Complex Variables 9798985972429

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Complex Variables

Complex Variables Sixth Edition

Robert P. Massé

Lorem Dolor Facilisis

Ipsum

Also by Robert P. Massé

Copyright © 2022 Robert P. Massé

Physics: Nature of Physical Fields and Forces

All rights reserved.

Physics: Where It Went Wrong Sixth Edition Vectors and Tensors of Physical Fields Number Theory

ISBN: 979-8-9859724-2-9

Laplace Transforms Matrices and Determinants

Complex Variables

by Robert P. Massé

Massé, Robert P.

Kay E. Massé

Physics: Principle of Least Action i

Dedication This book is dedicated to my beloved wife Kay whose loving and insightful support made it possible

ii

Preface Mathematics is an essential tool in the continuing struggle to understand the physical world.

iii

!

Complex variable theory constitutes a very elegant and

extremely useful branch of mathematics. The range of applications of complex variables in the sciences, engineering,

!

First Edition – January 2, 2020

!

Second Edition – January 20, 2020

!

Third Edition – December 9, 2020

!

Fourth Edition – June 19, 2021

!

Fifth Edition – March 15, 2022

!

Sixth Edition – June 5, 2022

and mathematics is broad and far-reaching. The use of complex variable theory in studying functions arising from physical problems often leads to a deeper insight into the general properties of these functions. !

The material in this book is organized and presented in a

manner designed to provide a comprehensive first look at complex variable theory. Most of the sources for this book are given in the references at the end of the book. !

The research that forms the foundation of this book could

not

have

been

accomplished

without

the

outstanding

interlibrary loan programs of Florida made available through the good resources of the Lee County Library System and the Tampa Bay Library Consortium.

!

Robert P. Massé

iv

Contents 1!

Complex Numbers and the Complex Plane

!

1.1!

Complex Numbers

!

1.2!

Complex Conjugates

!

1.3!

Algebraic Properties of Complex Numbers

!

1.4!

Properties of Complex Conjugates

!

1.5!

Multiplicative Inverse of a Complex Number

!

1.6!

Complex Plane

!

1.7!

Modulus of Complex Number

!

1.8!

Moduli Inequalities

!

1.9!

Square Roots of Complex Numbers

!

1.10! Polar Form of Complex Numbers

!

1.11! Euler’s Formula

!

1.12! de Moivre’s Theorem

!

1.13! Exponential Form of Complex Numbers

!

1.14! Integer Roots of Complex Numbers

!

1.15! de Moivre’s Theorem for Integer Roots

!

1.16! Rational Powers of Complex Numbers

!

1.17! Complex Polynomials

2!

Complex Functions

!

2.0!

Function Classifications

!

2.1!

Mapping of Complex Variables

!

2.2!

Point Sets in the Complex Plane

!

2.3!

Limits of Complex Functions

!

2.4!

Continuity of a Complex Function

!

2.5!

Infinity in Complex Numbers

!

2.6!

Riemann Sphere

!

2.7!

Limits Tending to Infinity

!

2.8!

Bounded Complex Functions

!

2.9!

Bolzano-Weierstrass Theorem

3! Complex Differentiation !

3.1!

Derivative of a Complex Function

!

3.2!

Existence of the Derivative

!

3.3!

Differentiation Rules

!

3.4!

Cauchy-Riemann Equations

!

3.5!

Exponential Form of the Cauchy-Riemann Equations

!

3.6!

Wirtinger Derivatives

!

3.7!

Holomorphic Functions

!

3.8!

Derivatives of Complex Functions of a Real-Valued

!

!

Parameter

!

3.9!

L’Hôpital’s Rule v

4! Elementary Complex Functions

6! Cauchy’s Integral Formula

!

4.1!

Branches, Branch Cuts, and Branch Points

!

6.1!

Cauchy’s Integral Formula for Holomorphic

!

4.2!

Complex Exponential Function

!

!

Functions

!

4.3!

Complex Trigonometric Functions

!

6.2!

Cauchy’s Integral Formula for Derivatives

!

4.4!

Complex Hyperbolic Functions

!

6.3!

Higher Order Derivatives of a Holomorphic Function

!

4.5!

Complex Logarithmic Function

!

6.4!

Cauchy’s Inequality

!

4.6!

Complex Power Function

!

6.5!

Liouville’s Theorem

!

4.7!

Inverse Complex Trigonometric Functions

!

6.6!

Gauss’ Mean-Value Theorem

!

4.8!

Inverse Complex Hyperbolic Functions

!

6.7!

Modulus Theorems

!

4.9!

Complex-Base Logarithmic Function

!

6.8!

Fundamental Theorem of Algebra

!

6.9!

Harmonic Functions

5! Complex Integration !

5.1!

Integration of Real-Valued Functions

7! Complex Sequences

!

5.2!

Integration of Complex Functions

!

7.1!

Sequences of Complex Numbers

!

5.3!

Contour Integration

!

7.2!

Complex Sequence Operations

!

5.4!

Cauchy-Goursat Theorem for Simply Connected

!

7.3!

Sequences of Complex Functions

!

!

Domains

!

7.4!

Uniform Convergence of Sequences of Complex

!

5.5!

Cauchy-Goursat Theorem for Multiply Connected

!

!

Functions

!

!

Domains

!

5.6!

Winding Number

8! Complex Series !

8.1!

Series of Complex Numbers

!

8.2!

Convergent and Divergent Complex Series

vi

!

8.3!

Tests for Convergence and Divergence of a Complex

10! Residues

!

!

Series

!

10.1! Residues

!

8.4!

Series of Complex Functions

!

10.2! Residue Integration

!

8.5!

Convergence of Complex Function Series

!

10.3! Evaluating Integrals of Real-Valued Functions Using

!

8.6!

Uniform Convergence of Complex Function Series

!

!

!

8.7!

Power Series

!

10.4! Summation of Infinite Series by Residues

!

8.8!

Taylor Series

!

10.5! Mittag-Leffler’s Expansion Theorem

!

8.9!

Maclaurin Series

!

8.10! Properties of Taylor and Maclaurin Series

11! Conformal Mapping

!

8.11! Laurent Series

!

11.1! Definition of Mapping

!

11.2! Conformal Mapping

9! Singularities and Zeros

!

11.3! Möbius Transformations

!

9.1!

Singularities

!

11.4! Automorphisms of a Unit Disk

!

9.2!

Zeros

!

11.5! Some Mappings

!

9.3!

Relation Between Zeros and Poles

!

11.6! Schwarz-Christoffel Transformation

!

9.4!

Argument Principle

!

11.7! Riemann Surfaces

!

9.5!

Rouché’s Theorem

!

9.6!

Hurwitz’s Theorem

12! Infinite Products

!

9.7!

Open Mapping Theorem

!

12.0! Finite Product of Complex Numbers

!

9.8!

Analytic Continuation

!

12.1! Infinite Product of Complex Numbers

!

9.9!

Schwarz’s Reflection Principle

!

12.2! Convergence and Divergence of Infinite Products

!

12.3! Uniform Convergence of an Infinite Product of

!

!

Residues

Functions vii

!

12.4!

Entire Function Expansion as an Infinite Product

! Appendix A!

The Greek Alphabet

! Appendix B!

Complex Exponential Function

! Appendix C!

Differentiation of the Cauchy Integral

!

Formula

! Appendix D!

Summary of Propositions

! Appendix E!

Identities of Elementary Complex Functions

!

E.1! Complex Exponential Function

!

E.2! Complex Trigonometric Functions

!

E.3! Complex Hyperbolic Functions

!

E.4! Complex Logarithmic Function

!

E.5! Inverse Complex Trigonometric Functions

!

E.6! Inverse Complex Hyperbolic Functions

! References

viii

Chapter 1 Complex Numbers and the Complex Plane

z = x+iy = re

iθ

1

!

In this chapter we will define complex numbers and

therefore represents an extension of the real number system.

review their algebraic properties in rectangular, polar, and

This extension results in the set of complex numbers (denoted

exponential form. We will present geometric representations of

by ! ). The creation of the complex number system was

complex numbers in the complex plane. For the moduli of

motivated by the need to be able to solve certain algebraic

complex numbers, we will develop inequalities. We will also

equations which could not be solved using only the real

determine the roots of complex numbers.

number system (see Section 1.1.1).

1.1! !

!

COMPLEX NUMBERS

where x and y are real numbers. Every complex number

A complex number z in rectangular or Cartesian form is

defined to be: !

z = x + i y!

z = x + y i!

z = x + i y is composed of two independent parts: a real part which is x , and an imaginary part which is y (not i y ). These

(1.1-1)

parts are written as:

(1.1-2)

!

equation:

!

i 2 = −1 !

(1.1-3)

The number i is taken to be the positive square root of −1 : ! !

i = −1 !

(1.1-4)

Since there is no real number whose square is a negative

y = Im z !

(1.1-5)

so that:

where x and y are real numbers and where i is defined by the

!

x = Re z !

!

or equivalently: !

Every complex number has a unique representation x + i y

z = Re z + i Im z = x + i y !

(1.1-6)

Note that both the real part and the imaginary part of a

complex number are real numbers. Therefore once we have defined

−1 = i , we can represent all complex numbers using

only real numbers together with i . In algebraic analysis the symbol i acts like any other ordinary algebraic symbol, and it can be treated as an algebraic entity in an expression.

number, the set of real numbers (denoted by ! ) does not

!

If z is an element of the set of complex numbers ! , this

include any square roots that are negative. The number i

fact is indicated by z ∈! , where the symbol ∈ means is an 2

element of or is a member of. If a complex number z can assume

which variable is the imaginary part of the complex number.

different complex numerical values from a set of complex

Since complex numbers ! constitute a set of ordered pairs of

numbers, then z is known as a complex variable.

real numbers, they are also denoted by ! 2 .

!

A number having a zero imaginary part, z = x + i 0 = x , is a

Example 1.1-1

pure real number (called a real complex number or simply a

Show that for z = x + i y :

real number). A number having a zero real part, z = 0 + i y = i y ,

( )

!Re i z = − Im z !

is a pure imaginary number. If y = 1 then by convention a complex number z is written as z = x + i . The number z = i is

Solution:

the most fundamental imaginary number and is known as the

We have i z = i x + i 2 y = − y + i x . Therefore:

imaginary unit. For a complex number having a negative

( )

( )

imaginary part such as z = 2 + −5 i , the number is written in

!Re i z = − y = − Im z !

the form z = 2 −5 i . The set of complex numbers ! includes all numbers (even those with zero imaginary part), and so the set of real numbers ! is a proper subset of the set of complex numbers ! (just as the set of integers ! is a proper subset of the set of real numbers ! , and the set of nonnegative integers

! is a proper subset of ! ). !

The real part x and the imaginary part y of any complex

variable z = x + i y can be written uniquely as an ordered pair of

(

)

real variables ( x, y ) , where ‘ordered’ means that x, y is not the same as

( y, x )

unless x = y . Using this complex number

notation, the real variables x and y must be ordered within the parentheses so that it is clear which variable is the real part and

( )

Im i z = Re z

1.1.1! !

( )

Im i z = x = Re z

NEED FOR COMPLEX NUMBERS

The need for complex numbers in mathematics arose

because certain cubic and quadratic equations were found to be unsolvable using only real numbers. Complex numbers were first employed by Girolamo Cardano (1501-1576) in the course of searching for certain cubic equation roots that were found to be not real. Some quadratic equations such as x 2 + 2 = 0 (where

x 2 = −2 and so x = ± −2 ) also have roots that are not real. The square root of a positive real number is real and of a negative real number is a pure imaginary number.

3

!

To obtain complete solutions of equations having some

the 1600s by Descartes conveyed the prevailing impression that

−1 must be used in

such numbers were fictitious or unreal. In fact, if imaginary

conjunction with real numbers, and so an enlarged number

numbers were encountered in an attempt to find roots of an

system ! is required that includes complex numbers. In the

equation, this was taken as proof that the equation had no

example x 2 = −2 we then have:

solution. Later on Gauss (1777-1855) attributed the unsuitable

roots that are imaginary, the number

!

x = ± −2 = ±

( −1)( 2) = ±

−1 2 = ± i 2 !

(1.1-7)

terminology of imaginary number for much of the mystery surrounding complex numbers.

Using the complex number system, all polynomial equations

!

with complex coefficients can be determined to have as many

can be represented as points on a two-dimensional plane (see

roots as the degree of the polynomial (see Section 6.8).

Section 1.6), and so should be considered as real as all other real

!

For a polynomial equation of degree two, a z 2 + b z + c = 0 ,

the roots are given by the quadratic formula: !

−b ± b2 − 4 a c ! z= 2a

Green, 1976; and Nahin, 1998). (1.1-8) Example 1.1-2

the discriminant b 2 − 4a c :

b2 − 4 a c > 0 ! roots are real and different!

numbers. Complex numbers now constitute a very important area of mathematics with applications in many fields (see

The nature of the roots of a quadratic equation is determined by

!

As time went on it became evident that complex numbers

(1.1-9)

Determine the roots of 2 z 2 + 4 z + 10 = 0 . Solution: Using the quadratic formula:

!

b2 − 4 a c = 0 ! roots are real and equal!

(1.1-10)

!

b2 − 4 a c < 0 ! roots are complex!

(1.1-11)

!

For many years mathematicians were very uncomfortable

− b ± b2 − 4 a c !z = 2a we find that the discriminant b2 − 4 a c is negative:

with complex numbers. The term imaginary number coined in 4

!z =

− 4 ± 16 − 80 − 4 ± − 64 = = 4 4

−4 ±

( −1)(64) 4

=

− 4 ± 8 −1 4

and so the roots are two complex numbers: !z = −1± 2 −1 = −1± 2 i

!

The operations of addition and multiplication of complex

numbers can more clearly be formulated using the algebraic operational rules for real binomials, where each complex number is treated as a binomial. The imaginary unit i of a complex number z = x + i y is considered as a “variable” when these rules are applied, and the definition i 2 = −1 is exploited

1.1.2! !

DEFINITIONS OF COMPLEX NUMBERS

We have defined complex numbers as numbers having the

form z = x + i y where x and y are real numbers, and where 2

i = −1 . This definition emphasizes the important role played by the imaginary unit i , and is consistent with the concept that real numbers are a subset of complex numbers. !

Complex numbers can also be defined in terms of the set

of ordered pairs

( x, y )

of real numbers ! 2 by specifying

(see Section 1.3). 1.1.2.1! !

EQUALITY OF COMPLEX NUMBERS

Two complex numbers z1 = x1 + i y1 and z2 = x2 + i y2 are

equal ( z1 = z2 ) if and only if their real parts are equal to each other ( x1 = x2 ) and their imaginary parts are equal to each other ( y1 = y2 ): !

Re z1 = Re z2 !

Im z1 = Im z2 !

(1.1-12)

operational rules for addition and multiplication that do not

In any equation involving complex numbers therefore, the real

involve the introduction of the imaginary unit i (see Section

parts on both sides of the equation must be equal and the

1.3.6). With this definition any ordered pair of real numbers

imaginary parts on both sides of the equation must be equal. A

following these operational rules is a complex number. This

complex equation is then the equivalent of two real equations.

approach aims at establishing an axiomatic foundation for the set of complex numbers ! as an algebraic field. Nevertheless this approach suffers from being less intuitive, and so more obscure when it comes to applying complex numbers to physical problems, and so we will not use it.

Example 1.1-3 Solve the complex equation:

(

)

!x + 2 y + i x + y = 4 where x and y are real numbers. 5

(

)

( )(

2

! − −1 = −1

Solution:

)

2

2

−1 = i 2 = −1

Equating real parts to each other, and imaginary parts to each other, we have: !x + 2 y = 4 !

Example 1.1-5

x+ y=0

Determine:

and so:

1.! i 4

!− y + 2 y = y = 4 !

2.! i 7 3.! i 18

NONNEGATIVE INTEGER POWERS OF i

1.1.2.2! !

x = −4

Integer powers of i are equal to either i , −1 , − i , or 1 : 0

4

8

12

!

i =i =i =i =i

!

4k

= 1!

k = 0, 1, 2, 3,! !

(1.1-13)

i 1 = i 5 = i 9 = i13 = i 4 k+1 = i !

k = 0, 1, 2, 3,! !

(1.1-14)

!

i 2 = i 6 = i10 = i14 = i 4 k+2 = −1!

k = 0, 1, 2, 3,! !

(1.1-15)

!

i 3 = i 7 = i11 = i15 = i 4 k+3 = − i !

k = 0, 1, 2, 3,! !

(1.1-16)

4.! i 21 Solution:

( )( ) i 7 = i 4 i 2 i = (1) ( −1) i = − i

1.! i 4 = i 2 i 2 = −1 −1 = 1 2.!

( ) ( ) ( −1) = −1 = ( i ) i = (1) i = i 4

3.! i18 = i 4 i 2 = 1 4.! i 21

4

5

4

5

Example 1.1-4

(

Show that: − −1 Solution: We have:

) = −1. 2

Example 1.1-6 Show that any polynomial in i having real coefficients: !z = a0 + a1 i + a2 i 2 +!+ an i n can be represented by a single complex number. 6

Solution: Using equations (1.1-13) through (1.1-16) we can rewrite the polynomial as:

(

) (

)

z = 5x 2 − i 2 y 2

3.!

z = −1

Solution:

!z = a0 − a2 + a4 −! + a1 − a3 + a5 −! i which has the form of a single complex number z = x + i y .

1.2!

2.!

COMPLEX CONJUGATES

1.! Re z = 3!

Im z = 5 !

z = 3− 5 i

2.! Re z = 5x 2 !

Im z = − 2 y 2 !

z = 5x 2 + i 2 y 2

3.! Re z = 0 !

Im z = 1 !

z = −i

If z = x + i y , then z = x − i y is a complex number known as

!

the complex conjugate or simply the conjugate of z . A complex number z will have the same real part as its conjugate z , but the negative of the imaginary part of z . The complex conjugate

z of a complex number z is obtained by replacing i → − i . The process of changing a complex number z to its conjugate z is referred to as conjugation. We see then that: !

z=z!

if and only if z is real!

(1.2-1)

!

z = −z !

if and only if z is pure imaginary!

(1.2-2)

1.3! !

Determine the real part, imaginary part, and complex conjugate of the following complex numbers: 1.!

z = 3+ 5 i

Complex numbers have the same operations of addition,

subtraction, multiplication, and division as do real numbers, and so form an algebraic number field. The procedures for performing each of these operations are different, however, from those of real numbers. ! !

Example 1.2-1

ALGEBRAIC PROPERTIES OF COMPLEX NUMBERS

A zero or null complex number is given by:

z = 0 + 0 i = 0!

(1.3-1)

and is usually written simply as z = 0 . From equation (1.3-1) we see that the complex equation z = 0 is equivalent to the two real equations: x = 0 and y = 0 . A complex number cannot equal zero unless both its real and imaginary parts equal zero. The 7

only complex number that can be considered both real and imaginary is zero.

1.3.1! !

of numbers is the unique number 0:

!

!

)

(

)

(1.3-6)

(

)

(1.3-7)

Im z1 + z2 = Im z1 + Im z2 !

(1.3-2)

numbers is the unique real number 1:

z = 1+ 0 i = 1 !

) (

and !

The unit identity element or unity for the complex set of

!

) (

Re z1 + z2 = Re z1 + Re z2 !

!

IDENTITY ELEMENTS

z +0 = 0+ z = z!

) (

We then have:

The additive identity element or zero for the complex set

!

(

z1 + z2 = x1 + i y1 + x2 + i y2 = x1 + x2 + i y1 + y2 ! (1.3-5)

!

!

The addition of complex numbers can be shown to be

commutative and associative by using the commutativity and associativity of their real parts and their imaginary parts

(1.3-3)

Note that in this book the numbers 0 and 1 can represent

either the real numbers 0 and 1, respectively, or the complex

separately. Example 1.3-1

numbers 0 + i0 and 1+ i0 , respectively. Both these numbers

Verify the commutativity of addition: z1 + z2 = z2 + z1 .

have a value that is real.

Solution:

!

The multiplicative identity element for the complex set of

numbers is the unique real number 1: !

(1)( z ) = ( z )(1) = z !

1.3.2!

ADDITION

!

From the definition of addition of complex numbers, we have:

(1.3-4)

Complex numbers are added by adding their real parts

(

) (

) (

) (

!z1 + z2 = x1 + i y1 + x2 + i y2 = x1 + x2 + i y1 + y2

)

From the commutative law for real numbers, we have:

(

) (

) (

) (

!z1 + z2 = x1 + x2 + i y1 + y2 = x2 + x1 + i y2 + y1

)

and their imaginary parts separately. In rectangular form, we

From the definition of addition of complex numbers, we

have:

have: 8

(

) (

) (

) (

)

!z1 + z2 = x2 + x1 + i y2 + y1 = x2 + i y2 + x1 + i y1 = z2 + z1

!

1.3.3! SUBTRACTION !

Two complex numbers are subtracted by subtracting their

form, we have:

(

) (

) (

) (

)

z1 − z2 = x1 + i y1 − x2 + i y2 = x1 − x2 + i y1 − y2 !

(

(

(1.3-8)

)

(1.3-9)

and

(

!

together in pairs and then adding:

)

(1.3-10)

We see that if z = x + i y , then the negative of z is

!

complex number − z such that z + −z = 0 . Also if for any z1 ∈!

!

we have: !

z + z1 = 0 !

! (1.3-11)

then z1 = − z , and − z is called either the inverse of z with respect to addition or the additive inverse.

) (

)

(1.3-13)

Re z1 z2 = Re z1 Re z2 − Im z1 Im z2 !

(

)

(1.3-14)

(

)

(1.3-15)

and

− z = −x − i y . For any complex number z , there exists a unique

( )

(

z1 z2 = x1 x2 − y1 y2 + i x1 y2 + y1 x2 !

We then have: !

Im z1 − z2 = Im z1 − Im z2 !

)( x2 + i y2 ) = x1 x2 + i y1 x2 + i x1 y2 + i 2 y1 y2 ! (1.3-12)

Using i 2 = −1 we obtain: !

Re z1 − z2 = Re z1 − Re z2 !

!

Two complex numbers are multiplied by multiplying all parts ! z1 z2 = x1 + i y1

We then have: !

Complex numbers follow the associative and distributive

laws of multiplication. Therefore we have z = x + i y = x + y i .

real parts and their imaginary parts separately. In rectangular

!

1.3.4! MULTIPLICATION

Im z1 z2 = Re z1 Im z2 + Im z1 Re z2 ! If k is a real constant, we have:

(

)

k z = k x + i y = k x + ik y !

(1.3-16)

and z is said to be scaled by k . ! !

If z = i , then we can write z 2 as:

(

)(

) (

) (

)

z 2 = z z = 0 + i 0 + i = 0 − 1 + i 0 + 0 = −1 !

(1.3-17)

9

!

The product of a complex number z and its complex

conjugate z is always a real number and is nonnegative:

(

)(

)

2

2

2

2

2

! z z = x + i y x − i y = x + i x y − i x y − i y = x + y ! (1.3-18) and so: ! !

!

Since complex numbers follow the associative and

distributive

(

zn = x + i y

)

n

laws

of

multiplication,

we

can

evaluate

using the binomial theorem (see Example 1.3-6).

Example 1.3-2

z z = x2 + y2 !

(1.3-19)

From algebraic properties of the real and imaginary parts

Show that 1 is the multiplicative identity. Solution:

of complex numbers, we obtain the addition and multiplication

From the definition of multiplication of complex numbers,

properties of complex numbers presented in Table 1.3-1 (see

we have:

Examples 1.3-1 and 1.3-5).

( )( ) ( ) (

) (

)(

! 1 z = 1 x + i y = 1+ i 0 x + i y

)

or

(1)( z ) = (1) ( x ) + i (1) ( y ) + i (0) ( x ) + i2 (0) ( y ) = x + i y = z Example 1.3-3 If z = 0 + i determine z 2 . Table 1.3-1! Complex number algebraic properties. !

We can specify a complex number z raised to an integer

power n by using mathematical induction with multiplication of complex numbers. We have z1 = z . We can then define z n as

Solution: From the definition of multiplication of complex numbers, we have:

(

)(

)

() ()

!z 2 = z 2−1 z1 = 0 + i 0 + i = 0 + i 0 + 0 i + i 2 = i 2 = −1

z n = z n−1 z1 = z n−1 z for n > 1 . 10

or

Example 1.3-4

(

)

(

2

)3

! 2 + i = 8 + 12i − 6 − i = 2 + 11i

Show that 1+ i = 2 i . Solution:

(

) ( 2

)(

)

2

! 1+ i = 1+ i 1+ i = 1+ i + 2 i = 1− 1+ 2 i = 2 i Example 1.3-5

! !

Show that z1 z2 = z2 z1 .

!z1 z2 = x1 + i y1

Complex numbers can be divided (except by zero):

z1 z2

=

( x1 + i y1 ) ! ( x2 + i y2 )

z2 ≠ 0 !

(1.3-20)

To separate the real and imaginary parts of the quotient z1 z2

Solution:

(

1.3.5! DIVISION

)( x2 + i y2 ) = x1 x2+ i x1 y2 + i y1 x2+ i 2 y1 y2

and place it in the form x + i y , the denominator is converted to a real number by multiplying the numerator and denominator by the complex conjugate of the denominator z2 :

or !z1 z2 = x2 x1+ i x2 y1 + i y2 x1 + i y2 y1 = ( x2 + i y2 ) ( x1 + i y1 ) 2

and so:

!

! Example 1.3-6

(

)

3

using the binomial theorem.

Solution:

(

z2

=

z1 z2 z2 z2

=

z1 z2 z2 z2

z2 ≠ 0 !

!

(1.3-21)

or

!z1 z2 = z2 z1

Determine 2 + i

z1

)3 ( )3 ( )2 ( i ) + 3 ( 2) ( i )2 + ( i )3

! 2+i = 2 + 3 2

x1 + i y1 ) ( x2 − i y2 ) ( ! = z2 ( x2 + i y2 ) ( x2 − i y2 ) z1

z2 ≠ 0 !

(1.3-22)

z2 ≠ 0 !

(1.3-23)

so we obtain: !

z1 z2

=

x1 x2 + y1 y2 x22 + y22

+i

x2 y1 − x1 y2 x22 + y22

!

We then have: 11

⎛ z ⎞ Re z1 Re z2 + Im z1 Im z2 Re ⎜ 1 ⎟ = ! 2 2 z ⎝ 2⎠ Re z2 + Im z2

!

(

) (

)

z2 ≠ 0 !

(1.3-24)

!

x2 − y2 y2

= x22 + y22 ≠ 0

x2

Solving the two equations for x and y , we have:

and

⎛ z ⎞ Re z2 Im z1 − Re z1 Im z2 ! Im ⎜ 1 ⎟ = 2 2 z ⎝ 2⎠ Re z2 + Im z2

!

(

) (

)

z2 ≠ 0 !

!x =

(1.3-25)

x1 x2 + y1 y2 x22

+

y22

!

y=

x2 y1 − x1 y2 x22

+

y22

z2 ≠ 0

!

These are the real and imaginary parts of z1 z2 as given in

Example 1.3-7 Determine z1 z2 in rectangular form x + i y without using

equation (1.3-23).

the complex conjugate of z2 .

Example 1.3-8

Solution:

If z1 = 6 + 8 i , z2 = 2+ i , and z3 = 1+ i determine the following

We have:

complex numbers in rectangular form x + y i :

x1 + i y1 ) ( ! = = x + iy z2 ( x2 + i y2 ) z1

or

(

)(

) (

) (

!x1 + i y1 = x + i y x2 + i y2 = x2 x − y2 y + i y2 x + x2 y Equating real and imaginary parts: !x2 x − y2 y = x1 !y2 x + x2 y = y1 These two simultaneous equations have a solution if:

)

1.!

z1 + z2

2.!

z1 − z2

3.!

z1 z2

4.!

( z1 + z2 ) z3

5.!

z1 z2

Solution: 1.!

(

) (

) (

) (

)

z1 + z2 = 6 + 8 i + 2 + i = 6 + 2 + i 8 +1 = 8 + 9 i 12

(

) (

) (

) (

)

2.!

z1 − z2 = 6 + 8 i − 2 + i = 6 − 2 + i 8 − 1 = 4 + 7 i

3.!

z1 z2 = 6 + 8 i 2 + i = 12 + 16 i + 6 i + 8i 2 = 12 + 22 i − 8

(

)(

)

! = 4 + 22 i 4.! 5.!

( z1 + z2 ) z3 = (8 + 9 i ) (1+ i ) = 8 − 9 + i (9 + 8) = −1+ 17 i z1 z2

=

6 + 8 i 6 + 8 i 2 − i 12 + 8 16 − 6 20 10 = = +i = +i 2+i 2 + i 2 − i 4 +1 4 +1 5 5

1.!

1 i

2.!

1 i3

Solution: Multiplying the numerator and denominator by the complex conjugate of the denominator, we have:

!= 4 + 2 i

1.!

1 1 ⎛ −i⎞ −i = = = −i i i ⎜⎝ − i ⎟⎠ 1

Example 1.3-9

2.!

1 1 1 1 ⎛ i⎞ i i = = = = = =i −1 i − i ⎜⎝ i ⎟⎠ − i 2 1 i3 i2 i

2

( )

2

Determine Re z and Im z . Solution:

(

)(

(

)

) ( )

!z 2 = x + i y x + i y = x 2 + 2 i x y + i 2 y 2 = x 2 − y 2 + i 2 x y Therefore: !Re z 2 = x 2 − y 2 !

Example 1.3-11

(

)

Determine Im ⎡⎣ 1+ i z ⎤⎦ . Solution:

Im z 2 = 2 x y

Example 1.3-10 Determine the following complex numbers in rectangular

(

) (

)(

)

(

) (

! 1+ i z = 1+ i x + i y = x + i y + i x − y = x − y + i x + y

)

Therefore:

(

)

!Im ⎡⎣ 1+ i z ⎤⎦ = x + y

form x + y i : 13

Equations (1.3-26) and (1.3-27) can be compared with equations Example 1.3-12

(1.3-5) and (1.3-13), respectively. Using the ordered pair

Determine in rectangular form x + y i :

representation of a complex number z = x + i y , we can write:

5+ i ! 5+ i

!

( ) ( ) ( )( )

z = x, y = x, 0 + 0,1 y,0 !

1.3.7! OPERATING WITH THE IMAGINARY UNIT

Solution:

!

We have:

5 + i 5 + i 5 + i 5 + i 25 + 10 i − 1 24 + 10i 12 + 5i ! = = = = = 5 − i 5 − i 5 + i 25 + 1 26 13 5+ i

Algebraic operations involving

! However:

!

!

The operational rules for addition and multiplication of

complex numbers in terms of ordered pairs as referred to in Section 1.1.2 are listed here for completeness only, and will not be used in this book. For two complex numbers z1 = x1 + i y1 and

z2 = x2 + i y2 we have the following algebraic rules that do not

!

) (

) (

)

z1 + z1 = x1 , y1 + x2 , y2 = x1 + x2 , y1 + y2 !

(

)(

) (

! !

−1 = 1

1 ! −1

−1

1

1

≠

−1

!

or!

i = i!

(1.3-29)

since!

1 i ≠ = −i! i

(1.3-30)

Similarly:

−1 = −1 −1 ≠

( −1)( −1) =

1 = 1!

(1.3-31)

since

include the imaginary unit i :

(

−1 must be consistent

with the definition i 2 = −1 . For example we have:

1.3.6! ORDERED PAIR OPERATIONAL RULES FOR COMPLEX NUMBERS

!

(1.3-28)

)

z1 z1 = x1 , y1 x2 , y2 = x1 x2 − y1 y2 , x1 y2 + y1 x2 !

(1.3-26)

!

−1 = i 2 ≠

( −1)( −1) =

1 = 1!

(1.3-32)

(1.3-27) 14

1.4! !

!

PROPERTIES OF COMPLEX CONJUGATES

product of complex numbers is equal to the sum, difference, or

If z, z1 , z2 ∈! , the following are some properties of the

complex conjugates z , z1 and z2 : !

z1 ± z 2 = z1 ± z2 !

The complex conjugate of any finite sum, difference, or

product, respectively of the complex conjugates of the numbers. In an equation involving only the arithmetic operations of addition, subtraction, multiplication, or division, the complex

(1.4-1)

numbers can all be replaced by their complex conjugates, and the equation will still be a valid equation.

!

−z = −z !

(1.4-2)

!

z1 − z2 = z1 − z2 !

(1.4-3)

!

z1 z2 = z1 z2 !

(1.4-4)

Example 1.4-1 Determine the complex conjugates of the following complex numbers: 1.! i 2.! 3+ 2i

!

zn = z n !

!

⎛ 1⎞ 1 ⎜⎝ z ⎟⎠ = z !

z ≠ 0!

(1.4-6)

⎛ z1 ⎞ z1 ⎜z ⎟= z ! ⎝ 2⎠ 2

z ≠ 0!

(1.4-7)

!

(1.4-5)

(

3.! − 3+ 2i

)

4.! 3+ 2i 5.! − −5 Solution: 1.!

!

z = z!

(1.4-8)

!

i z = −i z !

(1.4-9)

i =−i

2.! 3+ 2i = 3− 2i

(

)

(

3.! − 3+ 2i = − 3− 2i

) 15

(

) (

) (

!z1 z2 = x1 x2 − y1 y2 − i x1 y2 + x2 y1 = x1 − i y1

4.! 3+ 2i = 3+ 2i

)( x2 − i y2 )

and so:

5.! − −5 = − i 5 = i 5

!z1 z2 = z1 z2

Example 1.4-2 Show that z1 + z2 = z1 + z2 .

Example 1.4-4

Solution:

⎛ 1⎞ 1 Show that ⎜ ⎟ = . ⎝ z⎠ z

(

) (

) (

) (

!z1 + z2 = x1 + i y1 + x2 + i y2 = x1 + x2 + i y1 + y2

)

Therefore by definition of a complex conjugate:

(

) (

) (

) (

!z1 + z2 = x1 + x2 − i y1 + y2 = x1 − i y1 + x2 − i y2

)

and so:

Solution:

⎞ ⎛ x−iy ⎞ x+iy ⎛ 1⎞ ⎛ 1 ⎞ ⎛ x−iy !⎜ ⎟ = ⎜ = = ⎜ ⎟= ⎝ z ⎠ ⎝ x + i y ⎟⎠ ⎝ x + i y x − i y ⎠ ⎜⎝ x 2 + y 2 ⎟⎠ x 2 + y 2

(

)(

)

We also have:

!z1 + z2 = z1 + z2

1 1 x+iy x+iy ! = = = 2 2 z x−iy x +y x−iy x+iy

(

Example 1.4-3

)(

)

Therefore:

Show that z1 z2 = z1 z2 .

⎛ 1⎞ 1 !⎜ ⎟ = ⎝ z⎠ z

Solution:

(

!z1 z2 = x1 + i y1

)( x2 + i y2 ) = ( x1 x2 − y1 y2 ) + i ( x1 y2 + x2 y1 )

Therefore by definition of a complex conjugate:

Example 1.4-5 Show that z = z . 16

Solution:

!

!z = x + i y = x − i y = x + i y = z

Example 1.4-6

We can write:

(

) (

)

(1.4-10)

(

) (

)

(1.4-11)

!

z + z = x + i y + x − i y = 2 x = 2 Re z !

!

z − z = x + i y − x − i y = 2 i y = 2 i Im z !

We see that the sum of a complex number and its conjugate is

Show that i z = − i z .

a real number. Solution:

!

(

)

(

)

!i z = i x + i y = − i x − i y = − i z

We then have:

Re z = x =

!

Show that the equation

(2 + i)

2

= 3+ 4i remains a valid

equation if all the complex numbers are replaced by their conjugates.

(

(

)2 = 3− 4i

is a valid equation. We

have:

)2 = 4 − 4i − 1 = 3− 4i (

(1.4-12)

)

(

)

(1.4-13)

and as shown in Example 1.4-9:

2x y z x2 − y2 ! = 2 + i z x + y2 x2 + y2

!

We need to show that 2 − i

Therefore 2 − i

z−z ! 2i

z1 z2 + z1 z2 = 2 Re z1 z2 = 2 Re z1 z2 !

!

Solution:

(

Im z = y =

As shown below in Example 1.4-8, we also have:

Example 1.4-7

! 2−i

z+z ! 2

(1.4-14)

Example 1.4-8

(

)

(

)

Show that z1 z2 + z1 z2 = 2 Re z1 z2 = 2 Re z1 z2 . Solution:

(

)(

) (

)(

!z1 z2 + z1 z2 = x1 + i y1 x2 − i y2 + x1 − i y1 x2 + i y2

)2 = 3− 4i is a valid equation.

)

or 17

(

) (

(

)

(

) (

)

+ ⎡⎣ x1 x2 + y1 y2 + i x1 y2 − x2 y1 ⎤⎦

!

) (

)(

) ( )

!z 2 = x + i y x + i y = x 2 − y 2 + i 2 x y

!z1 z2 + z1 z2 = ⎡⎣ x1 x2 + y1 y2 − i x1 y2 − x2 y1 ⎤⎦

If z 2 = z 2 we have:

(

) ( ) (

) ( )

! x2 − y2 − i 2 x y = x2 − y2 + i 2 x y

and so:

(

!z1 z2 + z1 z2 = 2 x1 x2 + y1 y2

)

This equation is only true if y = 0 , in which case z is real, or if x = 0 , in which case z is pure imaginary.

or

(

)

(

!z1 z2 + z1 z2 = 2 Re z1 z2 = 2 Re z1 z2

)

Solve the quadratic equation z 2 − i z + 6 = 0 .

Example 1.4-9

2x y z x2 − y2 Show that = 2 . + i 2 2 2 z x +y x +y Solution:

2x y z z 2 x + i y x + i y x 2 − y 2 + 2i x y x 2 − y 2 ! = = = = + i z z z x−iy x+iy x2 + y2 x2 + y2 x2 + y2

Show that if z 2 = z 2 , then z is either real or pure imaginary. Solution:

)(

Solution: Using the quadratic formula:

− b ± b2 − 4 a c !z = 2a we have:

i ± i 2 − 24 i ± − 25 i ± 5i !z = = = 2 2 2

Example 1.4-10

(

Example 1.4-11

) (

) ( )

!z 2 = x − i y x − i y = x 2 − y 2 − i 2 x y

and so the roots are: !z = 3i !

or!

z = − 2i

and we have:

(

)(

)

! z − 3i z + 2i = z 2 − i z + 6 = 0 18

1.5! !

MULTIPLICATIVE INVERSE OF A COMPLEX NUMBER For any complex number z ≠ 0 there exists a unique

complex number z −1 such that: −1

n

⎛ 1⎞ 1 =⎜ ⎟ = n! ⎝ z⎠ z

z ≠ 0!

(1.5-6)

Example 1.5-1 Show that the multiplicative inverse of a complex number z is unique.

−1

z z = z z = 1!

!

( )

z − n = z −1

!

n

(1.5-1)

where z −1 = 1 z is called the inverse of z with respect to

Solution:

multiplication, the reciprocal of z , or the multiplicative

We will assume that both z −1 and z1−1 are multiplicative

inverse of z .

inverses of z . We can then write:

!

()

This multiplicative inverse is easily verified:

1 z x y ! = = 2 − i 2 2 2 z zz x +y x +y

!

z ≠ 0!

(1.5-2)

We can also write:

!

( z1 z2 ) ( z1−1 z2−1 ) = ( z1 z1−1 ) ( z2 z2−1 ) = 1!

(1.5-3)

z1 , z2 ≠ 0 !

(1.5-4)

)

()

( z1 z2 )

−1

=

(

z1−1

If n ∈! we have:

z2−1

)!

z1 , z2 ≠ 0 !

and so our assumption was wrong. The multiplicative inverse of a complex number z is unique, and z −1 = z1−1 . Example 1.5-2 Determine the inverse of z = 3− 5 i .

Therefore:

!

or

(

1 z z z x2 + y2 z =z = = 2 = 1! 2 z zz zz x +y

!

!

)

!z −1 = z1−1 z z −1 = z1−1 z z −1 = z1−1 1

and so: !

(

!z −1 = 1 z −1 = z1−1 z z −1

Solution: (1.5-5)

!z −1 =

1 z 3+ 5i 3+ 5i 3+ 5i = = = 2 2= z zz 34 3− 5i 3+ 5i 3 + 5

(

)(

)

19

Checking this inverse, we have:

from the set ! 2 , and the complex plane is a Euclidean plane.

3+ 5i 34 !z z = 3− 5i = =1 34 34

The identification of complex numbers as points on a plane by

(

−1

)

Wessel, Argand, and Gauss provided the missing foundation for complex numbers. !

Example 1.5-3

equal, z1 = z2 , then x1 = x2 and y1 = y2 . Therefore if and only if

Determine the inverse of z = 1+ i .

two complex numbers are equal will they correspond to the same point in the complex plane. Each point in the complex

Solution: !z −1 =

1 1 1− i 1− i 1 i = = = = − z 1+ i 1+ i 1− i 1+ 1 2 2

(

)(

)

Checking this inverse, we have:

(

plane then represents a unique complex number. As a result there exists a one-to-one correspondence between points of the complex plane and the set of complex numbers. For this reason the terms complex number and point in a complex plane are used

⎛1 i⎞ 1 i i 1 !z z = 1+ i ⎜ − ⎟ = − + + = 1 ⎝ 2 2⎠ 2 2 2 2 −1

)

interchangeably. The set of complex numbers ! can be said to form the complex plane, Argand diagram, or z-plane. !

1.6! !

A real number x is one-dimensional since it has just one

independent coordinate

( x) ,

With respect to a given Cartesian coordinate system in the

complex plane, the real part x of a complex number z = x + i y is

COMPLEX PLANE and so it can be represented

geometrically as a point on a line known as a number line. Similarly, a complex number z = x + i y is two-dimensional since

( )

it has two independent coordinates x, y , and so it can be represented geometrically as a point on a plane. The points

( x, y )

If two complex numbers z1 = x1 + i y1 and z2 = x2 + i y2 are

on a complex plane are ordered pairs of real numbers

plotted as the x coordinate, and the imaginary part y (not i y ) is plotted as the y coordinate (see Figure 1.6-1). A point on the x-axis will then represent a complex number having no

( )

imaginary part: z = x + i 0 = x or x, 0 , and a point on the y-axis will represent a complex number having no real part:

z = 0 + i y = i y or

(0, y ) .

For these reasons, the x-axis of a

Cartesian coordinate system in the complex plane is referred to as the real axis, and the y-axis is referred to as the imaginary 20

axis. The complex number zero is represented by the origin of

the x and y components of the vector are simply the

the coordinate system.

orthogonal projections of the radius vector onto the real and imaginary axis, respectively (see Figure 1.6-1). Complex numbers and the components of vectors in a rectangular coordinate system follow the same algebraic addition rule, and so complex numbers can be added graphically using the parallelogram rule of vectors.

Figure 1.6-1!

!

Geometric representations of a complex number z = x + i y as a point x, y and as a line vector (red line) in the Cartesian coordinate system in the z-plane.

( )

We see that two ways of representing complex numbers

are: algebraically as z = x + i y , and geometrically as a point

( )

specified by the coordinates x, y in the complex plane. Since all complex numbers are composed of two independent

Figure 1.6-2!

Argand diagram of complex numbers z = 3+ 2i , z = 3− 2i , and −z = − 3− 2i .

coordinates, it is also possible to interpret a complex number

z = x + i y as a two-dimensional line vector called a radius

!

vector or position vector. For a rectangular coordinate system,

complex plane: z = 3+ 2i , its conjugate z = 3− 2i , and its

In Figure 1.6-2 are shown three complex numbers in the 21

negative −z = − 3− 2i . For any complex number z we have the following geometrically: z and its conjugate z are reflections of each other with respect to the real axis (the x-axis), while z and its negative − z are reflections of each other through the origin.

1.7! !

MODULUS OF A COMPLEX NUMBER The modulus or magnitude of a complex number

z = x + i y is denoted by z and is defined as: !

z = x + i y ≡ x2 + y2 = z z ≥ 0 !

z

2

(

)2 (

)2

= z z = x 2 + y 2 = Re z + Im z !

(1.7-1)

(1.7-2)

is a real number, in which case the bars indicate the absolute value of z . From equation (1.7-2) we know that z z is real and non-

1.7.1! DISTANCE IN THE COMPLEX PLANE !

Since the set of complex numbers ! can be represented as

(

points on a Euclidean plane ! 2 , the Euclidean metric d z1 , z2

)

that is the distance between the points z1 and z2 exists in the complex plane, making it a metric space. From equation (1.7-1)

( )

the distance d 0, z from the origin of a rectangular coordinate

magnitude of the radius vector to the point

z = x + i y. !

nonnegative real number. Moreover the modulus z

of a

complex number z = x + i y is determined uniquely given x and

y. From equation (1.7-1) it follows that z = 0 if and only if

( x, y )

where

The modulus of the difference z2 − z1 of two complex

(

)

numbers z1 and z2 is equal to the distance d z1 , z2 between the two points z1 and z2 in the complex plane: !

(

)

d z1 , z2 =

( x2 − x1 )2 + ( y2 − y1 )2

negative. The modulus z of a complex number z is always a

x = 0 and y = 0 , and so:

(1.7-3)

be negative. In vector analysis the modulus z is known as the

understood to stand for the modulus of the number z unless z

!

z = 0!

system to the point z .(see Figure 1.7-1). This distance can never

The vertical bars z enclosing a number z should always be

!

if and only if!

and the Pythagorean Theorem we see that the modulus z is

where we have used equation (1.3-19). We also have: !

z = 0!

!

=

!

= z2 − z1

( x1 − x2 )2 + ( y1 − y2 )2

= z1 − z2 ! (1.7-4)

We can use the modulus to define regions in the complex plane. For example if we will let: !

z − z0 = r !

(1.7-5) 22

and ! !

( )2

z = + x2 + y2 = + x2 + − y

= z !

(1.7-9)

For the modulus of a complex conjugate, we have from

equations (1.7-8), (1.7-9), and (1.7-1): !

z = − z = z = + z z = + x2 + y2 ≥ 0 !

(1.7-10)

and

Figure 1.7-1!

2

2

Modulus z = + x + y .

!

2

2

(

r = z − z0 = x − x0

) + ( y − y0 ) 2

2

(1.7-6)

A circle with radius r centered at the origin is given by: !

r = z = x2 + y2 !

(1.7-7)

z1 − z2 = z2 − z1 !

!

2

z =+ x +y =+

( −x ) + ( − y ) 2

2

= −z !

(1.7-12)

As shown in Example 1.7-2, we also have:

z1 z2 = z1 z2 !

(1.7-13)

and by repeated application of equation (1.7-13): !

z1 z2 ! z n = z1 z2 ! zn !

(1.7-14)

From equation (1.7-14) we see that any product of complex number factors in the product equals zero. !

We have: 2

(1.7-11)

numbers can only equal zero if one or more of the complex

1.7.2! PROPERTIES OF THE MODULUS OF A COMPLEX NUMBER !

= zz = zz!

!

! !

2

= z

z

! We then have the circle:

2

!

! (1.7-8)

If n is an integer, from equation (1.7-14) we can write: n

zn = z !

(1.7-15) 23

We also have: !

z1 z2

=

z1 z2

z2 ≠ 0 !

!

(1.7-16)

! !

From equation (1.4-4) we have:

( z1 z2 ) ( z1 z2 ) = + ( z1 z1 )( z2 z2 )

! z1 z2 = +

From equation (1.5-1) we can write: !

( z1 z2 ) ( z1 z2 )

! z1 z2 = +

From equation (1.7-11) we then obtain:

1 z z z = = = 2! z zz z −1

z ≠ 0!

(1.7-17)

From equations (1.7-13) and (1.7-9) we can write:

z1 z2 = z1 z2 = z1 z2 = z1 z2 = z1 z2 = z1 z2 ! (1.7-18)

2

! z1 z2 = z1

2

z2

2

Since the modulus of a complex number is nonnegative, we can write: ! z1 z2 = z1 z2

Example 1.7-1 Determine z if z = 4 − 3i .

Example 1.7-3

Solution:

Show that 2

( )

! z = 4 − 3i = + 4 + − 3

2

= + 16 + 9 = + 25 = 5

z1 z2

=

z1 z2

where z2 ≠ 0 .

Solution: Example 1.7-2 Show that z1 z2 = z1 z2 . Solution:

From equation (1.7-13) we can write: ! z2

z1 z2

=

z2 z1 z2

= z1

and so:

From equation (1.7-1) we have: 24

!

z1 z2

=

z1

Solution:

z2

From equation (1.7-2) we have: !z

Example 1.7-4 2

Show that z1 + z2

(

2

= z1 + z1 z2 + z2 z1 + z2

Solution:

2

(

2

! z1 + z2

2

(

= z1 + z2

(

!z 2 = x + i y

) ( z1 + z2 )

Example 1.7-6

) ( z1 + z2 )

Determine z if z =

(

)

Therefore: ! z1 + z2

(

= z1 + z1 z2 + z2 z1 + z2

)2 = ( x + i y ) ( x + i y ) = x 2 − y 2 + i 2 x y

2

= z1 z1 + z1 z2 + z2 z1 + z2 z2

2

)

We also have:

or

2

)(

! z ≠ z2

From equation (1.4-1) we have: ! z1 + z2

).

(

= z z = x + i y x − i y = x2 + y2

Therefore:

From equation (1.7-2) we have: ! z1 + z2 = z1 + z2

2

2

2

)

5i . 3− i

Solution: Obtaining a real denominator:

( ) ( )

−5 + i 15 5i 3+ i 1 3 !z = = =− +i 3− i 3+ i 9 +1 2 2 and so:

Example 1.7-5 Does z

2

= z2 ?

⎛ 1 3⎞ ⎛ 1 3⎞ 1 9 + 10 !z = + z z = + ⎜− + i ⎟ ⎜− − i ⎟ = + + = 2⎠ ⎝ 2 2⎠ 4 4 2 ⎝ 2

25

Example 1.7-7

1 Show that z = 1 if and only if = z . z

(

! x − x0

)2 + ( y − y0 )2 = 1

Therefore z − z0 = 1 describes all points on a unit circle

Solution:

(

)

centered at point z0 : that is, a circle with center x0 , y0 and

We have:

radius 1 . We see that a unit circle centered at the origin is

1 1z z ! = = z z z z Therefore

or

1 = z only if z = 1 . z

Conversely, if z = 1 , then: 2

!z =1= z z

1 and so = z . z

then given by z = 1 .

Example 1.7-9

( )

Show that Re z , and Re 1 z have the same sign for all z . Solution: We have:

1 1z z z ! = = = 2 z z z zz z

( )

Example 1.7-8

Since Re z = Re z , we see that the Re z , and Re 1 z have the

Describe the set of points determined by z − z0 = 1.

same sign for all z .

Solution:

Example 1.7-10

We have: ! z − z0 =

( x − x0 )2 + ( y − y0 )2 = 1

Show that if z1 z2 = 0 and z2 ≠ 0 , then z1 = 0 .

26

Solution:

2

2

! z1 + z2 + z1 − z2 = 2

Dividing z1 z2 = 0 by z2 : 2

z1 z2 z1 z2 z2 z1 z2 ! = = = z1 = 0 2 z2 z2 z2 z2

(

2

z1 + z2

2

)

Example 1.7-12

(

)2 (

)2

2

2

Show that z1 z2 + z1 z2 − z1 z2 − z1 z2 = 4 z1 z2 . Solution:

Example 1.7-11 2

2

(

2

Show that z1 + z2 + z1 − z2 = 2 z1 + z2

2

).

We have:

(

! z1 z2 + z1 z2

Solution:

)2 = z12 z22 + z12 z22 + 2 z1 z2 z1 z2

Similarly, we have:

(

We can write: 2

(

)(

! z1 + z2 = z1 + z2 z1 + z2

! z1 z2 − z1 z2

)

and so:

(

From equation (1.4-1) we then have: 2

(

)(

)

)(

)2

or from equation (1.7-2):

(

Similarly, we have:

(

)2 (

! z1 z2 + z1 z2 − z1 z2 − z1 z2 = 4 z1 z2 z1 z2

! z1 + z2 = z1 + z2 z1 + z2 = z1 z1 + z2 z2 + z1 z2 + z1 z2

2

)2 = z12 z22 + z12 z22 − 2 z1 z2 z1 z2

)2 (

)2

2

! z1 z2 + z1 z2 − z1 z2 − z1 z2 = 4 z1 z2

)

2

! z1 − z2 = z1 − z2 z1 − z2 = z1 z1 + z2 z2 − z1 z2 − z1 z2

1.8!

Adding these two equations: 2

2

(

! z1 + z2 + z1 − z2 = 2 z1 z1 + z2 z2 Using equation (1.7-11) we obtain:

)

!

MODULI INEQUALITIES Real numbers are one-dimensional and can be ordered

since they can be arranged according to their value on a number line. Given x1 , x2 ∈! , then either x1 > x2 , x1 = x2 , or 27

x1 < x2 . All real numbers satisfy these conditions and are said to be trichotomous. !

Complex numbers are two-dimensional and have no

natural order since they cannot be arranged on a number line,

− z ≤ Re z ≤ z !

!

− z ≤ Im z ≤ z !

(1.8-3)

We also have:

z =

!

but must be represented on a plane. Complex numbers can be neither greater than nor less than other complex numbers (see

or

Proposition 1.8-1). The set of real numbers ! differs then from

!

( Re z )2 + ( Im z )2 ≤ ( Re z + Im z )

2

z ≤ Re z + Im z !

!

(1.8-4)

(1.8-5)

the set of complex numbers ! in that real numbers can be ordered, while complex numbers cannot.

Example 1.8-1

!

Show that z ≤ Re z + Im z ≤ 2 z .

Inequalities for complex numbers apply only to the

moduli of complex numbers (which are real numbers); not to the complex numbers themselves. Given z1, z2 ∈! , statements

Solution:

such as z1 > z2 have no meaning, but statements such as

From equation (1.8-1) we have:

z1 > z2 do have meaning. Therefore any inequality statement

2

such as M > 0 where M is not a modulus implies that M must be a real number. Statements such as Re z1 > Re z2

and

Im z1 > Im z2 are meaningful since the real and imaginary parts

!

Rewriting equation (1.7-2) as 2

(

)2 (

)2

2

2

z = Re z + Im z = Re z + Im z !

z ≥ Re z ≥ Re z !

)2

2

2

! Re z + Im z

( (

)

2

= Re z + Im z + 2 Re z Im z

)

2

= z + 2 Re z Im z

2

2

or (1.8-1)

we see that: !

)2 (

Since:

of a complex number are both real numbers. !

(

! z = Re z + Im z = Re z + Im z

! Re z + Im z

2

we have equation (1.8-5):

z ≥ Im z ≥ Im z !

(1.8-2)

! z ≤ Re z + Im z 28

We can also write:

(

! Re z − Im z

)

2

1.! One and only one of the three relations: z > 0 , z = 0 , or 2

− z > 0 will be valid for every z ∈! .

2

= Re z + Im z − 2 Re z Im z ≥ 0

2.! For z > 0 and w > 0 where z, w ∈! , we will have

or 2

z w > 0.

2

! Re z + Im z ≥ 2 Re z Im z

3.! For z > 0 and w > 0 where z, w ∈! , we will have

and so:

(

! Re z + Im z

z + w > 0.

)

2

2

2

= Re z + Im z + 2 Re z Im z

!

(

2

≤ 2 Re z + Im z

)

2

!

=2 z

2

or ! Re z + Im z ≤ 2 z Therefore: ! z ≤ Re z + Im z ≤ 2 z

Proposition 1.8-1, Complex Numbers Cannot be Ordered: The set of complex numbers ! cannot be ordered.

( i ) ( i ) = i2 > 0 . This is not true, however, since i2 = −1 . !

!

If the set of complex numbers ! could be ordered, we

would expect the following axioms to be followed:

If we assume that i < 0 , then we should still have

( i ) ( i ) = i2 > 0 . This also is not true, however, since i2 = −1 . !

Finally, since 1 > 0 we should have i 2 + 1 > 0 , but this is not

true since i 2 + 1 = 0 . !

Therefore i does not conform to any ordering process, and

so complex numbers cannot be ordered. !

Note that the symbol

■

■

signifies the end of a proof in this

book. !

Proof:

Since i ≠ 0 , if we assume that i > 0 , then we should have

It is the imaginary unit i that requires complex numbers to

be two-dimensional, and so prohibits them from being ordered on a number line. Complex numbers should never be referred to as either positive or negative. 29

and

1.8.1!

TRIANGLE INEQUALITY FOR MODULI

Proposition 1.8-2, Triangle Inequality:

triangle inequality as defined by:

z1 + z2 ≤ z1 + z2 !

(1.8-6)

!

)(

(

)

z1 + z2 = z1 + z2 z1 + z2 !

(1.8-7)

!

(

)(

)

z1 + z2 = z1 + z2 z1 + z2 !

(1.8-8)

(1.8-13)

2

2

2

z1 + z2 ≤ z1 + 2 z1 z2 + z2 !

(1.8-14)

2

(

z1 + z2 ≤ z1 + z2

!

)! 2

(1.8-15)

z1 + z2 ≤ z1 + z2 !

!

(1.8-16)

■ 2

z1 + z2 = z1 z1 + z1 z2 + z1 z2 + z2 z2 !

(1.8-9)

2

2

(

)

2

z1 + z2 = z1 + 2 Re z1 z2 + z2 !

2

2

(

!

By mathematical induction we can obtain the generalized

triangle inequality: (1.8-10)

)

! !

From equation (1.8-2) we can write: !

2

have the triangle inequality:

Using equation (1.4-13) we have: !

2

From equation (1.7-9) we obtain:

or !

2

Since z1 + z2 ≥ 0 and z1 + z2 ≥ 0 , from equation (1.8-15) we

From equation (1.4-1) we have: 2

(1.8-12)

and so:

We can write: 2

2

z1 + z2 ≤ z1 + 2 z1 z2 + z2 !

!

!

Proof: !

2

Using equation (1.7-13) we can write:

Moduli of any two complex numbers z1 and z2 obey the !

2

z1 + z2 ≤ z1 + 2 z1 z2 + z2 !

!

2

z1 + z2 ≤ z1 + 2 Re z1 z2 + z2 !

z1 + z2 + z3 + ! + zn ≤ z1 + z2 + z3 + ! + zn ! (1.8-17) The triangle inequality is important in the proof of many

propositions in complex analysis. To interpret the triangle (1.8-11)

inequality geometrically, we will consider the triangle shown in 30

red in Figure 1.8-1. The lengths of the three sides of this triangle are z1 , z2 and z3 = z1 + z2 . We can see that we will always

Proposition 1.8-3:

have z1 + z2 ≤ z1 + z2 . That is, the sum of the lengths of any

Moduli of any two complex numbers z1 and z2 obey the

two sides of a triangle will always be equal to or greater than

inequality defined by:

the length of the third side. This is the rational for calling equation (1.8-16) the triangle inequality.

z1 − z2 ≤ z1 + z2 !

!

(1.8-18)

Proof: Replacing z2 with − z 2 in the triangle inequality for

!

complex number moduli, we have:

( )

z1 + −z2 = z1 − z2 ≤ z1 + − z2 !

!

(1.8-19)

or from equation (1.7-8):

z1 − z2 ≤ z1 + z2 !

!

(1.8-20)

■

Proposition 1.8-4: Moduli of any two nonzero complex numbers z1 and z2 obey the law:

z1 + z2 = z1 + z2 !

! Figure 1.8-1!

Triangle inequality z1 + z2 ≤ z1 + z2 can be visualized using the red lines.

(1.8-21)

if and only if z1 = c z2 where c > 0 . Proof: !

If z1 = c z2 where c > 0 , then since c ∈! we can write: 31

(

)

(

)

z1 + z2 = c z2 + z2 = c + 1 z2 = c + 1 z2

! !

= c z2 + z2 = c z2 + z2 = z1 + z2 !

!

Conversely, if z1 + z2 = z1 + z2 , we have using equation

(1.8-22)

(1.7-13):

(

z1 z2 =

!

)

2

2

2

2

(

+ z2

1

(

z1 z2

!

)= 2

2

2

z1 + 2 z1 z2 + z2 !

(1.8-24)

)

+ z2

2

z2

2

2

2

(

2

(1.8-28)

)

(1.8-29)

)

2

− 2 Re z1 z2 − z2 ! (1.8-25)

(z

1

+ z2

)

2

2

(

!

)

(1.8-26)

)

We also have by definition:

z2 z2

=

z2

> 0!

(1.8-30)

where c > 0 and c ∈! .!

(1.8-31)

OTHER MODULI INEQUALITIES

We can derive other inequalities for moduli of complex

Proposition 1.8-5, Reverse Triangle Inequality: Moduli of any two complex numbers z1 and z2 obey the reverse

Since z1 + z2 = z1 + z2 we have from equation (1.8-26):

(

=

z1

numbers by using the triangle inequality.

− z1 + z2 = 2 z1 z2 − 2 Re z1 z2 !

z1 z2 = Re z1 z2 !

2

z1 z2

■

1.8.2!

or

!

)

)

z1 = c z2 !

!

− z1 + z2 = z1 + 2 z1 z2 + z2 − z1

!

!

(

and so we must have:

Using equations (1.8-24) and (1.8-10) we can now write: 1

)

z1 z2 = Re z1 z2 = z1 z2 > 0 !

!

2

= z1 + 2 z1 z2 + z2 = z1 + 2 z1 z2 + z2 ! (1.8-23)

(z (z

(

and so we see that Im z1 z2 = 0 , and we then have:

or from equation (1.7-18):

!

)

2

Therefore using equation (1.7-11):

! z1 + z2

!

(

2

⎡⎣ Re z1 z2 ⎤⎦ + ⎡⎣ Im z1 z2 ⎤⎦ = Re z1 z2 !

(1.8-27)

triangle inequality as defined by: !

z1 − z2 ≥ z1 − z2 !

(1.8-32)

Proof: 32

!

We can write the triangle inequality for complex number Proposition 1.8-6:

moduli given in equation (1.8-16) in the form:

( z1 − z2 ) + z2

!

≤ z1 − z2 + z2 !

triangle inequality as defined by:

or

z1 ≤ z1 − z2 + z2 !

!

(1.8-34)

!

z1 − z2 ≥ z1 − z2 !

(1.8-35)

We can also write the triangle inequality for complex

number moduli given in equation (1.8-16) in the form:

(

)

z1 + z2 − z1 ≤ z1 + z1 − z2 !

!

(1.8-36)

z2 ≤ z1 + z1 − z2 !

!

(1.8-37)

z1 − z2 ≥ z2 − z1 !

!

(1.8-40)

We write the triangle inequality for complex number

moduli given in equation (1.8-16) in the form:

( z1 + z2 ) + ( − z2 ) ≤

!

(1.8-38)

From equations (1.8-35) and (1.8-38) we see that we must

have:

z1 − z2 ≥ z1 − z2 !

which is known as the reverse triangle inequality.

(1.8-39) ■

z1 + z2 + − z2 !

(1.8-41)

or

z1 ≤ z1 + z2 + z2 !

(1.8-42)

z1 + z2 ≥ z1 − z2 !

(1.8-43)

and so: !

and so:

!

!

!

or

!

z1 + z2 ≥ z1 − z2 !

! Proof:

and so: !

Moduli of any two complex numbers z1 and z2 obey the reverse

(1.8-33)

!

We can also write the triangle inequality for complex

number moduli given in equation (1.8-16) in the form: !

( z1 + z2 ) + ( − z1 ) ≤

z1 + z2 + − z1 !

(1.8-44)

or !

z2 ≤ z1 + z2 + z1 !

(1.8-45) 33

and so:

z1 + z2 ≥ z2 − z1 !

! !

(

(1.8-46)

(

have: (1.8-47)

!

Summarizing equations (1.8-6) and (1.8-18):

z1 ± z2 ≤ z1 + z2 !

!

z1 ± z2 ≥ z1 − z2 !

(1.8-49)

and so from equations (1.8-48) and (1.8-49): !

(1.8-50)

(

)2 = 4 ( Re ( z1 z2 ))

(

)

2

( (

2

= 4 i 2 Im z1 z2

≥0

))

2

( (

= − 4 Im z1 z2

))

2

≤0

SQUARE ROOTS OF COMPLEX NUMBERS All complex numbers have square roots. Moreover, it is

possible to calculate the square root of any complex number using the algebraic properties of complex numbers that have already been discussed in previous sections.

Example 1.8-2 Show that z1 z2 + z1 z2

2i

(

1.9! !

z1 − z2 ≤ z1 ± z2 ≤ z1 + z2 !

z1 z2 − z1 z2

! z1 z2 + z1 z2 ! z1 z2 − z1 z2

(1.8-48)

Summarizing equations (1.8-32) and (1.8-40):

2

and so:

■

!

)

!Im z1 z2 =

z1 + z2 ≥ z1 − z2 !

z1 z2 + z1 z2

Similarly, we have:

From equations (1.8-43) and (1.8-46) we see that we must

!

)

!Re z1 z2 =

)

2

(

≥ 0 and z1 z2 − z1 z2

)

2

≤ 0.

!

We will let z = x + i y be an arbitrary complex number and

w = u + i v be given by:

Solution:

!

We see that z1 z2 is the complex conjugate of z1 z2 . From

so that:

equation (1.4-12) we then have:

!

(

w = u + i v = z1 2 = x + i y

(

w2 = u + i v

)1 2 !

)2 = z = x + i y !

(1.9-1)

(1.9-2) 34

or !

2

u − v + 2 iu v = x + i y !

u2 − v 2 = x !

(1.9-4)

Since u 2 and v 2 each have two values, together they

suggest that there exist four solutions for w = z 1 2 . But there can

2 uv = y!

(1.9-5)

apply to the signs of u and v . Specifically, the signs of u and v must be chosen to satisfy equation (1.9-5). Taking u as positive

(

)

x 2 + y 2 = u2 − v 2

2

(

)

we can make v positive or negative according to the sign of y . 2

+ 4 u2 v 2 = u2 + v 2 !

(1.9-6)

From equations (1.9-1), (1.9-9), and (1.9-10) we can write

w = u + i v = z 1 2 as:

2

2

2

2

u +v = x + y !

(1.9-7)

!w = z

1 2

Using equation (1.9-4) with equation (1.9-7), we have: 2

2

2

2

u +u − x = x + y !

(1.9-8)

2

u =

x + x2 + y2 2

!

and similarly we can obtain:

(1.9-9)

⎡ x + x2 + y2 ⎢ =± + i sgn y ⎢ 2 ⎣

( )

⎤ − x + x2 + y2 ⎥ ! (1.9-11) ⎥ 2 ⎦

where !

or !

(1.9-10)

means that u and v are both real numbers.

or

!

!

only be two roots for w = z 1 2 , and so some constraint must

We then can write:

!

2

Both u 2 and v 2 will be nonnegative since ± x ≤ x 2 + y 2 . This !

and

!

− x + x2 + y2

(1.9-3)

equation:

!

v =

!

2

Therefore, equating the real and imaginary parts of this

!

2

⎧⎪ 1 if y ≥ 0 ! sgn y = ⎨ −1 if y < 0 ⎪⎩

( )

(1.9-12)

Equation (1.9-11) can also be written as: !

w= z

1 2

⎡ =±⎢ ⎢ ⎣

z +x 2

( )

+ i sgn y

z −x ⎤ ⎥! 2 ⎥ ⎦

(1.9-13) 35

Example 1.9-1

!w = z

Determine the square roots of z = 3+ 4i .

( )

⎡ 1 !w = ± ⎢ +i 2 ⎢⎣

From equation (1.9-11) we have: !w = z

⎡ 0 + 0 +1 =±⎢ + i sgn +1 2 ⎢ ⎣

0 + 0 +1 ⎤ ⎥ 2 ⎥ ⎦

or

Solution:

1 2

1 2

⎡ 3+ 9 + 16 =±⎢ + i sgn + 4 2 ⎢ ⎣

( )

− 3+ 9 + 16 ⎤ ⎥ 2 ⎥ ⎦

or

The Cartesian coordinate system can be transformed into a

x = r cosθ !

!

y = r sin θ !

(1.10-1)

where r ≥ 0 and θ is in radians (see Figure 1.10-1). We then

(

)

!w 2 = ⎡⎣ 2 + i ⎤⎦ ⎡⎣ 2 + i ⎤⎦ = 4 − 1 + 4i = 3+ 4i = z

)(

⎤ i⎥ ⎦

polar coordinate system using the transformation equations:

2 ⎤ ⎥ = ± ⎡⎣ 2 + i ⎤⎦ 2 ⎥⎦

Checking this result:

(

⎤ ⎡ 1 1 + ⎥=±⎢ ⎣ 2 ⎥⎦ 2

1.10! POLAR FORM OF COMPLEX NUMBERS !

⎡ 8 !w = ± ⎢ +i 2 ⎢⎣

1 2

) (

)

!w 2 = − ⎡⎣ 2 + i ⎤⎦ − ⎡⎣ 2 + i ⎤⎦ = 4 − 1 + 4i = 3+ 4i = z

have:

(

)

! x 2 + y 2 = r 2 cos 2θ + r 2 sin 2θ = r 2 cos 2θ + sin 2θ = r 2 ! !

(1.10-2)

Using equation (1.10-1) we can express a complex number

z in terms of the polar coordinates r and θ : Example 1.9-2

!

Determine the square roots of z = i .

or

Solution: From equation (1.9-11) we have:

!

z = x + i y = r cosθ + i r sin θ !

(

)

z = r cosθ + isin θ !

(1.10-3)

(1.10-4)

Equation (1.10-3) is an algebraic representation of a complex

( )

number z in polar form z r, θ . 36

In vector analysis r is the magnitude of the radius vector

We see from this equation that r is equal to the modulus or

to the point z = x + i y , and the x and y components of the

magnitude of the complex number z . The modulus r is

vector given in equation (1.10-1) are the orthogonal projections

measured from the coordinate system origin to the point z (see

of the radius vector onto the real and imaginary axes,

Figure 1.10-1). Note that r is unique, and is never a negative

!

respectively (see Section 1.6).

number. A complex number z equals zero if and only if its modulus equals zero: r = x 2 + y 2 = z = 0 . We can see this from equation (1.10-4) since cosθ and sin θ are never zero for the same value of θ . !

The angle θ in equation (1.10-4) is called the argument or

phase of z and is written as:

θ = arg ( z ) !

!

(1.10-6)

The angle θ is measured in radians from the positive x-axis to the line segment joining the origin and the point z , with the positive direction of rotation being in the counterclockwise direction (see Figures 1.10-1 and 1.10-2). Unlike the modulus r

()

of a complex number, the argument arg z Figure 1.10-1!

Geometric representation of a complex number z = x + i y in polar form, where r = z .

of a complex

number is not unique (see Section 1.10.2).

1.10.1! COMPLEX CONJUGATE IN POLAR FORM !

From equations (1.10-2) and (1.7-1) we can write r in the

form: !

!

In the polar coordinate system, the complex conjugate of

(

)

z = r cosθ + isin θ is given by: r = x2 + y2 = z = x + i y = z z !

(1.10-5)

!

(

)

z = r cosθ − i sin θ !

(1.10-7) 37

so that z and z have the same modulus r = z . We can write: !

( ( )

( ))

z = r cos − θ + i sin − θ !

(1.10-8)

From equation (1.10-8) we see that for z and z we have: !

( )

()

arg z = − arg z !

0 < arg z ≤ 2 π !

(1.10-9)

Therefore the complex conjugate z of z is a reflection of z about the real axis as shown in Figures 1.10-2 and 1.6-2.

!

(

) (

)

z z = r cosθ + i sin θ r cosθ − i sin θ !

(1.10-10)

or !

(

)

z z = r 2 cos 2θ + sin 2θ = r 2 !

(1.10-11)

Therefore: !

z z = z = r = x2 + y2 !

(1.10-12)

as given in equation (1.10-5).

1.10.2! ARGUMENT OF A COMPLEX NUMBER ! !

From equations (1.10-1) and (1.10-5) we can write:

cosθ =

x = r

x 2

x +y

2

!

sin θ =

y = r

y 2

x +y

2

! (1.10-13)

and from equations (1.10-13) and (1.10-12) we obtain: !

cosθ =

x Re z ! = r z

sin θ =

y Im z ! = r z

(1.10-14)

We then have: ! Figure 1.10-2!

!

Geometric representation of a complex number z and its complex conjugate z in polar form.

From equations (1.10-4) and (1.10-7) we have:

tan θ =

sin θ y Im z ! = = cosθ x Re z

(1.10-15)

and so the argument θ is given by: !

y θ = arg ( z ) = tan −1 ! x

(1.10-16) 38

!

()

If z = 0 = 0 + i0 , then θ = arg z is undefined since equation

(1.10-16) becomes: !

θ = arg ( z ) = tan

−1 0

0

!

(1.10-17)

There is an ambiguity in the determination of the angle θ

using equation (1.10-16) since: !

θ = tan

−1

(

)

Express in polar form z = r cosθ + isin θ :

Any real number can then be used as a valid argument. !

Example 1.10-1

y −y ! = tan −1 x −x

(1.10-18)

The correct value of the angle θ can be found using the signs of

1.! i 2.! 1+ i 3.! − i 4.! 1+ i 3 5.! −1+ i

both the real part x and the imaginary part y of the complex

Solution:

number to determine in which quadrant z is positioned (see

1.! For z = i we have:

Table 1.10-1)

r = z = x 2 + y 2 = 0 + 12 = 1

!z = x + i y = 0 + i !

θ = tan −1

!

(

)

!z = r cosθ + isin θ = cos

y 1 π = tan −1 = x 0 2

π π + i sin = 0 + i 2 2

2.! For z = 1+ i we have: Table 1.10-1! Quadrants of z .

!z = x + i y = 1+ i ! !

r = z = x 2 + y 2 = 12 + 12 = 2

θ = tan −1

y 1 π = tan −1 = x 1 4 39

⎛ π π⎞ !z = r cosθ + isin θ = 2 ⎜ cos + i sin ⎟ 4 4⎠ ⎝

(

)

⎛ 1 1 ⎞ = 2⎜ +i ⎟⎠ = 1+ i ⎝ 2 2

!

!θ = tan −1

( )2

r = z = x 2 + y 2 = 0 + −1 = 1

θ = tan −1

!

y −1 3 π = tan −1 = x 0 2

3π 3π !z = r cosθ + isin θ = cos + i sin = 0−i 2 2

(

)

For z = 1+ i 3 we have: !z = x + i y = 1+ i 3 !

r = z = x 2 + y 2 = 12 + 3 = 2 y 3 π = tan −1 = x 1 3

⎛ π π⎞ !z = r cosθ + isin θ = 2 ⎜ cos + i sin ⎟ 3 3⎠ ⎝

(

)

! 5.!

For z = −1+ i we have: !z = x + i y = −1+ i

( )

⎛ 2 2⎞ = 2⎜− +i ⎟ = −1+ i 2 2 ⎝ ⎠

!

Since cosθ and sin θ are periodic functions with period

2π :

(

)

(

sin θ = sin θ + 2 π !

)

cosθ = cos θ + 2 π !

(1.10-19)

a complex number z in polar form can be written as:

θ = tan −1

!

y 1 3π = tan −1 = tan −1 −1 = x −1 4

!

! 4.!

2

⎛ 3π 3π ⎞ !z = r cosθ + i r sin θ = 2 ⎜ cos + i sin ⎟ 4 4 ⎠ ⎝

3.! For z = − i we have: !z = x + i y = 0 − i !

( −1)2 + ( +1)2 =

!r = z = x 2 + y 2 =

⎛1 3⎞ = 2 ⎜ + i ⎟ = 1+ i 3 2 ⎠ ⎝2

!

(

)

(

)

z = r cos θ + 2 k π + i r sin θ + 2 k π !

(1.10-20)

where k is any integer including zero. The argument of a complex number z is not unique then, but is determined up to an integer multiple of ± 2 π : ! !

()

arg z = θ + 2 k π !

k = 0, ± 1, ± 2, ! !

(1.10-21)

()

Therefore arg z is a multivalued function, and a complex

number z in polar form can have an infinity of possible arguments that differ from each other by some multiple of 2 π . Equation (1.10-4) then becomes: 40

( (

)

(

z = r cos θ + 2 k π + i sin θ + 2 k π

!

)) !

k = 0, ± 1, ± 2, !

! !

(1.10-22)

Example 1.10-3 Determine z for the following arguments:

Example 1.10-2

1.! θ = 0

Show that two complex numbers z1 and z2 are equal if and

2.! θ =

only if their moduli are equal and their arguments differ by

π 2

some multiple of 2 π .

3.! θ = π

Solution:

4.! θ =

3π 2

Let: !z1 = r1 cosθ1 + i r1 sin θ1 !

z2 = r2 cosθ 2 + i r2 sin θ 2

If z1 = z2 and θ1 = θ 2 + 2 k z where k is an integer, then: !Re z1 = Re z2 !

In polar form we have: z = r cosθ + i r sin θ . 1.! For θ = 0 we have:

Im z1 = Im z2

!z = r cos0 + i r sin0 = r

and so: !z1 = z2

2.! For θ =

Conversely, if z1 = z2 , then

z1 = z2

and so r1 = r2 = r .

Equating real parts of z1 and z2 , and equating imaginary parts of z1 and z2 , we have: !r cosθ1 = r cosθ 2 !

Solution:

r sin θ1 = r sin θ 2

π we have: 2

!z = r cos

π π + i r sin = i r 2 2

3.! For θ = π we have: !z = r cos π + i r sin π = − r !

(see Figure 1.10-3)

Therefore θ1 = θ 2 + 2 k z . 41

4.! For θ =

()

3π we have: 2

!z = r cos

!arg 1 = 0 + 2 k π = 2 k π !

k = 0, ± 1, ± 2, !

2.! If z = i then θ = π 2 . From equation (1.10-21) we have:

3π 3π + i r sin = −ir 2 2

()

!arg i =

π + 2kπ ! 2

k = 0, ± 1, ± 2, !

3.! If z = − i then θ = − π 2 . From equation (1.10-21) we have:

( )

!arg − i = −

π + 2kπ ! 2

k = 0, ± 1, ± 2, !

4.! If z = −1− i then θ = − 3π 4 . From equation (1.10-21) we have: Figure 1.10-3!

( )

Polar representation of the number z = − r .

!arg − i = −

3π + 2kπ ! 4

k = 0, ± 1, ± 2, !

Example 1.10-4 Determine arg z for the following:

1.10.3! PRINCIPAL VALUE OF ARGUMENT !

()

The ambiguity associated with arg z can be eliminated by

1.!

z =1

2.!

z=i

3.!

z = −i

the principal value of arg z or the principal argument of z ,

4.!

z = −1− i

and is written θ = Arg z . We then have:

specifying a certain range of values for θ ≤ 2 π . The value of θ that falls within the range −π < θ ≤ π is generally referred to as

Solution:

!

1.! If z = 1 then θ = 0 . From equation (1.10-21) we have:

and

()

()

−π < Arg z ≤ π !

()

!

(1.10-23)

42

()

!

()

arg z = Arg z + 2 k π !

!

k = 0, ± 1, ± 2, ! ! (1.10-24)

()

The exact range of values chosen for Arg z is arbitrary,

but must not encompass more than 2 π if ambiguity is to be eliminated. The range of angles for the principal argument of z

()

is sometimes chosen to be 0 ≤ Arg z ≤ 2 π , the same range that

()

is commonly used for polar angles. While arg z always has an

()

associated set of values, Arg z has only one unique value. We

()

()

see the that arg z is a multivalued function, while Arg z is a

!

the fourth quadrant is − π 2 < θ ≤ 0 . Therefore tan −1 y x provides

()

the principal value of arg z when z is in either the first or fourth quadrants. !

Since tan θ has a period of π , we can convert the range of

tan −1 y x to the range π 2 < θ ≤ π by adding π : !

1.10.4! DISCONTINUITY IN ARGUMENT

!

()

The function −π < Arg z ≤ π is not continuous, but has a

discontinuity on the negative real axis. Approaching the

When z is in second quadrant (which is π 2 < θ ≤ π ),

tan −1 y x will have a negative value in the range − π 2 < θ ≤ 0 .

single-valued function.

!

For a complex plane, the first quadrant is 0 < θ ≤ π 2 and

(− π

)

can convert the range of tan −1 y x to the range −π < θ ≤ −π 2 by

approaches π ; approaching the negative real axis from the

!

()

()

approaches − π . Therefore

Arg z jumps by 2 π across the negative real axis. When z is a

!

()

negative real number, by convention Arg z is taken to have the value π and not − π , which is why the range is written as

−π < θ ≤ π . !

−1

Using tan y x as in equation (1.10-16) to determine

() arg ( z ) has

Arg z , we must take into account that the principal value of the range −π < θ ≤ π , while the principal value

tan −1 y x has the range − π 2 < θ ≤ π 2 .

(1.10-25)

tan −1 y x will have a positive value in the range 0 < θ ≤ π 2 . We subtracting π :

lower half-plane ( y → 0− ), Arg z

)

When z is in third quadrant (which is −π < θ ≤ −π 2 ),

negative real axis from the upper half-plane ( y → 0+ ), Arg z

()

(

2 0 !

(1.10-27)

x < 0, y < 0

43

!

Note that if x = 0 , then y x becomes infinitely large. The

Since the complex conjugate z of z is a reflection of z

!

complex plane quadrants corresponding to the coordinates in

about the real axis, we have from equation (1.10-9):

equation (1.10-27) are shown in Figure 1.10-4.

!

( )

()

Arg z = − Arg z !

(1.10-29)

Example 1.10-5 Determine the principal argument for the following complex numbers:

Figure 1.10-4! Complex plane quadrants. !

!

()

z=3

2.!

z=

3.!

z = −1

4.!

z = −i

i 5

Solution:

()

1.! Arg 3 = 0

We also have:

⎧ π ⎪ ⎪ 2 ⎪ π Arg z = ⎨ − ⎪ 2 ⎪ π ⎪ ⎩

1.!

⎛ i⎞ π 2.! Arg ⎜ ⎟ = ⎝ 5⎠ 2

x = 0, y > 0 x = 0, y < 0 x < 0, y = 0

!

(1.10-28)

( )

3.! Arg −1 = π

( )

4.! Arg − i = −

π 2

44

Example 1.10-6 If z = 2 − i 2 3 determine:

1.10.5! MULTIPLICATION OF COMPLEX NUMBERS IN POLAR FORM

1.! r

!

() arg ( z )

two complex numbers z1 and z2 can be written:

2.! Arg z 3.!

(

()

Solution:

1.! r = x 2 + y 2 =

( 2 )2 + ( − 2

3

)

2

= 4 + 12 = 16 = 4

)

or !

(

z1 z2 = r1 r2 cosθ1 cosθ 2 − sin θ1 sin θ 2

(

)

)

+ r1 r2 i sin θ1 cosθ 2 + cosθ1 sin θ 2 !

!

From equations (1.10-5) and (1.10-16) we have:

)(

z1 z2 = r1 cosθ1 + i r1 sin θ1 r2 cosθ 2 + i r2 sin θ 2 ! (1.10-30)

!

4.! Express z in polar form using Arg z .

2.! tan θ =

Using the polar form of complex numbers, the product of

( )

(1.10-31)

( )

where θ1 = arg z1 and θ 2 = arg z2 . !

Using trigonometric identities for the sum of two angles,

we obtain:

−2 3 − 3 = =− 3 2 1

(

)

(

)

z1 z2 = r1 r2 ⎡⎣cos θ1 + θ 2 + isin θ1 + θ 2 ⎤⎦ !

!

(1.10-32)

We then have from equations (1.10-32) and (1.10-21):

and so:

(

)

( )

( )

()

!arg z1 z2 = arg z1 + arg z2 + 2 k π !

()

!

!Arg z = θ = − π 3 3.! arg z = − π 3+ 2 k π !

k = 0, ± 1, ± 2, !

()

4.! Using Arg z we can write:

⎛ π⎞ ⎛ π⎞ π π !z = 4cos ⎜ − ⎟ + 4i sin ⎜ − ⎟ = 4 cos − 4i sin 3 3 ⎝ 3⎠ ⎝ 3⎠

k = 0, ± 1, ± 2, ! ! (1.10-33)

Multiplication of two complex numbers results, therefore,

in their moduli being multiplied and their arguments being added. The product of two complex numbers is a new complex number that is scaled and rotated relative to the two factors (see Figure 1.10-5). 45

We have:

( )

!arg z1 =

3π ! 2

( )

arg z2 = π

and so: !arg ( z1 z2 ) = arg ( z1 ) + arg ( z2 ) + 2 k π =

3π 5π +π + 2kπ = + 2kπ 2 2

where k = 0, ± 1, ± 2, ! .

1.10.6! DIVISION OF COMPLEX NUMBERS IN POLAR FORM !

Using the polar form of complex numbers, the ratio of two

complex numbers z1 and z2 can be written: Figure 1.10-5!

!

Polar representation of the multiplication of two complex numbers z1 and z2 .

If a complex number is multiplied by a real number, the

product will have the same argument as the complex number

z1

!

z2

=

r1 cosθ1 + i r1 sin θ1 r2 cosθ 2 + i r2 sin θ 2

!

z2 ≠ 0 !

(1.10-34)

Multiplying the numerator and denominator by z2 : !

since the argument of all real numbers is zero.

z1 z2

=

( )( )! r2 ( cosθ 2 + i sin θ 2 ) ( cosθ 2 − i sin θ 2 ) r1 cosθ1 + i sin θ1 cosθ 2 − i sin θ 2

z2 ≠ 0 !

(1.10-35)

z2 ≠ 0 !

(1.10-36)

or Example 1.10-7

(

)

z1

=

If z1 = −5i and z2 = − 3 determine arg z1 z2 .

!

Solution:

and so:

z2

(

)(

r1 cosθ1 + i sin θ1 cosθ 2 − i sin θ 2 r2

cos θ 2 + sin θ 2 2

2

)!

46

!

z1 z2

=

r1

Solution:

(cosθ1 cosθ 2 + sinθ1 sinθ 2 )

r2

+

!

r1 r2

(

We have for 1+ i 3 :

)

i sin θ1 cosθ 2 − cosθ1 sin θ 2 !

z2 ≠ 0 !

!r = 12 + 3 = 2 !

(1.10-37)

we obtain: !

z2

=

r1 r2

⎡⎣cos (θ1 − θ 2 ) + i sin (θ1 − θ 2 ) ⎤⎦ !

z2 ≠ 0 !

!

z2

=

r1 r2

=

z1 z2

!

tan −1

3 1 π = tan −1 = 3 3 6

We have for 3+ i 3 :

(1.10-38)

!r = 32 + 3 = 12 = 2 3 !

We then have:

z1

3 π = 1 3

⎛ π π⎞ !1 + i 3 = 2 ⎜ cos + i sin ⎟ 3 3⎠ ⎝

Using trigonometric identities for the difference of two angles,

z1

tan −1

⎛ π π⎞ !3+ i 3 = 2 3 ⎜ cos + i sin ⎟ 6 6⎠ ⎝

(1.10-39)

and so from equation (1.10-38) we can write:

and

⎡π π ⎤ ⎡ π π ⎤⎞ 1+ i 3 2 ⎛ ! = ⎜⎝ cos ⎢ 3 − 6 ⎥ + i sin ⎢ 3 − 6 ⎥⎟⎠ ⎣ ⎦ ⎣ ⎦ 3+ i 3 2 3

⎛z ⎞ ! arg ⎜ 1 ⎟ = arg z1 − arg z2 + 2 k π ! k = 0, ± 1, ± 2, ! ! (1.10-40) ⎝ z2 ⎠

( )

( )

or

Division of two complex numbers results, therefore, in their

1+ i 3 1 ⎛ π π⎞ ! = cos + i sin ⎜ 6 6 ⎟⎠ 3+ i 3 3⎝

moduli being divided and their arguments being subtracted. Example 1.10-8 Determine

1+ i 3 3+ i 3

in polar form.

!

Using equations (1.10-40) and (1.10-21) with z1 = 1 and

z2 = z we can write:

47

⎛1⎞ arg ⎜ ⎟ = − arg z + 2 k π ! ⎝z⎠

()

!

k = 0, ± 1, ± 2, ! ! (1.10-41)

From equation (1.10-9) we then have:

⎛1⎞ arg ⎜ ⎟ = arg z ! ⎝z⎠

( )

!

0 < arg z ≤ 2 π !

1.10.7! PRINCIPAL ARGUMENT FOR PRODUCTS AND RATIOS !

(1.10-42)

For the product and ratio of two complex numbers z1 and

z2 we have:

(

)

( )

( )

Arg z1 z2 = Arg z1 + Arg z2 + 2 k π !

!

(1.10-43)

(

)

Example 1.10-9

where k is a unique integer such that −π < Arg z1 z2 ≤ π , and

z Show that = cos 2θ + isin 2θ . z

!

Solution:

!

We have:

z cosθ + isin θ cosθ + isin θ cosθ + isin θ ! = = z cosθ − isin θ cosθ − isin θ cosθ + isin θ or

z cos θ − sin θ + 2icosθ sin θ ! = z cos 2 θ + sin 2 θ 2

2

and so using trigonometric identities:

z ! = cos 2θ + isin 2θ z See equation (1.13-48) for another derivation of this relation.

(

)

( )

( )

Arg z1 z2 = Arg z1 − Arg z2 + 2 k π !

(1.10-44)

(

)

where k is a unique integer such that −π < Arg z1 z2 ≤ π . Note that in general:

(

)

( )

( )

( )

( )

!

Arg z1 z2 ≠ Arg z1 + Arg z2 !

!

Arg z1 z2 ≠ Arg z1 − Arg z2 !

(

)

(1.10-45) (1.10-46)

While principal arguments added together or subtracted from each other may not result in a principal value, it is possible that non-principal arguments added together or subtracted from each other may result in a principal value. Example 1.10-10

( )

( )

If z1 = −5 and z2 = i , determine Arg z1 + Arg z2 and

(

)

Arg z1 z2 . 48

!

We have:

and so:

( )

( )

( )

()

!Arg z1 = Arg −5 = π !Arg z2 = Arg i =

!

π 2

!

(

)

( )

!Arg z1 z2 = Arg −5i = −

( )

( )

!

π 2

( )

()

!Arg z1 + Arg z2 = Arg −5 + Arg i = π +

π 3π = 2 2

(cosθ + i sinθ )2 = (cos 2θ + i sin 2θ ) !

(1.10-49)

Using mathematical induction as shown in Example

)

( )

!

( )

(

z n = r n cosθ + i sin θ

unless we add − 2 π to Arg z1 + Arg z2 . In this case we

Example 1.10-11

have k = −1 in equation (1.10-43).

Show that cosθ + i sin θ

(

1.10.8! INTEGER POWERS IN POLAR FORM power can be written as:

( )( ) (

)n = r n (cos nθ + i sin nθ ) !

(1.10-51)

)n = (cos nθ + i sin nθ ) .

Solution:

In polar form a complex number z raised to the second

z 2 = z z = r cosθ + i r sin θ

(1.10-50)

z raised to an integer power n :

( ) ( )

(cosθ + i sinθ )n = (cos nθ + i sin nθ ) !

theorem (see Section 1.12). We then have for a complex number

!Arg z1 z2 ≠ Arg z1 + Arg z2

!

(1.10-48)

where n is an integer. This equation is known as de Moivre’s

We see that:

(

)

z 2 = r 2 cos 2θ + isin 2θ !

1.10-11, we obtain:

and so:

!

(

Solution:

We will use mathematical induction. We will assume that the equation:

)2 = r 2 (cosθ + isinθ )2 !

Using equation (1.10-32) we have:

(1.10-47)

(cosθ + i sinθ )k = (cos k θ + i sin k θ ) is valid for some integer k ≥ 1 . We then have: 49

(cosθ + i sinθ )k+1 = (cos k θ + i sin k θ ) (cosθ + i sinθ ) or

(cosθ + i sinθ )k+1= (cos k θ cosθ − sin k θ sinθ ) ! + i ( sin k θ cosθ + cos k θ sin θ ) and so using trigonometric identities:

(cosθ + i sinθ )k+1 = cos ( k + 1) θ + i sin ( k + 1) θ This equation is obviously valid for k = 1 as given in equation (1.10-49). Therefore it is valid for all k ∈! , and we can let n = k + 1 to obtain:

(cosθ + i sinθ )n = (cos nθ + i sin nθ )

Determine w = z 2 using z in polar form if z = 1+ i . Solution:

Example 1.10-13 Determine w = z 3 using z in polar form if z = 1+ i 3 . Solution: !z = x + i y = 1+ i 3 From equations (1.10-12) and (1.10-16) we have: 2

2

From equations (1.10-12) and (1.10-16) we have: 2

2

r = x + y = 1 +1 = 2

2

r= x +y = 1 +

( 3)

2

=2

y 3 = = 3 x 1

Therefore:

θ = tan −1 3 =

!z = x + i y = 1+ i

2

⎛ ⎛ π π⎞ π π⎞ w = z 2 = 2 ⎜ cos 2 + i sin 2 ⎟ = 2 ⎜ cos + i sin ⎟ = 2i 4 4⎠ 2 2⎠ ⎝ ⎝

!tan θ =

Example 1.10-12

2

Using equation (1.10-51) we obtain:

π 3

Using equation (1.10-51) we obtain:

⎛ π π⎞ !w = z 3 = 23 ⎜ cos3 + i sin3 ⎟ = 8 cos π + i sin π = −8 3 3⎠ ⎝

(

)

1 π θ = tan −1 = 1 4 50

Example 1.10-14 Determine w = z10 using z in polar form if z = 1+ i . Solution:

!

The Maclaurin series for the function e z where e is the

base of the natural logarithms is:

!z = x + i y = 1+ i From equations (1.10-12) and (1.10-16) we have:

r = x 2 + y 2 = 12 + 12 = 2 !tan θ =

1.11! EULER'S FORMULA

y 1 = =1 x 1

π 4

If z is a real number x , then this Maclaurin series is simply the expected series expansion for e x :

x 2 x 3 x 4 x5 x 6 x 7 e = 1+ x + + + + + + + !! 2 3! 4! 5! 6! 7! x

(1.11-2)

is also valid as will be shown in Proposition 8.8-8 and Section

( ) 2

(1.11-1)

If z is a complex number, this Maclaurin series expansion for e z

Using equation (1.10-51), we obtain: !w = z10 =

!

!

Therefore:

θ = tan −1 1 =

z 2 z 3 z 4 z5 z 6 z 7 e = 1+ z + + + + + + + ! ! 2 3! 4! 5! 6! 7! z

10 ⎛

π π⎞ cos10 + i sin10 ⎜⎝ 4 4 ⎟⎠

8.9. For now we will assume the validity of equation (1.11-1) when z is a complex number. If z is pure imaginary z = iθ , then equation (1.11-1) becomes: ! eiθ = 1+ iθ +

or

⎛ ⎛ ⎛ π⎞ π ⎞⎞ !w = 25 ⎜ cos ⎜ 2π + ⎟ + i sin ⎜ 2π + ⎟ ⎟ 2⎠ 2⎠⎠ ⎝ ⎝ ⎝ and so:

⎛ ⎛π⎞ ⎛ π ⎞⎞ !w = 32 ⎜ cos ⎜ ⎟ + i sin ⎜ ⎟ ⎟ = 32i ⎝ 2⎠ ⎝ 2⎠⎠ ⎝

( ) +( ) +( ) +( ) +( ) iθ

2

2

iθ

3!

3

iθ

4!

4

iθ

5!

5

iθ

6!

6

+! !

(1.11-3)

where θ is a real number. We therefore have: !

θ2 θ3 θ4 θ5 θ6 θ7 e = 1+ iθ − − i + + i − − i + ! ! 2 3! 4! 5! 6! 7! iθ

(1.11-4)

and so: 51

⎛ θ2 θ4 θ6 ⎞ ⎛ ⎞ θ3 θ5 θ7 ! e = ⎜ 1− + − +!⎟ + i ⎜ θ − + − +!⎟ ! (1.11-5) 2 4! 6! 3! 5! 7! ⎝ ⎠ ⎝ ⎠ iθ

!

The two bracket terms are just the Maclaurin series for the

real cosine and sine functions of a real variable θ . We then

derivative, we find that Euler's formula is consistent with this definition (see Section 4.2). !

The moduli of eiθ and e−iθ are given by:

!

eiθ = cosθ + i sin θ = cos 2θ + sin 2θ = 1 !

(1.11-9)

!

e−iθ = cosθ − i sin θ = cos 2θ + sin 2θ = 1!

(1.11-10)

obtain the very elegant and important Euler's formula: !

iθ

e = cosθ + i sin θ !

(1.11-6)

!

where θ is in radians and where: !

( )!

cosθ = Re e

iθ

of complex exponential functions. As shown in Example 1.11-1,

( )!

sin θ = Im e

iθ

(1.11-7)

We also have: ! !

( )

we have: !

( )

e−iθ = cos −θ + i sin −θ = cosθ − i sin θ = e iθ !

(1.11-8)

iθ

Euler’s formula reveals that the functions e , cosθ , and

sin θ are all related. More specifically, the complex exponential function of a pure imaginary number can be expressed in terms of real cosine and sine functions. !

Real cosine and sine functions can be expressed in terms

The derivation of Euler's formula given above cannot be

considered a proof since we have not yet demonstrated that the Maclaurin series converges for imaginary exponents. This will be done in Chapter 8 (see Examples 8.9-1 and 8.3-3). Euler's

eiθ + e−iθ ! cosθ = 2

eiθ − e−iθ ! sin θ = 2i

(1.11-11)

We can write: !

sin θ 1 eiθ − e−iθ ! tan θ = = iθ −iθ cosθ i e + e

(1.11-12)

Example 1.11-1

eiθ + e−iθ eiθ − e−iθ Show that cosθ = and sin θ = where θ is a 2 2i real number.

formula is often simply taken as the definition of eiθ . If we

Solution:

define the exponential function as a function that is its own

Using Euler's formula, we have: 52

( )

( )

!eiθ + e−iθ = cosθ + i sin θ + cos − θ + i sin − θ or iθ

!e + e

−iθ

= cosθ + i sin θ + cosθ − i sin θ = 2cosθ

and so:

eiθ + e−iθ !cosθ = 2

θ eiθ 2 + e−iθ !cos = 2 2

2

or squaring: 2

⎛ θ⎞ 1 1 !⎜ cos ⎟ = ⎡ eiθ + e−iθ + 2eiθ 2 e−iθ 2 ⎤ = ⎡ eiθ + e−iθ + 2 ⎤ ⎦ 4⎣ ⎦ 2⎠ 4⎣ ⎝ and so: !cos 2

Similarly:

( )

( )

!eiθ − e−iθ = cosθ + i sin θ − ⎡⎣cos − θ + i sin − θ ⎤⎦ or !eiθ − e−iθ = cosθ + i sin θ − cosθ + i sin θ = 2 i sin θ and so:

θ 1+ cosθ = 2 2

Similarly:

θ eiθ 2 − e−iθ !sin = 2 2i

2

or squaring: 2

eiθ − e−iθ !sin θ = 2i

⎛ θ⎞ 1 1 !⎜ sin ⎟ = − ⎡⎣ eiθ + e−iθ − 2eiθ 2 e−iθ 2 ⎤⎦ = − ⎡⎣ eiθ + e−iθ − 2 ⎤⎦ 4 4 ⎝ 2⎠ and so from equation (1.11-11):

Example 1.11-2 Show that cos 2

θ 1+ cosθ θ 1− cosθ and sin 2 = . = 2 2 2 2

!sin 2

θ 1− cosθ = 2 2

Example 1.11-3 Solution: Using equation (1.11-11) we can write:

3 1 Show that sin 3 θ = sin θ − sin3θ . 4 4 53

Solution:

1.12!

Using equation (1.11-11) we can write:

⎡ eiθ − e−iθ ⎤ !sin θ = ⎢ ⎥ 2i ⎣ ⎦

!

3

induction. We can also derive this theorem using Euler's formula.

Using the binomial theorem

( ) ( )(

1 ⎡ iθ e 8i ⎢⎣

3

+ 3 eiθ

Using the polar form of complex numbers, de Moivre’s

theorem was derived in Example 1.10-11 using mathematical

3

!sin 3 θ = −

DE MOIVRE'S THEOREM

2

) ( )(

− e−iθ + 3 eiθ

− e−iθ

) ( 2

)

3⎤ + − e−iθ ⎥ ⎦

Proposition 1.12-1, de Moivre’s Theorem: de Moivre’s theorem states that:

We then have:

1 !sin θ = − ⎡ e3 iθ − 3e2 iθ e−iθ + 3eiθ e−2 iθ − e−3 iθ ⎤ ⎦ 8i ⎣

(cosθ + i sinθ )n = (cos nθ + i sin nθ ) !

!

3

or

(1.12-1)

where n is an integer and θ is a real variable. Proof:

!sin 3 θ = −

1 ⎡ 3 iθ iθ −iθ −3 iθ ⎤ e − 3e + 3e − e ⎦ 8i ⎣

and so:

3 ⎛ eiθ − e−iθ ⎞ 1 ⎛ e3 iθ − e−3 iθ ⎞ !sin θ = ⎜ − ⎟ 4⎝ 2 i ⎟⎠ 4 ⎜⎝ 2i ⎠ 3

Therefore from equation (1.11-11):

3 1 !sin 3 θ = sin θ − sin3θ 4 4

!

Equation (1.12-1) follows from Euler's formula given in

equation (1.11-6):

eiθ = cosθ + i sin θ !

!

(1.12-2)

since

(cosθ + i sinθ ) = ( e ) n

!

iθ

n

i nθ = e ( ) = cos nθ + i sin nθ ! (1.12-3)

■

!

de Moivre’s theorem is also valid where n is a negative

integer (see Example 1.12-4). 54

Expanding this equation using the binormal theorem:

Example 1.12-1

!cos3θ + i sin3θ = cos3θ + 3icos 2θ sinθ − 3 cos θ sin 2θ − i sin 3θ

Determine cos 2θ and sin 2θ as a function of powers of cosθ

Equating the real and imaginary parts of this equation

and sin θ .

separately, we have:

Solution:

!cos3θ = cos3θ − 3 cos θ sin 2θ !

Using de Moivre's theorem, we can write:

(

!cos 2θ + i sin 2θ = cosθ + i sin θ

)

sin3θ = 3cos 2θ sinθ − sin 3θ

We also have:

2

(

)

!cos3θ = cos3θ − 3 cos θ 1− cos 2θ = 4cos3θ − 3cos θ

or

(

!cos 2θ + i sin 2θ = cos 2θ − sin 2θ + 2 i sin θ cos θ Equating the real and imaginary parts of this equation separately, we have:

)

!sin3θ = 3 1− sin 2θ sinθ − sin 3θ = 3sinθ − 4 sin 3θ

Example 1.12-3 n

!cos 2θ = cos θ − sin θ ! 2

2

and!

sin 2θ = 2 sin θ cos θ

⎛ 1+ i tan θ ⎞ 1+ i tan nθ Show that ⎜ . = 1− i tan nθ ⎝ 1− i tan θ ⎟⎠

Example 1.12-2

Solution:

Determine cos3θ and sin3θ as a function of powers of cosθ

We can write:

and sin θ .

1+ i tan θ cosθ + i sin θ ! = 1− i tan θ cosθ − i sin θ

Solution:

or

Using de Moivre's theorem, we can write:

(

!cos3θ + i sin3θ = cosθ + i sin θ

)3

(

)2

cosθ + i sin θ 1+ i tan θ cosθ + i sin θ cosθ + i sin θ ! = = 1− i tan θ cosθ − i sin θ cosθ + i sin θ cos 2 θ + sin 2 θ

55

Therefore:

1+ i tan θ ! = cosθ + i sin θ 1− i tan θ

(

(

! cosθ + i sin θ

)

2

1+ i tan nθ ! = cos nθ + i sin nθ 1− i tan nθ

(

)2

(

1 cos nθ + i sin nθ

)

2

cos nθ − i sin nθ cos nθ − i sin nθ

! cosθ + i sin θ

(

)− n = cos nθ +1 i sin nθ

(

cos nθ − i sin nθ = cos nθ − i sin nθ )− n = cos 2 2 nθ + sin nθ

! cosθ + i sin θ

)2 n = cos 2 nθ + i sin 2 nθ

1+ i tan nθ ! = cos nθ + i sin nθ 1− i tan nθ

(

(cosθ + i sinθ )n

=

or

Using de Moivre's theorem, we can write: n

1

Therefore:

Replacing θ with nθ :

⎛ 1+ i tan θ ⎞ !⎜ = cosθ + i sin θ ⎝ 1− i tan θ ⎟⎠

)− n =

and so:

(

! cosθ + i sin θ

= cos 2 nθ + i sin 2 nθ

Therefore: n

⎛ 1+ i tan θ ⎞ 1+ i tan nθ !⎜ = 1− i tan nθ ⎝ 1− i tan θ ⎟⎠

)− n = cos ( − nθ ) + i sin ( −nθ )

1.13! EXPONENTIAL FORM OF COMPLEX NUMBERS !

We have seen that it is possible to represent complex

numbers in rectangular and polar form. We will now show that Example 1.12-4

it is also possible to represent complex numbers in a form

Show that de Moivre's theorem is valid for negative values

known as the exponential form or the polar exponential form.

of n .

!

Using Euler's formula, we can rewrite the polar form of a

complex number given in equation (1.10-4) in exponential form: Solution: We can write using de Moivre's theorem:

!

(

)

z = r cosθ + isin θ = r eiθ !

(1.13-1)

56

As we will show, the exponential form of complex numbers

!

simplifies the algebraic operations of multiplication, division,

center at the coordinate origin in the complex plane (see Figure

and raising to a power. With equations (1.10-12) and (1.10-16),

1.13-1). Equation (1.13-6) and Figure 1.13-1 are useful for

equation (1.13-1) can also be written as:

exploring the values of eiθ .

! !

z = r eiθ = z e

i arg ( z )

!

Equation (1.13-6) is the equation of a unit circle having its

(1.13-2)

The complex conjugate of z in exponential form can be

obtained using equations (1.13-1) and (1.11-8): ! !

(

) ( ) = r ( cos ( − θ ) + i sin ( − θ )) = r e− iθ !

z = r eiθ = r cosθ + i sin θ = r cosθ − i sin θ

(1.13-3)

We see again that the complex conjugate of a number z can be obtained by replacing i with − i wherever if occurs in z . From equations (1.13-1) and (1.13-3) we have: !

2

z z = r eiθ r e−iθ = r 2 = z !

(1.13-4)

We also have using equation (1.11-9): ! ! !

z = r eiθ = r eiθ = r !

(1.13-5)

Figure 1.13-1!

If r = 1 , equation (1.13-1) becomes:

z = cosθ + isin θ = eiθ !

and we have z = eiθ = 1.

(1.13-6)

! !

Geometric representation of z = eiθ in the complex plane.

From Figure 1.13-1 we see that:

e0i = 1!

(1.13-7) 57

!

ei π 2 = i !

(1.13-8)

!

!

e π i = −1!

(1.13-9)

and so:

!

ei3π 2 = e−i π 2 = − i !

(1.13-10)

!

e 2π i = 1 !

(1.13-11)

Equation (1.13-9) written in the form e π i + 1 = 0 is known

!

as Euler’s equation. It is a remarkable equation, connecting the constants e , i , π , the integers 0 , 1 , the operation of addition

( + ) , and the equality relation ( = ) .

!

(1.10-21) we also have:

k = 0, ± 1, ± 2, ! !

(1.13-12)

= i!

k = 0, ± 1, ± 2, ! !

(1.13-13)

i π +2 k π ) e ( = −1 !

k = 0, ± 1, ± 2, ! !

(1.13-14)

!

e

!

i π e (

! !

i 3π e(

!

2+2 k π )

= 1!

2+2 k π )

−i π =e (

2+2 k π )

= −i !

k = 0, ± 1, ± 2, ! !

(1.13-15)

From equation (1.11-8) we have the complex conjugate of

eiθ : ! !

iθ

e =e

− iθ

()

! eiθ = eiθ 1 = eiθ ei 2 k π = eiθ +i 2 k π !

k = 0, ± 1, ± 2, ! !

(1.13-17)

k = 0, ± 1, ± 2, ! !

(1.13-18)

k = 0, ± 1, ± 2, ! !

(1.13-19)

Since we have:

()

arg z = θ + 2 k π !

!

()

we see that arg z is not unique for the exponential form of complex numbers, just as it is not unique for the polar form of complex numbers. We have in exponential form:

From equations (1.13-7) through (1.13-11), and equation 2kπ i

()

arg 1 = 2 k π !

!

From equation (1.13-12) we have:

(1.13-16)

! !

i θ +2 k π ) ! z = r eiθ +i 2 k π = r e (

k = 0, ± 1, ± 2, ! !

(1.13-20)

Two complex numbers z1 = r1 eiθ1 and z2 = r2 eiθ 2 are equal if

and only if: ! r1 = r2 !

and!

θ1 = θ 2 + 2 π k !

k = 0, ± 1, ± 2, ! !

(1.13-21)

Example 1.13-1 If eiθ = −1, determine all values of θ . Solution: Using equation (1.13-6) we have: !e iθ = cosθ + i sin θ = −1 Therefore: 58

(

)

!cosθ = −1 !

θ = π + 2kπ !

k = 0, ± 1, ± 2, !

and so r = 1 and, using equation (1.13-12), i n + 1 θ = 2 k π i .

!sin θ = 0 !

θ = π ± 2kπ !

k = 0, ± 1, ± 2, !

We then have θ = 2 k π

( n + 1) where k = 0, ± 1, ± 2, !, n , and

so:

Therefore: !θ = π + 2 k π !

k = 0, ± 1, ± 2, !

!z = e

i 2 k π ( n+1)

!

k = 0, ± 1, ± 2, !, n

as given in equation (1.13-14). Example 1.13-3 Example 1.13-2

Express in exponential form z = r eiθ the following complex

If z n = z where n is a positive integer, determine all values of z ≠ 0 .

numbers:

Solution:

2.! 1− i

Using equations (1.13-1) and (1.13-3) we have:

3.! −1− i

!z = r eiθ !

z = r e− iθ

( )

!z = r e

iθ

n

= r e− iθ

!r n−1 ei nθ = e− iθ Therefore: n−1 i ( n+1) θ

e

Solution: 1.! For 1+ i :

or

!r

The points corresponding to these complex numbers are shown in Figure 1.13-2.

Therefore: n

1.! 1+ i

=1

!r = 12 + 12 = 2 and

1 π θ! = tan −1 = + 2 k π ! 1 4

k = 0, ± 1, ± 2, !

Therefore: 59

i π 1! + i = 2 e (

4+2 k π )

!

k = 0, ± 1, ± 2, !

Therefore: i 7π 1! − i = 2 e (

4+2 k π )

k = 0, ± 1, ± 2, !

!

3.! For −1− i : !r =

( −1)2 + ( −1)2 =

2

and

θ! = tan −1

−1 5π = + 2kπ ! −1 4

k = 0, ± 1, ± 2, !

Therefore: i 5π !−1− i = 2 e (

! Figure 1.13-2!

Geometric representation of the points 1+ i , 1− i , and −1− i in the complex plane.

!

k = 0, ± 1, ± 2, !

The polar representation of a circle of radius r centered at

a point z0 is given by: !

z = z0 + r eiθ !

(1.13-22)

(see Figure 1.13-3 in which z0 = 1+ i and r = 1 ). From equation

2.! For 1− i :

(1.13-22) we have:

( )2

!r = 12 + −1 = 2

!

and

θ! = tan −1

4+2 k π )

z − z0 = eiθ !

(1.13-23)

Using equation (1.11-9) we can write:

−1 7 π = + 2kπ ! 1 4

k = 0, ± 1, ± 2, !

!

z − z0 = 1!

(1.13-24) 60

!

The product of two complex numbers z1 = r1 e

z1 = r1 e

i θ1

and

is given by:

z1 z2 = r1 e

!

i θ1

i θ1

r2 e

i θ2

iθ

= r1 r2 e 1 e

i θ2

i θ +θ = r1 r2 e ( 1 2 ) !

(1.13-27)

From equation (1.13-2) we then obtain: i θ +θ z1 z2 = z1 z2 e ( 1 2 ) !

!

(1.13-28)

Using equation (1.10-21) we can write: !

(

) (

) (

)

arg z1 z2 = θ1 + 2 k1 π + θ 2 + 2 k 2 π !

(1.13-29)

where k1 = 0, ± 1, ± 2, ! and k 2 = 0, ± 1, ± 2, ! . Letting k = k1 + k 2 , we have:

(

)

Polar representation of the circle z = 1+ i + eiθ .

Figure 1.13-3!

!

(

)

arg z1 z2 = θ1 + θ 2 + 2 k π !

k = 0, ± 1, ± 2, ! !

We then see that:

(

)

mod 2 π ! (1.13-31) ( ) ( ) Note that for Arg ( z1 z2 ) equations (1.13-30) and (1.13-31) do not arg z1 z2 = arg z1 + arg z2 !

1.13.1! MULTIPLICATION OF COMPLEX NUMBERS IN EXPONENTIAL FORM

!

!

We have:

necessarily apply (see Section 1.10.7).

!

e

i θ1 i θ 2

e

i θ +θ = e ( 1 2) !

(1.13-25)

as shown in Example 1.13-4. If n is an integer we also have: !

(e ) iθ

n

= e inθ !

(1.13-26)

(1.13-30)

Example 1.13-4 Show that: i θ +θ !e ( 1 2 ) = ei θ1 ei θ 2

Solution: 61

We have from Euler's formula: i θ +θ !e ( 1 2 ) = cos θ1 + θ 2 + i sin θ1 + θ 2

(

)

(

Solution:

)

Letting z2 = i , from equation (1.13-8), we have:

From trigonometric identities we have !

i θ +θ e ( 1 2 ) = cosθ1 cosθ 2 − sin θ1 sin θ 2 + i sin θ1 cosθ 2 + sin θ 2 cosθ1

(

or

)

!z2 = i = ei π

2

Since z2 = 0 + 1i we have: !r2 = 1

i θ +θ !e ( 1 2 ) = cosθ1 + i sin θ1 cosθ 2 + i sin θ 2

(

)(

)

Therefore from equation (1.13-27): i θ +θ i θ +π 2 !z1 z2 = z1 i = r1 e ( 1 2 ) = r1 e ( 1 )

Therefore: i θ +θ !e ( 1 2 ) = ei θ1 ei θ 2

Multiplying a complex number z1 by i therefore rotates the complex number by π 2 .

Example 1.13-5 Show using z in exponential form that if z = ei π , then we have z

2

= 1.

If z1 = r1 eiθ1 and z2 = r2 eiθ 2 are both negative real numbers, show using their arguments that their product is a positive

Solution: We have using equations (1.10-5) and (1.13-3): 2

Example 1.13-7

iπ

!z = z z = e e

iπ

iπ

=e e

−i π

0

= e =1

Example 1.13-6 If z1 = r1 eiθ1 , determine z1 multiplied by i .

real number. Solution: Since z1 and z2 are both negative real numbers, we have: !θ1 = π !

θ2 = π

Therefore using equations (1.13-27) and (1.13-11): 62

i θ +θ !z1 z2 = r1 r2 e ( 1 2 ) = r1 r2 e2 π i = r1 r2

Therefore arg z 2 has values that include + 2 k π while 2 arg z does not. We see then that arg z 2 ≠ 2 arg z .

and so the product z1 z2 is a positive real number.

Example 1.13-10

Example 1.13-8

Given z1 = 1+ i and z2 = 1+ i 3 , find z1 z 2 .

Using the exponential form z = r eiθ of complex numbers, show that z z is real.

Solution:

Solution:

From equation (1.10-15) we have:

From equation (1.13-3) we have:

( )( r e ) = r

!z z = r e

iθ

−iθ

2

e

i (θ −θ )

=r

!tan θ1 = 2

!z1 z2 =

(

1 +1 e

!z1 z2 =

(

2 ei π

Example 1.13-9

Solution: From equation (1.13-30) we have:

( )

!

k = 0, ± 1, ± 2, !

() (

)

y2 x2

=

3 = 3 1

2

2

iπ 4

)

⎛ 2 ⎜ 1 + ⎝

( )

⎞ 3 ei π 3 ⎟ ⎠ 2

k = 0, ± 1, ± 2, !

4

) (2 e ) = 2 iπ 3

i π 2e(

4+ π 3)

= 2 2 ei 7 π

12

The effect of multiplying a complex number z by eiθ can

be seen by writing: !

We also have from equation (1.10-21): !2 arg z = 2 θ + 2 k π = 2 θ + 4 k π !

tan θ 2 =

or

Show that arg z 2 ≠ 2 arg z .

( )

1 = = 1! x1 1

Therefore θ1 = π 4 and θ 2 = π 3 . We can then write:

and so z z is real.

arg z 2 = arg z z = θ + θ + 2 k π = 2θ + 2 k π !

y1

i arg z + θ ) ! z eiθ = r ei arg z eiθ = r e (

(1.13-32)

and so multiplication of any complex number z by eiθ results in a rotation of the number z by θ . This is illustrated in Figure 63

1.13-4 for eiθ = ei π 2 = i . Multiplication of any complex number

z by i then corresponds to a rotation of z by θ = π 2 . If z is a real number and so is on the x-axis, then the product of z and i

!

z ei3π

!

z ei 2 π = z !

will fall on the y-axis, consistent with the Argand diagram

2

= z e− i π 2 = − z i ! rotates z by − π 2 ! rotates z by 2 π !

(1.13-36) (1.13-37)

Example 1.13-11

representation of pure imaginary numbers.

Show that e π i = −1 using ei π 2 = i . Solution: We can write:

(

! ei π

2

)( e ) = ( i ) ( i ) iπ 2

or !e π i = i 2 = −1

Figure 1.13-4! Representation of a complex number z multiplied by ei π 2 = i .

1.13.2! DIVISION OF COMPLEX NUMBERS IN EXPONENTIAL FORM !

!

From equations (1.13-7) through (1.13-11) we have the

following rotations:

!

!

z ei0 = z !

does not rotate z !

(1.13-33)

!

z ei π 2 = z i !

rotates z by π 2 !

(1.13-34)

!

( )

z ei π = z −1 = −z !

rotates z by π !

(1.13-35)

We have from equation (1.10-38):

z1 z2

=

r1

(

)

(

)

⎡⎣cos θ1 − θ 2 + i sin θ1 − θ 2 ⎤⎦ ! r2

z2 ≠ 0 !

(1.13-38)

From equation (1.13-1) we then have: !

z1 z2

=

r1 r2

i θ −θ e ( 1 2)!

z2 ≠ 0 !

(1.13-39)

Therefore: 64

⎛z ⎞ arg ⎜ 1 ⎟ = arg z1 − arg z2 ! ⎝ z2 ⎠

( )

!

( )

z2 ≠ 0 !

From equation (1.13-39) we can write: (1.13-40)

1 ei0 1 i 0−θ 1 z = = iθ = e ( ) = e−iθ ! z re r r −1

!

Using equation (1.10-21) we can write: !

We also have from equations (1.13-46) and (1.13-3):

⎛ z1 ⎞ arg ⎜ ⎟ = θ1 + 2 k1 π − θ 2 + 2 k 2 π ! ⎝ z2 ⎠

(

) (

)

(1.13-41)

where k1 = 0, ± 1, ± 2, ! and k 2 = 0, ± 1, ± 2, ! . Letting k = k1 − k2 , we have: !

k = 0, ± 1, ± 2, ! !

(1.13-42)

⎛ z1 ⎞ arg ⎜ ⎟ = arg z1 − arg z2 ! ⎝ z2 ⎠

( )

(

Note that for Arg z1 z2

( )

)

mod 2 π !

(1.13-43)

!

equations (1.13-40) and (1.13-42) do

⎛ z1 ⎞ arg ⎜ ⎟ = arg z1 z2−1 = arg z1 − arg z2 ! ⎝ z2 ⎠

)

( )

( )

z2 ≠ 0 ! (1.13-44)

( )

z ≠ 0 ! (1.13-48)

as was shown in Example 1.10-9.

( )

arg z2−1 = − arg z2 !

Show that arg z −1 = − arg z = arg z .

We have from equation (1.13-46):

1 1 1 = iθ = e−iθ ! z re r

z2 ≠ 0 !

and so!

arg z −1 = − θ

We have from equation (1.13-3): !z = r e − iθ !

we see from equations (1.13-31) and (1.13-44) that: !

z r eiθ = −iθ = e 2iθ = cos 2θ + isin 2θ ! z re

and

!z −1 =

Since

(

z ≠ 0 ! (1.13-47)

Solution:

not necessarily apply (see Section 1.10.7). !

!

Example 1.13-12

We then see that: !

1 −iθ r e−iθ z z = e = 2 = 2! r r r −1

!

⎛z ⎞ arg ⎜ 1 ⎟ = θ1 − θ 2 + 2 k π ! ⎝ z2 ⎠

z ≠ 0 ! (1.13-46)

and so arg z = − θ

We also have: (1.13-45)

!z = r e iθ !

and so − arg z = − θ 65

Therefore arg z −1 = arg z = − arg z .

!

If r = 1 we have z = eiθ from equation (1.13-1). We then can

write using equation (1.11-11): Example 1.13-13 Determine

!

1 in polar form. z

z+

1 1 = eiθ + iθ = eiθ + e−iθ = 2 cosθ ! z e

(1.13-49)

z−

1 1 = eiθ − iθ = eiθ − e−iθ = 2 i sin θ ! z e

(1.13-50)

and

Solution:

!

We have from equation (1.13-46) and Euler's formula:

1 1 e−iθ 1 1 ! = iθ = = ⎡⎣cos − θ + i sin − θ ⎤⎦ = ⎡⎣cosθ − i sin θ ⎤⎦ z re r r r

( )

( )

we then have: !

as illustrated in Figure 1.13-5.

cosθ =

1⎛ 1⎞ ! z + ⎜ ⎟ 2⎝ z⎠

sin θ =

1 ⎛ 1⎞ ! z − ⎜ ⎟ 2i ⎝ z⎠

(1.13-51)

As is shown in Example 1.13-14, we can also write: !

zn +

1 = 2 cos nθ ! zn

(1.13-52)

zn −

1 = 2 i sin nθ ! n z

(1.13-53)

and ! or !

Figure 1.13-5!

cos nθ =

1 2

⎛ n 1⎞ ⎜⎝ z + n ⎟⎠ ! z

sin nθ =

1 2i

⎛ n 1⎞ ⎜⎝ z − n ⎟⎠ ! z

(1.13-54)

Geometric representation of z = r eiθ and 1 z . 66

Example 1.13-14

Solution:

1 1 Show that z + n = 2 cos nθ and z n − n = 2 i sin nθ if z = eiθ . z z

Using equation (1.13-51) we can write:

n

1 !cos 2θ = 2 2

Solution: Using Euler's formula, with z = eiθ we can write:

( ) ( ) (

1 !z ± n = eiθ z n

n

± e

iθ

= cosθ + i sin θ ± cosθ + i sin θ

)− n

!cos 2θ =

)

(

)

!

or !z n ±

1 = cos nθ + i sin nθ ± cos nθ − i sin nθ zn

(

) (

)

1 = 2 cos nθ ! n z

zn −

1 = 2 i sin nθ n z

Example 1.13-15 Using z n +

)

!

Since z is a real number, we can write:

e

log e z

= z !

(1.13-55)

where log e is the natural logarithm. From equations (1.13-2)

Therefore: !z n +

(

!cos 2θ = 2 cos 2θ − 1

1 !z ± n = ⎡⎣cos nθ + i sin nθ ⎤⎦ ± ⎡⎣cos −nθ + i sin −nθ ⎤⎦ z

(

1 1 2 cos 2θ + 4 2

Therefore:

From de Moivre's theorem we then have: n

⎤ ⎤ 1 ⎡⎛ 2 1 ⎞ = z + + 2 ⎥ ⎥ 4 ⎢⎜⎝ ⎟ z2 ⎠ ⎦ ⎣ ⎦

or using equation (1.13-54):

)n (

−n

2

⎡ 1⎤ 1⎡ 2 1 1 z + = z + 2 z + 2 ⎢ ⎥ ⎢ z 4 z z ⎣ ⎦ ⎣

1 = 2 cos nθ find an expression for cos 2 θ in zn

terms of cos θ . 2

and (1.13-55) we then have: !

z= z e

iarg ( z )

=e

log e z + i arg ( z )

!

(1.13-56)

1.13.3! COMPLEX NUMBERS IN EXPONENTIAL FORM RAISED TO INTEGER POWERS !

Since the argument of a complex number is multivalued as

given in equation (1.10-21), the integer power of a complex number z is given by: 67

(

n

! w= z = re

i (θ +2 k π )

) =r e n

n i n (θ +2 k π )

!

r = x 2 + y 2 = 12 + 12 = 2

k = 0, ± 1, ± 2, ! !(1.13-57)

1 π θ = tan −1 = 1 4

where n is an integer. From equation (1.13-12) we have:

ei 2 n k π = 1 !

!

k = 0, ± 1, ± 2, ! !(1.13-58)

Using equation (1.13-59), we obtain:

and so equation (1.13-57) becomes:

2

(

w= z =

)

w = z n = r n ei n θ = r n cos nθ + isin nθ !

!

(1.13-59)

( 2)

2

2i π 4 i π 2 e ( ) = 2e ( ) = 2i

See Example 1.10-12.

We then have: n

z = r

!

n

n

!

n

=r = z !

where z

n

!

arg z n = n θ = n arg z !

(1.13-60) !

is real and positive. We also have:

( )

()

(1.13-61)

()

( )

!

so that z is a positive real number. We also see that

or

n i nθ

w= z =r e

is a single-valued function.

Example 1.13-16

( )

z n = z −1

m

!

(1.13-62)

where m = − n . Using equation (1.13-46), we have:

From equation (1.13-61) we see that arg z = arg z n only if θ = 0 n

If n is negative we can write:

!

( )

z n = z −1

m

m

−n

m

⎛ 1⎞ ⎛ 1⎞ ⎡1 ⎤ = ⎢ e−iθ ⎥ = ⎜ ⎟ e−i m θ = ⎜ ⎟ e i n θ ! (1.13-63) ⎝ r⎠ ⎝ r⎠ ⎣r ⎦

z n = r n e i nθ !

(1.13-64)

and so equation (1.13-59) is also valid for negative integer

Determine w = z 2 if z = 1+ i .

values of n . For all integer values of n we then have:

Solution:

!

!z = x + i y = 1+ i From equations (1.10-12) and (1.10-15) we have:

(

)

z n = r n ei nθ = r n cos nθ + i sin nθ !

(1.13-65)

( z1 z2 )n = z1n z2n !

(1.13-66)

and !

68

!

If k is an integer, we have: n k

!

z z =z

!

(z )

n k

n+k

!

!

(1.13-67)

= z nk !

(1.13-68)

!

z=w !

(

)n

(1.14-5)

Using de Moivre's theorem: !

!

(

)

(

)

z = r cosθ + i sin θ = ρ n cos nφ + i sin nφ !

(1.14-6)

r = ρn!

n

ρ = r1 n = r !

(1.14-7)

where the nth root function of r always produces a nonnegative

If a complex number z is equal to: n

)

The moduli of both sides of this equation must be equal:

1.14! INTEGER ROOTS OF COMPLEX NUMBERS !

(

z = r cosθ + i sin θ = ρ n cosφ + i sin φ !

(1.14-1)

where n ≥ 2 is an integer and w is a complex number, then w is

real number. The arguments must also be equal: !

nφ = θ + 2 k π !

k = 0, 1, 2, !, n − 1 !

(1.14-8)

considered to be the nth root of z . To find all n roots of a

where k ranges from k = 0 to k = n − 1 since there are n distinct

complex number z ≠ 0 we must solve the equation:

nth roots. We then have:

!

w = z1 n!

(1.14-2)

For n > 2 the method for finding quadratic roots given in Section 1.9 will not work, and the following methods using polar coordinates must be employed. ! ! !

(

)

z = r cosθ + isin θ = r e !

(

)

w = ρ cosφ + isin φ = ρ eiφ !

!

φ=

θ + 2 kπ ! n

k = 0, 1, 2, !, n − 1 !

(1.14-9)

The n roots all be different since for these roots n > k , and

so each φ will differ from the previous one by less than 2 π . For

k > n − 1 the same roots as given in equation (1.14-9) will simply

Writing z and w in polar form, we have: iθ

!

repeat plus some multiple of 2 π . We can see this for k = n : (1.14-3) ! (1.14-4)

φ=

θ + 2 nπ θ θ = + 2π = ! n n n

(1.14-10)

which is the same as for k = 0 .

Equation (1.14-1) then becomes: 69

! !

We can write equation (1.14-9) in the form:

( )

arg wk =

()

arg z + 2 k π n

!

Example 1.14-1

k = 0, 1, 2, !, n − 1 ! (1.14-11)

!

The n roots wk of a complex number z = r eiθ are given by:

!

wk = z 1 n = z

1 n

i θ +2 k π ) n i θ +2 k π ) e( = r1 n e (

or !

wk = z 1 n = r 1 n e iθ n e i 2 k π n ! n

k = 0, 1, 2, !, n − 1 ! (1.14-13)

is the nth root of r . Each distinct root wk has the

same modulus r 1

n

( )

but a different argument arg wk . The n

roots are all found, therefore, on the same circle of radius of r

1 n

centered at the origin, with the roots equally spaced on the circle, and separated by 2 π n radians (see Figures 1.14-1, 1.14-2, 1.14-3, and 1.14-4). These roots constitute the vertices of a regular n-sided polygon.

! !

We have z = −1 where r = −1 = 1 , θ = π , and n = 2 . From i π + 2k π ) 2 !wk = z 1 2 = 11 2 e ( !

k = 0, 1

w0 = e i π 2 = i !

where r 1

!

Solution:

equation (1.14-12) we find the two square roots of z = −1:

n

k = 0, 1, 2, !, n − 1 ! (1.14-12)

!

Determine the square roots of −1 .

We can write equation (1.14-12) in polar coordinates:

wk = z1

n

⎡ ⎛θ + 2 kπ ⎞ ⎛θ + 2 kπ ⎞ ⎤ = r 1 n ⎢cos ⎜ + i sin ⎟⎠ ⎜⎝ ⎟⎠ ⎥ n n ⎝ ⎣ ⎦ k = 0, 1, 2, !, n − 1 !

i π +2 π ) 2 w1 = e ( = e i3π 2 = − i

Using equation (1.14-14), we can also write the square roots of −1 in polar form:

⎛ π + 2 kπ ⎞ ⎛ π + 2 kπ ⎞ !wk = cos ⎜ ! + i sin ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠

k = 0, 1

Therefore the square roots of −1 are:

⎛π ⎞ ⎛π⎞ !w0 = cos ⎜ ⎟ + i sin ⎜ ⎟ = i ! ⎝2⎠ ⎝ 2⎠

k=0

⎛ 3π ⎞ ⎛ 3π ⎞ ⎛ π⎞ ⎛ π⎞ !w1 = cos ⎜ ⎟ + i sin ⎜ ⎟ = cos ⎜ − ⎟ + i sin ⎜ − ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2⎠ ⎝ 2⎠ ⎛π⎞ ⎛π⎞ != cos ⎜ ⎟ − i sin ⎜ ⎟ = − i ! ⎝ 2⎠ ⎝ 2⎠

k =1

(1.14-14)

70

Example 1.14-2

Example 1.14-3

Determine the square roots of i .

Determine the cube roots of 1 .

Solution:

Solution:

We have z = i where r = i = 1 , θ = π 2 , and n = 2 . From equation (1.14-12) we find the two square roots of z = i : i π wk = z1 2 = 11 2 e (

w0 = e i π 4 = i π w1 = e (

1 2

2 + 2π ) 2

+i

2 + 2k π ) 2

k = 0, 1

!

We have z = 1 where r = 1 = 1 , θ = 0 , and n = 3 . From equation (1.14-12) we find the three cube roots of z = i : i 0+2 k π ) 3 !wk = z 1 3 = 11 3 e ( = ei 2 k π 3 !

1

k = 0, 1, 2

and so the roots all lie on a unit circle separated by 2 π 3 .

2

The cube roots of 1 are:

1

1

!w0 = 1 !

k=0

Using equation (1.14-14), we can also write the square roots

!w1 = e i 2 π 3 !

k =1

of i in polar form:

!w2 = e i 4 π 3 !

k=2

= e i 3π 4 = −

2

−i

2

⎛ π 2+ 2 kπ ⎞ ⎛ π 2+ 2 kπ ⎞ !wk = cos ⎜ + i sin ⎟⎠ ⎜⎝ ⎟⎠ ! 2 2 ⎝

k = 0, 1

Therefore the square roots of i are:

as shown in Figure 1.14-1. As a check we can cube

w2 = e i 4 π 3 :

( ) ( 3

! w2 = e i 4 π

⎛π ⎞ ⎛π⎞ 1 1 !w0 = cos ⎜ ⎟ + i sin ⎜ ⎟ == ! +i ⎝4⎠ ⎝ 4⎠ 2 2

k=0

⎛ 5π ⎞ ⎛ 5π ⎞ 1 1 !w1 = cos ⎜ ⎟ + i sin ⎜ ⎟ = − ! −i ⎝ 4 ⎠ ⎝ 4 ⎠ 2 2

k =1

This example can be compared with Example 1.9-2.

3

) =e 3

i 4π

=1

Using equation (1.14-14), we can also write the cube roots of 1 in polar form:

⎛ 2 kπ ⎞ ⎛ 2 kπ ⎞ !wk = cos ⎜ + i sin ⎜⎝ 3 ⎟⎠ ! ⎝ 3 ⎟⎠

k = 0, 1, 2 71

Therefore the cube roots of 1 are:

()

()

w0 = cos 0 + i sin 0 = 1 !

k=0

⎛ 2π ⎞ ⎛ 2π ⎞ 1 3 w1 = cos ⎜ + i sin = − + i! ⎟ ⎜ ⎟ 2 2 ⎝ 3 ⎠ ⎝ 3 ⎠

k =1

⎛ 4π ⎞ ⎛ 4π ⎞ 1 3 w2 = cos ⎜ + i sin = − − i! ⎜⎝ 3 ⎟⎠ 2 2 ⎝ 3 ⎟⎠

k=2

Example 1.14-4 Determine the 4th roots of 1 . Solution: We have z = 1 where r = 1 = 1 , θ = 0 , and n = 4 . From equation (1.14-12) we find the four fourth roots of z = 1: !wk = z 1 4 = 11 4 e i 2 k π 4 = e i k π 2 !

k = 0, 1, 2, 3

and so the roots all lie on a unit circle separated by π 2 . The

4th roots of 1 are: w0 = e0i = 1 !

w1 = e i π 2 = i

w2 = e i 2 π 2 = e i π = −1 !

w3 = e i 3π 2 = − i

as shown in Figure 1.14-2. We can check that these are the roots of the polynomial w4 −1 = 0 :

(

)(

)(

)(

) (

)(

) (

! w − 1 w − i w + 1 w + i = w2 − 1 w2 + 1 = w 4 − 1

)

Using equation (1.14-14), we can also write the 4th roots of 1 in polar form: Figure 1.14-1!

Cube roots of z = 1.

⎛ kπ ⎞ ⎛ kπ ⎞ !wk = cos ⎜ + i sin ⎜⎝ 2 ⎟⎠ ! ⎝ 2 ⎟⎠

k = 0, 1, 2, 3

Therefore the 4th roots of 1 are: 72

()

()

w0 = cos 0 + i sin 0 = 1 !

k=0

⎛π⎞ ⎛π⎞ w1 = cos ⎜ ⎟ + i sin ⎜ ⎟ = i ! ⎝ 2⎠ ⎝ 2⎠

k =1

⎛ 2π ⎞ ⎛ 2π ⎞ w2 = cos ⎜ + i sin ⎜⎝ 2 ⎟⎠ = − 1 ! ⎝ 2 ⎟⎠

k=2

⎛ 3π ⎞ ⎛ 3π ⎞ w2 = cos ⎜ ⎟ + i sin ⎜ ⎟ = − i ! ⎝ 2 ⎠ ⎝ 2 ⎠

k=3

Example 1.14-5 Determine the 6 th roots of 1 . Solution: We have z = 1 where r = 1 , θ = 0 , and n = 6 . From equation (1.14-12) we find the six sixth roots of z = 1 : i 0+2 k π ) 6 !wk = z 1 6 = 11 6 e ( = ei k π 3 !

k = 0, 1, 2,!, 5

and so the roots all lie on a unit circle separated by π 3 . The

6 th roots of 1 are: 1 i + 3 2 2

w0 = e0i = 1 !

w1 = ei π 3 =

1 i w2 = e i 2 π 3 = − + 3! 2 2

w3 = ei π = −1

1 i w4 = e i 4 π 3 = − − 3! 2 2

w5 = ei 5π 3 =

1 i − 3 2 2

as shown in Figure 1.14-3. The 6 th roots are positioned at the equally spaced vertices of a regular 6-sided polygon inscribed inside a circle centered at the origin and having a radius of 1 . Using equation (1.14-14), we can also write the

6 th roots of 1 in polar form: Figure 1.14-2!

4

th

roots of 1 .

⎛ kπ ⎞ ⎛ kπ ⎞ !wk = cos ⎜ + i sin ⎜⎝ 3 ⎟⎠ ! ⎝ 3 ⎟⎠

k = 0, 1, 2,!, 5 73

The n th roots are positioned at the equally spaced vertices of a regular n-sided polygon inscribed inside a circle centered at the origin and having a radius of 1 . Example 1.14-7 Determine the cube roots of i . Solution: We have z = i where r = i = 1 , θ = π 2 , and so from equation (1.14-12) we find the three cube roots of z = i : i π wk = z 1 3 = 11 3e (

2+2 k π ) 3

i π =e(

2+2 k π ) 3

!

k = 0, 1, 2

and so the roots all lie on a unit circle separated by 2 π 3 . Figure 1.14-3!

6 th roots of 1 .

The cube roots of i are:

Example 1.14-6

w0 = ei π 6 !

k=0

Determine the n th roots of 1 .

w1 = e i 5π 6 !

k =1

Solution:

w2 = e i 3π 2 = e− i π 2 = − i !

k =2

We have z = 1 where r = 1 and θ = 0 . From equation (1.14-12)

Using equation (1.14-14), we can also write the cube roots of

we find the n th roots of z = 1 :

i in polar form:

i 0+2 k π ) n !wk = 11 n e ( = ei 2 kπ n !

k = 0, 1, 2,!, n − 1

⎛ π 2+ 2 kπ ⎞ ⎛ π 2+ 2 kπ ⎞ wk = cos ⎜ + i sin ⎟⎠ ⎜⎝ ⎟⎠ ! 3 3 ⎝

k = 0, 1, 2

74

Therefore the cube roots of i are:

⎛π⎞ ⎛π⎞ 3 1 w0 = cos ⎜ ⎟ + i sin ⎜ ⎟ = + i! 2 2 ⎝ 6⎠ ⎝ 6⎠ ⎛ 5π ⎞ ⎛ 5π ⎞ 3 1 w1 = cos ⎜ ⎟ + i sin ⎜ ⎟ = − + i! 2 2 ⎝ 6 ⎠ ⎝ 6 ⎠ ⎛ 3π ⎞ ⎛ 3π ⎞ w2 = cos ⎜ ⎟ + i sin ⎜ ⎟ = − i ! ⎝ 2 ⎠ ⎝ 2 ⎠ as shown in Figure 1.14-4.

Example 1.14-8

k=0

Determine the cube roots of − 27 . Solution:

k =1

For z = − 27 we have r = − 27 = 27 , θ = π , and n = 3. From equation (1.14-12) we find:

k =2

i π + 2k π ) 3 i π + 2k π ) 3 !wk = z 1 3 = 27 1 3 e ( ! = 3e (

k = 0, 1, 2

and so the roots all lie on a circle of radius 3 separated by

2 π 3 . The cube roots of − 27 are: w0 = 3e i π 3 !

i π +2 π ) 3 w1 = 3e ( = 3 e iπ = − 3

i π +4π ) 3 w2 = 3e ( = 3e i 5π 3 = 3e− i π

3

Example 1.14-9 Determine the cube roots of z = 1− i . Solution:

(

)

For z = 1− i we have r = 12 + 12 = 2 , θ = tan −1 −1 1 = − π 4 , and n = 3. From equation (1.14-14) we find:

Figure 1.14-4!

Cube roots of z = i .

⎡ ⎛ −π 4 + 2 k π ⎞ ⎛ −π 4 + 2 k π ⎞ ⎤ wk = 2 1 6 ⎢cos ⎜ + i sin ⎟⎠ ⎜⎝ ⎟⎠ ⎥ ! k = 0, 1, 2 3 3 ⎝ ⎣ ⎦ 75

and so the roots all lie on a circle of radius 21 2 π 3 . The cube roots of z = 1− i are:

w0 = 2 w1 = 2

w2 = 2

k=0

()

When k = 0 and θ = Arg z , the n th root of a complex

number z = r eiθ is referred to as the principal n th root, and is a referred to as a principal n th root function. For a principal n th

k =1

⎡ ⎛ 5π ⎞ ⎛ 5π ⎞ ⎤ cos + i sin ⎢ ⎜ ⎟ ⎜⎝ 4 ⎟⎠ ⎥ ! 4 ⎝ ⎠ ⎣ ⎦

1 6

1.14.1! PRINCIPAL ROOT FUNCTION

unique value. A function restricted to its principal n th root is

⎡ ⎛ 7π ⎞ ⎛ 7π ⎞ ⎤ cos + i sin ⎢ ⎜ ⎟⎠ ⎜⎝ 12 ⎟⎠ ⎥ ! 12 ⎝ ⎣ ⎦

1 6

separated by

!

⎡ ⎛ π⎞ ⎛ π ⎞⎤ cos − + i sin ⎢ ⎜ ⎟⎠ ⎜⎝ − 12 ⎟⎠ ⎥ ! 12 ⎝ ⎣ ⎦

1 6

6

root function w = z 1 n , we have: !

k=2

w= z

1 n i Arg ( z ) n

e

= r1 n e

!

(1.14-15)

−π < θ < π !

(1.14-16)

Equation (1.14-12) then becomes:

w = r 1 n ei θ n !

Example 1.14-10

!

Determine the 4 th roots of −1 .

and equation (1.14-15) becomes:

Solution:

!

For z = −1 We have r = − 1 = 1 , θ = π , and n = 4 . From equation (1.14-12) we find the 4 th roots of z = −1 :

wk = z

1 4

1 4

=1

e

i (π +2k π ) 4

k = 0, 1, 2, 3

!

and so the roots all lie on a unit circle separated by π 2 . The

4

th

roots of z = −1 are:

w0 = e w2 = e

iπ 4

w1 = e

!

i (π + 4 π ) 4

= e i 5π

4

= e− i 3π 4 ! w3 = e

i Arg ( z ) n

i (π +2 π ) 4

i (π + 6 π ) 4

=e

i 3π 4

= e i 7π

4

⎡ ⎛θ ⎞ ⎛θ ⎞⎤ w = r 1 n ⎢cos ⎜ ⎟ + i sin ⎜ ⎟ ⎥ ! ⎝ n⎠ ⎦ ⎣ ⎝ n⎠

−π < θ < π ! (1.14-17)

1.14.2! PRINCIPAL SQUARE ROOT FUNCTION The principal square root function w = z 1 ! by: 1 2 i Arg ( z ) 2 i Arg z 2 ! w= z e = z e () !

2

is then given (1.14-18)

Equation (1.14-16) then becomes: ,

= e− i π

!

w = r 1 2 ei θ 2 !

−π < θ < π !

(1.14-19)

4

and equation (1.14-17) becomes: 76

w= r

! where r

1 2

1 2

⎡ ⎛θ ⎞ ⎛θ ⎞⎤ ⎢cos ⎜ ⎟ + i sin ⎜ ⎟ ⎥ ! −π < θ < π ! ⎝ 2⎠ ⎦ ⎣ ⎝ 2⎠

(1.14-20)

or taking the n th root: m

⎛ θ θ⎞ cos + i sin = cos mθ + isin mθ ⎜⎝ n n ⎟⎠

!

is taken as the positive square root of r .

!

Example 1.14-11

Example 1.15-1

Solution:

Show that:

)

For z = 1− i we have r = 12 + 12 = 2 , θ = tan −1 −1 1 = − π 4 ,

!cos

and so from equation (1.14-20), we find: 1 4

!w = 2

1.15! !

4

(

! cosθ + i sin θ

!

2

and so:

If m and n are both integers and if n > 0 , we can write

θ θ = cos m n + i sin m n = cos m θ + i sin m θ n n !

)1 2 = cos θ2 + i sin θ2

⎛ θ θ⎞ !cosθ + i sin θ = ⎜ cos + i sin ⎟ 2 2⎠ ⎝

using de Moivre's theorem: !

θ 1− cosθ = 2 2

or

DE MOIVRE'S THEOREM FOR INTEGER ROOTS

mn

sin

From equation (1.15-2) with m = 1 and n = 2 we have:

⎡ ⎛π⎞ ⎛π ⎞⎤ cos − i sin ⎢ ⎜ ⎟ ⎜⎝ 8 ⎟⎠ ⎥ ⎣ ⎝ 8⎠ ⎦

⎛ θ θ⎞ cos + i sin ⎜⎝ n n ⎟⎠

θ 1+ cosθ ! = 2 2

(1.15-2)

Solution:

⎡ ⎛ −π 4⎞ ⎛ −π 4⎞ ⎤ cos + i sin ⎢ ⎜ ⎟ ⎜⎝ 2 ⎟⎠ ⎥ ⎣ ⎝ 2 ⎠ ⎦

Therefore the principal square root of z = 1− i is: !w = 21

)1 n

!

Determine the principal square root of z = 1− i .

(

(

(1.15-1)

cosθ + i sin θ = cos 2

θ θ θ θ − sin 2 + i 2 sin cos 2 2 2 2

Equating the real parts of this equation, we have:

cosθ = cos 2

θ θ − sin 2 2 2 77

For the principal mth root, k = 0 , and we have:

Using:

cos 2

θ θ + sin 2 = 1 2 2

we obtain: !cosθ = 2cos 2

θ −1! 2

cosθ = 1− 2sin 2

θ 1+ cosθ !cos = ! 2 2

θ 2

=r

! and so z n

m

de Moivre's theorem:

⎡ ⎛ nθ + 2 k π ⎞ ⎛ nθ + 2 k π ⎞ ⎤ cos + i sin ⎢ ⎜ ⎟⎠ ⎜⎝ ⎟⎠ ⎥ m m ⎣ ⎝ ⎦ k = 0, 1, 2, !, m − 1 ! (1.16-1)

denotes the n th power of each of the m roots of z .

zn

m

( )

≠ zn

1 m

(

! cosθ + i sin θ

)n = cos nθ + i sin nθ

and m

⎛ nθ nθ ⎞ !⎜ cos + i sin ⎟ = cos nθ + i sin nθ = cosθ + i sin θ m m⎠ ⎝

(

)n

Taking the mth root:

⎛ nθ nθ ⎞ !⎜ cos + i sin ⎟ = cosθ + i sin θ m m⎠ ⎝

(

)n m

We then have de Moivre's theorem for rational powers of complex numbers:

In general we have: !

(1.16-3)

Let α = n m where n and m are integers. We can write using

If m and n are relatively prime integers and if m > 0 , then

z

⎡ ⎛ nθ ⎞ ⎛ nθ ⎞ ⎤ cos + i sin ⎢ ⎜ ⎟⎠ ⎜⎝ m ⎟⎠ ⎥ ! m ⎝ ⎣ ⎦

Solution:

θ 1− cosθ sin = 2 2

from equations (1.13-65) and (1.14-14) we have: !

m

of complex numbers.

1.16! RATIONAL POWERS OF COMPLEX NUMBERS

n m

= rn

Show that de Moivre's theorem is valid for rational powers

This example can be compared with Example 1.11-2.

n m

m

Example 1.16-1

or

!

zn

!

!

(1.16-2)

(

! cosθ + i sin θ

)α = cos α θ + i sin α θ 78

Solution:

1.17! COMPLEX POLYNOMIALS !

We are given:

A complex polynomial is an expression having the form:

()

2

P z = a0 + a1 z + a2 z +!+ an−1 z

!

n−1

n

+ an z !

!P ( z ) =

(1.17-1)

and where n is a nonnegative integer. A polynomial will have

()

A complex rational function f z is a function that is the

!

()

()

quotient of two complex polynomials P z and Q z having no common divisor, and so has the form: !

a0 + a1 z + a2 z 2 + !+ an−1 z + an z ( ) ! f (z) = = Q ( z ) b0 + b1 z + b2 z 2 +!+ bm−1 z m−1 + bm z m n−1

P z

!P ( z ) =

k =0

n

∑

(

)

ak r k cos k θ + i sin k θ = 0

Equating the real and imaginary parts to zero:

()

n

∑

ak r cos k θ = 0 ! k

k =0

is

)k = 0

k =0

∑

(1.17-2)

)

ak r k sin k θ = 0

k =0

Multiplying the second equation by i and subtracting it from the first: n

(

)

If z = r cosθ + i sin θ is a root of a polynomial equation

P z = 0 with real coefficients, such that:

k

k

k

k =0

Using de Moivre’s theorem again: n

∑

!

n

∑a z

∑ a r (cos k θ − isin k θ ) = 0

!

Example 1.17-1

!P ( z ) =

∑

(

ak r k cosθ + i sin θ

From de Moivre’s theorem we can write:

!

max m, n .

()

ak z k =

n

n

where an ≠ 0 and bm ≠ 0 , and where the degree of f z

(

∑

n

k =0

where the coefficients a0 , a1 , !, an−1 , an are complex constants only a finite number of terms.

n

k

=0

k =0

show that z is also a solution.

(

ak r k cosθ − i sin θ

)k = 0

k =0

Since the coefficients are real, ak = ak , and we have: 79

!P ( z ) =

n

∑a z k

k

=0

k =0

and so if z is a solution to a polynomial equation having real coefficients, so is its conjugate z . We see this when we determine complex roots using the quadratic formula.

80

Chapter 2 Complex Functions

() ()

()

w = f z = u z + iv z

81

!

In this chapter we will consider the nature of complex-

valued functions. We will explore the topology of set of points

2.0.2!

in complex planes where complex functions are defined, and

!

we will examine the limits and continuity of complex functions.

single real variable t has the form:

We will define the Riemann sphere, and we will discuss the

!

concept of infinity for complex numbers.

2.0! !

()

()

Mathematical functions can be classified as real-valued

variables, and complex-valued functions of complex variables.

REAL FUNCTIONS OF A REAL VARIABLE

()

A real function (real-valued function) f x of a single

real variable x has the form: !

() ()

w= f t =u t +iv t !

(2.0-2)

()

u t and the imaginary part v t of w are real-valued functions

functions of real variables, complex-valued functions of real

!

()

A complex function (complex-valued function) f t of a

where w = u + i v is a complex number, and both the real part

FUNCTION CLASSIFICATIONS

2.0.1!

COMPLEX FUNCTIONS OF A REAL VARIABLE

of the real variable t :

()

!

()

()

u t = Re ⎡⎣ f t ⎤⎦ !

!

()

v t = Im ⎡⎣ f t ⎤⎦ !

()

A complex function w = f t

(2.0-3)

dependent upon a real

()

variable t is specified then in terms of two real functions u t

()

()

and v t . The complex function w = f t defines the rule for obtaining a complex number w from a real variable t . Such functions can be represented graphically on a complex plane by

()

y= f x !

(2.0-1)

()

where the function f x defines the rule for obtaining a real number y from a real number x . The real number y is known

()

as the value of f x for x . Real functions are defined on a set of real numbers, and they can be represented graphically on a real

()

plane by plotting the curve y = f x .

()

plotting the curve w = f t .

2.0.3! !

COMPLEX FUNCTIONS OF A COMPLEX VARIABLE

()

A complex function f z of a single complex variable

z = x + i y has the form: !

() ()

() ( )

( )

w = f z = u z + i v z = u x, y + i v x, y !

(2.0-4) 82

()

where w = u + i v is a complex number, and both the real part

represent the complex function w = f z : a z-plane and a w-

u x, y

plane. The function w = f z provides the link between points

( )

( )

and the imaginary part v x, y

of w are real-valued

functions of the two real variables x and y :

( )

!

()

( )

u x, y = Re ⎡⎣ f z ⎤⎦ !

!

()

in an area of the z-plane and image points in an area of the w-

()

v x, y = Im ⎡⎣ f z ⎤⎦ !

(2.0-5)

()

A complex function w = f z dependent upon a complex

plane. !

( )

We can consider the complex numbers z = x, y as being

( )

transformed to complex numbers w = u,v

()

by the function

variable z is specified then in terms of two real functions

w = f z . Such a transformation is called a mapping of the

u x, y

complex variable z = x, y into a complex variable w = u,v by

( )

( )

and v x, y , which in turn depend upon the two

independent real variables x and y , such that: !

()

(

)

w = u + iv = f z = f x + i y !

()

The complex function f z

() variables from ( x, y )

(2.0-6)

defines the rule for obtaining a

functions are defined on a set of complex numbers. The

by:

function f z

!

!

is defined in its domain of definition in the

( )

( )

Since both z x, y and w u,v are complex numbers that

!

to specify z = x + i y , and two dimensions are required to specify

!

w = u + i v . The complex function w = f z

and

is then a four-

dimensional function, and so cannot be represented by either a two-dimensional or even a three-dimensional diagram. This

()

makes the function w = f z

very difficult to visualize. To

resolve this problem, two complex planes are generally used to

( )

( )

()

The complex conjugate of the function w = f z is given

() ( )

( )

w = f z = u x, y − i v x, y !

(2.0-7)

We can write:

have a real and an imaginary part, two dimensions are required

()

( )

to u,v where u = u x, y and v = v x, y

are two real functions. !

complex plane.

( )

the function f z . This mapping represents a change in real

complex number w from a complex number z . Complex

()

( )

!

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

w + w = ⎡⎣u x, y + i v x, y ⎤⎦ + ⎡⎣u x, y − i v x, y ⎤⎦ = 2 u x, y ! (2.0-8)

w − w = ⎡⎣u x, y + i v x, y ⎤⎦ − ⎡⎣u x, y − i v x, y ⎤⎦ = 2 i v x, y

!!

(2.0-9)

Therefore we have: 83

()

( )

Re ⎡⎣ f z ⎤⎦ = u x, y =

!

Solution:

w+ w ! 2

(2.0-10)

In rectangular form w = z 2 is:

(

w− w ! Im ⎡⎣ f z ⎤⎦ = v x, y = 2i

()

!

( )

Therefore:

(2.0-11)

( )

v x, y = 2 x y

()

If the function w = f z is expressed in exponential form

z = r eiθ , we have:

Example 2.0-2

(

()

) ( )

( )

w = f z = f r eiθ = u r, θ + i v r, θ !

!

()

( )

( )

Express w = f z = z 2 in the form w = u r, θ + i v r, θ .

(2.0-12)

Solution:

where !

( )

!u x, y = x 2 − y 2 !

as in equation (1.4-12). !

)

!w = u ( x, y ) + i v ( x, y ) = z 2 = ( x + i y ) 2 = x 2 − y 2 + i ( 2 x y )

and

( )

( )

u r, θ = Re ⎡ f r eiθ ⎤ ! ⎣ ⎦

()

( )

( )

v r, θ = Im ⎡ f r eiθ ⎤ ! (2.0-13) ⎣ ⎦

In exponential form w = z 2 is:

A complex function w = f z is then specified in terms of a pair

( )

( )

( )

( )

( )

!w = u r, θ + i v r, θ = z = r e 2

iθ

2

= r 2 ei 2 θ

of real functions u r,θ and v r,θ , which in turn depend upon

Using Euler's formula given in equation (1.11-6), we can

the real variables r and θ .

write:

!

In chapter 1 we have already seen a number of complex

functions of a complex variable. For example: w = z 1 Section 1.9, w = z n in Section 1.13.3, and w = z 1

n

in

in Section 1.14.

( )

( )

( ( )

( ))

( )

v r, θ = r 2 sin 2θ

( )

( )

Therefore:

( )

!u r, θ = r 2 cos 2θ !

Example 2.0-1

()

2

!w = u r, θ + i v r, θ = r 2 cos 2θ + isin 2θ

( )

( )

Express w = f z = z 2 in the form w = u x, y + i v x, y . 84

Therefore:

Example 2.0-3

()

Express w = f z = z

2

( )

( )

( )

( )

!u x, y = e x cos y !

in the form w = u x, y + i v x, y .

v x, y = e x sin y

Solution: 2

In rectangular form w = z

Example 2.0-5

is:

!w = u ( x, y ) + i v ( x, y ) = z = z z = ( x + i y ) ( x − i y ) = x 2 + y 2 2

()

Express w = f z =

1 in the form w = u r, θ + i v r, θ . z

( )

( )

Solution:

Therefore:

( )

( )

!u x, y = x 2 + y 2 !

()

and so w = f z = z

v x, y = 0 2

In exponential form w =

is a real-valued function of the

1 is: z

1 1 e− iθ !w = u r, θ + i v r, θ = = iθ = z re r

( )

complex variable z .

( )

Using Euler's formula given in equation (1.11-6), we can Example 2.0-4

write:

()

( )

( )

Express w = f z = e z in the form w = u x, y + i v x, y . Solution: In rectangular form w = e is:

( )

( )

!w = u x, y + i v x, y = e = e

x+i y

x

=e e

iy

Using Euler's formula given in equation (1.11-6), we can write:

( )

( )

(

( ))

( )

1 cos −θ + isin −θ r

( )

( )

1 cosθ − isin θ r

or

z

z

( ( )

( )

!w = u r, θ + i v r, θ =

!e z = u x, y + i v x, y = e x cos y + isin y

)

!w = u r, θ + i v r, θ =

(

)

Therefore:

( )

!u r, θ =

1 cosθ ! r

( )

v r, θ = −

1 sin θ r 85

2.0.4! !

PARAMETRIC FORM OF A COMPLEX VARIABLE

!

definitions:

If the real and imaginary parts of a complex variable

z = x + i y are given in terms of a real parameter t :

()

()

x=x t !

!

y= y t !

(2.0-14)

() ()

()

()

The functions x = x t

(2.0-15)

()

and y = y t

are known as parametric

curves in the z-plane, where x = Re z and y = Im z , and where the parameter t takes on some range of real values.

2.1! !

by z ∈ A . Point Set: Any collection of points in the complex plane constitutes a point set. A given point is determined to

z t = x t +iy t !

!

Element: Each object in a set is referred to as an element of the set. If z is an element of a set A , this is denoted

then the complex variable z can be expressed in parametric form:

If A and B are two sets, then we have the following set

MAPPING OF COMPLEX VARIABLES We will begin our discussion of the mapping of complex

variables by first discussing the concept of sets. We will then describe the mapping of points and of sets of points from one

belong to a specific point set by some definite rule. Intersection: The intersection of two sets A and B consists of all elements that are in both of the two sets. The intersection of sets A and B is denoted by A ∩ B (see Figure 2.1-1). Union: The union of two sets A and B consists of all elements that are in at least one of the two sets. The union of sets A and B is denoted by A ∪ B . Empty Set: An empty set has no elements (technically this

complex plane to another.

is not a set, but it is a useful concept). An empty set

2.1.1!

A is denoted by A = ∅ . For a set B ≠ ∅ we have

!

SETS

A set can be defined as a collection or aggregate of distinct

B ∪ ∅ = B and B ∩ ∅ = ∅ .

objects. Each object in a set is referred to as an element of the

Infinite Set: An infinite set contains infinitely many

set. A set can consist of finitely or infinitely many elements. The

elements. An infinite set is not enumerable (its

number of elements in a set is referred to as its cardinality.

elements cannot be counted). 86

Equal: Two sets A and B are equal if they have all elements in common. We then have A = B , where

A ⊆ B and B ⊇ A . Proper Subset: If the set B is a subset of the set A , but not all elements of A are elements of B so that A ≠ B , then B is called a proper subset of A . When B is a proper subset of A this can be denoted by A ⊃ B or

B ⊂ A . Every set has ∅ as a proper subset. Complement: The complement of a set A is the set of all similar elements in the complex plane that are not in

A . The complement (or relative complement) of a point set A with respect to a point set B consists of all the points of B that are not in A . The complement Figure 2.1-1! Relation of sets of points A and B . The subset in red is the result.

( )

of A is denoted by AC . We also have AC

C

= A.

Difference: The difference A − B of two sets A and B is the set of elements in A that are not members of B .

Disjoint: Two sets A and B are disjoint if they have no elements in common. We then have A ∩ B = ∅ . Subset: The set B is a subset of the set A if all elements of

B are also elements of A . If z ∈B when B is a subset of A , we also have z ∈ A . When B is a subset of A , this is denoted by either A ⊇ B or B ⊆ A .

We then have A − B = A ∩ BC . Class: A collection of sets that all share a common distinct property is called a class. Each set of the class is called a member of the class. Cover or Covering: A class of sets E covers (is a covering of) a set A if every element of A belongs to some 87

member set of E . That is, a covering of a set A is a

and range refer to the same set of points Ω . The inverse image

class of sets E whose union contains A .

z = f −1 w consists of the set of points S where w = f z . Both

Subcover or Subcovering: This is a subset of a class of sets

( )

S and Ω are subsets of the set ! of complex numbers. !

E covering A that also covers the set A .

()

Mappings are referred to as either a mapping of one set

into or onto another set. The mapping of a set S by a function

2.1.2!

MAPPING OF POINTS IN A COMPLEX PLANE

( )

()

w = f z is referred to as S mapped into Ω if not every point w

into a

of Ω is the image of some point z of S . If every point in Ω is an

complex variable w = u,v by a function f z determines how

image point of at least one point in S , then the mapping of a set

a set of points in the z-plane is transformed into a set of points

S by w = f z is referred to as S mapped onto Ω .

!

A mapping of the complex variable z = x, y

( )

()

in the w-plane. We need to be concerned then about which points in the z-plane are mapped into which points in the wplane. Both the z-plane and the w-plane are complex planes. !

()

A function f z is often defined for only some subset of

all the points in the z-plane. We will let S be the restricted set of

()

points in the z-plane for which the function f z is defined. This set of points S is known as the domain of definition or

()

simply as the domain of the function f z . !

When a point z0 in the z-plane is mapped into a point w0

in the w-plane, the point w0 is known as the image of z0 , and

z0 is called the inverse image or pre-image of w0 . Mapping of

()

the set of points S in the z-plane by a function w = f z defines an image set of points Ω in the w-plane. This set of points Ω in

()

the w-plane is known as the range of f z . Therefore image

()

Example 2.1-1 Determine the domain of definition of the following complex functions:

()

1.!

f z = z2

2.!

f z =

()

4 z −1

3.!

f z =

()

4 z+z

Solution:

() 2 w = u + i v = ( x + i y) = x2 − y2 + i 2 x y .

1.! The domain of f z = z 2 is the entire z-plane. We have

88

()

2.! The domain of f z =

4 is all complex numbers z −1

()

except z = 1 where f z is not defined.

()

()

()

!

z-plane can be mapped into any given point w0 in the w-plane by a function f z , then the mapping is one-to-one, as shown

4 3.! Since z + z = 2 Re z , the domain of f z = is all z+z

2.1.3!

function of z . In other words, if only one unique point z0 in the

only if, for any two points z1 and z2 in the domain of definition

()

complex numbers except Re z = 0 where f z is not

of f z , we have:

defined.

!

MAPPING WITH A SINGLE-VALUED FUNCTION

()

()

in Figure 2.1-2. A function w = f z will be one-to-one if and

( )

( )

f z1 = f z2 !

only if z1 = z2 !

(2.1-1)

A one-to-one function is called an injective or univalent function.

If w = f z is a single-valued function of z , then one and

()

only one value of w in the range of f z will correspond to

()

each value of z in the domain of definition of f z . That is, only one point w will correspond to any one or more points z . Therefore only one point in the w-plane can be the image of any

()

point in the z-plane. If a function f z maps every point z into

()

the same point w0 , then f z is a constant function with a range of one point.

2.1.4! !

MAPPING WITH A ONE-TO-ONE FUNCTION

()

If w = f z is a single-valued function of z , and if any one

()

value w0 in the range of f z corresponds to one and only one

()

()

value z0 in the domain of f z , then f z

is a one-to-one

Figure 2.1-2! Complex point z0 in the z-plane mapped into point w0 in the w-plane by a function w = f z .

()

89

one-to-one for the same reason. It is possible to make

2.1.5! MAPPING WITH A MULTIVALUED FUNCTION

w = z 2 into a one-to-one function by restricting its

()

If w = f z is a multivalued function of z , then more than

!

domain of definition.

one value of w will correspond to some values of z in the

()

domain of f z . Therefore a given point z0 in the set of points

3.!

two w values ( ± w ) correspond to a single value of z .

S in the z-plane can be mapped into more than one point w in

This mapping is also not one-to-one. Note that w = z

the set of points Ω in the w-plane. A number of points in the w-

is the inverse function of w = z 2 , and so the inverse

plane can then be the images of a single point z0 in the z-plane.

function w = z of the single-valued function w = z 2 is

Example 2.1-2 Determine if the following functions are single-valued or multivalued functions of z .

w = z is equal to w2 = z , which is multivalued since

multivalued. 4.

w = z is not one-to-one since it maps the entire z plane onto the nonnegative x-axis of the w plane.

1.!

w= z

2.!

w = z2

Example 2.1-3

w= z

If z = 2 eiθ where 0 ≤ θ ≤ π 2 is mapped with the function

3.! 4.!

w= z

Solution: 1.!

w = z is single-valued and one-to-one since any distinct value of z will be mapped onto only one distinct value of w .

2.!

()

()

w = f z = z 2 , what is the range of w = f z ? Solution:

z = 2eiθ where 0 ≤ θ ≤ π 2 is a quarter of a circle having a radius of 2 as shown in Figure 2.1-3. We then have:

()

2

( )

!w = f z = z = 2 e

iθ

2

= 4e2iθ !

0 ≤θ ≤ π 2

w = z 2 is single-valued since only one value of w

Letting φ = 2 θ , we see that the quarter of a circle is mapped

corresponds to two z values: ± z . This mapping is not

into a semicircle: 90

!w = 4 eiφ !

()

0 ≤φ < π

(

)

(

) (

!f z = 2 z −1 = 2 x + i y −1 = 2 x −1 + 2 i y

()

and so the range of w = f z = z 2 is a semicircle of radius 4. We then obtain the mapping shown in Figure 2.1-3.

)

Therefore: !u = 2 x − 1 !

v = 2y

We then obtain −5 < Re w < 3 and − 2 < Im w < 2 . Mapping is as shown in Figure 2.1-4.

Figure 2.1-3! Mapping of z = 2eiθ with 0 ≤ θ ≤ π 2 using the

()

function f z = z 2 . Example 2.1-4 If Re z < 2 and Im z < 1 , what is the image of z in the

()

w-plane when mapped with the function w = f z = 2 z − 1 ?

Figure 2.1-4!

Mapping of Re z < 2 , Im z < 1 using the

()

function f z = 2 z − 1 .

Solution: 91

2.1.6! INVERSE IMAGE OF A COMPLEX NUMBER !

()

()

If w = f z is a one-to-one mapping, then in the range of

f z there will exist one and only one inverse image z0 of a point w0 . The inverse image of a given point w0 in a set Ω is

( )

then the unique point z0 in the set S for which w0 = f z0 . The

metric space, this terminology is easily obtained from the theory of metric spaces. We will begin our discussion of point sets by considering open and closed disks. We will then discuss neighborhoods and the classification of points. We will define open sets, closed sets, connected sets, domains, and regions.

inverse image of the entire set of points Ω is the domain of

2.2.1!

definition of the function f z . We then have:

!

()

( )

z = f −1 w !

!

(2.1-2)

( ( ))

f −1 f z = z !

!

(2.1-3)

and

( ( ))

f f −1 w = w !

!

(2.1-4)

()

any continuous curve in the z-plane onto a continuous curve in the w-plane.

!

z − z0 < δ !

(2.2-1)

where δ is any given positive real number. According to this equation, all the points z in the set S lie within a distance from the point z0 that is less than δ . These points therefore lie inside

If w = f z is a continuous one-to-one function, then it will map

2.2!

Let z0 be a point in the z-plane, and S be the set of all

points z ∈! such that: !

such that:

OPEN DISKS

POINT SETS IN THE COMPLEX PLANE To describe the mapping of points from one complex

plane to another, some terminology is necessary regarding sets of points in a complex plane. Since a complex plane forms a

a circle of radius r = δ , and so form a disk centered at z0 having a radius less than δ (see Section 1.7.1). This disk is referred to as an open δ -disk because it does not include the boundary circle of radius r = δ (see Figure 2.2-1). !

( )

We will use the notation Dδ z0 for open disks centered

at a point z0 and having a radius of δ , where the boundary

( )

circle z − z0 = δ is not included. An open unit disk D1 z0

centered at point z0 and having a radius δ = 1 is specified by the equation z − z0 < 1 . 92

2.2.2! ! !

CLOSED DISKS

We will now let S be the set of all points z ∈! such that:

z − z0 ≤ δ !

(2.2-4)

where δ is any given positive real number. According to this equation, all the points z in S lie within a distance from z0 that is less than or equal to δ . These points therefore include all

( )

points of the open disk Dδ z0 together with all points forming the boundary circle of radius r = δ .

Figure 2.2-1! Graphical representation of the set of points S in an open δ -disk Dδ z0 about the point z0 . The disk’s boundary circle is dashed to indicate that points on this circle are not in the set S .

( )

!

Using equation (1.7-4) we can rewrite equation (2.2-1) as:

!

( x − x0 )2 + ( y − y0 )2 < δ !

(2.2-2)

or !

( x − x0 )2 + ( y − y0 )2 < δ 2 !

(2.2-3)

which is a disk of radius less than r = δ centered at the point

( x0 , y0 ) .

Figure 2.2-2! Graphical representation of the set of points in a closed δ -disk Dδ z0 about the point z = z0 . The disk’s boundary circle is solid to indicate that the disk is closed.

( )

93

!

The disk is called a closed δ -disk because all points of the

boundary circle of radius r = δ belong to the set of points that

2.2.4!

form the disk (see Figure 2.2-2). We will use the notation

!

Dδ z0 for closed disks centered at a point z0 and having a

neighborhood of z = z0 without including the point z0 itself.

( )

DELETED NEIGHBORHOODS

Sometimes it is useful to consider a set of points in a δ

radius of δ . While real numbers can be discussed in terms of

Such a δ neighborhood is known as a deleted δ neighborhood

open and closed intervals on a line, the comparable entities for

of a point z0 , and is given by:

complex numbers are open and closed disks on a complex

0 < z − z0 < δ !

!

plane.

(2.2-6)

An open δ -disk centered at a point z0 that does not include the

2.2.3!

point z0 is called a punctured δ -disk. We will use the notation

NEIGHBORHOODS

( )

We will define the neighborhood of a point to be any

Dδ* z0 for an open punctured disk centered at a point z0 and

open disk centered at that point. Using the concept of an open

having a radius of δ . The set of points in a deleted δ

δ -disk, we can define the δ neighborhood of a point z0 by:

neighborhood of z0 then form a punctured δ -disk Dδ* z0 .

!

!

z − z0 < δ !

(2.2-5)

( )

Example 2.2-1

where the set of all points contained in the δ -disk Dδ z0 are

Determine the neighborhoods of the following sets of points:

then said to be in the δ neighborhood of z = z0 . A point z = z0

1.!

z 1 2.! 1 < z < 2

Proposition 2.2-3:

3.! 1 ≤ z ≤ 2

A closed set contains all its limit points.

4.! 1 < z ≤ 2

Proof: !

From Proposition 2.2-2 we know that a finite point set

Let S be any closed set, and let z0 be a limit point of S . If

( )

z0 is an exterior point of S , then there must exist a Dδ z0 with

δ > 0 such that Dδ ( z0 ) lies entirely outside S . But this is in

contradiction to the definition of z0 being a limit point of S .

Solution: The point sets defined by: Re z > 1 , 1 < z < 2 , 1 ≤ z ≤ 2 , and 1 < z ≤ 2 are shown in Figure 2.2-5. 98

1.! The top left-hand diagram of Figure 2.2-5 shows the

()

point set Re z = x > 1 . This set is open since all of its points are interior points. None of its points fall on a boundary. This set includes most of the right half-plane except for the narrow strip 0 < Re z ≤ 1 . 2.! The top right-hand diagram of Figure 2.2-5 shows the point set 1 < z < 2 . This set is open since all of its points are interior points. None of its points fall on the boundary circles. This set forms an open circular annulus. Note that the point set 0 < z < 2 is an open circular annulus in the form of a punctured disk centered at the point z = 0 . 3.! The bottom left-hand diagram of Figure 2.2-5 shows the point set 1 ≤ z ≤ 2 . This set is closed

Figure 2.2-5!

Representations of the sets of points: Re z > 1 , 1 < z < 2 , 1 ≤ z ≤ 2 , and 1 < z ≤ 2 .

since it contains all points on the boundary circles

z = 1 and z = 2 . 4.! The bottom right-hand diagram of Figure 2.2-5 shows the point set 1 < z ≤ 2 . This set is neither open nor closed since it contains all points on the outer boundary circle z = 2 , but does not contain any points on the inner boundary circle z = 1 .

2.2.7! HALF-PLANES !

An open half-plane is defined as the set of points located

on only one side of an infinite straight line in a plane. If that line is the y-axis, the open half plane is either a right or left halfplane (see Figure 2.2-6). A closed half-plane is defined as the set of points located in an open half-plane together with its boundary line (see Figure 2.2-6). 99

boundary points. We will now determine if the intersection has only interior points.

Figure 2.2-6!

Open right half-plane set (upper diagram), and a closed right half-plane set (lower diagram). Figure 2.2-7!

Proposition 2.2-5: The intersection of two open sets of points in the complex plane

!

is an open set.

z0 ∈S1 ∩ S2 . Since S1 and S2 are open, we will have Dδ z0 as an

Proof: !

Largest open disks (circular areas) centered at point z0 within the intersection of sets S1 and S2 .

Let S1 and S2 be two open point sets that intersect. The

intersection is not closed since open sets do not have any

If the point z0 is a point within the intersection, we have:

( )

1

( )

open disk within S1 and Dδ z0 as an open disk within S2 . By 2

choosing the smallest radius of the two radii δ1 and δ 2 , and

( )

setting it equal to δ , we have Dδ z0 ⊂ S1 ∩ S2 as an open disk 100

within the intersection of S1 and S2 (see Figure 2.2-7). Therefore the intersection S1 ∩ S2 has interior points and no boundary points, and so is an open set.

The empty point set ∅ is both open and closed.

■

Proof:

Proposition 2.2-6: A point set S is closed if and only if its complement S C is open. Proof: plane that are not contained in S .

Since by definition the point set ∅ has no points, it has no

points which are not interior points, and so ∅ is open. But ∅ is then the complement of the set of all points z ∈!

which is open (as shown in Proposition 2.2-8), and so ∅ must be closed (see Proposition 2.2-6).

If S C is open and if z0 is a limit point of S , then every disk

!

! !

By definition S C will contain all points in the complex

!

Proposition 2.2-7:

( )

Dδ z0 contains a point of S . Therefore z0 is not an interior

■

Proposition 2.2-8:

point of S C . Since S C is open, z0 is also not a boundary point of

The set of points z ∈! is both open and closed.

C

S . This means that z0 must belong to S . Because S will then contain all its limit points, S is closed.

Proof: !

C

()

For any point z ∈! there will exist an open δ -disk Dδ z

Conversely, if S is closed and if z0 is a point in S , then z0

that lies within the z-plane. Therefore by definition all points

cannot be a limit point for S which contains all of its limit

z ∈! are interior points. The set of points z ∈! is then an open

points. This means that some open disk Dδ z0 exists within S C

set.

!

( ) that contains no points of S . Therefore Dδ ( z0 ) ⊂ S C , and so

z0

!

Since z ∈! includes all possible points in the z-plane, the

is an interior point of S C . Since z0 can be any point in S C , by

complement set of points to z ∈! is ∅ . But ∅ is open (as

definition S C is open.

shown in Proposition 2.2-7), and so z ∈! must be closed (see

■

Proposition 2.2-6).

■

101

2.2.8! !

all points along each open interval of the curve (see

PATH DEFINITIONS

curves C1 and C4 in Figure 2.2-8). A smooth curve

A path in the complex plane can consist of one segment or

of a finite chain of segments. We will now establish some definitions that describe paths in the complex plane.

except possibly at two endpoints. Piecewise-smooth Curve: A piecewise-smooth curve

Path Segment: A path segment in the complex plane is a continuous curve or arc.

consists of a finite number of connected path segments each of which is a smooth curve. At the

Continuous Curve or Arc: A continuous curve or arc in the

()

juncture between any two segments, z ′ t

has a

complex plane is defined as a set of points that can be

discontinuity (see curve C2 in Figure 2.2-8 which has

represented by a continuous complex function. A

a cusp between two segments where z ′ t

curve is then both a continuous function and a set of

continuous).

()

is not

points. The continuous function can be given in the

Simple Curve: A simple curve (or Jordan curve) is a curve

form of a parameterization (such as z t , where t is a

or arc that does not cross itself (see curves C1 and C4

()

real parameter). From equation (2.0-15) we have !

will then possess tangents at all points of the curve

() ()

()

()

()

z t = x t +iy t ! where x t

and y t

(2.2-8) are real-valued continuous

functions. The parameter t

has definite limits

()

in Figure 2.2-8). The function z t defining a simple curve will then be one-to-one. Closed Curve: A closed curve is a curve that has no endpoints (see curves C3 and C4 in Figure 2.2-8).

a ≤ t ≤ b such that the curve proceeds from a point

Simple Closed Curve: A simple closed curve is a simple

z1 = z a to a point zn = z b . The parameterization of

curve that is closed. Such a curve is called a closed

the curve provides a direction to the curve.

Jordan curve. The area of the complex plane enclosed

()

()

Smooth Curve: A smooth curve is a continuous curve or

()

arc that will have a continuous derivative z ′ t ≠ 0 at

by a closed Jordan curve is called the interior of the Jordan curve (see curve C4 in Figure 2.2-8). A simple 102

closed curve divides the complex plane into two parts: an interior part bounded by the curve; and an

Contour: A contour is a piecewise-smooth curve in the complex plane along which integration can be

exterior part that is unbounded.

performed. Polygonal path or polygonal line: A polygonal path consists of a finite number of connected straight line segments (see Figure 2.2-9). In a polygonal path the junction points of the segments are points where a

()

discontinuity in z ′ t exists.

Figure 2.2-8! ! ! ! ! !

Curves in the complex plane: ! Curve C1 is smooth, simple, and not closed. ! Curve C2 is piecewise-smooth, simple, and ! ! ! not closed. ! Curve C3 is closed but not simple. ! Curve C4 is smooth, simple, and closed.

Figure 2.2-9! A piecewise-smooth path consisting of connected straight line segments forming a polygonal line from a point z1 to a point zn . Example 2.2-4 Parameterize the semicircular curve C shown in Figure 2.2-10. 103

Solution: Since r = 1 the curve C can be parameterized as: !

()

z θ = eiθ !

0 ≤ θ ≤ 2π

Figure 2.2-10! Semicircular path.

2.2.9! !

STRUCTURE OF POINT SETS

We will now establish some definitions that describe the

structure of point sets in the complex plane. Connected Set: The internal structure of an open set S in the complex plane is described as connected if any

Figure 2.2-11! S1 – connected set of points showing points z1 ! ! and z2 connected by a polygonal path. ! S2 – starlike set of points. Convex Set: A set S is described as convex if each pair of points of the set can be joined by a single straight line segment where every point in the line segment is also in S . Every convex set in the complex plane is connected.

two points of S can be joined by a polygonal path all

Starlike or Star-shaped Set: A set S is called starlike with

of whose segments lie within S (see set S1 in Figure

respect to a point z0 in S if every other point z in S

2.2-11).

is visible from z0 . The point z0 is called a star center 104

of the set S . A convex set will be starlike with respect

connected (see domain S2 in Figure 2.2-12). A simple

to the point z0 (see set S2 in Figure 2.2-11).

closed curve surrounding a hole in a domain cannot be shrunk to a point while remaining within the

2.2.10! DOMAINS AND REGIONS !

domain.

Point sets in the complex plane can be categorized as

either domains or regions: Domain: An open connected set of points is called a domain. All neighborhoods are therefore domains. A domain cannot contain any of its boundary points. The domain of definition of a function (see Section 2.1.2) is different from domain as defined here. The domain of definition of a function may or may not be a domain as defined here. Simply Connected Domain: A domain is considered to be simply connected if every point inside of any closed curve in the domain lies within the domain. In a simply connected domain, any simple closed curve

Figure 2.2-12! S1 – simply connected domain. ! S2 – multiply connected domain.

can be shrunk to a point while remaining within the

Region: A domain together with all, some, or none of its

domain during the contraction. A simply connected

boundary points is called a region. Therefore region is

domain is then a domain without holes (see domain

a more general designation for a set of points than

S1 in Figure 2.2-12).

domain. A domain is always a region, but a region is

Multiply Connected Domain: If a connected domain is not simply connected, it is considered to be multiply

not always a domain. Open Region: An open region is a domain. 105

Closed Region: A domain together with all its boundary points is known as a closed region or as a region.

3.! The set 1 ≤ z ≤ 2 is closed since it contains points z = 1 and z = 2 which are all its boundary points (see lower

Example 2.2-5 Determine if the following sets of points are domains.

left diagram of Figure 2.2-5). Therefore this set is not a domain, but is a closed region.

1.! Re z > 1 2.! 1 < z < 2 3.! 1 ≤ z ≤ 2 4.! Re z 2 > 0 5.!

Re z > 1

Solution: 1.! The set Re z > 1 is open since all of its points are interior points, and the set is simply connected (see upper left

Figure 2.2-13! Open sets of points that are not connected and so are not domains.

diagram of Figure 2.2-5). Therefore it is a simply connected domain. 2.! The set 1 < z < 2 is open since all of its points are

4.! The set Re z 2 > 0 is an open set that is not connected since the point 0 is not included in the set, and so no continuous path exists between the +x and −x parts of

interior points, and the set is connected (see upper right

the set (see right diagram of Figure 2.2-13). Therefore

diagram of Figure 2.2-5). Therefore it is a domain, but is

this set is neither a domain nor a region.

not a simply connected domain. 106

5.! The set Re z > 1 is an open set that is not connected since points Re z ≤ 1 are not included in the set, and so no continuous path exists between the +x and −x parts of the set (see left diagram of Figure 2.2-13). Therefore this set is neither a domain nor a region.

2.2.11! BOUNDED, UNBOUNDED, AND COMPACT SETS !

Point sets in the complex plane can also be categorized as

bounded or unbounded. Bounded Set: A set of points S in the complex plane is considered bounded if there exists a circle of

Figure 2.2-14! Graphical representation of the bounded set S that is bounded with z ≤ M .

constant radius M such that z ≤ M for every point in S (see Figure 2.2-14). A bounded set of points S is

Unbounded Set: A set of points S in the complex plane is

therefore contained within a closed disk of radius M .

called unbounded if it is not bounded. Therefore a

The real and imaginary parts of a set of bounded

set of points S is unbounded if there exist points of

complex numbers must also be bounded. A set of

S which lie outside a circle of any radius no matter

points S described by a function f z is bounded if

how large. The set of all points in the complex plane

for some M > 0 we have:

is unbounded.

()

!

()

f z ≤ M!

for all z in S !

(2.2-9)

Compact Set: A set of points S is compact if and only if it is closed and bounded. A region specified by 107

z − z0 ≤ M where M > 0 is compact. All closed disks

( )

Dδ z0 are compact. !

Proposition 2.2-9: If the upper bound of a point set S is not an element of S , then

If a set of points S is bounded by M , the upper bound of

it is a limit point of S .

S is defined to be at least one point z of S such that z > M − ε , where ε > 0 can be arbitrarily small. The largest value of a set of

Proof:

real numbers (such as moduli) is often specified using a concept

!

known as the supremum or superior value. The supremum is

S , then by definition of an upper bound, there must exist a

also known as the least upper bound, and is the smallest of all

number of points of S that lie between z and z − ε , where

possible upper bounds for a set. The notation used for this is

ε > 0 can be arbitrarily small. If the number of such points were

sup (as for example sup z ). !

If a set of numbers S has a largest value, there is no

difference between the maximum and the supremum of the set. If a set of numbers S does not have a maximum, then the supremum is the smallest number not in S that has a modulus greater than any number in S. Therefore the maximum of a set always belongs to the set, while the supremum may or may not belong to the set. !

A bounded set of points S on the real number line will

have a least upper bound M > 0 . This is known as the least

If the upper bound z of a point set S is not an element of

to be only finite, then the largest of them would be the upper bound of S instead of z . Therefore an infinity of such points must exist, making z a limit point of S .

■

Example 2.2-6 Determine if the following sets of points are bounded, compact, both, or neither: 1.!

z 0 there

must exist a real number δ > 0 such that:

()

0 < f x − y0 < ε !

!

!

Equation (2.3-4) describes an open set of points within a

disk of radius δ that is centered at point z0 , with only the point

when 0 < x − x0 < δ !

(2.3-2)

()

( )

z0 excluded. This disk is therefore the punctured disk Dδ* z0 , and so equation (2.3-4) describes a deleted neighborhood of the

For any given function f x , the number δ usually depends on

point z0 .

ε (see Figure 2.3-1).

!

2.3.2! !

()

()

( )

defined at all points of the punctured disk Dδ* z0 . For the limit

LIMIT OF A COMPLEX FUNCTION

For a complex function w = f z

( )

For the limit w0 = f z0 to exist, the function f z must be

( )

()

w0 = f z0 to exist, f z need not be defined at the point z = z0

of a single complex

since that point is excluded: 0 < z − z0 . It is necessary, however,

variable z , if w can become arbitrarily close to some w0 as z

that z0 be a limit point of the point set S which is the domain

approaches but do not necessarily reach z0 , we define w0 as the

of definition of f z . This assures that points exist arbitrarily

complex limit of the function w = f z :

close to z = z0 so that f z

()

!

()

( )

lim f z = f z0 = w0 !

z → z0

(2.3-3)

()

exist a real number δ > 0 such that:

!

()

when 0 < z − z0 < δ !

(2.3-4)

()

For any given function f z , the number δ generally depends on ε . The limits in equation (2.3-4) can be expressed in terms of the distance between two points using equation (1.7-4): !

( ()

)

d f z − w0 < ε !

(

)

when 0 < d z − z0 < δ ! (2.3-5)

() f ( z0 ) itself.

()

( )

of a function f z depends upon values of f z close to f z0 , and not upon

f z − w0 < ε !

will be defined in a deleted

neighborhood of z0 . As the point z0 is approached, the limit w0

If the limit w0 exists, then for any real number ε > 0 there must !

()

The deleted neighborhood 0 < z − z0 < δ in the z-plane is

()

mapped by the function w = f z into a deleted neighborhood in the w-plane described by

()

()

f z − w0 < ε where we have

lim f z = w0 . The circle of radius ε in the w-plane will

z → z0

encompass the deleted neighborhood 0 < z − z0 < δ as mapped

()

onto the w-plane by f z . The neighborhoods shown in Figure 2.3-2 then represent the limit process given in equation (2.3-4). 110

Proof: !

Seeking a contradiction, we will assume that two different

() f ( z ) = L1 !

limits exist for f z at the point z = z0 :

lim

!

z → z0

()

lim f z = L2 !

z → z0

(2.3-6)

where L1 ≠ L2 . We then must have for any real number ε > 0 :

ε ! 2

when 0 < z − z0 < δ 1 !

(2.3-7)

ε f z − L2 < ! 2

when 0 < z − z0 < δ 2 !

(2.3-8)

()

f z − L1
0 such that when 0 < z − 2 + i < δ then:

(

) (

)

! 3 z + 2i − 6 + 5 i < ε

■

To determine if such a real number δ exists, we will begin

Proposition 2.3-2:

(

()

) ( ) z − ( 2 + i ) . We start by factoring out 3:

If a complex function f z has a limit at a given point z = z0 ,

by transforming 3 z + 2 i − 6 + 5 i < ε to obtain an equation

the limit must be independent of the direction of approach to the

expressed in terms of

point z0 .

⎛ 2 ⎞ ⎛ 5 ⎞ !3 ⎜ z + i ⎟ − ⎜ 2 + i ⎟ = 3 z − 2 + i < ε 3 ⎠ ⎝ 3 ⎠ ⎝

(

Proof: !

know that the limit: z→2+i

L2 = L1 !

!

From the definition of the limit of a complex function, we

()

Since there cannot be more than one limit for f z

as

z → z0 (see Proposition 2.3-1), the limit must be independent of the direction of approach to the point z = z0 . This conclusion derives directly from the definition of a limit of a complex function.

■

)

Therefore:

(

)

!z − 2+ i
0 :

()

lim f z = f z0 = w0 !

!

Since:

!

()

If w = f z = u z + i v z is defined in a domain D except

If the limit lim f z = w0 exists, then lim f z = w0 .

!

If equations (2.3-20) and (2.3-21) are true, then for any real

number ε > 0 we have: ε ! u − u0 < ! 2

when 0 < z − z0 < δ 1 !

(2.3-22)

and !

(2.3-18)

ε v − v0 < ! 2

when 0 < z − z0 < δ 2 ! (2.3-23)

We have from the triangle inequality (Proposition 1.8-2): !

() ( ) (

) (

)

f z − f z0 = u − u0 + i v − v0 ≤ u − u0 + v − v0 !

(2.3-24) 113

(

)

Let δ = min δ 1 , δ 2 . We then have using equations (2.3-22) and (2.3-23): !

() ( )

f z − f z0
0 we have:

!

z → z0

when 0 < z − z0 < δ ! (2.3-25)

lim f z = f z0 = w0 !

z → z0

()

(2.3-32)

and so if lim f z = f z0 = w0 we must have:

Therefore since ε can be arbitrarily small: !

when 0 < z − z0 < δ !

v − v0 < ε !

!

Proof: (2.3-31)

!

Since: 114

()

lim f z = w0 !

!

z → z0

(2.3-35)

()

f z − w0 < ε !

when 0 < z − z0 < δ !

(2.3-36)

Therefore from Proposition 2.3-4 and equation (2.3-36) we have

() ()

()

since w = f z = u z − i v z :

()

f z − w0 < ε !

!

when 0 < z − z0 < δ !

()

lim f z = w0 !

z → z0

(2.3-37)

(2.3-38)

when 0 < z − z0 < δ

!

(2.3-41)

()

()

lim c f z = c lim f z = c A !

!

z → z0

z → z0

(2.3-42)

Proposition 2.3-7: z → z0

!

()

If lim f z = A and if c is a complex constant, then:

()

()

lim c f z = c lim f z = c A ! z → z0

(2.3-39)

()

()

If lim f z = A and lim g z = B , then

Proposition 2.3-6:

Proof:

()

■

functions.

z → z0

()

c f z − A = c f z −cA < ε!

! !

We can state a number of properties of limits of complex

!

when 0 < z − z0 < δ ! (2.3-40)

and since ε can be arbitrarily small:

2.3.4! PROPERTIES OF LIMITS

z → z0

ε ! c

Therefore:

■

!

()

f z −A
0 :

we must have for any real number ε > 0 : !

()

Since lim f z = A , for some 0 < z − z0 < δ we must have

!

z → z0

() ()

()

()

lim ⎡⎣ f z ± g z ⎤⎦ = lim f z ± lim g z = A ± B ! z → z0 z → z0

z → z0

(2.3-43)

Proof: !

()

z → z0

z → z0

real number ε > 0 : !

()

Since lim f z = A and lim g z = B , we must have for any

()

f z −A
0 , some real number δ > 0

a continuous function.

must exist such that:

!

()

■

From equation (2.4-9) we see that if a complex function

()

g z is continuous at a point z0 , then 1 g z is also continuous

()

at z0 (providing g z ≠ 0 ).

( ( )) ( ( )) < ε ! for all z ∈D ( z ) ! and so f ( g ( z )) is continuous at z . ■ !

() and if complex function f ( w) is continuous at a point w = w0 where w0 = g ( z0 ) , then f ( g ( z )) is also continuous at z0 . Proof: From the condition for continuity of

( )

f w

given in

equation (2.4-4), we know that for every real number ε > 0 , some real number δ > 0 must exist such that:

( )

( )

(2.4-13)

when w − w0 < δ !

(2.4-10)

()

If a complex function w = f z is continuous at a point z = z0 ,

()

then w = f z is continuous at z0 . Proof: !

f w − f w0 < ε !

0

0

If a complex function w = g z is continuous at a point z = z0 ,

!

δ

Proposition 2.4-5:

Proposition 2.4-4:

!

f g z − f g z0

() ()

()

Since w = f z = u z + i v z is continuous at a point z = z0 ,

() () continuous at the point z0 . Since w = f ( z ) = u ( z ) − i v ( z ) then from equation (2.4-7) we can conclude that f ( z ) must also be from Proposition 2.4-2 we know that u z and v z are both

continuous at the point z0 .

■

or 121

will let:

()

If f z is given by:

⎧⎪ z 2 !f z = ⎨ ⎪⎩ 0

()

()

We are given that w = f z is continuous at a point z0 . We

!

Example 2.4-1

ε=

!

z≠i z=i

()

()

Solution:

( )

2

!

when z − z0 < δ !

(2.4-15)

()

()

For z = i + δ we have:

( )

f z0
0 such that:

or

(2.4-24)

z − z0 < δ !

lim g f z = g w0 = g f z0

!

w0 we must have for any real number ε > 0 a real number

!

when ( ( )) ( ) Therefore g ( f ( z )) is continuous at z :

g f z − g w0 < ε !

!

0

⎛ ⎞ lim g f z = g ⎜ lim f z ⎟ ! z → z0 ⎝ z → z0 ⎠

Proof: !

when z − z0 < δ !

0

( ( )) is continuous at z , and we have:

then g f z !

()

f z − w0 < ε !

!

From equations (2.4-23) and (2.4-24) we then have:

lim f z = f z0 = w0 !

z → z0

when w − w0 < δ w ! (2.4-23)

Since w0 is a limit for f z as z → z0 , for any real number ε > 0

Proposition 2.4-7:

!

( ( )) ( )

g f z − g w0 < ε !

!

( ) ( )

g w − g w0 < ε !

when w − w0 < δ w ! (2.4-22)

All complex polynomials:

()

P z = a0 + a1 z + a2 z 2 +!+ an−1 z n−1 + an z n !

!

(2.4-27)

are continuous over the finite complex plane. Proof: 123

()

From Proposition 2.4-1 we know that the function f z = z

!

is continuous for z ∈! .

()

We will now assume that f z = z n is continuous for some

!

n for all z ∈! . We then have:

()

n

f z =z z=z

!

n+1

Proof: !

Follows from Proposition 2.4-8 and equation (2.4-9).

2.5!

!

(2.4-28)

()

!

■

INFINITY IN COMPLEX NUMBERS Real numbers extend to both + ∞ and − ∞ . Since complex

From equation (2.4-8) we see that f z = z n+1 is continuous for

numbers cannot be ordered, however, the concept of ± ∞ has no

all z ∈! . From mathematical induction, we see then that

meaning for complex numbers. Only z

()

and not z can be

f z = z n is continuous for all z ∈! . Moreover since all

considered to approach infinity, and z extends only to + ∞ .

complex constants are continuous, from equation (2.4-8) we see

There is no − ∞ in complex numbers. Moreover, since only z

that each product of a complex constant coefficient and a power

can approach ∞ , only one point at + ∞ is considered to exist for

of z is continuous. Finally, using equation (2.4-7) we see that

complex numbers, irrespective of the direction from which this

the sum of all the terms in equation (2.4-27) is continuous.

infinity is approached. For a complex number, neither the real

Therefore we can conclude that every complex polynomial is

and imaginary parts nor the argument have any meaning at

continuous in the point set ! , and so is continuous over the

infinity.

finite complex plane.

!

■

Proposition 2.4-9, Continuity of Complex Rational Functions:

infinity: !

z ± ∞ = ∞!

!

( z ) (∞) = ∞ !

z ≠ 0!

(2.5-2)

!

z = ∞! 0

z ≠ 0!

(2.5-3)

All complex rational functions: !

()

f z =

a0 + a1 z + a2 z 2 + !+ an−1 z n−1 + an z n 2

b0 + b1 z + b2 z +!+ bm−1 z

m−1

+ bm z

m

!

(2.4-29)

()

where an ≠ 0 and bm ≠ 0 , and where the degree of f z is

(

)

A complex number z has the following operations with

(2.5-1)

max m, n , are continuous over the finite complex plane. 124

z = 0! ∞

! !

z ≠ ∞!

(2.5-4)

When the point at infinity is not included in the set of

complex numbers, the set ! is known as the finite complex

through the middle of a sphere having a unit radius with a center at the origin of a coordinate system in R 3 (point O in Figure 2.6-1). The north pole (denoted by N in Figure 2.6-1) has

(

)

coordinates 0, 0, 1 in R 3 .

number system or the finite complex plane. When we refer to the complex plane, we will always mean the finite complex plane. When the point at infinity is included in the set of complex numbers, the set !∞ is known as the extended complex number system or the extended complex plane. The extended complex plane is considered to be a closed surface consisting of the finite complex plane together with the single point at infinity: !∞ = ! ∪ ∞ . !

If a region is unbounded, then the region is said to

contain the point at infinity. The region z > R > 0 where R can be arbitrarily large is known as a neighborhood of infinity. That is, the region outside a disk centered at the origin and having a radius of arbitrarily large size is termed a neighborhood of the point z = ∞ .

2.6! !

RIEMANN SPHERE The extended complex plane can be visualized using a 3

Figure 2.6-1!

!

Stereographic projection of complex numbers onto a Riemann sphere.

Any point z in the complex plane can be projected onto

the Riemann sphere by drawing a straight line from the north

three-dimensional surface in R known as a Riemann sphere.

pole of the sphere to the point z . The line will intersect the

This sphere is constructed by passing the complex plane

sphere at only one point. The point P where this line intersects 125

the sphere is the projection of the point z onto the sphere. The

parallel to the extended complex plane. For a point at infinity

point z is then mapped onto the Riemann sphere at the point P.

then, this line will not intersect the sphere.

This type of mapping from a plane to a sphere (or from a

!

sphere to a plane) is known as stereographic projection.

are points on the sphere close to N . The set of points z > 1 δ

For a very small positive number δ , we see that z = 1 δ

Points of the complex plane on the circle z = 1 (where the

at the north pole of the Riemann sphere can be considered a

Riemann sphere intersects the complex plane) are mapped onto

neighborhood of the point at infinity. Letting δ → 0 , we will

the equator of the Riemann sphere. Points of the complex plane

have z → ∞ . The point N at the north pole then corresponds

with z < 1 (inside the sphere) are

to the point at infinity. Only one point on the Riemann sphere

mapped onto the Riemann southern hemisphere. The origin

corresponds to infinity, regardless of the direction in the

z = 0 of the complex plane is mapped onto the south pole of the

extended complex plane. The Riemann sphere allows us to

Riemann sphere. Points of the complex plane outside the circle

visualize the meaning of infinity for complex numbers.

!

()

in the open disk D1 0

z > 1 (outside the sphere) are mapped onto the Riemann northern hemisphere.

2.7!

!

!

Only one point P on the Riemann sphere will correspond

on the Riemann sphere. The mapping from the complex plane No point in ! corresponds to the north pole of the

Riemann sphere. The point at infinity in the extended complex

()

If z0 is a point on the complex plane, then: !

plane is taken to be the north pole of the Riemann sphere. As

z → ∞ , the line from the north pole N to z will become

()

Proposition 2.7-1:

to the Riemann sphere is therefore one-to-one. !

For complex numbers we will define z → ∞ to mean

z → ∞ and f z → ∞ to mean f z → ∞ .

to any point z in the complex plane ! . Conversely, only one point z in the complex plane ! will correspond to any point P

LIMITS TENDING TO INFINITY

()

lim f z = ∞ !

z → z0

(2.7-1)

if and only if: !

lim

z → z0

1

()

f z

= 0!

(2.7-2) 126

Proof:

Proof:

!

!

For equation (2.7-1) to be true we must have for any real

number ε > 0 a real number δ > 0 such that:

1 f z > ! ε

()

!

when 0 < z − z0 < δ !

number ε > 0 a real number δ > 0 such that: (2.7-3)

1 −0 0 can be arbitrarily small.

()

f z − w0 < ε !

!

■

⎛ 1⎞ f ⎜ ⎟ − w0 < ε ! ⎝ z⎠

!

(2.7-7)

when 0 < z − 0 < δ !

(2.7-8) ■

Proposition 2.7-3: If z0 is a point on the complex plane, then:

If z0 is a point on the z-plane and w0 is a point on the w-plane,

()

lim f z = w0 !

z →∞

z → z0

(2.7-9)

if and only if: (2.7-5)

⎛ 1⎞ lim f ⎜ ⎟ = 0 ! z → z0 ⎝ z ⎠

! if and only if:

⎛ 1⎞ lim f ⎜ ⎟ = w0 ! z→0 ⎝ z ⎠

()

lim f z = ∞ !

!

then:

!

1 ! δ

we have equation (2.7-6) since ε can be arbitrarily small.

Proposition 2.7-2:

!

when 0 < z
0 a real number δ > 0 such that: 127

1 f z > ! ε

()

!

when z >

1 ! δ

2.8!

(2.7-11)

!

Rewriting equation (2.7-9) with 1 z replacing z :

⎛ 1⎞ f ⎜ ⎟ −0 0 be a real constant. If f x

is a real-valued

()

function that is continuous in a region R , and if f x ≤ M for ! ■

in agreement with Proposition 2.7-1.

()

Proposition 2.8-1:

()

()

If the limit lim f z = w0 exists where w0 is finite, then f z z → z0

is a bounded function in some neighborhood of the point z = z0 . Proof: ! !

We also have:

z −1 !lim =0 z →1 5z − 2

()

considered to be bounded in R (see Section 2.2.11).

Solution:

5z − 2 !lim =∞ z →1 z − 1

()

Similarly, if f z is a complex function that is continuous

in a region R , and if f z ≤ M for all points in R , then f z is

5z − 2 z −1

We have:

()

all points in R , then f x is considered to be bounded in R .

(2.7-12)

we have equation (2.7-10) since ε > 0 can be arbitrarily small.

!f z =

BOUNDED COMPLEX FUNCTIONS

Since:

()

lim f z = w0 !

z → z0

()

then f z

(2.8-1)

is continuous in some neighborhood of the point

z = z0 , and for any real number ε > 0 we have: !

()

f z − w0 < ε !

when 0 < z − z0 < δ !

(2.8-2)

We can write for any neighborhood of z0 : 128

!

()

()

()

f z = f z − w0 + w0 ≤ f z − w0 + w0 < ε + w0 !

(2.8-3)

Proposition 2.9-1, Bolzano-Weierstrass Theorem for Point Sets in the Complex Plane:

Let:

M = ε + w0 !

!

If S is a bounded infinite set of points in the complex plane, then

(2.8-4)

S has a limit point.

We then have:

()

f z < M!

!

(2.8-5)

()

Proof: !

Since the infinite set of points S is bounded, we will

and so f z is a bounded function in some neighborhood of the

enclose it within a finite-diameter square A0 having sides L

point z0 .

that are parallel to the coordinate axes. We can now partition

■

A0 into four equal squares using lines parallel to the sides L .

Proposition 2.8-2:

Since S contains an infinite number of points, at least one of the

()

If a complex function f z has a limit at the point z = z0 , then

()

f z must be bounded in a neighborhood of z0 .

Follows from Proposition 2.8-1.

2.9! !

obtain square A2 , also containing infinitely many points. ■

BOLZANO-WEIERSTRASS THEOREM If an infinite set of points S is bounded, then we can

expect that these points will have a point about which they will gather.

this square A1 (see Figure 2.9-1). We then follow the same procedure by partitioning square A1 into four equal squares to

Proof: !

four squares also contains infinitely many points. We will label

!

Repetitively following this procedure, we will obtain an

infinite sequence of nested squares: A0 ⊃ A1 ⊃ A2 ⊃ ! ⊃ An ! , with each square An only containing points also existing in the square An−1 . The side of each square An is L 2 n . Since each of the nested squares will have a finite diameter, the points of S within each of the squares is bounded. In the limit as the number of squares n → ∞ , we will obtain a square that is infinitesimally small containing infinitely many points of S . 129

Figure 2.9-1! Nested squares (in green). Large green square A0 has sides of length L . !

Let z = z0 be a point in the nth square as n → ∞ . For any

given ε > 0 and for some integer N where n > N , all the points

( )

in the nth square will be contained within a disk Dδ z0 radius δ = L

(2 2 ) < ε . n

of

This disk then contains infinitely

many points. For every nth square then, every δ neighborhood of z0 contains a limit point for the set of points within the disk. Since z0 is a point within all the nested squares, z0 is therefore a limit point for the infinite set of points S .

■

130

Chapter 3 Complex Differentiation

∂u ∂v ! = ∂x ∂ y

∂u ∂v =− ∂y ∂x

131

!

In this chapter we will use the concepts of limit and

continuity of complex functions to extend the calculation of derivatives of real-valued functions to derivatives of complex functions. We will determine the necessary and sufficient conditions for the derivative of a complex function to exist. In so doing we will develop the Cauchy-Riemann differential equations which provide special relations that must hold between the real and imaginary parts of a differentiable complex function. Finally, we will define an important class of complex functions known as holomorphic functions.

3.1! !

DERIVATIVE OF A COMPLEX FUNCTION Just as the concept of limits is essential to the differential

calculus of real-valued functions, so it is with complex valued functions. Since there are differences in the limiting processes of real and complex functions, so too differences exist between the

Δx = x − x0 !

!

(3.1-1)

()

so that x = x0 + Δx . If f x

is a real function defined in the

()

neighborhood of x0 , then the change in the function f x due

() ! Δ f ( x ) = f ( x ) − f ( x0 ) = f ( x0 + Δx ) − f ( x0 ) ! (3.1-2) ! By definition, the derivative f ′ ( x0 ) at a given point x0 of a real-valued function f ( x ) is the limit: f ( x ) − f ( x0 ) f ( x0 + Δx ) − f ( x0 ) Δ f ( x) x f ′ ( ) = lim = lim = lim to the change from point x0 to the point x is Δ f x , where:

0

Δx →0

Δx

x − x0

x → x0

Δx

Δ x →0

!!

(3.1-3)

() be differentiable at the point x0 . The derivative of f ( x ) is then

provided that the limit exists. If this limit exists, f x is said to specified with reference to the given point x0 . We must have

()

() ( ) ( ) ( ) f ′ ( x0 ) to exist at the point x0 . Therefore f ( x )

Δ f x = f x − f x0 = f x0 + Δx − f x0 → 0 as Δx → 0 for the limit

must be

differential calculus of real and complex functions.

continuous at point x0 .

3.1.1!

point x0 as Δx → 0 . For the limit f ′ x0 to exist, it is necessary

!

DERIVATIVE OF REAL-VALUED FUNCTIONS OF A REAL VARIABLE

Given some point x0 on the x-axis, we will consider

another point x which is located near x0 where we will define:

!

There are only two directions of approach to the given

( )

that the same limit obtain as x approaches the point x0 from

(

)

(

)

either the plus side x > x0 or the minus side x < x0 of x0 . The limiting process for real functions of a real variable is therefore one-dimensional. 132

3.1.2!

DERIVATIVE OF COMPLEX FUNCTIONS OF A COMPLEX VARIABLE

Given some point z0 in an open set in the complex plane,

!

we will consider another point z which is located within a neighborhood of z0 , where we will define: (3.1-4)

(

) (

)

Δz = x − x0 + i y − y0 = Δx + i Δ y !

()

()

limit

(3.1-5)

()

z0 , then the change in the function f z due to the change from

() Δ f ( z ) = f ( z ) − f ( z0 ) = f ( z0 + Δz ) − f ( z0 ) !

point z0 to the point z is Δ f z , where:

(3.1-6)

( )

By definition, the derivative f ′ z0 at a given point z0 of

!

()

()

For a function f z to be complex differentiable at a point

z0 , there must then exist for any real number ε > 0 a real

()

( )

Δz →0

( ) = lim f ( z ) − f ( z0 ) = lim f ( z0 + Δz ) − f ( z0 )

Δf z Δz

z → z0

z − z0

Δz

Δ z→ 0

! !

(3.1-7)

provided that the limit exists. For the limit to exist we know

()

from Proposition 2.8-2 that f z

must be bounded in a

()

neighborhood of z0 . If the limit exists, f z

is said to be

( )− f′

f z − f z0

!

z − z0

( z0 ) < ε !

when z − z0 < δ ! (3.1-8)

or

( )− f′

Δf z

!

Δz

( z0 ) < ε !

when Δ z < δ !

(3.1-9)

The number δ will generally depend on ε , but the derivative

( )

f ′ z0

a complex function f z is the limit:

f ′ z0 = lim

must be

continuous at point z0 (see Proposition 3.3-1).

If f z is a complex function defined in the neighborhood of

!

() ( ) ( ) ( ) f ′ ( z0 ) to exist at the point z0 . Therefore f ( z )

Δ f z = f z − f z0 = f z0 + Δz − f z0 → 0 as Δz → 0 for the

number δ > 0 such that:

so that z = z0 + Δz . From equation (3.1-4) we have: !

specified with reference to the given point z0 . We must have

!

Δz = z − z0 !

!

()

differentiable at the point z0 . The derivative of f z is then

is independent of both ε and Δz since the derivative

()

represents the actual spatial rate of change in the function f z at the given point z0 . From equation (3.1-9), we have: !

()

( )

Δ f z = f ′ z0 Δ z + ε Δ z !

(3.1-10)

which gives us the mean value theorem: !

()

( ) (

) ( ) (

)

f z = f z0 + z − z0 f ′ z0 + ε z − z0 !

(3.1-11) 133

From equations (3.1-7), (3.1-4), and (3.1-6) with Δz → 0 , we

!

( )

can write the derivative f ′ z0 at a given point z0 in the form:

()

df z

( )

f ′ z0 =

!

dz

= lim z = z0

( ) = lim f ( z0 + Δ z ) − f ( z0 ) !

Δf z

Δz → 0

Δz

Δz

Δz → 0

3.2! !

EXISTENCE OF THE DERIVATIVE Since Δz = Δx + i Δ y is a complex number, for Δz → 0 we

must have both Δx → 0 and Δ y → 0 . The limiting process for complex functions of a complex variable then involves both Δx

! !

(3.1-12)

and Δ y . How Δz → 0 is dependent upon how both Δx → 0 and

We will now omit the subscript 0 from z0 , and we will use

Δ y → 0 . The limiting process for complex functions of a

simply z to represent the point at which we wish to obtain the

complex variable is therefore two-dimensional. There are an

derivative:

infinity of directions of approach in the complex plane to a

!

()

f′ z =

( ) = lim

df z dz

Δz → 0

( ) = lim (

Δf z Δz

)

point z0 as Δz → 0 .

( )!

f z + Δz − f z

Δz → 0

Δz

(3.1-13)

!

( )

For the limit f ′ z0

to exist at a point z0 , it is necessary

that the same limit be determined as z0 is approached from any direction in the z-plane (see Proposition 2.3-2). That is, the same

Example 3.1-1

()

()

Determine the derivative f ′ z of the function f z = 4 z .

limit must be obtained independent of the argument of Δz . !

()

Since the derivative of a complex function f z at a point

Solution:

is independent of the direction along which the derivative is

We can write:

taken, f z will be differentiable at the point only if the spatial

()

!f ′ z =

()

( ) = lim f ( z + Δ z ) − f ( z ) = lim 4 ( z + Δ z ) − 4 z

df z dz

Δz → 0

or

Δz

Δz → 0

Δz

()

rate of change in f z at the point is the same no matter from what direction the point is approached. !

The two-dimensional direction-independent requirement

for the existence of a derivative of a complex function provides

()

! f ′ z = lim

Δ z→0

4 Δz =4 Δz

a much more stringent condition that must be satisfied than does the analogous one-dimensional existence requirement for 134

a derivative of a real-valued function. As a result, the number of complex functions that are differentiable is greatly restricted. Many relatively simple complex functions are not differentiable

Therefore:

()

! f ′ z = lim

(

)

( ) = lim

f z + Δz − f z Δz

Δz → 0

3Δx + i Δ y Δ x → 0 Δx + i Δ y Δy→0

(see Examples 3.2-1 and 3.2-3), and those that are differentiable have properties very different from their real counterparts.

Letting Δz → 0 along a path parallel to the y-axis so that

!

Δx = 0 and Δz = Δy , we have:

In particular, complex functions have one property not

()

commonly found in real functions. If a complex function f z

has one derivative at a point, it will have an infinity of higher

()

! f ′ z = lim

Δy → 0

i Δy =1 i Δy

order derivatives at the same point (see Proposition 6.2-1). By

Letting Δz → 0 along a path parallel to the x-axis so that

comparison, if a real function f x

Δ y = 0 and Δz = Δx , we have:

()

has one derivative at a

point, it may have no, some, or many higher order derivatives

()

3Δx =3 Δx → 0 Δx

! f ′ z = lim

at the same point. Example 3.2-1

()

Show that the continuous function f z = 3x + i y is not differentiable at any point in the complex plane. Solution:

(

()

)

()

Since the limit f ′ z = lim ⎡⎣ f z + Δz − f z ⎤⎦ Δz depends Δz → 0 upon the direction of approach to a point z , no derivative

()

of the function f z = 3x + i y exists at any point in the complex plane despite the fact that this function is

We will choose two paths to compute a limit of the function

() ! f ( z + Δz ) − f ( z ) = ⎡⎣3 ( x + Δx ) + i ( y + Δ y ) ⎤⎦ − ⎡⎣3x + i y ⎤⎦

continuous at all points in the complex plane.

f z = 3x + i y as we approach a point z . We have:

Example 3.2-2

()

Show that the function f z = z 2 is differentiable at all

or

(

)

()

! f z + Δz − f z = 3Δx + i Δ y

points in the complex plane. 135

Solution:

Letting Δz → 0 along a path parallel to the y-axis so that

We have:

Δx = 0 and Δz = Δ y , we have:

()

! f ′ z = lim

Δz → 0

(

)

( ) = lim

f z + Δz − f z Δz

( z + Δz )2 − z 2

()

! f ′ z = lim

Δz

Δz → 0

or

Δx = 0 Δy → 0

Δx − i Δ y −iΔy = lim = −1 Δx + i Δ y Δ y → 0 i Δ y

Letting Δz → 0 along a path parallel to the x-axis so that

()

! f ′ z = lim

Δz → 0

z 2 + 2 z Δz + Δz 2 − z 2 Δz

(

)

= lim 2 z + Δz = 2 z Δz → 0

Since the limit is independent of the path of approach Δz to

()

Δ y = 0 and Δz = Δx , we have:

()

Δy = 0

2

a point z , the derivative of the function f z = z exists at

Δx − i Δ y Δx = lim =1 Δ x → 0 Δx + i Δ y Δ x → 0 Δx

! f ′ z = lim

Since the limits depend upon the path of approach to a

()

all points in the complex plane.

point z , no derivative of the function f z = z exists at any point in the complex plane, despite the fact that this

Example 3.2-3

function is continuous at all points in the complex plane.

()

Show that the function f z = z is not differentiable at any

The dependence of the limit on direction of approach can

point in the complex plane.

also be shown by expressing Δz in exponential form

Δz = Δr eiθ . We then have:

Solution: We can write:

z + Δz ) − ( z ) z + Δz ) − ( z ) ( ( Δz ! f ′ ( z ) = lim = lim = lim Δz →0

!

Δz

Δx − i Δ y = lim Δ x → 0 Δx + i Δ y

Δz →0

Δz

Δz →0

Δz

Δz Δr e− iθ − 2iθ ! f ′ z = lim = lim = e Δ z → 0 Δz Δr → 0 Δr eiθ

()

where the dependence of the limit lim on the direction of approach θ is clear.

Δz →0

Δy→0

136

3.3! !

DIFFERENTIATION RULES

!

n−1 n d ⎡⎣ f z ⎤⎦ = n ⎡⎣ f z ⎤⎦ f ′ z ! dz

!

d ⎡ f z ⎤ g z f ′ z − f z g′ z ! ⎢ ⎥= 2 dz ⎢⎣ g z ⎥⎦ ⎡⎣ g z ⎤⎦

!

d d ⎡ ⎤ = f g′ z ! f g z ⎦ dg dz ⎣

The existence of a limit of a complex function requires the

existence of limits for the real and imaginary parts of the function (see Proposition 2.3-4). The operational rules for limits then apply equally to real-valued functions and complex functions. Therefore the sum, product, quotient, and chain differentiation rules that apply for real-valued functions also

()

() ()

()

()

(3.3-6)

() () () () ()

⎡⎣ g z ⎤⎦ ≠ 0 ! (3.3-7)

( ( ))

()

2

()

(3.3-8)

apply for complex functions (assuming all the derivatives exist at the given point). If c is a complex constant, n is an integer,

()

()

and f z and g z are complex functions, we then have: !

!

!

dc = 0! dz

(3.3-1)

dz = 1! dz

(3.3-2)

Example 3.3-1 Show that

d ⎡⎣ f z g z ⎤⎦ = f z g ′ z + f ′ z g z . dz

() ()

() ()

() ()

Solution:

()

(3.3-3)

Δz → 0

d ⎡⎣ f z ± g z ⎤⎦ = f ′ z ± g ′ z ! dz

(3.3-4)

!

d ⎡ f z g z ⎤⎦ = g z f ′ z + f z g ′ z ! dz ⎣

(3.3-5)

() ()

) (

)

() ()

() (

)

Adding and subtracting lim f z g z + Δz Δz :

!

() ()

(

f z + Δz g z + Δz − f z g z d ! ⎡⎣ f z g z ⎤⎦ = lim Δz → 0 dz Δz

() ()

d ⎡⎣ c f z ⎤⎦ = c f ′ z ! dz

()

We can write:

()

() ()

()

() ()

(

) ()

g z + Δz − g z d ! ⎡⎣ f z g z ⎤⎦ = lim f z Δz → 0 dz Δz

() ()

()

!+ lim

Δz → 0

(

)

( )g

f z + Δz − f z Δz

( z + Δz ) 137

and we have: ! !

Using Proposition 2.3-8 we have:

d ⎡⎣ f z g z ⎤⎦ = f z g ′ z + f ′ z g z dz

!

The continuity of a complex function at a given point does

or

() ()

() ()

() ()

not assure that the function will be differentiable at the point (see examples 3.2-1 and 3.2-3). For a complex function to be differentiable at a point, however, the function must be continuous at the point.

()

()

If f z is differentiable at a point z = z0 , then f z is

!

z → z0

( )

z → z0

Δz

z → z0

!

( )}

()

()

(

■

is differentiable at a point z = z0 , then the real and

()

(

)

Proof:

()

Since f z is differentiable at point z = z0 , we have from

Proposition 3.3-1 that

( )

( )

Proposition 3.3-2:

can write:

( )

( )

(3.3-13)

!

()

(3.3-12)

!

(3.3-9)

z − z0

⎡ f z − f z0 lim ⎡⎣ f z − f z0 ⎤⎦ = lim ⎢ z → z0 z → z0 z − z0 ⎢⎣

(3.3-11)

imaginary parts of f z are continuous at z0 .

Δf z

z → z0

()

( )( )

{( )

()

If f z

( ) = lim f ( z ) − f ( z0 ) !

( z − z0 ) !

lim f z = lim f z0 + ⎡⎣ f z − f z0 ⎤⎦ = f z0 + 0 = f z0 z → z0

z → z0

and so from Proposition 2.3-9, the limit lim z − z0 exists. We

!

z → z0

at a point z0 , as given in equation (2.4-3).

()

( )

z − z0

()

Since f z is differentiable at point z0 , the limit given in

f ′ z0 = lim

( ) lim

Therefore from Proposition 2.3-7:

equation (3.1-7) exists: !

()

()

f z − f z0

This is just the definition of a function f z that is continuous

continuous at z0 . Proof:

( )

lim ⎡⎣ f z − f z0 ⎤⎦ = f ′ z 0 = 0 !

!

!

Proposition 3.3-1, Differentiable Functions are Continuous:

()

lim ⎡ f z − f z0 ⎤⎦ = lim z → z0 ⎣ z → z0

⎤ z − z0 ⎥ ! (3.3-10) ⎥⎦

)

()

f z

is continuous at

z0 . From

Proposition 2.4-2 we know that the real and imaginary parts of

()

f z must then be continuous at z0 .

■

138

Proposition 3.3-3:

()

If f z

is differentiable at a point z = z0 , then the real and

()

!

( )

f ′ z0 = lim

z → z0

imaginary parts of f z are differentiable at z0 . Proof: !

( ) ( ( )) f ( z ) − f ( z0 ) ! g ( f ( z )) − g ( f ( z0 ))

Using the inverse function z = g w = g f z , we have:

or

()

Since f z is differentiable at point z0 , the limit given in

!

( )

f ′ z0 = lim

equation (3.1-13) exists. From Proposition 2.3-4 we know that

()

z → z0

1

( ( )) ( ( ))

g f z − g f z0

()

()

at z0 . Therefore the real and imaginary parts of f z

are

!

( )

( )

function z = g w exists and is differentiable at w0 = f z0 with

( )

g ′ w0 ≠ 0 , then we have:

( )

f ′ z0 =

!

1

( )

g ′ w0

( )

f ′ z0 = lim

(3.3-14)

! !

( )

By definition of a derivative f ′ z0 is given by:

( )

f ′ z0 = lim

z → z0

()

( )!

f z − f z0 z − z0

( ) ( )

g w − g w0

!

(3.3-18)

Letting:

( )

h w =

( ) ( )!

g w − g w0

(3.3-19)

w − w0

In the limit as w → w0 , we obtain: !

Proof:

z → z0

1

w − w0

!

!

( )

()

Proposition 3.3-4:

()

(3.3-17)

With w = f z we can write:

■

If w = f z is differentiable at a point z = z0 , and the inverse

!

f z − f z0

the limits of the real and imaginary parts of f z will also exist differentiable at z0 .

(3.3-16)

( )

( )

h w0 = g ′ w0 !

!

(3.3-20)

and so equation (3.3-18) can be written: (3.3-15)

!

( )

f ′ z0 = lim

z → z0

1 1 ! = lim z → z h w 0 h f z

( )

( ( ))

(3.3-21) 139

()

( )

Since f z is differentiable at z0 , and g w is differentiable at

( ( ))

w0 , then z = g f z

is continuous at z0 and we can use

Proposition 2.4-7 to write:

( )

f ′ z0 =

!

()

f′ z ≡

!

⎛ ⎞ h ⎜ lim f z ⎟ ⎝ z → z0 ⎠

()

=

1

( )

h w0

!

(3.3-22)

( )

f ′ z0 =

( )

g ′ w0

!

!

(3.3-23)

CAUCHY-RIEMANN EQUATIONS

(

()

(3.4-2)

()

)

(

)

()

()

(

()

()

()

()

point z = x + i y in a domain D . Writing the function w = f z in the form:

() ( )

( )

w = f z = u z + i v z = u x, y + i v x, y !

the derivative is given by equation (3.1-7):

(3.4-1)

)

( )

3.4.1!

derivative of a complex function at a point. We will consider the derivative f ′ z of a complex function w = f z at a given

(

!

( )

⎡⎣u x, y + i v x, y ⎤⎦ ! (3.4-4) − Δlim x→ 0 Δ x + iΔ y Δy→0

!

the necessary and sufficient conditions for the existence of a

)

⎡⎣u x + Δx, y + Δy + i v x + Δx, y + Δy ⎤⎦ lim ! f′ z = Δx→ 0 Δx + iΔy Δy → 0

We will now develop two partial differential equations

() ()

Δz

Δz → 0

or

known as the Cauchy-Riemann equations that together provide

!

Δz

⎡⎣u z + Δz + i v z + Δz ⎤⎦ − ⎡⎣ u z + i v z ⎤⎦ ! (3.4-3) f ′ z = lim Δz → 0 Δz

!

■

3.4!

Δz →0

the real and imaginary parts of w = f z must exist:

and so: !

dz

( ) = lim f ( z + Δz ) − f ( z ) !

Δf z

If this derivative exists, then from Proposition 2.3-4 the limits of

1

1

( ) = lim

df z

NECESSARY CONDITIONS FOR THE EXISTENCE OF A DERIVATIVE

()

If the derivative f ′ z given in equation (3.4-4) exists at a

( )

point z x, y in a domain D , the limits in this equation cannot

( )

depend upon the path of approach to z x, y as Δz → 0 since each limit must be unique (see Proposition 2.3-1). To examine the necessary conditions for the existence of the derivative

()

f ′ z , we will choose two different paths of approach within

( )

the domain D to the point z x, y as shown in Figure 3.4-1: a 140

path parallel to the real axis and a second path parallel to the imaginary axis.

()

f′ z =

! !

∂u ∂v +i ! ∂x ∂x

(3.4-6)

Now lettings Δz → 0 parallel to the y-axis so that Δx = 0

and Δz = Δ y → 0 (see Figure 3.4-1), equation (3.4-4) becomes:

(

) ( )

(

) ( )

⎡⎣u x, y + Δ y − u x, y ⎤⎦ + i ⎡⎣v x, y + Δ y − v x, y ⎤⎦ f ′ z = lim Δy →0 i Δy ! ! (3.4-7)

()

or from the definition of real partial derivatives

()

f′ z =

! !

1 ∂u ∂v ∂v ∂u ! + = −i i ∂y ∂y ∂y ∂y

()

Since the function w = f z

( )

(3.4-8)

is differentiable at a point

z x, y , the derivatives given in equations (3.4-6) and (3.4-8), Figure 3.4-1! !

( )

Two paths of approach to a point z x, y .

Letting Δz → 0 parallel to the x-axis so that Δ y = 0 and

Δz = Δx → 0 (see Figure 3.4-1), equation (3.4-4) becomes:

(

) ( )

(

) ( )

which were calculated for two different directions of approach, must exist and be equal. We then can write: !

()

f′ z =

( ) = ∂u + i ∂v = ∂v − i ∂u !

df z dz

∂x

∂x

∂y

∂y

(3.4-9)

⎡⎣u x + Δ x, y − u x, y ⎤⎦ + i ⎡⎣v x + Δ x, y − v x, y ⎤⎦ f ′ z = lim Δx → 0 Δx ! ! (3.4-5) Since the derivative f ′ z exists, the limits of the real and

Equating the real and imaginary parts, we obtain:

imaginary parts of equation (3.4-5) exist. From the definition of

These two linear partial differential equations are known as the

real partial derivatives, we then have:

Cauchy-Riemann equations.

()

()

!

∂u ∂v ! = ∂x ∂ y

∂u ∂v =− ! ∂y ∂x

(3.4-10)

141

!

()

() ( )

( )

If the derivative f ′ z of a function f z = u x, y + i v x, y

( ) of u ( x, y ) and v ( x, y ) with respect to

exists at a point z x, y , then the first order partial derivatives

x and y must exist and

must satisfy the Cauchy-Riemann equations at this point. The Cauchy-Riemann equations therefore constitute necessary conditions for a complex function to be differentiable at a point.

3.4.2! !

Given that we considered only two paths of approach to

( )

the point z x, y , it is not obvious that the Cauchy-Riemann equations constitute sufficient conditions for any complex

() ()

() ( )

( )

w = f z = u z + i v z = u x, y + i v x, y !

!

(3.4-11)

()

to have a derivative f ′ z at a given point so that we have:

∂u ∂v ∂v ∂u ! f′ z = +i = −i ∂x ∂x ∂ y ∂y

()

! !

(3.4-12)

We will now show that if the partial derivatives ∂u ∂x ,

Since we are assuming that the partial derivatives exist

( )

and are continuous in a neighborhood of the point z x, y , we can write Δw = Δu + i Δv using: !

⎛ ∂u ⎞ ⎛ ∂u ⎞ Δu = ⎜ + ε1 ⎟ Δx+ ⎜ + η1 ⎟ Δ y ! ⎝ ∂x ⎠ ⎝ ∂y ⎠

(3.4-13)

⎛ ∂v ⎞ ⎛ ∂v ⎞ Δv = ⎜ + ε 2 ⎟ Δx+ ⎜ + η2 ⎟ Δ y ! ⎝ ∂x ⎠ ⎝ ∂y ⎠

(3.4-14)

and !

SUFFICIENT CONDITIONS FOR THE EXISTENCE OF A DERIVATIVE

function:

!

where ε1 , ε 2 → 0 and η1 , η2 → 0 as Δx → 0 and Δ y → 0 . We then

( )

have at the point z x, y : !

⎛ ∂u ⎛ ∂u ∂v ⎞ ∂v ⎞ Δw = Δu + i Δv = ⎜ + i ⎟ Δx+ ⎜ + i ⎟ Δy ∂x ⎠ ∂y⎠ ⎝ ∂x ⎝ ∂y

(

where ε = ε1 + i ε 2 and η = η1 + i η 2 .

z x, y

equation (3.4-16) as:

()

equations hold at this point, then the derivative f ′ z given in equation (3.4-12) exists.

)

⎛ ∂u ⎛ ∂u ∂v ⎞ ∂v ⎞ ! Δw = ⎜ + i ⎟ Δx + ⎜ + i ⎟ Δ y+ ε Δx+ η Δ y ! (3.4-16) ∂x ⎠ ∂y⎠ ⎝ ∂x ⎝ ∂y

!

within a domain D , and if the Cauchy-Riemann

(

or

∂u ∂ y , ∂v ∂x , and ∂v ∂ y exist and are continuous at a point

( )

)

+ ε1 + i ε 2 Δx+ η1 + i η2 Δ y ! (3.4-15)

!

Using the Cauchy-Riemann equations, we can rewrite

⎛ ∂u ⎛ ∂v ∂v ⎞ ∂u ⎞ ! Δw = ⎜ + i ⎟ Δx+ ⎜ − + i ⎟ Δ y+ ε Δx+ η Δ y ! (3.4-17) ∂x ⎠ ∂x ⎠ ⎝ ∂x ⎝ ∂x 142

⎛ ∂u ∂v ⎞ Δw = ⎜ + i ⎟ Δx+ i Δ y + ε Δx+ η Δ y ! ∂x ⎠ ⎝ ∂x

(

!

)

()

and is unique if f z has continuous first partial derivatives (3.4-18)

equations therefore constitute sufficient conditions for a

Δw ⎛ ∂u ∂v ⎞ Δx Δy ! =⎜ +i ⎟ +ε +η Δz ⎝ ∂x ∂x ⎠ Δz Δz

!

(3.4-19)

and so:

lim

Δz →0

Δw ⎛ ∂u ∂v ⎞ Δx Δy ! − ⎜ + i ⎟ = lim ε +η Δ z → 0 Δz ⎝ ∂x ∂x ⎠ Δz Δz

( )

point z x, y .

DERIVATIVE OF A COMPLEX FUNCTION IN RECTANGULAR FORM

(3.4-20) !

Δw ⎛ ∂u ∂v ⎞ Δx Δy ! lim − ⎜ + i ⎟ ≤ lim ε +η Δ z → 0 Δz ∂x ⎠ Δ z → 0 Δz Δz ⎝ ∂x

From equations (3.4-23) and (3.4-10), we then have the

Δz ≤ 1 and Δy

( )

complex function at a point z x, y : (3.4-21) !

Δz ≤ 1 . Since ε → 0 and η → 0

as Δz → 0 , we have:

Δw ∂u ∂v ! lim = +i Δ z → 0 Δz ∂x ∂x

()

()

! !

(3.4-24)

(3.4-22) !

dw ∂u ∂v ! f′ z = = +i dz ∂x ∂x

dw ∂u ∂v ∂v ∂u ∂u ∂u ∂v ∂v = f′ z = +i = −i = −i = +i ! dz ∂x ∂x ∂y ∂y ∂x ∂y ∂y ∂x

We have from equation (3.2-24):

or !

()

following equivalent expressions for the derivative of a

We have Δx

!

()

complex function f z to have a derivative f ′ z at a given

3.4.3!

or !

( ) satisfies the Cauchyz ( x, y ) . The Cauchy-Riemann

with respect to x and y , and if f z Riemann equations at the point

Using Δz = Δ x + i Δ y , we have:

!

()

independent of how Δz → 0 . The derivative f ′ z then exists

or

dw dz

2

()

= f′ z

2

2

2

(3.4-23)

!

2

2

⎛ ∂u ⎞ ⎛ ∂v ⎞ ⎛ ∂u ⎞ ⎛ ∂v ⎞ =⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ +⎜ ⎟ ⎝ ∂x ⎠ ⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎝ ∂y ⎠ 2

2

2

2

⎛ ∂u ⎞ ⎛ ∂u ⎞ ⎛ ∂v ⎞ ⎛ ∂v ⎞ = ⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ + ⎜ ⎟ ! (3.4-25) ⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎝ ∂x ⎠ ⎝ ∂ y ⎠

143

!

We can summarize the results of the last two sections in

Proposition 3.4-1.

3.4.4!

Determine where the derivative of the complex function

()

f z = z 2 exists.

CAUCHY-RIEMANN EQUATIONS IN RECTANGULAR FORM

Solution:

Proposition 3.4-1, Cauchy-Riemann Equations:

()

( ) ( ) ( ) to be differentiable z ( x, y ) are that the Cauchy-Riemann equations: w = f z = u x, y + i v x, y

∂u ∂v = ! ∂x ∂ y

∂u ∂v =− ! ∂y ∂x

at a point

() (

)2 = x 2 − y 2 + 2 x y i

Therefore: !u = x 2 − y 2 !

v = 2x y

and so: (3.4-26)

( )

hold at the point z x, y , and that the partial derivatives ∂u ∂x ,

∂u ∂ y , ∂v ∂x , and ∂v ∂ y all exist and are continuous in a

( )

neighborhood of the point z x, y .

∂u ! = 2x! ∂x

∂u = −2 y! ∂y

!

Given in Sections 3.4.1 and 3.4.2.

!

From the Cauchy-Riemann equations we see that, if a

■

∂v = 2 y! ∂x

∂v = 2x ∂y

These partial derivatives are continuous everywhere. We then have:

∂u ∂v ! = 2x = ! ∂x ∂y

Proof:

()

()

! f z = z2 = u z + i v z = x + i y

The necessary and sufficient conditions for a complex function

!

Example 3.4-1

∂u ∂v = −2 y = − ∂y ∂x

These equations show that the Cauchy-Riemann equations

()

hold everywhere in the complex plane. The derivative f ′ z

()

complex function f z is differentiable at a point z , and if the

of f z = z 2 then exists everywhere in the complex plane

Re f z = u or the Im f z = v is a constant, then f z

(see Example 3.2-2).

()

()

()

is a

constant. 144

Example 3.4-3 Example 3.4-2

Determine where the derivative of the complex function

Determine where the derivative of the complex function

()

f z = z

2

exists.

()

f z = x 2 + i y 2 exists. Solution:

()

Solution:

()

!f z = z

2

()

()

(

)(

For f z = x 2 + i y 2 we have:

)

= u z + i v z = z z = x + i y x − i y = x2 + y2

Therefore:

v=0

∂u ! = 2x! ∂x

and so:

∂u = 2 y! ∂y

∂v = 0! ∂x

∂v =0 ∂y

These partial derivatives are continuous everywhere. Only at the point z = 0 do the Cauchy-Riemann equations hold, however, so that:

∂u ∂v ! = 2x = 0 = ! ∂x ∂y

()

∂u ∂v = 2y = 0= − ∂y ∂x

()

The derivative f ′ z of f z = z the point z = 0 .

v = y2

and so:

!u = x 2 + y 2 !

∂u ! = 2x! ∂x

!u = x 2 !

2

∂v = 0! ∂x

∂u = 0! ∂y

∂v =2y ∂y

These partial derivatives are continuous everywhere. Only at points x = y do the Cauchy-Riemann equations hold, however, so that:

∂u ∂v ! = 2x = 2 y = ! ∂x ∂y

()

∂u ∂v =0=− ∂y ∂x

()

The derivative f ′ z of f z = x 2 + i y 2 therefore exists only at points x = y .

therefore exists only at Example 3.4-4

()

Show that the complex function f z = z is not differentiable anywhere in the complex plane. 145

Equation (3.4-25) and the Cauchy-Riemann equations give

Solution:

()

()

us:

()

!f z = z = u z + iv z = x − i y

2

Therefore: !u = x !

3.5! ∂u = 0! ∂y

∂v = 0! ∂x

∂v = −1 ∂y

∂u ∂v ! = 1 ≠ −1 = ! ∂x ∂y

()

()

()

differentiable at any point of ! (see Example 3.2-3).

∂u ∂v ∂u ∂v dw ! = − dz ∂x ∂ y ∂ y ∂x Solution:

()

( )

() ()

() ( )

( )

w = f z = u z + i v z = u r,θ + i v r,θ !

!

anywhere in the complex plane, and so f z = z is not

Show that:

()

We will now consider a complex function w = f z that

w = f z = f r eiθ in the form:

∂u ∂v =0=− ∂y ∂x

The Cauchy-Riemann equations for f z = z do not hold

Example 3.4-5

!

EXPONENTIAL FORM OF THE CAUCHY-RIEMANN EQUATIONS

has a derivative f ′ z at a point z = r eiθ . Writing the function

We then have:

2

2

v=−y

and so:

∂u ! = 1! ∂x

2

⎛ ∂u ⎞ ⎛ ∂v ⎞ dw ∂u ∂u ∂v ∂v ∂u ∂v ∂u ∂v ! =⎜ ⎟ +⎜ = + = − dz ∂x ∂x ∂x ∂x ∂x ∂ y ∂ y ∂x ⎝ ∂x ⎠ ⎝ ∂x ⎟⎠

() ⎡⎣u ( z + Δz ) + i v ( z + Δz )⎤⎦ − ⎡⎣ u ( z ) + i v ( z )⎤⎦

(3.5-1)

the derivative f ′ z as given in equation (3.4-3) is:

()

! f ′ z = lim

Δz

Δz → 0

! (3.5-2)

where Δz = Δr eiθ + i r eiθ Δθ . We then have:

(

)

(

)

⎡⎣u r + Δ r, θ + Δθ ⎤⎦ + i ⎡⎣v r + Δ r, θ + Δθ ⎤⎦ ! f ′ z = lim Δr → 0 Δr eiθ + i r eiθ Δθ

()

Δθ → 0

!

− lim

Δr → 0 Δθ → 0

( )

( )!

u r, θ + i v r, θ

Δr eiθ + i r eiθ Δθ

(3.5-3) 146

Since the derivative exists, this limit cannot depend upon the path of approach as Δz → 0 .

⎡ ∂u ∂v ⎤ f ′ z = e− iθ ⎢ + i ⎥ ! ∂r ⎦ ⎣ ∂r

()

!

(3.5-5)

Now lettings Δz → 0 along an arc so that Δr = 0 and

!

Δz = i r eiθ Δθ (see Figure 3.5-1), we have:

(

) ( )

u r , θ + Δθ − u r, θ e−iθ f′ z = lim r Δθ → 0 i Δθ

()

!

(

) ( )

v r , θ +Δθ − v r, θ i e−iθ + lim r Δθ → 0 i Δθ

!

!(3.5-6)

or e−iθ ⎡ ∂u ∂v ⎤ e−iθ ⎡ ∂v ∂u ⎤ ! f′ z = + = − i ⎢ ⎥ ⎢ ⎥ r ⎣ i ∂θ ∂θ ⎦ r ⎣ ∂θ ∂θ ⎦

()

! Figure 3.5-1! !

( )

Two paths of approach to a point z r, θ .

Letting Δz → 0 along a ray so that Δθ = 0 and Δz = Δr eiθ

(see Figure 3.5-1), we have: !

! or

()

f′ z = e

− iθ

lim

+ ie

(

) ( )

u r + Δ r, θ − u r, θ

lim

( )

is differentiable at a point

z r, θ , the derivatives given in equations (3.5-5) and (3.5-7), which were calculated for two different directions of approach,

!

Δr

Δr→ 0

()

Since the function w = f z

must exist and be equal. We then can write:

Δr→ 0

− iθ

!

(3.5-7)

(

) ( )

v r + Δ r, θ − v r, θ Δr

!

(3.5-4)

!

()

f′ z =e

− iθ

⎡ ∂u ∂v ⎤ e− iθ ⎡ ∂v ∂u ⎤ + i = − i ⎢ ∂r ⎥ ⎢ ∂θ ⎥! ∂r r ∂ θ ⎣ ⎦ ⎣ ⎦

(3.5-8)

Using Euler’s formula e−iθ = cosθ − i sin θ , we can write

equation (3.5-8) as:

⎡ ∂u ∂v ⎤ ∂w ! f ′ z = cosθ − i sin θ ⎢ + i ⎥ = cosθ − i sin θ ! (3.5-9) ∂r ∂r ∂r ⎣ ⎦

() (

)

(

)

147

and as:

cosθ − i sin θ f′ z = r

()

! !

!

(

)

!

3.5.1!

Proposition 3.5-2:

⎡ ∂v ∂u ⎤ − i cosθ − i sin θ ∂w − i ⎢ ∂θ ⎥= ∂ θ r ∂θ ⎣ ⎦

CAUCHY-RIEMANN EQUATIONS IN POLAR FORM

()

derivative is:

()

f′ z =e

!

Proposition 3.5-1, Polar Form of Cauchy-Riemann Equations:

( )

If the complex function f z = u r,θ + i v r,θ has a derivative

()

f′ z

in some neighborhood of a point z = r eiθ , then the

( )

Cauchy-Riemann equations in polar form at the point z r, θ are:

∂u 1 ∂v ! = ∂r r ∂θ

!

1 ∂u ∂v =− ! r ∂θ ∂r

(3.5-11)

Equating the real parts and equating the imaginary parts

of equation (3.5-8), we have equation (3.5-11). These are the

3.5.2! !

⎡ ∂u ∂v ⎤ e− iθ ⎡ ∂v ∂u ⎤ + i = − i ! ⎢ ∂r ∂r ⎥⎦ r ⎢⎣ ∂θ ∂θ ⎥⎦ ⎣

(3.5-12)

Proof: !

Given in the derivation of equation (3.5-8).

!

Using the Cauchy-Riemann equations together with

■

■

DERIVATIVE OF A COMPLEX FUNCTION IN POLAR EXPONENTIAL FORM

The derivative of a complex function in polar exponential

⎛ ⎛ ∂u 1 ∂u ⎞ ∂v 1 ∂v ⎞ f ′ z = ⎜ cosθ − sin θ + i cos θ − sin θ ⎜⎝ ∂r r ∂θ ⎟⎠ ∂r r ∂θ ⎟⎠ ⎝

()

! !

Cauchy-Riemann equations in polar form.

− iθ

Euler’s formula, equation (3.5-12) can be written in the form:

Proof: !

( )

has a derivative f ′ z , then the polar exponential form of the

(3.5-10)

() ( )

() ( )

If at a point z a complex function w = f z = u r, θ + i v r, θ

!

(3.5-13)

or

⎛ ⎛ ∂v 1 ∂v ⎞ ∂u 1 ∂u ⎞ f ′ z = ⎜ sinθ + cosθ − i sin θ + cos θ ⎜⎝ ∂r r ∂θ ⎟⎠ ∂r r ∂θ ⎟⎠ ⎝

()

! ! !

!

(3.5-14)

() ( )

( )

For w = f z = u r, θ + i v r, θ we can write:

form has now been obtained in equation (3.5-8). 148

()

f′ z =

! !

∂f ∂ f ⎛ ∂u ∂v ⎞ ⎛ ∂u ∂v ⎞ +i = ⎜ +i ⎟ +i⎜ +i ⎟! ∂r ∂θ ⎝ ∂r ∂r ⎠ ⎝ ∂θ ∂θ ⎠

Let z = r ei θ . Using Euler’s formula we have: (3.5-15)

We then have from equation (3.5-12):

∂ f ∂u ∂v = +i = f ′ z eiθ ! ∂r ∂r ∂r

()

!

(3.5-16)

∂ f ∂u ∂v = +i = i f ′ z r eiθ ! ∂θ ∂θ ∂θ

()

(3.5-17)

∂v ⎞ ∂u ∂v ⎛ ∂u ! + i i r = + i ⎜⎝ ∂r ∂r ⎟⎠ ∂θ ∂θ

(3.5-18)

Equating the real and imaginary parts, we obtain the CauchyRiemann equations in polar form: !

() (

)

= r 2 ei 2θ = r 2 cos 2θ + i r 2 sin 2θ

!u = r 2 cos 2θ !

v = r 2 sin 2θ

and so:

From equations (3.5-16) and (3.5-17) we have: !

()

2

Therefore:

and !

()

! f z = z 2 = u z + i v z = r eiθ

∂u 1 ∂v ! = ∂r r ∂θ

1 ∂u ∂v =− ! r ∂θ ∂r

(3.5-19)

∂v ! = 2 r sin 2θ ! ∂r

∂v = 2 r 2 cos 2θ ∂θ

These partial derivatives are continuous everywhere. We then have:

∂u 1 ∂v ! = 2 r cos 2θ = ! ∂r r ∂θ

1 ∂u ∂v = − 2 r sin 2θ = − r ∂θ ∂r

()

f ′ z exists everywhere (see Example 3.4-1). From equation (3.5-12) we have:

() f ′ ( z ) and where it exists.

Expressing the complex function f z = z 2 in exponential

Solution:

∂u = − 2r 2 sin 2θ ∂θ

The Cauchy-Riemann equations hold everywhere, and so

Example 3.5-1

form, determine its derivative

∂u ! = 2r cos 2θ ! ∂r

⎡ ∂u ∂v ⎤ ! f ′ z = e− iθ ⎢ + i ⎥ = e− iθ ⎡⎣ 2r cos 2θ + i 2r sin 2θ ⎤⎦ ∂r ⎦ ⎣ ∂r

()

Using Euler’s formula we have:

()

! f ′ z = 2 r e− iθ e2 iθ = 2 r eiθ = 2 z 149

3.6! !

WIRTINGER DERIVATIVES

Proposition 3.6-1:

For a complex variable z = x + i y we have from equation

The complex variables z and z are independent.

(1.4-12): !

z+z x= ! 2

z−z y= ! 2i

(3.6-1)

These equations can be viewed as transformations from the conjugate coordinates

( )

( x, y ) . We can now write:

z, z

to the rectangular coordinates

!

∂ ∂ ∂x ∂ ∂ y ∂ 1 ∂ 1 ! = + = + ∂z ∂x ∂z ∂ y ∂z ∂x 2 ∂ y 2i

!

∂ ∂ ∂x ∂ ∂ y ∂ 1 ∂ 1 ! = + = − ∂z ∂x ∂z ∂ y ∂z ∂x 2 ∂ y 2i

(3.6-2)

!

Using equations (3.6-4) and (3.6-5) we have:

(

)

(

)

(

(

)

(

)

(

!

∂ x − iy ⎞ 1 ∂z 1 ⎛ ∂ x − i y = ⎜ −i = 1+ i 2 = 0 ! ⎟ ∂z 2 ⎝ ∂x ∂y ⎠ 2

!

∂ x + iy ⎞ 1 ∂z 1 ⎛ ∂ x + i y = ⎜ +i = 1+ i 2 = 0 ! ⎟ ∂z 2 ⎝ ∂x ∂y ⎠ 2

) )

(3.6-6)

(3.6-7)

Therefore z and z are independent even though, given z , it is (3.6-3)

or

always possible to determine z .

■

Example 3.6-1

!

∂ 1⎛ ∂ ∂⎞ = ⎜ −i ⎟ ! ∂z 2 ⎝ ∂x ∂y⎠

!

∂ 1⎛ ∂ ∂⎞ = ⎜ +i ⎟! ∂z 2 ⎝ ∂x ∂y⎠

!

Proof:

(3.6-4)

∂ −z2 Calculate . e ∂z Solution:

(3.6-5)

Equations (3.6-4) and (3.6-5) are known as the Wirtinger

derivatives or Wirtinger operators, and equation (3.6-5) is

Since z and z are independent, we have:

∂ − ! e ∂z

z

2

∂ −zz − = e = − z e− z z = − z e ∂z

z

2

called the d-bar operator. 150

!

From equations (3.6-4) and (3.6-5) we also have:

! !

⎛ ∂ ∂ ∂⎞ =i⎜ − ⎟! ∂y ⎝ ∂z ∂z ⎠

∂ ∂ ∂ = + ! ∂x ∂z ∂z

or (3.6-8)

Equations (3.6-4) and (3.6-5) can be written in terms of

!

!

!

⎛ ∂ ∂⎞ ! r − i ⎜⎝ ∂r ∂θ ⎟⎠

∂ 1 = ∂z 2 z

⎛ ∂ ∂⎞ ! r + i ⎜⎝ ∂r ∂θ ⎟⎠

( ) = 1 ⎛ ∂u − ∂v ⎞ + i ⎛ ∂v + ∂u ⎞ !

(3.6-14)

2 ⎜⎝ ∂x

∂ y ⎟⎠

2 ⎜⎝ ∂x

∂ y ⎟⎠

and so:

∂f z

!

∂z

(3.6-9) ! (3.6-10)

(3.6-13)

∂z

polar coordinates as:

∂ 1 = ∂z 2 z

( ) = 1 ⎛ ∂u + i ∂u ⎞ + 1 ⎛ i ∂v − ∂v ⎞ !

∂f z

2 ⎜⎝ ∂x

∂ y ⎟⎠

2 ⎜⎝ ∂x

∂ y ⎟⎠

If we have:

( ) = 0!

∂f z

!

(3.6-15)

∂z

then we must have: Proposition 3.6-2:

()

For a complex function w = f z to be differentiable at a point

z , it is necessary that:

( ) = 0!

(3.6-11)

∂z

! !

Proof:

() ( ) use equation (3.6-5) to calculate ∂ f ( z ) ∂z : ∂ f ( z ) 1 ⎛ ∂u ∂u ⎞ i ⎛ ∂v ∂v ⎞ ! ! = +i + +i !

( )

For a complex function w = f z = u x, y + i v x, y , we can

∂z

2 ⎜⎝ ∂x

∂u ∂v − = 0! ∂x ∂ y

∂v ∂u + = 0! ∂x ∂ y

(3.6-16)

∂u ∂v ! = ∂x ∂ y

∂u ∂v =− ! ∂y ∂x

(3.6-17)

or

∂f z

!

!

∂ y ⎟⎠

2 ⎜⎝ ∂x

∂ y ⎟⎠

(3.6-12)

These are just the Cauchy-Riemann equations. Therefore ∂ f z ∂z = 0 is equivalent to the Cauchy-Riemann equations. If a function f z is differentiable, therefore, it must be independent of z . ■ !

()

()

()

We see that any function f z that is independent of z

will satisfy the Cauchy-Riemann equations; moreover, any 151

()

function f z

∂u ∂v ! = 4x = ! ∂x ∂y

that satisfies the Cauchy-Riemann equations

cannot be a function of z . Proposition 3.6-3:

Example 3.6-3

()

Any complex function f z will satisfy the Cauchy-Riemann equations if and only if it is independent of z .

()

Determine if the function f z = 2 x 2 + i 4 x y is differentiable using equation (1.4-12):

Proof: !

∂u ∂v =8y ≠ − = −4y ∂y ∂x

Follows from Proposition 3.6-2.

!x =

■

z+z ! 2

y=

z−z 2i

Solution:

Example 3.6-2

() (

)

Determine if f z = 2 x 2 + 4 y 2 + i 4 x y is dependent on z ,

We have using equation (1.4-12):

()

! f z = 2 x2 + i 4 x y = 2

and if it satisfies the Cauchy-Riemann equations. Solution:

z+z z+z z−z + i4 2 2 2i

or

We have:

() (

)

()

(

(

! f z = 2 x2 + 4 y2 + i 4 x y = 2 x2 + 4 y y + i x

)

()

(

)(

)

! f z = z + z + z + z z − z = z + z + z2 − z 2

()

Since f z is dependent of z , it is not differentiable.

or

)

! f z = 2 x2 + i 4 y x − i y = 2 x2 + i 4 y z

()

Example 3.6-4

() (

Since f z is dependent of z it does not satisfy the Cauchy-

Determine if the function f z = x + i y

Riemann equations as can be seen:

using equation (1.4-12):

)2

is differentiable

152

z+z !x = ! 2

z−z y= 2i

Solution:

!

∂2 1 ∂ ⎛1⎛ ∂ ∂ ⎞⎞ 1 ∂ ⎛1⎛ ∂ ∂ ⎞⎞ = + i − i + i ! ∂z ∂z 2 ∂x ⎜⎝ 2 ⎜⎝ ∂x ∂ y ⎟⎠ ⎟⎠ 2 ∂ y ⎜⎝ 2 ⎜⎝ ∂x ∂ y ⎟⎠ ⎟⎠

!

!

(3.6-20)

or

We have: ⎛ z+z

z−z⎞

2

!f ( z) = ( x + i y) = ⎜ +i = z2 ⎟ 2i ⎠ ⎝ 2 2

∂2 1 ⎛ ∂2 ∂2 ∂2 ∂2 ⎞ = ⎜ 2 +i −i + 2⎟ ! ∂z ∂z 4 ⎝ ∂x ∂x ∂ y ∂ y ∂x ∂ y ⎠

!

()

(3.6-21)

Therefore:

Since f z is independent of z , it is differentiable.

∂2

∂2 ! + =4 ∂z ∂z ∂x 2 ∂ y 2

!

Proposition 3.6-4:

∂2

(3.6-22)

■

The Laplace operator can be written in terms of Wirtinger

3.7!

derivatives:

∂2

∂2 ! + 2 =4 2 ∂z ∂z ∂x ∂y

!

∂2

(3.6-18)

Using equations (3.6-4) and (3.6-5) to calculate the

derivatives ∂ ∂z and ∂ ∂z , we have: ! and so:

We will now consider a more restrictive set of functions

that are differentiable not only at some single point z = z0 , but also at every point z of a δ neighborhood of z0 . A complex

Proof: !

!

HOLOMORPHIC FUNCTIONS

∂2 ∂ ⎛1⎛ ∂ ∂ ⎞⎞ = + i ⎟⎟ ! ⎜ ⎜ ∂z ∂z ∂z ⎝ 2 ⎝ ∂x ∂y⎠⎠

()

function w = f z is said to be holomorphic at a point z0 if

()

f z is defined and differentiable in an open disk z − z0 < δ

()

centered at z0 , where δ > 0 . For f z to be holomorphic at a (3.6-19)

()

point z0 in a domain D requires then that the function f z be

( )

differentiable in a whole neighborhood Dδ z0

that exists

within D . Therefore z0 must be an interior point of D . 153

!

() f ( z ) to be holomorphic

While it is possible for a derivative f ′ z to exist at just a

single point, it is then not possible for

3.7.1! HOLOMORPHIC NECESSARY CONDITIONS

just at a single point. For this reason holomorphicity is

!

considered to be a global property of a function, while

not sufficient conditions for a complex function f z to be

differentiability is considered to be a local property of a

holomorphic at a point z0 . It is possible for a function f z to

function. We can conclude then that, for a function to be

satisfy the Cauchy-Riemann equations at z0 , but not be

holomorphic along a curve, the function must be differentiable

differentiable at any other point in a neighborhood of z0 , and so

not only at points on the curve, but at points in some region

not be holomorphic.

surrounding the curve. !

()

A complex function w = f z is said to be holomorphic on

()

The Cauchy-Riemann equations constitute necessary but

()

3.7.2!

()

HOLOMORPHIC SUFFICIENT CONDITIONS

a domain if f z is defined and differentiable at every point in

!

the domain. If f z

w = f z = u x, y + i v x, y to be holomorphic at a point z0 is

()

is holomorphic over the entire finite

The sufficient conditions for a complex function

() ( )

( )

( )

( )

complex plane, the function is referred to as an entire function.

that the first order partial derivatives of u x, y and v x, y

Functions such as e z , sin z , and cos z are differentiable

satisfy the Cauchy-Riemann equations at z0 :

everywhere in the finite complex plane (see Chapter 4), and so are entire functions. Polynomials are also entire functions (see Section 3.7.4). !

()

If a function f z is holomorphic at a point z0 , then z0 is

()

()

called a regular point of the function f z . If a function f z is

∂u ∂v = ! ∂x ∂ y

!

∂v ∂u =− ! ∂x ∂y

(3.7-1)

and that these partial derivatives are continuous in some neighborhood of z0 .

not holomorphic at a point z0 , but is holomorphic at some point

Example 3.7-1

in every neighborhood of z0 , then z0 is called a singular point

Is f z = z 2 + z entire? Determine the derivative of f z .

()

or singularity of f z , as discussed in Section 9.1.

()

()

Solution: 154

(

)2

i x+i y !ei z = e ( ) = ei x e− y = e− y cos x + i sin x

(

!z 2 + z = u + i v = x + i y + x + i y or

(

) (

!u + i v = x 2 − y 2 + i 2 x y + x + i y = x 2 − y 2 + x + i 2 x y + y

)

!u = e− y cos x !

∂u = −2 y ∂y

∂v ! = 2 y! ∂x

∂v = 2x +1 ∂y

∂u ∂v = −2 y = − ∂y ∂x

()

()

f z = z 2 + z is entire. The derivative of f z = z 2 + z is: df z ∂u ∂v ! = +i = 2x +1 + i 2 y = 2 x + i y +1= 2 z +1 dz ∂x ∂x

(

) ( )

(

)

Example 3.7-2

() f (z).

Is f z = e i z an entire function? Determine the derivative of

Solution:

∂v = − e− y sin x ∂y

∂u ! = − e− y cos x ! ∂y

∂v = e− y cos x ∂x

∂u ∂v ! = − e− y sin x = ! ∂x ∂y

∂u ∂v = − e− y cos x = − ∂y ∂x

and the partial derivatives are continuous. Therefore

and the partial derivatives are continuous. Therefore

()

∂u ! = − e− y sin x ! ∂x

The Cauchy-Riemann equations then hold for all z :

The Cauchy-Riemann equations then hold for all z :

∂u ∂v ! = 2x +1= ! ∂x ∂y

v = e− y sin x

and so:

and so:

∂u ! = 2 x + 1! ∂x

)

() df ( z ) ∂u ∂v = +i = − e− y sin x + i

()

f z = ei z is entire. The derivative of f z = ei z is:

dz

∂x

∂x

(e

−y

)

(

cos x = e− y − sin x + i cos x

)

or df ( z ) i x+i y ! = i e− y ( cos x + i sin x ) = i e− y ei x = i ei x− y = i e ( ) dz

and so: df ( z ) ! = i ei z dz 155

Proof:

Example 3.7-3

()

Show that the complex function f z = 3x + i x y is not holomorphic at any point in the complex plane.

!

then defined and exist everywhere in D . Therefore the function

()

f z must be continuous in D .

We have:

v=xy

■

Proposition 3.7-2, Operations for Holomorphic Functions:

and so:

∂u ! = 3! ∂x

() f ( z ) are

Proposition 3.4-1, we know that if a complex function f z is holomorphic on a domain D , the partial derivatives of

Solution:

!u = 3x !

From the definition of a holomorphic function and from

() () f ( z ) ± g ( z ) , f ( z ) g ( z ) , and f ( z ) g ( z )

If the complex functions f z and g z are both holomorphic

∂u = 0! ∂y

∂v = y! ∂x

∂v =x ∂y

We then have:

∂u ∂v ! ≠ ! ∂x ∂ y

∂u ∂v ≠− ∂y ∂x

on a domain D , then

()

with g z ≠ 0 are also all holomorphic on D . Proof: !

From the definition of a holomorphic function, and from

the differentiation rules given in equations (3.3-4), (3.3-5), and

() ()

() ()

( ) g (z)

Since the Cauchy-Riemann equations do not hold for any

(3.3-7) we know that f z ± g z , f z g z , and f z

point, the complex function f z = 3x + i x y is then not

with g z ≠ 0 are all holomorphic on D .

()

()

■

holomorphic at any point in the complex plane. Proposition 3.7-3, Composition for Holomorphic Functions: Proposition 3.7-1, Continuous Holomorphic Functions

()

If a complex function f z is holomorphic on a domain D , then

()

f z is continuous in D .

() () on a domain D , then g ( f ( z )) is also holomorphic on D .

If the complex functions f z and g z are both holomorphic

Proof: 156

!

Follows from the definition of a holomorphic function,

and from the differentiation rule given in equation (3.3-8).

■

!

Since the domain D is connected, from equation (3.7-4) we

see that u is constant in D . Similarly, from equation (3.7-5) we

()

see that v is constant in D . Therefore f z = u + i v is a constant Proposition 3.7-4, Zero Derivative Theorem:

()

in D .

Let f z be a holomorphic function in a domain D . Then if and

()

()

only if f ′ z = 0 for all z ∈D will f z be a constant in D .

!

constant in D for all z ∈D , then we will have

( ) both f ′(z) = 0 . ■

() ( )

( )

If for a function w = f z = u x, y + i v x, y we have:

()

f ′ z = 0!

(3.7-2)

( )

∂u ∂v ∂v ∂u ! f′ z =0= +i = −i ∂x ∂x ∂ y ∂y

()

()

(3.7-3)

!

( )

∂v = 0! ∂x

∂v = 0! ∂y

() ( )

( )

For a function w = f z = u x, y + i v x, y we have using

equation (3.4-24): (3.7-4) !

and !

()

is a constant in D .

have for all points z x, y in D : !

()

or imaginary part of f z is constant for all z ∈D , then f z

Proof:

∂u = 0! ∂y

equal to constants. Therefore we must have

If f z is a holomorphic function in a domain D , and if the real

Equating real parts and imaginary parts with zero, we then

∂u = 0! ∂x

( ) is u ( x, y ) and

Proposition 3.7-5:

at all points z x, y in D , equation (3.4-24) gives us: !

() ( )

Conversely, if a function w = f z = u x, y + i v x, y

v x, y

Proof: !

!

()

f′ z =

∂u ∂u ∂v ∂v −i = +i = 0! ∂x ∂y ∂y ∂x

(3.7-6)

for all z ∈D since either u or v is constant. Therefore by (3.7-5)

()

Proposition 3.7-4 f z is a constant in D .

■

157

Proposition 3.7-6:

()

()

If f z and its conjugate f z are holomorphic functions in a

()

domain D , then f z is a constant in D .

() ( )

( )

w = f z = u x, y + i v x, y !

(3.7-12)

()

■

()

() ( )

( )

f z = u x, y − i v x, y !

(3.7-8)

are holomorphic at all points z in D . Using the Cauchy-

()

Riemann equations, we have for f z : !

∂u ∂v ! = ∂x ∂ y

∂u ∂v =− ! ∂y ∂x

(3.7-9)

()

()

is constant for all z ∈D , then f z is a constant in D .

!

Proof: !

() ( )

()

f z = u + iv = c!

!

! and so:

(3.7-13)

()

where c is a real constant in the domain D . If f z = 0 in D ,

()

()

then f z is clearly constant in D . If f z ≠ 0 in D , we can

∂u ∂v ! = ∂ y ∂x

(3.7-10)

!

()

f z

∂v = 0! ∂x

2

= u2 + v 2 = c2 !

(3.7-14)

where c ≠ 0 . Taking partial derivatives, we then have:

Therefore we must have:

∂u = 0! ∂x

( )

For a function w = f z = u x, y + i v x, y we are given:

write:

and for f z :

∂u ∂v =− ! ∂x ∂y

()

If f z is a holomorphic function in a domain D , and if f z

(3.7-7)

and !

is a

Proposition 3.7-7, Constant Modulus Theorem:

We are given that the functions:

!

∂u ∂v +i = 0! ∂x ∂x

everywhere in D . Therefore by Proposition 3.7-4 f z constant in D .

Proof: !

()

f′ z =

!

(3.7-11)

!

u

∂u ∂v +v = 0! ∂x ∂x

u

∂u ∂v +v = 0! ∂y ∂y

(3.7-15)

With the Cauchy-Riemann equations, these equations become: 158

u

!

∂u ∂u −v = 0! ∂x ∂y

(3.7-16)

at all points z in D . From the Cauchy-Riemann equations:

∂u ∂v = ! ∂x ∂ y

!

v

!

∂u ∂u +u = 0! ∂x ∂y

(3.7-17)

Squaring and adding equations (3.7-16) and (3.7-17), we obtain:

(u + v ) 2

!

2

⎡ ⎛ ∂u ⎞ 2 ⎛ ∂u ⎞ 2 ⎤ ⎢⎜ ⎟ +⎜ ⎟ ⎥ = 0! ⎢⎣ ⎝ ∂x ⎠ ⎝ ∂ y ⎠ ⎥⎦

c

!

()

f′ z

2

∂u = 0! ∂x

∂u = 0! ∂y

(3.7-22)

∂v = 0! ∂x

∂v = 0! ∂y

(3.7-23)

we then have: !

(3.7-18) ! !

= 0!

(3.7-19)

()

()

Therefore f ′ z = 0 in D , and so by Proposition 3.7-4 f z is a constant in the domain D .

(3.7-21)

and

or using equations (3.7-14) and (3.4-25): 2

∂u ∂v =− ! ∂y ∂x

Therefore as shown in the proof of Proposition 3.7-4 we

() ( )

( )

see that w = f z = u x, y + i v x, y is a constant in D .

■

Proposition 3.7-9, Constant Argument Theorem:

■

()

If f z is a holomorphic function in a domain D , and if a

()

()

()

a real function for all z ∈D , then f z is a constant in D .

! !

Proof: !

() ( )

( )

For a function w = f z = u x, y + i v x, y we are given:

( )

v x, y = 0 !

()

constant in D .

If f z is a holomorphic function in a domain D , and if f z is

Proof:

()

branch of arg f z is constant for all z ∈D , then f z is a

Proposition 3.7-8:

(3.7-20)

() ( )

( )

For a function w = f z = u x, y + i v x, y we are given that

( ) is constant for all z ∈D . Therefore arg f ( z ) is also constant, and we can write: arg f z

the tangent of

159

v = c! u

!

(3.7-24)

We do not have to consider the case where u = 0 since

()

from Proposition 3.7-5 we know that if the real part of f z is

()

constant for all z ∈D , then f z is a constant in D . !

Using equation (3.7-24) we have:

()

(3.7-25)

We now multiply equation (3.7-25) by the complex constant

1− i c : !

(1− i c ) f ( z ) = (1− i c )(u + i cu ) = (1− i c ) u (1+ i c ) !

(3.7-26)

3.7.3!

!

(1− i c ) f ( z ) = (1+ c ) u !

(3.7-27)

!

1+ c

2

(z) = u !

()

(

!

(

)

⎡ 1− i c ⎤ Im ⎢ f z ⎥ = 0! 2 1+ c ⎢⎣ ⎥⎦

()

The sum, product, quotient, and chain differentiation rules

()

()

and g z are holomorphic functions we have: !

dc = 0! dz

(3.7-30)

!

dz = 1! dz

(3.7-31)

!

d ⎡ c f z ⎤⎦ = c f ′ z ! dz ⎣

!

d ⎡ f z ± g z ⎤⎦ = f ′ z ± g ′ z ! dz ⎣

(3.7-33)

!

d ⎡⎣ f z g z ⎤⎦ = g z f ′ z + f z g ′ z ! dz

(3.7-34)

()

()

(3.7-32)

(3.7-28)

)(

) ()

is holomorphic, ⎡⎣ 1− i c 1+ c 2 ⎤⎦ f z = u is also holomorphic. From equation (3.7-28) we have for all z ∈D :

Since f z

DIFFERENTIATION RULES FOR HOLOMORPHIC FUNCTIONS

functions. If c is a complex constant, n is an integer, and f z

and so:

(1− i c ) f

()

that apply for real-valued functions also apply for holomorphic

or 2

) ()

⎡ 1− i c 1+ c 2 ⎤ f z is a constant in D . We can conclude that ⎣ ⎦ the function f z is also a constant for all z ∈D . ■

!

f z = u + i v = u + i cu !

!

)(

(

where c is a real constant in the domain D . !

Therefore from Proposition 3.7-5 we know that the function

(3.7-29)

() ()

() ()

()

() ()

()

() ()

160

n n−1 d ⎡⎣ f z ⎤⎦ = n ⎡⎣ f z ⎤⎦ f ′ z ! dz

()

!

!

() ()

()

()

(3.7-35)

() () () () ()

⎡⎣ g z ⎤⎦ ≠ 0 ! (3.7-36)

d ⎡ f z ⎤ g z f ′ z − f z g′ z ! ⎢ ⎥= 2 dz ⎢⎣ g z ⎥⎦ ⎡⎣ g z ⎤⎦

( ( ))

()

2

( ( )) g ′ ( z ) !

(3.7-37)

functions are holomorphic functions.

! !

DIFFERENTIABILITY OF COMPLEX POLYNOMIALS AND RATIONAL FUNCTIONS

()

P z = a0 + a1 z + a2 z 2 +!+ an−1 z n−1 + an z n !

(3.7-38)

where the coefficients a0 , a1 , !, an−1 , an are complex constants and where n is a nonnegative integer. All complex polynomials are continuous (see Proposition 2.4-8), and have derivatives over the entire finite complex plane, and so are holomorphic over the entire finite complex plane. All polynomials are !

(

()

)

is

()

Since somewhere in the finite complex plane there will be

()

points at which Q z = 0 , complex rational functions cannot be entire functions.

3.8!

Complex polynomials have the form:

therefore entire functions.

()

where an ≠ 0 and bm ≠ 0 , and where the degree of f z

domain having no points at which the denominator Q z = 0 .

The sums, differences, products, and quotients of holomorphic

3.7.4!

(3.7-39)

max m, n . Complex rational functions are holomorphic on any

d ⎡ f g z ⎤⎦ = f ′ g z ⎣ dz

!

!

a0 + a1 z + a2 z 2 + !+ an−1 z n−1 + an z n ( ) f (z) = = ! Q ( z ) b0 + b1 z + b2 z 2 +!+ bm−1 z m−1 + bm z m P z

A complex rational function f z is a function that is the

()

()

quotient of two complex polynomials P z and Q z having no

!

DERIVATIVES OF COMPLEX FUNCTIONS OF A REAL-VALUED PARAMETER If the x and y coordinates in the complex plane can be

specified by some function of a real-valued parameter t , we then have: !

()

() ()

()

w t = f t = u t + iv t !

()

(3.8-1)

()

()

where f t is a complex function, and u t and v t are realvalued functions of t . The derivative dw dt is then given by: !

( ) = du (t ) + i dv (t ) !

dw t dt

dt

dt

(3.8-2)

common divisor, and so has the form given in equation (1.17-2): 161

!

The same differentiation rules apply for complex functions

() w = f (z)

w = f t of a real-valued parameter t as for complex functions of a complex variable (assuming all the derivatives

()

exist). If c is a complex constant, n is an integer, and f t and

()

3.9!

Proposition 3.9-1, L’Hôpital’s Rule:

()

( ) ( ) f (z) f ′(z) = lim ! g ( z ) z → z g′ ( z )

( )

a point z = z0 , and if f z0 = g z0 = 0 and g ′ z0 ≠ 0 , then:

dc = 0! dt

(3.8-3)

lim

! !

()

If f z and g z are complex functions that are holomorphic at

g t are complex functions we have: !

L’HÔPITAL’S RULE

d ⎡ c f t ⎤⎦ = c f ′ t ! dt ⎣

()

()

z → z0

(3.8-4)

(3.9-1)

0

Proof: !

!

d ⎡⎣ f t ± g t ⎤⎦ = f ′ t ± g ′ t ! dz

() ()

()

()

d ⎡⎣ f t g t ⎤⎦ = g t f ′ t + f t g ′ t ! dt

() ()

() ()

() ()

!

n n−1 d ⎡⎣ f t ⎤⎦ = n ⎡⎣ f t ⎤⎦ f ′ t ! dz

!

() ()

!

(3.8-5)

()

d ⎡f t ⎤ ⎢ ⎥= dt ⎢⎣ g t ⎥⎦

()

() ()

() ()

()

( ( ))

2

!

!

We can write:

()

f z z − z0

() ! lim = lim z→ z g (z) z→ z g z () f z

0

(3.8-7)

()

( )

2

⎡⎣ g t ⎤⎦ ≠ 0 ! (3.8-8)

( )

Since f z0 = g z0 = 0 , we can also write:

!

lim

z → z0

()

f z z − z0

( ) = lim g (z) z→ z g (z) f z

0

z − z0

d ⎡ ⎤ = df g ′ t ! f g t ⎦ dg dt ⎣

()

(3.9-2)

0

z − z0

()

g t f ′ t − f t g′ t ⎡⎣ g t ⎤⎦

(3.8-6)

!

(3.8-9)

()

= lim

z → z0

( )

f z − f z0 z − z0

() ( )

g z − g z0 z − z0

!

(3.9-3)

Using the definition of a derivative, this equation becomes: 162

! !

lim

z → z0

( ) = lim f ′ ( z ) ! g ( z ) z → z g′ ( z ) f z

0

()

( )

g ′ z0 ≠ 0 !

(3.9-4)

()

Because f z and g z are holomorphic at point z0 , their

derivatives exist in some neighborhood of z0 , and so the limits in equation (3.9-4) exist. This equation, known as L’Hôpital’s rule, is very useful for evaluating limits that involve an indeterminate form.

■

Example 3.9-1 Determine:

z6 + 1 !lim 2 z→i z +1 Solution: Since the limit has an indeterminate form we can use L’Hôpital’s rule to obtain:

( )( ) ( )( )

z6 + 1 6 z5 !lim 2 = lim = lim 3 z 4 = 3 i 2 i 2 = 3 −1 −1 = 3 z→i z +1 z→i 2 z z→i

163

Chapter 4 Elementary Complex Functions

(

ln z = log e r + i θ + 2 k π

)

164

!

In this chapter we will consider a number of complex

!

It is important to specify the branch being employed when

functions including the exponential functions, trigonometric

presenting results obtained from a multivalued function. If a

functions, hyperbolic functions, logarithmic functions, power

multivalued function is restricted to a range corresponding to

functions,

its principal values, the associated branch is referred to as the

inverse

trigonometric

functions,

and

inverse

hyperbolic functions. We will show that each of these complex functions is an extension to the complex plane of a real-valued function. We will begin by discussing concepts relating to the branches of multivalued complex functions.

4.1!

BRANCHES, BRANCH CUTS, AND BRANCH POINTS

4.1.1! !

principal branch of the function.

4.1.2! !

BRANCH CUTS

A branch cut of a multivalued function is a curve in the

complex plane across which the function is discontinuous. The purpose of a branch cut is to demarcate branches, with each edge of the curve belonging to different branches. The branch cut acts to transform a complex function that is multivalued

BRANCHES

()

If w = f z is a multivalued function of a complex variable

z , then more than one value of w will correspond to some

()

values of z in the domain of f z (see Section 2.1.5). A number

into one that is single-valued and continuous within each of its branches. A branch cut need not be a straight line although such a line is generally selected for branch cuts. Also there is no unique branch cut for any given multivalued function.

of points in the w-plane can then be the images of a single point in the z-plane.

4.1.3!

!

!

If we consider a restricted part of the range of a

BRANCH POINTS

A point z0 is designated a branch point for a complex

()

()

multivalued function in which the function assigns just one

multivalued function f z if f z does not return to its initial

value w for each z , and in which the function is continuous,

value after once traversing a closed circle of arbitrarily small

we refer to this part of the range as a branch of the multivalued

radius centered at z0 . Therefore a multivalued function will

function. Within any given branch of a multivalued function,

never be single-valued in any δ neighborhood Dδ z0

the function will be single-valued and continuous.

branch point z0 , where z − z0 < δ . If a closed contour does not

( )

of a

165

The function w = z1

2

encircle a branch point of a multivalued function, that function

!

will be single-valued when traversing such a contour.

function w = z 2 . The function w = z 1

!

only when a branch of the function is specified.

Every branch cut of a multivalued function will join two

is the inverse of the two-to-one 2

From equation (1.14-13) we know that the n roots wk of a

(and only two) branch points of the function. The location of

!

any branch point depends on the function, and is independent

complex number z = r e iθ are given by:

of the chosen branch cut. Therefore the location of branch

!

points is unique for a given multivalued function. Branch

()

If a branch cut for a function f z

extends from 0 to

()

infinity, the point at infinity is also a branch point for f z . We can see this by writing:

z=

!

1 ! ζ

i θ + 2k π ) n ! wk = z1 n = r 1 n e (

k = 0, 1, 2, !, n − 1!

(4.1-4)

where the modulus r 1

! (4.1-2)

n

of the nth root always consists of a

n

()

()

()

The complex square root function w = f z = z

!

! is a very

simple multivalued function, having only two possible values of w for each z value. For this reason we will now use w = z1

()

For the complex square root w = f z = z1

as discussed in 2

where z = r eiθ

wk = z 1 2 = r 1 2 e iθ 2 e i k π !

k = 0, 1 !

(4.1-5)

i θ +2 k π ) 2 ! wk = z 1 2 = r 1 2 e (

k = 0, 1 !

(4.1-6)

or

COMPLEX SQUARE ROOT FUNCTION 1 2

( )

but a different argument arg wk

we then have:

point of f z , then z = ∞ must also be a branch point of f z .

as an example of a multivalued function.

!

Section 1.14.

When z = 0 we have ζ = ∞ . We see then that if z = 0 is a branch

!

(4.1-3)

modulus r 1

(4.1-1)

ln z = − ln ζ !

4.1.4!

k = 0, 1, 2, !, n − 1!

nonnegative real number. Each distinct root wk has the same

and so: !

wk = z1 n = r 1 n ei θ n ei 2 k π n !

or

points of a function can be located at infinity. !

becomes single-valued

2

!

As z traverses a circle C about the origin, the argument

θ = arg ( z ) will vary continuously from 0 to 2 π . Upon

()

completing one circuit of C , we will have θ = arg z = 2 π , but

166

( )

arg w will change only by θ 2 = π , and so w will not return to

to return to the value w0 . The principal complex square root

its initial value. Therefore the origin must be a branch point for

function has a discontinuity along the negative real axis as is

the function f z = z1 2 . The complex square root function is

demonstrated in Example 4.1-1.

()

not single-valued in any neighborhood of the origin z = 0 . !

()

The other branch point for the f z = z1

()

that although f z = z1

2

2

is infinity. Note

does not have a finite value at the

Example 4.1-1 Show that the principal complex square root function

()

w = f z = z1

!

negative real axis.

()

The principal complex square root function is obtained

when k = 0 in equations (4.1-3) through (4.1-6):

Solution:

⎛ θ θ⎞ ! w0 = z1 2 = r 1 2 eiθ 2 = r 1 2 ⎜ cos + i sin ⎟ ! − π < θ ≤ π ! (4.1-7) 2 2⎠ ⎝ The other complex square root function is obtained when k = 1 :

⎛ θ θ⎞ w1 = z1 2 = r 1 2 e i θ 2 e i π = r 1 2 ei π ⎜ cos + i sin ⎟ ! 2 2⎠ ⎝

!

(4.1-8)

w1 = z1 2 = − r 1 2 e i θ

2

From equation (1.14-18) we have for the principal complex square root function: !z 1 2 = z

1 2 i Arg ( z ) 2

e

and from equation (1.14-19): !z 1 2 = r1 2 ei θ 2 !

or using equation (1.13-9): !

2

where − π < θ ≤ π is discontinuous along the

point z = ∞ , nevertheless z = ∞ is a branch point for f z = z1 2 .

⎛ θ θ⎞ = − r 1 2 ⎜ cos + i sin ⎟ = − w0 ! 2 2⎠ ⎝

(4.1-9)

−π < θ ≤ π

()

Since Arg z has a discontinuity along the negative real axis (see Section 1.10.4), the limit lim z1 z → z0

2

will not exist for any

and so w0 and w1 are not identical, but give us the ± square

point z0 on the negative real axis as we will now show. We

roots.

can write:

!

The complex square root w = z1

2

will continuously change

over a closed curve that encloses the origin. After one complete

! lim z1 2 = lim z z → z0

z → z0

1 2 i Arg ( z ) 2

e

cycle w0 becomes w1 = −w0 . Another complete cycle is required 167

We first approach the point z0 along the upper half of a circle of radius r = z0 as shown in Figure 4.1-1. We then have: ! lim z1 2 = lim z z → z0

z → z0

1 2 i Arg ( z ) 2

e

= lim r 1 2 e i θ

2

θ →π

and so:

⎛ π π⎞ ! lim z 1 2 = r 1 2 ⎜ cos + i sin ⎟ = r 1 z → z0 2 2⎠ ⎝

2

(0 + i ) = i r 1 2

We next approach the point z0 along the lower half of a circle of radius r = z0 as shown in Figure 4.1-1. We then have: ! lim z1 2 = lim z z → z0

z → z0

1 2 i Arg ( z ) 2

e

= lim r 1 2 ei θ

2

Figure 4.1-1!

θ → −π

and so:

Paths of approach for determining the limit lim z 1 2 . z → z0

⎛ −π −π ⎞ ! lim z1 2 = r 1 2 ⎜ cos + i sin = r1 ⎟ z → z0 2 2 ⎠ ⎝

2

!

(0 − i ) = − i r1 2

From Example 4.1-1 we see that both the principal

complex square root function

()

f z = z1

2

()

and Arg z

are

Since the limit obtained is dependent upon the path of

discontinuous along the negative real axis for a branch defined

approach, no true limit exists for the principal complex

by −π < θ ≤ π . The negative real axis is then a branch cut for the

()

()

along the negative real axis.

principal square root function f z = z 1 2 . This branch cut is

Therefore the principal complex square root function is

represented by the purple line in Figure 4.1-2. Immediately

square root function f z = z1 discontinuous

along

the

2

negative

real

()

discontinuity is just the discontinuity in Arg z .

axis.

This

()

above this branch cut Arg z = + π ; immediately below this

()

branch cut Arg z = − π . The principal complex square root 168

()

function f z = z 1

2

is then not defined on the negative real

()

axis. Both the complex square root function and Arg z are single-valued and continuous on any path that does not cross the branch cut.

where

θ !u = r 1 2 cos ! 2

v = r 1 2 sin

θ 2

From equation (3.5-11) the Cauchy-Riemann equations are:

∂u 1 ∂v ! = ! ∂r r ∂θ

1 ∂u ∂v =− r ∂θ ∂r

We then have:

∂u 1 1 θ 1 ∂v ! = ! cos = ∂r 2 r 1 2 2 r ∂θ

()

Therefore f z = z 1

2

1 ∂u 1 1 θ ∂v = − 1 2 sin = − r ∂θ 2r 2 ∂r

is holomorphic within any one branch

except at the point z = 0 where r = 0 . Figure 4.1-2! Branch cut and branch point for the principal square root function. The branch point is z = 0 . Example 4.1-2

()

Show that f z = z 1

2

is holomorphic within any one branch

except at the point z = 0 . Solution: Within any one branch we have:

()

!w = u + i v = f z = z1 2 = r 1 2 ei θ

Figure 4.1-3! 2

Branch cut for w = z 1 0 ≤ θ < 2π .

2

for the branch defined by

169

!

While the location of branch cuts is not unique, any

!

branch cut must end at its branch points which are unique. For

() (

)(

)

w = f z = ⎡⎣ z − z1 z − z2 ⎤⎦

1 2

= r11 2 r21 2 eiφ ei kπ !

(4.1-13)

example, the branch cut for the square root function with

0 ≤ θ < 2 π is different from that for the principal square root function for which −π < θ ≤ π . The branch cut for 0 ≤ θ < 2 π is along the positive real axis (see Figure 4.1-3). For both branch cuts the origin is a branch point, as is the point at infinity.

4.1.5! !

TWO BRANCH POINTS IN THE COMPLEX PLANE

Some functions have two branch points in the complex

plane. For example, the square root function: !

() (

)

w = f z = z2 − 1

1 2

!

(4.1-10)

has two branch points: z = ±1 . Two possible branch cuts for this function are shown in Figure 4.1-4. !

Writing equation (4.1-10) in a more general form, we have:

() (

)(

)

! w = f z = ⎡⎣ z − z1 z − z2 ⎤⎦

1 2

=

r11 2 r21 2 e

i (θ1 +θ 2 ) 2+i k π

Figure 4.1-4!

! (4.1-11)

where k = k1 + k2 , 0 < θ1 < 2 π , and 0 < θ 2 < 2 π . We will define: !

φ=

θ1 + θ 2 ! 2

so that equation (4.1-11) becomes:

(4.1-12)

!

(

2

)

1 2

Two possible branch cuts for w = z − 1 and their associated branch points. In top diagram branch points of each cut are also located at infinity.

If a closed contour encircles only the point z1 , then by

moving once around the contour, θ1 will change by 2 π while

θ 2 will return to its initial value. Similarly, if a closed contour 170

encircles only the point z2 , then by moving once around the contour, θ 2 will change by 2 π while θ1 will return to its initial

()

(

)

value. In either case f z will change by i θ1 + θ2 / 2 = i π and will move from one branch to the next one. !

If a closed contour encircles both points, z1 and z2 , then

() (

2

)

w1 = f z = z − 1

!

1 2

= r11 2 r21 2 e 2 π i ei π = − w0 !

(4.1-17)

and so again there is no discontinuity in going from θ = 0 to

θ = 2π .

moving once around the contour, both θ1 and θ 2 will change by

4.2!

2 π . In this case f z will return to its initial value.

!

!

defined as that unique function which is its own derivative,

()

We see that by choosing the branch cut shown in the lower

() (

2

)

z = ±1 with a closed contour, w = f z = z − 1 single-valued function. !

() (

)

w0 = f z = z 2 − 1

= r11 2 r21 2 !

() (

)

1 2

w0 = f z = z 2 − 1

= r11 2 r21 2 e 2 π i = r11 2 r21 2 ! (4.1-15)

and so there is no discontinuity in going from θ = 0 to θ = 2 π . ! !

For k = 1 and θ1 = θ 2 = 0 , equation (4.1-13) becomes:

() (

2

)

w1 = f z = z − 1

1 2

=

r11 2 r21 2

0

iπ

e e = − w0 !

(4.1-16)

and for k = 1 and θ1 = θ 2 = 2 π , equation (4.1-13) becomes:

d x e = ex = f x ! dx

()

()

f′ x =

4.2.1!

DEFINITION OF THE COMPLEX EXPONENTIAL FUNCTION

(4.1-14)

and for k = 0 and θ1 = θ 2 = 2 π , equation (4.1-13) becomes: !

()

!

Let z1 = +1 and z2 = −1 . For k = 0 and θ1 = θ 2 = 0 , equation 1 2

()

f′ x = f x :

becomes a

(4.1-13) becomes: !

()

The real exponential function, f x = e x ≡ exp x , can be

()

diagram of Figure 4.1-4, and by encircling both branch points 1 2

COMPLEX EXPONENTIAL FUNCTION

!

()

f 0 = 1!

(4.2-1)

Ideally we will want the complex exponential function

f z = e z ≡ exp z to have properties like those of the real exponential function, and to agree with the real exponential function when z is real. We will begin by defining the complex exponential function as:

(

)

e z = e x+i y = e x ei y = e x cos y + isin y !

!

(4.2-2)

()

If Im z = y = 0 so that z is real, we see that we have: !

e z = e x+i0 = e x !

(4.2-3) 171

which is important if e z is to be the complex equivalent to the

()

real exponential function e x . If Re z = x = 0 , we see that we have: iy

e = cos y + i sin y !

!

(4.2-4)

which is Euler's formula as given in equation (1.11-6). We also see that if z is pure imaginary then equation (4.2-2) becomes

e z = ei y , and so e z is actually oscillatory.

! e z = e x+i y = e x ei y = e x ei y = e x = e x = 1 and so x = 0 . We then have:

(

)

!e z = e x+i y = e x ei y = e0 cos y + isin y = cos y + i sin y = 1 Equating the real and imaginary parts of this equation, we obtain: !cos y = 1 !

sin y = 0

Therefore we must have:

Example 4.2-1

!y = 2 k π !

Determine e if z = 3− i π . z

and so:

Solution: Using equation (4.2-2) we have:

( )

k = 0, ± 1, ± 2,!

( )

!e3−i π = e3 cos −π + i e3 sin −π = e3 cos π − i e3 sin π = − e3 The complex exponential can have a negative value unlike the real exponential. Example 4.2-2 z

If e = 1 determine z . Solution: We can write using equation (1.11-9):

!z = x + i y = 0 + i 2 k π = i 2 k π !

k = 0, ± 1, ± 2,!

Example 4.2-3 If e z = −1+ i determine z . Solution: From Example 4.2-2 we have: ! e z = e x+i y = e x ei y = e x ei y = e x = e x = 2 Therefore: !x = log e 2 =

1 log e 2 2 172

π

where log e is the natural logarithm. We then have:

(

)

(

1.! For ee we have:

)

!e z = e x cos y + isin y = 2 cos y + isin y = −1+ i

π

!ee = ecos π +isin π = e−1 =

and so equating real and imaginary parts: ! 2 cos y = −1!

e5 i − e−5 i 2.! For we have: 2i

2 sin y = 1

( )

!cos y =

−1 2

=−

2 ! 2

sin y =

1 2

=

2 2

or

e5 i − e−5 i cos5 + i sin5 − cos5 + i sin5 2i sin5 ! = = = sin5 2i 2i 2i

and so: !y =

3π + 2 kπ ! 4

k = 0, ± 1, ± 2,! !

Therefore:

⎛3 ⎞ 1 !z = x + i y = log e 2 + ⎜ + 2 k ⎟ π i ! 2 ⎝4 ⎠

k = 0, ± 1, ± 2,!

In analogy to the real exponential function, we can also

define the complex exponential function to be that function which is its own derivative. !

Example 4.2-4

e −e 2i

Solution:

d z e = ez = f z ! dz

()

()

f 0 = 1!

(4.2-5)

definition. We can write equation (4.2-2) as:

eπ 5i

()

f′ z =

We will now determine if equation (4.2-2) is consistent with this

Evaluate 1.! e

( )

e5 i − e−5 i cos5 + i sin5 − cos −5 − i sin −5 ! = 2i 2i

or

2.!

1 e

−5i

!

()

w = f z = u + i v = e z = e x cos y + i e x sin y !

(4.2-6)

u = e x cos y !

(4.2-7)

where !

v = e x sin y !

173

We then have from equation (3.4-24):

(

)

(

)

x x ∂ e cos y ∂ e sin y d z ∂u ∂v e = +i = +i ! dz ∂x ∂x ∂x ∂x

!

Therefore: (4.2-8)

∂u ∂v = e x cos y = ! ∂x ∂y

∂u ∂v = − e x sin y = − ! (4.2-12) ∂y ∂x

and so the Cauchy-Riemann equations are satisfied for e z

and so:

d z e = e x cos y + i e x sin y = e x cos y + i sin y = e z ! dz

(

!

!

)

(4.2-9)

everywhere in the complex plane. The complex exponential function e z is then differentiable for all z ∈! and, since it is also

for all z ∈! . We see that equation (4.2-2) is consistent with the

continuous for all z ∈! , it is holomorphic for all z ∈! .

definition of the complex exponential function as a function

Therefore the complex exponential function is an entire

which is its own derivative: f ′ z = f z . This relation between

function.

a function and its derivative does assure that the derivative

!

exists.

function is the only entire complex function that has the

()

()

In Appendix B we show that the complex exponential

following properties:

4.2.2! NATURE OF THE COMPLEX EXPONENTIAL FUNCTION !

( )

( )

Both Re e z = e x cos y and Im e z = e x sin y are continuous

( )

1.! It is its own derivative: !

()

f′ z =

and have continuous first-order derivatives for all x, y . The complex exponential function e z is then continuous for all

d z e = ez = f z ! dz

()

()

f 0 = 1!

(4.2-13)

2.! It agrees with the real exponential function when z is

z ∈! . Using equation (4.2-7), we have:

real:

!

∂u = e x cos y ! ∂x

∂v = e x sin y ! ∂x

(4.2-10)

!

∂u = − e x sin y ! ∂y

∂v = e x cos y ! ∂y

(4.2-11)

!

()

f′ z =

d z e = ex = f x ! dx

()

()

f 0 = 1!

(4.2-14)

174

4.2.3! !

DERIVATIVES OF COMPLEX EXPONENTIAL FUNCTIONS

()

If g z is a holomorphic function, we have:

d g (z) g z e = e ( ) g′ z ! dz

()

!

!

(4.2-15)

From the definition of the complex exponential function

(

)

()

Solution:

(

)

e0 = e0+i 0 = e0 cos 0 + i sin0 = 1!

! !

Determine the derivative of f z = z 3 e z +2i .

(4.2-16)

For any two complex numbers z1 = x1 + i y1 and z2 = x2 + i y2 ,

we have: z

z +z2

z

e 1e 2 = e 1

!

!

(4.2-17)

as shown in Example 4.2-7. !

)

d ! z 3 e z+2i = 3 z 2 e z+2i + z 3 e z +2i dz Example 4.2-6

()

ez

Determine the derivative of f z = e .

Since e z is holomorphic, we have from equation (4.2-15):

( ) ( ez

n

= e x cos y + i e x sin y

)

n

(

)

= en x cos n y + i sin n y !

(4.2-18)

Therefore we have: !

( ) =e

!

Using equation (4.2-17) we can write:

Solution:

z z d ! ee = ee e z dz

If n is an integer, using de Moivre's theorem given in

equation (1.12-1) we can write: !

( )

ALGEBRAIC PROPERTIES OF THE COMPLEX EXPONENTIAL FUNCTION

e z = e x+i y = e x ei y = e x cos y + i sin y when z = 0 we obtain:

Example 4.2-5

(

4.2.4!

!

ez

n

nx

z − z2

e1

n x+i y ei n y = e ( ) = e n z !

z

z

e 2 = e 1!

n = 0, ± 1, ± 2,! ! (4.2-19)

(4.2-20)

or z

!

e1 e

z2

z − z2

=e1

!

(4.2-21) 175

From this equation with z1 = 0 and z2 = z we see that:

1 = e− z ! z e

!

(4.2-22)

Show that e

We can write:

z1 z2

=e e .

!e z+i π = e z ei π

Solution:

From equation (1.13-9) we have:

We can write: z + z2

!e 1

Show that e z + i π = − e z . Solution:

Example 4.2-7 z1+ z2

Example 4.2-8

!ei π = −1

x + iy + x +iy x + x +i y + y = e( 1 1 ) ( 2 2 ) = e( 1 2 ) ( 1 2 )

Therefore: !e z + i π = − e z

or since x1 and x2 are real: i y +y !e z1 + z2 = e x1 e x2 e ( 1 2 )

From equation (1.13-25) we have:

Show that if e z1 = e z2 then z1 = z2 + i 2 k π .

i y +y !e ( 1 2 ) = ei y1 ei y2

Solution:

Therefore: !e z1+ z2 = e x1 e x2 ei y1 ei y2 = e x1 ei y1 e 2 ei y2 = e x1+ i y1 e x2+ i y2 x

and we have finally: !e z1 + z2 = e z1 e z2

Example 4.2-9

If e z1 = e z2 we can write: z

e1

! z = e z1− z2 = 1 e2 From Example 4.2-2 we then have: !z1 − z2 = i 2 k π 176

or

Solution:

!z1 = z2 + i 2 k π

4.2.5!

We have: z

!e = e

MODULUS AND ARGUMENT OF THE COMPLEX EXPONENTIAL FUNCTION

!

Writing equation (4.2-2) in polar form:

!

e z = e x+i y = e x e i y = e x cos y + i sin y !

(

)

(4.2-23)

Therefore e z ≤ e

(4.2-24) !

where from equation (1.11-9): !

e

= cos y + isin y =

! 2

cos y + sin y = 1!

(4.2-25)

We then have for all z ∈! : !

e z = e x = e x = eRe z !

where e z = e

z

only if y = 0 so that z is

For all z ∈! we have:

ez ≠ 0 !

(4.2-28)

as shown in example 4.2-11. Example 4.2-11

(4.2-26)

Show that e z ≠ 0 for all z ∈! . Solution:

We also have: !

z

real.

e z = e x+i y = e x ei y !

2

=e

( Re z ) 2+ ( Im z ) 2

! e z = e x = e x = eRe z

we see that the modulus of e is given by:

iy

=e

x 2+ y 2

From equation (4.2-26) we have:

z

!

x+iy

z

ez ≤ e !

(4.2-27)

Let z = x + i y . We then have:

(

!e z = e x+i y = e x ei y = e x cos y + i sin y Example 4.2-10

)

If e z = 0 we must have: z

Show that e z ≤ e .

!e x cos y = 0 !

e x sin y = 0 177

But x is a real number, and e x > 0 for all values of x . For

The complex exponential function is periodic, therefore, with

e z = 0 then we must have cos y = 0 and sin y = 0 for the

a pure imaginary period of 2 π i . The real exponential function

same value of y . This never happens, however, and so

is not periodic. While we see that e z has a period of 2 π i , this

e z ≠ 0 for all z .

does not prove that no smaller period exists for e z . We will now show that 2 π i is the smallest period of e z .

The argument of e z is given by:

!

( ) ( ) Since arg ( e ) is multivalued, the complex exponential function

Proposition 4.2-1:

single-valued. The complex exponential function does not,

!

arg e z = arg e x+i y = y + 2 k π !

!

k = 0, ± 1, ± 2,!! (4.2-29)

The periodicity of the complex exponential function e z is 2 k π i

z

z

where k is an integer:

x

e is multivalued, whereas the real exponential function e is

therefore, have a defined inverse without restricting its domain of definition. If the argument of e z is restricted to its principal value, then k = 0 and we have: !

( ) z

Arg e = y !

(4.2-30)

!

From equation (1.13-12) we have:

e

i 2 kπ

= 1!

e z = e z ei 2 k π = e z+i 2 k π !

Proof: ! !

k = 0, ± 1, ± 2,!!

!

If T is the period of e z we will have:

e z+T = e z eT = e z !

(4.2-34)

(4.2-31)

T = Tx + i Ty !

(4.2-35)

where Tx and Ty are real numbers. !

Therefore: !

(4.2-33)

We can assume that T is a complex number:

4.2.6! PERIODICITY OF THE COMPLEX EXPONENTIAL FUNCTION !

e z = e z+ i 2k π !

!

From equation (4.2-34) we have:

eT = 1!

(4.2-36)

From the definition of the complex exponential given in

k = 0, ± 1, ± 2,!!

(4.2-32)

equation (4.2-2), we obtain: 178

!

T

(

)

e x cosTy + i sinTy = 1 !

(4.2-37)

or !

T

e x cosTy + i sinTy = 1!

(4.2-38)

cosTy + i sinTy = 1!

(4.2-39)

cosTy + i sinTy = 1 !

!

cosTy = 1 !

(4.2-40)

sinTy = 0 !

!

Since e z = e z+i 2 k π

■

the z-plane can be divided into

!

( 2 k − 1) π < y < ( 2 k + 1) π !

k = 0, ± 1, ± 2,! !

(4.2-42)

continuous, and so each strip is a branch of e z . Within each

From the real and imaginary parts of this equation we obtain: !

Follows from the proof of Proposition 4.2-1.

In each strip of width 2 π , the complex exponential function is

and since Tx is real, we must have Tx = 0 . Therefore: !

!

horizontal strips such that:

Since we have from equation (1.11-9): !

Proof:

branch the function e z is one-to-one.

4.2.7!

(4.2-41)

Since sinTy = 0 we must have Ty = n π where n is an

integer. Since cosTy = 1 we must have n = 2 k where k is an

!

PRINCIPAL VALUE OF THE COMPLEX EXPONENTIAL FUNCTION

If we restrict the argument of e z to its principal value so

that k = 0 in equation (4.2-42), then w = e z is called the

integer. Therefore the periodicity of e z is T = 0 + 2 k π i = 2 k π i

principal value of the complex exponential function. Equation

where k is an integer. The complex exponential function e z is

(4.2-42) becomes:

then periodic with the smallest period being a pure imaginary

!

period of 2 π i .

■

Proposition 4.2-2:

e z = 1 if and only if z = 2 k π i where k is an integer.

−π < y < π !

(4.2-43)

The resulting domain is known as the principal branch of e z , and the strip − π < y < π is referred to as the fundamental strip of e z .

179

4.2.8! !

and so e2 i x e2 y = 1 and e2 i x e2 y = 1 . From equation (1.11-9)

CONJUGATE COMPLEX EXPONENTIAL FUNCTION

we have e2 i x = 1 . Therefore e2 y = 1 and so y = 0 . We then

The conjugate complex exponential function can be

obtained from equation (4.2-2) using the properties of sine and cosine: z

x

x

!cos 2 x = 1!

sin 2 x = 0

and!

and so x = n π . Therefore:

e = e cos y + i e sin y = e cos y − i e sin y !

!

x

have cos 2x + isin 2x = 1 , or

x

(4.2-44)

!z = n π !

n = 0, ± 1, ± 2,!

or

( ( )

( ))

e z = e x cos − y + i sin − y !

!

(4.2-45)

e z = e x e−i y = e x−i y = e z !

(4.2-46)

()

! f z = e z = e x−i y = e x cos y − i e x sin y = u + i v

Example 4.2-12 iz

iz

For what values of a complex variable z is e = e ?

We are given ei z = e i z , which is: !ei ( x−i y ) = e−i ( x−i y ) or !e i x e y = e− i x e− y

Solution: We can write:

for all z ∈! .

Solution:

()

Determine if f z = e z is holomorphic.

Therefore: !

Example 4.2-13

We then have: !u = e x cos y !

v = − e x sin y

and so:

∂u ! = e x cos y ! ∂x

∂u = − e x sin y ∂y

∂v ! = − e x sin y ! ∂x

∂v = − e x cos y ∂y 180

Checking the Cauchy-Riemann equations:

∂u ∂v ! = e x cos y ≠ ! ∂x ∂y

∂u ∂v = − e x sin y ≠ − ∂y ∂x

()

Therefore f z = e z is not holomorphic anywhere. From

()

equation (4.2-46) we then see that f z = e z is also not holomorphic anywhere.

4.2.9! !

MAPPING PROPERTIES OF THE COMPLEX EXPONENTIAL FUNCTION

Mapping from the z-plane to the w-plane using the

complex exponential function: !

w = e z = e z+i 2 k π !

k = 0, ± 1, ± 2,!!

(4.2-47)

can be seen to be the mapping of many points to one point. Mapping using the complex exponential function is then not

Figure 4.2-1!

one-to-one. Since e z is periodic with period 2 π i , e z can be

Horizontal strips of width 2 π in the z-plane. The fundamental strip is highlighted in red.

On each such strip e z will perform in the same way. The

plotted on the z-plane as horizontal strips extending to infinity

!

in both the plus and minus x directions. By selecting a restricted

strips are all defined by:

domain of e z = e x+i y in the form of a horizontal strip such as:

!

!

−∞ < x < ∞!

−π < y ≤ π !

(4.2-48)

and

−∞ < x < ∞!

( 2 k − 1) π < y ≤ ( 2 k + 1)π !

(4.2-49)

k = 0, ± 1, ± 2, +! !

the mapping of w = e z becomes one-to-one. The width of these

!

strips is 2 π as measured along the imaginary axis (see Figure

The particular strip defined by equation (4.2-48) is called the

4.2-1).

fundamental strip as noted previously.

(4.2-50)

181

!

For the fundamental strip we have:

!

w = e z = e x+ i y = e x e i y = ρ eiφ !

!

axis in the z-plane will produce a set of rays emanating from (4.2-51)

To visualize mapping using the complex exponential

the origin in the w-plane. Points with x = 0 will be mapped onto a unit circle in the w-plane, while points with x < 0 will be mapped inside a unit circle in the w-plane.

function e z , we can rewrite equation (4.2-51) as:

w = u + i v = e z = e x ei y = ρ eiφ !

!

(4.2-52)

Letting y = θ , we can write:

w = u + i v = e x eiθ = ρ eiφ !

!

(4.2-53)

and so: ! !

eiθ = eiφ !

ρ = ex !

(4.2-54)

We see that w = e z maps a vertical line x = a in the z-plane

to a circle in the w-plane with center at the origin and radius

ρ = ea (see Figure 4.2-2): !

u 2 + v 2 = e2 a !

(4.2-55)

and w = e maps a horizontal line y = θ in the z-plane to a ray z

emanating from the origin in the w-plane having an angle φ = θ where −π < θ ≤ π (see Figure 4.2-2). !

z

Therefore, the mapping w = e of a set of lines parallel to

Figure 4.2-2! Mapping of lines x = a and y = θ using w = e z . In the w-plane ρ = ea and φ = θ . !

With each increase in y = θ by 2 π , the image point in the

w-plane moves completely around the circle. The range of w

the imaginary axis in the z-plane will produce a set of

includes all circles with center at the origin w = 0 . The origin

concentric circles having their center at the origin in the w-

itself is not included in the range of w .

z

plane. The mapping w = e of a set of lines parallel to the real

182

Example 4.2-14 Using w = e z = e x+i y , map the region defined by x0 < x < x1 and −π < y = θ < π 2 shown in Figure 4.2-3. Solution: !

In the z-plane we have x0 < x < x1 and −π < y = θ < π 2

4.3! !

COMPLEX TRIGONOMETRIC FUNCTIONS Using Euler’s formula eiθ = cosθ + i sin θ

equation (1.11-6), the following equations for real cosine and sine functions can be derived (see Example 1.11-1):

as the dimensions of a rectangle. !

In the w-plane we have ρ0 = e x0 and ρ1 = e x1 as the

radii of a 3 4 annulus.

as given in

eiθ + e−iθ cosθ = ! 2

!

eiθ − e−iθ sin θ = ! 2i

(4.3-1)

where θ is a real number in radians. !

We see then that real cosine and sine functions can be

expressed in terms of complex exponential functions. Real cosine and sine functions cannot, however, be expressed in terms of real exponential functions.

4.3.1! !

DEFINITION OF COMPLEX TRIGONOMETRIC FUNCTIONS

Based upon equation (4.3-1) we can define the complex

trigonometric functions for z ∈! as follows: ! !

For cos z and sin z :

ei z + e−i z ! cos z = 2

ei z − e−i z ! sin z = 2i

(4.3-2)

Figure 4.2-3! Mapping of the region defined by x0 < x < x1 and −π < y = θ < π 2 using w = e z . 183

! ! !

For tan z we then have:

sin z ei z − e−i z tan z = = − i i z −i z ! cos z e +e

Example 4.3-1

⎛ 1⎞ z ≠ ⎜ n + ⎟ π ! (4.3-3) 2⎠ ⎝

Solution:

For sec z , csc z , and cot z :

!

sec z =

1 2 = i z −i z ! cos z e + e

⎛ 1⎞ z ≠ ⎜ n + ⎟ π ! (4.3-4) 2⎠ ⎝

!

csc z =

1 2i = i z −i z ! sin z e − e

z ≠ nπ !

!

1 ei z + e−i z cot z = = i i z −i z ! tan z e −e

Determine cos i and sin i .

(4.3-5)

Using equation (4.3-2), we have: 2

2

ei + e−i e−1 + e e2 + 1 !cos i = = = 2 2 2e and 2

z ≠ nπ !

2

ei − e−i e−1 − e e2 − 1 !sin i = = =i 2i 2i 2e (4.3-6)

where n = 0, ± 1, ± 2, ± 3, ! . Since the complex exponential

Example 4.3-2

(

)

function is multivalued, all the complex trigonometric

Determine cos π − 2 i .

functions are also multivalued.

Solution:

!

If z is real, then the definitions for the complex cosine and

sine functions given in equation (4.3-2) become, respectively, the expressions for the real cosine and sine functions given in

Using equation (4.3-2), we have:

(

equation (4.3-1). This is important if complex cosine and sine functions are to be the complex equivalents, respectively, of the real cosine and sine functions.

)

!cos π − 2 i =

e(

i π −2 i )

−i π −2 i ) +e ( e2 ei π + e−2 e−i π = 2 2

or

(

)

!cos π − 2 i =

(

( ( )

)

( ))

e2 cos π + i sin π + e−2 cos −π + i sin −π 2

184

and so:

and so:

(

− e2 − e− 2 e2 + e−2 !cos π − 2i = =− = − cosh 2 2 2

(

)

(

)

! i − 1 ei z − i + 1 e−i z = 0

)

or

Example 4.3-3

!e2 i z =

Show that 2 sin z cos z = sin 2 z .

(1+ i ) (1+ i ) = 2 i = i (1− i ) ( 1+ i ) 2

i π From equation (1.13-13) we have e (

Solution:

⎛π ⎞ !2i z = i ⎜ + 2 k π ⎟ ! ⎝2 ⎠

Using equation (4.3-2), we have:

ei z − e−i z ei z + e−i z ei 2 z − e−i 2 z !2 sin z cos z = 2 = 2i 2 2i

2+2 k π )

= i . Therefore:

k = 0, ± 1, ± 2, ± 3, !

or !z =

and so:

π + kπ ! 4

k = 0, ± 1, ± 2, ± 3, !

!2 sin z cos z = sin 2 z (see Example 1.12-1). Example 4.3-4 Determine all complex numbers z such that cos z = sin z . Solution: Using equation (4.3-2), we have:

ei z + e−i z ei z − e−i z !cos z = = = sin z 2 2i

4.3.2! NATURE OF THE COMPLEX TRIGONOMETRIC FUNCTIONS !

Complex sine and complex cosine functions are entire

since they each are linear combinations of exponential functions that are entire. The other trigonometric functions given in equations (4.3-3) through (4.3-6) are not entire. Both tan z and

sec z are holomorphic on those domains that do not include points where cos z is zero, and both cot z and csc z are holomorphic on those domains that do not include points where sin z is zero. 185

4.3.3! !

DERIVATIVES OF COMPLEX TRIGONOMETRIC FUNCTIONS

Since the complex sine and complex cosine functions are

!

d tan z = sec 2 z ! dz

⎛ 1⎞ z ≠ ⎜n+ ⎟ π ! 2⎠ ⎝

(4.3-11)

!

d sec z = sec z tan z ! dz

⎛ 1⎞ z ≠ ⎜n+ ⎟ π ! 2⎠ ⎝

(4.3-12)

!

d csc z = − csc z cot z ! dz

z ≠ nπ !

(4.3-13)

!

d cot z = − csc 2 z ! dz

z ≠ nπ !

(4.3-14)

entire, their derivatives are also entire. These derivatives are given by: !

d d ⎛ ei z + e−i z ⎞ i ei z − i e−i z ei z − e−i z cos z = ⎜ =− ! (4.3-7) ⎟= dz dz ⎝ 2 2 2i ⎠

and so:

d cos z = − sin z ! dz

!

(4.3-8)

4.3.4! IDENTITIES OF COMPLEX TRIGONOMETRIC FUNCTIONS

Similarly: !

d d ⎛ ei z − e−i z ⎞ i ei z + i e−i z ei z + e−i z ! sin z = ⎜ = = dz dz ⎝ 2i ⎟⎠ 2i 2

(4.3-9)

!

!

Most of the identities that exist for real trigonometric

functions also apply for complex trigonometric functions. This is to be expected since both real and complex trigonometric

and so: !

where n = 0, ± 1, ± 2, ± 3, ! .

d sin z = cos z ! dz

functions can be expressed similarly in terms of complex (4.3-10)

It can be shown that the derivatives of the other four

complex trigonometric functions have the same form as their equivalent real functions:

exponential functions. !

From equations (4.3-2) and (4.3-3) we have:

( )

!

cos −z = cos z !

!

tan −z = − tan z !

( )

( )

sin −z = − sin z !

(4.3-15) (4.3-16) 186

!

cos 2 z + sin 2 z = 1 !

(4.3-17)

)

(4.3-18)

(

)

(4.3-19)

cos z1 ± z2 = cos z1 cos z2 ∓ sin z1 sin z2 !

!

sin z1 ± z 2 = sin z1 cos z2 ± cos z1 sin z2 !

!

tan z1 ± z 2 =

(

)

tan z1 ± tan z2 ! 1∓ tan z1 tanz2

(

) (

)

(

) (

)

!

2 sin z1 + z2 sin z1 − z2

!

2 cos z1 + z2 sin z1 − z2 = sin 2 z1 − sin 2 z2 !

(4.3-20)

= cos 2 z2 − cos 2 z1 !

(4.3-21) (4.3-22)

!

From equations (4.3-21) and (4.3-22), we have:

!

⎛ z2 + z1 ⎞ ⎛ z2 − z1 ⎞ ! cos z2 − cos z1 = − 2 sin ⎜ sin ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠

!

⎛ z2 + z1 ⎞ ⎛ z2 − z1 ⎞ ! sin z2 − sin z1 = 2 cos ⎜ cos ⎜ ⎝ 2 ⎟⎠ ⎝ 2 ⎟⎠

Letting z1 = z and z2 = π , we have from equations (4.3-18)

and (4.3-19):

(

!

!

(

)

(4.3-27)

(

)

(4.3-28)

!

cos z + π = − cos z !

!

sin z + π = − sin z !

!

Letting z1 = z and z2 = π 2 , we also have from equations

(4.3-18) and (4.3-19): !

⎛ π⎞ cos ⎜ z + ⎟ = sin z ! 2⎠ ⎝

(4.3-29)

!

⎛ π⎞ sin ⎜ z + ⎟ = cos z ! 2⎠ ⎝

(4.3-30)

Example 4.3-5 (4.3-23)

Show that cos 2 z + sin 2 z = 1 . Solution:

(4.3-24)

From equation (4.3-2), we have: 2

! ! !

Letting z1 = z2 = z in equations (4.3-18) and (4.3-19), we get:

cos 2 z = cos 2 z − sin 2 z ! sin 2 z = 2 sin z cos z !

(4.3-25) (4.3-26)

⎛ ei z + e−i z ⎞ e2 i z + e− 2 i z + 2 2 !cos z = ⎜ ⎟ = 2 4 ⎝ ⎠ 2

⎛ ei z − e−i z ⎞ e2 i z + e− 2 i z − 2 2 !sin z = ⎜ ⎟ =− 2i 4 ⎝ ⎠ 187

or

Therefore:

(

e2 i z + e− 2 i z + 2 − e2 i z − e− 2 i z + 2 2 2 !cos z + sin z = 4

)

!cos z1 + z2 =

(

Example 4.3-6

(

e

i z2

+e 2

−i z2

e

−

i z1

−e 2i

−i z1

e

i z2

−e 2i

−i z2

)

)

(

)

( )! !tan ( z1 + z2 ) = cos ( z1 + z2 ) sin z1 + z2

Adding and subtracting the term: −i z1 i z2

(

)

!tan z1 + z2 =

we obtain:

)

!cos z1 + z2 =

(

)

tan z1 − z2 =

( ) cos ( z1 − z2 ) sin z1 − z2

From equations (4.3-19) and (4.3-18) we have:

e

iz

1∓ tan z1 tan z2

.

From equation (4.3-3), we have:

i z +z −i z +z iz iz −i z −i z e ( 1 2 ) + e ( 1 2 ) e 1e 2 + e 1e 2 !cos z1 + z2 = = 2 2

+e 4

tan z1 ± tan z2

Solution:

From equation (4.3-2), we have:

(

)

Show that tan z1 ± z2 =

Solution:

(

−i z1

Example 4.3-7

Show that cos z1 + z2 = cos z1 cos z2 − sin z1 sin z2 .

−i z2

+e 2

!cos z1 + z2 = cos z1 cos z2 − sin z1 sin z2

!cos 2 z + sin 2 z = 1

iz

i z1

and so:

and so:

e 1e !

e

e 1e

i z2

+e iz

!+

−i z1 −i z2

e 1e

e

i z2

iz

+ e 1e 4

−i z1 −i z2

+e

e

−i z2

+e iz

−i z1 i z2

− e 1e 4

e

−i z 2

(

)

!tan z1 − z2 = −i z1

−e

e

i z2

sin z1 cos z2 + cos z1 sin z2 cos z1 cos z2 − sin z1 sin z2 sin z1 cos z2 − cos z1 sin z2 cos z1 cos z2 + sin z1 sin z2

and so: 188

(

)

!tan z1 + z2 =

(

)

!tan z1 − z2 =

sin z1 cos z2 + cos z1 sin z2 cos z1 cos z2 cos z1 cos z2 − sin z1 sin z2 cos z1 cos z2 sin z1 cos z2 − cos z1 sin z2 cos z1 cos z2 cos z1 cos z2 + sin z1 sin z2 cos z1 cos z2

and so:

(

Therefore: !e2 i z i tan z − e2 i z = − 1− i tan z or !e2 i z =

or

(

)

!tan z1 ± z2 =

tan z1 ± tan z2

If sin z = 2 find sin 2 z .

Example 4.3-8

Solution:

1+ i tan z . 1− i tan z

We can write: !cos 2 z = 1− sin 2 z = 1− 4 = −3

Solution:

and so:

From equation (4.3-3), we have:

ei z − e−i z !tan z = − i i z −i z e +e

)

!cos z = i 3 From equation (4.3-26) we have:

()

!sin 2 z = 2 sin z cos z = 2 2

or

(

1+ i tan z 1− i tan z

Example 4.3-9

1∓ tan z1 tan z2

Show that e2 i z =

)

! e2 i z + 1 i tan z = e2 i z − 1

(

! ei z + e−i z tan z = − i ei z − e−i z

3i=4 3i

) 189

!

In terms of hyperbolic sines and cosines, we have:

!

cos z = cos x cosh y − i sin x sinh y !

(4.3-31)

!

sin z = sin x cosh y + i cos x sinh y !

(4.3-32)

sin x cos x + i sinh y cosh y tan z = ! cos 2 x + sinh 2 y

!

(4.3-33)

!sin z =

(

Show that: !

cos z = cos x cosh y − i sin x sinh y

!

sin z = sin x cosh y + i cos x sinh y

(

)

2i

or

e ( !cos z = cos x e ( !sin z = sin x

Example 4.3-10

)

e− y cos x + i sin x − e y cos x − i sin x

y

y

+ e− y 2

+ e− y 2

) − i sin x ( e

) + i cos x ( e

y

y

− e− y

)

2 − e− y

)

2

Therefore: !cos z = cos x cosh y − i sin x sinh y !sin z = sin x cosh y + i cos x sinh y

Solution: Using z = x + i y , equation (4.3-2) becomes: !cos z =

e

i ( x+i y )

+e 2

−i ( x+i y )

sin z =

!

e

i ( x+i y )

−e 2i

−i ( x+i y )

Example 4.3-11 Show that tan z =

We then have:

e− y ei x + e y e−i x !cos z = ! 2

e− y ei x − e y e−i x sin z = 2i

and so: !cos z =

(

)

(

e− y cos x + i sin x + e y cos x − i sin x 2

)

sin x cos x + i sinh y cosh y . 2 2 cos x + sinh y

Solution: From equations (4.3-31) and (4.3-32) we have: !tan z =

sin x cosh y + i cos x sinh y cos x cosh y − i sin x sinh y

We can write: 190

sin x cosh y + i cos x sinh y ) ( cos x cosh y + i sin x sinh y ) ( !tan z = (cos x cosh y − i sin x sinh y )(cos x cosh y + i sin x sinh y ) or !tan z =

(

)

(

sin x cos x cosh 2 y − sinh 2 y + i sinh y cosh y sin 2 x + cos 2 x

)

cos 2 x cosh 2 y + sin 2 x sinh 2 y

!

sin z = sin 2 x + sinh 2 y !

(4.3-37)

!

cos z =

cosh 2 y − sin 2 x !

(4.3-38)

real cosine and sine functions. Therefore the complex cosine

)

cos 2 x cosh 2 y − sinh 2 y + sinh 2 y

and sine functions are unbounded.

Using cosh 2 y − sinh 2 y = 1 : !tan z =

(4.3-36)

both cos z and sin z can be larger than 1 unlike the moduli of

sin x cos x + i sinh y cosh y

(

cos z = cos 2 x + sinh 2 y !

Note that since sinh y is not bounded ( sinh y → ∞ as y → ∞ ),

or using cosh 2 y − sinh2 y = 1 and sin 2 x + cos 2 x = 1 : !tan z =

!

Example 4.3-12

sin x cos x + i sinh y cosh y

Show that:

cos 2 x + sinh 2 y

cos z = cos 2 x + sinh 2 y

4.3.5! !

MODULUS OF COMPLEX TRIGONOMETRIC FUNCTIONS

Using equations (4.3-31) and (4.3-32), the modulus of cos z

and sin z can be written as: 2

2

2

2

Solution: From the identity:

!

cos z = cos x cosh y + sin x sinh y !

(4.3-34)

!

sin z = sin 2 x cosh 2 y + cos 2 x sinh 2 y !

(4.3-35)

We also have:

sin z = sin 2 x + sinh 2 y

!cosh 2 y − sinh2 y = 1 we have from equations (4.3-34) and (4.3-35): ! cos z =

(

)

cos 2 x 1+ sinh2 y + sin 2x sinh 2 y 191

(

)

! sin z = sin 2 x 1+ sinh 2 y + cos 2 x sinh 2 y or

(

!

)

! cos z = cos 2 x + cos 2 x + sin 2x sinh 2 y ! sin z =

2

(

2

2

)

2

sin x + cos x + sin x sinh y

Therefore the modulus of cos z and sin z are given by: ! cos z =

cos 2 x + sinh 2 y

! sin z =

sin 2 x + sinh 2 y

Example 4.3-13 Show that cos z =

From equation (4.3-36) we have: ! cos z = cos 2 x + sinh 2 y =

cos 2 x + cosh 2 y − 1

or

Therefore:

cosh 2 y − sin 2 x

From equations (4.3-36), (4.3-37), and (4.3-38) we have the

inequalities: !

cos z ≥ cos x !

(4.3-39)

!

sin z ≥ sin x !

(4.3-40)

!

cos z ≥ sinh y !

(4.3-41)

!

sin z ≥ sinh y !

(4.3-42)

!

cos z ≤ cosh y !

(4.3-43)

!

sin z + cos z ≥ 1 !

cosh 2 y − sin 2 x .

Solution:

! cos z =

! cos z =

2

2

(4.3-44)

Example 4.3-14 Show that cos5i ≥ 1 . Solution:

(

cosh 2 y − 1− cos 2 x

)

From equation (4.3-36) we have with x = 0 and y = 5 : ! cos z = cos 2 x + sinh 2 y = 1+ sinh 2 5 > 1 192

(

sin z = sin z + 2 π

Example 4.3-15

)

Solution:

Show that: 2

2

sin z + cos z ≥ 1

From equations (4.3-2) and (1.13-12) we have: i z+2 π ) −i z+2 π ) e( +e ( ei z + e−i z !cos z + 2 π = = = cos z 2 2

(

Solution: From equations (4.3-40) and (4.3-39) we have: 2

2

2

2

2

! sin z + cos z ≥ sin x + cos x

)

i z+2 π ) −i z+2 π ) e( −e ( ei z − e−i z !sin z + 2 π = = = sin z 2i 2i

(

2

)

or !

! sin z + cos z ≥ sin x + cos x = 1 2

2

While we have shown that cos z and sin z have a period of

2 π , this does not prove that no smaller period exists for these functions. We will now show that 2 π is the smallest period of

4.3.6! ! ! !

PERIODICITY OF THE COMPLEX TRIGONOMETRIC FUNCTIONS

cos z and sin z . !

complex exponential function is 2 k π i :

As shown in Example 4.3-16 we have:

(

)

cos z = cos z + 2 π !

(

)

sin z = sin z + 2 π ! Example 4.3-16 Show that:

(

cos z = cos z + 2 π

(4.3-45)

!

e z = e z+i 2 k π !

(4.3-47)

where k is an integer. Since cos z and sin z are defined in terms (4.3-46)

of ei z and e−i z : !

)

From Proposition 4.2-1 we know that the periodicity of the

ei z + e−i z ! cos z = 2

ei z − e−i z ! sin z = 2i

(4.3-48)

both cos z and sin z must be periodic with a period of 2 k π . The smallest period of cos z and of sin z is then T = 2 π . We also see 193

We see then that all the zeros of cos z and of sin z are real.

that both complex secant and cosecant have a periodicity of

!

2 kπ .

The complex tan z , has only the same zeros as the complex sin z ,

!

since tan z = sin z cos z . Similarly, the complex cot z has the same

For the complex tangent we can use equations (4.3-27) and

zeros as the complex cos z .

(4.3-28) to write:

tan z =

!

( (

sin z + π

) )

sin z − sin z = = = tan z + π ! cos z − cos z cos z + π

(

)

(4.3-49)

and so the complex tangent and cotangent are periodic with a period of π .

4.3.7!

ZEROS OF COMPLEX TRIGONOMETRIC FUNCTIONS

Since cos 2 x ≥ 0 and sinh 2 y ≥ 0 , we see from equation

!

(

y

(4.3-36) that cos z = 0 if and only if sinh y = e − e

−y

)

2 = 0 and

cos x = 0 . We will then have cos z = 0 only if y = 0 and

(

)

x = 2 k + 1 π 2 where k = 0, ± 1, ± 2,! . Therefore the complex cos z has only the same zeros as the real cos x , and so all the zeros of cos z are real. !

Similarly, since sin 2 x ≥ 0 and sinh 2 y ≥ 0 , we see from

equation (4.3-37) that we will have sin z = 0 if and only if

(

sinh y = e y − e− y

)

2 = 0 and sin x = 0 Therefore we must have

y = 0 and x = k π where k = 0, ± 1, ± 2,! . Therefore the complex sin z has only the same zeros as the real sin x , and so all the zeros of sin z are real.

In summary, we have for k = 0, ± 1, ± 2,! :

! !

sin z = 0 !

when z = k π !

!

cos z = 0 !

when z =

!

tan z = 0 !

when z = k π !

!

cot z = 0 !

when z =

(4.3-50)

( 2 k + 1) π ! 2

(4.3-51) (4.3-52)

( 2 k + 1) π ! 2

(4.3-53)

Example 4.3-17 Determine z if cos z = 2 . Solution: This is a case where z cannot be real since z is greater than one. We have from equation (4.3-31): !cos z = cos x cosh y − i sin x sinh y = 2 or !cos x cosh y = 2 !

sin x sinh y = 0 194

Equating the real and imaginary parts of this equation, we

and so for sin x sinh y = 0 we must have: !x = k π !

k = 0, ± 1, ± 2, ! !

or!

have:

y=0

!e x cos y = −1!

If y = 0 then cosh y = 1 and so cos x = 2 . This is not possible,

Since e x > 0 for x ∈! we have from e x sin y = 0 :

however, since x is real. Therefore we must have: !x = k π !

e x sin y = 0

y = kπ !

!sin y = 0 !

k = 0, ± 1, ± 2, !

k = 0, ± 1, ± 2, !

Since e x > 0 and y = k π we have from e x cos y = −1 :

and so:

( )

!cos x = −1

( )k

!e x cos y = e x −1 = −1

k

( )

If k is even we have e x = −1 which has no solution.

k

We then have −1 cosh y = 2 . Since cosh y > 0 we must also

If k is odd we have e x = 1 , and so:

have: !k = 2 n !

n = 0, ± 1, ± 2, !!

)

n = 0, ± 1, ± 2, !

Therefore:

and so cos x = 1 . We then have x = 2 n π and y = cosh −1 2 .

(

)

!z = i 2 n + 1 π !

Therefore z = 2 nπ + i cosh −1 2 with n = 0, ± 1, ± 2, ! . Example 4.3-18

(

y = 2n +1 π !

!x = 0 !

4.3.8!

n = 0, ± 1, ± 2, !

PURE IMAGINARY TRIGONOMETRIC FUNCTIONS

Determine z if 1+ e z = 0 .

!

Solution:

z = i y where y is a real number, we have from equation (4.3-2):

Using Euler’s formula we can write: !1 + e z = 1+ e x+i y = 1+ e x ei y = 1+ e x cos y + i e x sin y = 0

!

If z is a pure imaginary number so that we can write 2

ei y + e−i cos i y = 2

( )

2

y

e y + e− y = = cosh y ! 2

(4.3-54)

and 195

( )

! sin i y =

e

i 2y

−e 2i

−i 2y

=−

y

1 e −e i 2

−y

=i

y

e −e 2

−y

= i sinh y ! (4.3-55)

!!

Therefore:

( )

cosh y = cos i y !

!

(4.3-56)

and !

!

−i z iz ei z − e−i z ei z − e−i z e − e ei z − e−i z sin z = = = = = sin z 2i − 2i 2i 2i

!

(4.3-58)

ei z + e−i z ei z + e−i z e−i z + e i z ei z + e−i z cos z = = = = = cos z ! 2 2 2 2

!!

( )

sinh y = − i sin i y !

(4.3-57)

Example 4.3-19

!

(4.3-59) We then have: 2

!

sin z = sin z sin z = sin z sin z !

!

cos z = cos z cos z = cos z cos z !

(4.3-60)

Show that cosh 2 x − sinh 2 x = 1 . Solution: From equations (4.3-56) and (4.3-57), we have: 2

2

( )

!cosh x = cos i x

( ) ( )

!cosh 2 x − sinh 2 x = cos 2 i x + sin 2 i x = 1

4.3.9! CONJUGATE COMPLEX SINE AND COSINE FUNCTIONS !

To map from the z-plane to the w-plane the complex sine

function:

Therefore from equation (4.3-17) we have:

( )

(4.3-61)

4.3.10! MAPPING PROPERTIES OF THE COMPLEX SINE FUNCTION !

!i 2 sinh 2 x = − sinh 2 x = sin 2 i x

2

The conjugate complex sine and cosine functions can be

obtained from equations (4.3-2) and (4.2-46):

!

w = sin z !

(4.3-62)

we will use equation (4.3-32): !

w = u + i v = sin z = sin x cosh y + i cos x sinh y !

(4.3-63)

We then have: !

u = sin x cosh y !

v = cos x sinh y !

(4.3-64)

and so we can write: 196

!

u2 v2 cosh y − sinh y = 2 − = 1! sin x cos 2 x 2

2

(4.3-65). This equation represents a hyperbola in the w-plane. A (4.3-65)

to the w-plane using equation (4.3-66). This equation represents

We also have: !

the upper half of an ellipse in the w-plane (see Figure 4.3-1).

u2 v2 sin x + cos x = + = 1! 2 2 sinh y cosh y 2

horizontal line y = b in the z-plane where 0 < b can be mapped

2

(4.3-66)

4.4! !

COMPLEX HYPERBOLIC FUNCTIONS The real hyperbolic cosine and sine functions are defined

in terms of real exponential functions:

e x + e− x cosh x = ! 2

!

e x − e− x sinh x = ! 2

(4.4-1)

where x is dimensionless. These real hyperbolic functions can easily be extended to the complex plane by defining them in terms of complex exponential functions.

4.4.1! Figure 4.3-1!

!

Mapping of lines x = a and y = b using w = sin z = sin x + i y .

(

)

To visualize the mapping of the complex sine function, we

will consider two lines in the z-plane: a vertical line and a horizontal line. A vertical line x = a in the z-plane where

0 < a ≤ π 2 can be mapped to the w-plane using equation

!

DEFINITION OF COMPLEX HYPERBOLIC FUNCTIONS

Based upon equation (4.4-1) we will define the complex

hyperbolic functions for z ∈! as follows: ! ! !

For cosh z and sinh z :

e z + e− z ! cosh z = 2

e z − e− z ! sinh z = 2

(4.4-2)

For tanhz : 197

!

sinh z e z − e− z tanh z = = ! cosh z e z + e− z

!

We then have:

!

sech z =

!

!

1 ! cosh z

1 ! csch z = sinh z 1 ! coth z = tanh z

⎛ 1⎞ z ≠ ⎜n+ ⎟ π i! 2⎠ ⎝

⎛ 1⎞ z ≠ ⎜n+ ⎟ π i! 2⎠ ⎝

(4.4-3)

! (4.4-4)

Both the complex hyperbolic cosine and hyperbolic sine

functions are entire since they are linear functions of complex functions given in equations (4.4-3) through (4.4-6) are not

z ≠ nπ i !

(4.4-5)

entire. Both tanh z and sech z are holomorphic on those domains that do not include points where cosh z is zero, and

z ≠ nπ i !

(4.4-6)

exponential function is multivalued, all the complex hyperbolic functions are also multivalued. If z is real, then the definitions for complex hyperbolic

cosine and hyperbolic sine functions given in equation (4.4-2) become, respectively, the expressions for real hyperbolic cosine

both coth z and csch z are holomorphic on those domains that do not include points where sinh z is zero.

4.4.3! DERIVATIVES OF COMPLEX HYPERBOLIC FUNCTIONS !

important if the complex hyperbolic cosine and sine functions

The derivatives of the complex hyperbolic cosine and

hyperbolic sine functions are given by: !

and hyperbolic sine functions given in equation (4.4-1). This is

d d ⎛ e z + e− z ⎞ e z − e− z cosh z = ⎜ = = sinh z ! dz dz ⎝ 2 ⎟⎠ 2

(4.4-7)

Similarly:

are to be the complex equivalent, respectively, of the real hyperbolic cosine and sine functions.

NATURE OF THE COMPLEX HYPERBOLIC FUNCTIONS

exponential functions which are entire. The other hyperbolic

where n = 0, ± 1, ± 2,! (see Section 4.4.8). Since the complex

!

4.4.2!

!

d d ⎛ e z − e− z ⎞ e z + e− z sinh z = ⎜ = = cosh z ! ⎟ dz dz ⎝ 2 ⎠ 2

(4.4-8)

198

!

It can also be shown that the derivatives of the other four

ei z − e−i z sinh i z = = i sin z ! 2

( )

!

complex hyperbolic functions have the same form as their equivalent real functions: !

!

!

!

(4.4-14)

or

d tanh z = sech2z ! dz

⎛ 1⎞ z ≠ ⎜n+ ⎟ π i! 2⎠ ⎝

d sechz = − sechz tanh z ! dz

⎛ 1⎞ z ≠ ⎜n+ ⎟ π i! 2⎠ ⎝

(4.4-10)

d csch z = − csch z coth z ! dz

z ≠ nπ i !

(4.4-11)

d coth z = − csch2z ! dz

z ≠ nπ i !

(4.4-9)

( )

cos z = cosh i z !

!

( )

sin z = − i sinh i z !

(4.4-15)

If in the definitions of cos z and sin z given in equation

!

(4.3-2) we replace z with i z , we obtain using equation (4.4-2):

(4.4-12)

!

e z + e− z cos i z = = cosh z ! 2

(4.4-16)

!

e z − e− z sin i z = − = i sinh z ! 2i

(4.4-17)

( )

( )

or

( )

( )

(4.4-18)

sinh z − i sin i z tanh z = = = − i tan i z ! cosh z cos i z

(4.4-19)

cosh z = cos i z !

!

where n = 0, ± 1, ± 2,! .

sinh z = − i sin i z !

We then have:

4.4.4!

!

RELATION OF COMPLEX HYPERBOLIC FUNCTIONS TO COMPLEX TRIGONOMETRIC FUNCTIONS

If in the definitions of cosh z and sinh z given in equation

(4.4-2) we replace z with i z , we obtain using equation (4.3-2): !

ei z + e−i z cosh i z = = cos z ! 2

( )

(4.4-13)

( ) ( )

!

( )

Example 4.4-1 Evaluate: 1.! 2.!

() sinh ( i ) cosh i

199

(

)

(4.4-23)

!

cosh z1 ± z2 = cosh z1 cosh z2 ± sinh z1 sinh z2 !

4.!

sin i

()

!

sinh z1 ± z2 = sinh z1 cosh z2 ± cosh z1 sinh z2 !

!

tanh z1 ± z2 =

!

sinh 2 z = 2 sinh z cosh z !

(4.4-25)

!

cosh 2 z = cosh2 z− sinh 2 z !

(4.4-26)

!

tanh 2 z =

2 tanh z ! 2 1+ tanh z

(4.4-27)

!

1+ coth 2 z ! coth 2 z = 2 coth z

(4.4-28)

!

sech 2 z = 1− tanh 2 z !

(4.4-29)

!

csch 2 z = coth 2 z − 1 !

(4.4-30)

!

sinh z1 + sinh z2 = 2 sinh

!

sinh z1 − sinh z2 = 2 sinh

() () !sinh ( i ) = i sin (1) !cosh i = cos 1

From equations (4.4-16) and (4.4-17), we have:

() () !sin ( i ) = isinh ( 1) !cos i = cosh 1

4.4.5!

IDENTITIES OF COMPLEX HYPERBOLIC FUNCTIONS

From equations (4.3-18) and (4.3-19) together with the

relations of the complex hyperbolic functions to the complex trigonometric functions given in equations (4.4-15) and (4.4-18), a number of identities can be derived. These include:

!

(4.4-22)

()

From equations (4.4-13) and (4.4-14), we have:

!

)

cos i

Solution:

!

(

3.!

( )

cosh −z = cosh z ! 2

2

cosh z − sinh z = 1 !

( )

sinh −z = − sinh z !

(4.4-20) (4.4-21)

(

)

tanh z1 ± tanh z2 1± tanh z1 tanh z2

z1 + z2 2 z1 − z2 2

!

cosh

cosh

(4.4-24)

z1 − z2 2 z1 + z2 2

!

(4.4-31)

!

(4.4-32) 200

!

!

cosh z1 + cosh z2 = 2 cosh

cosh z1 − cosh z2 = 2 sinh

z1 + z2 2 z1 + z2 2

cosh

sinh

z1 − z2

!

2 z1 − z2 2

!

(4.4-33)

(4.4-34)

Example 4.4-2

⎛ π Evaluate sinh ⎜ 1+ 2 ⎝

⎛ π !sinh ⎜ 1+ 2 ⎝

⎞ e + e−1 i ⎟ = i cosh1 = i 2 ⎠

Example 4.4-3 Show that cosh 2 z − sinh 2 z = 1 . Solution:

⎞ i⎟ . ⎠

From equation (4.4-18) we have:

(

)2 (

!cosh 2 z − sinh 2 z = cos i z − − i sin i z Solution:

π Taking z1 = 1 and z2 = i and using equation (4.4-23), we 2 have: ⎛ π !sinh ⎜ 1+ 2 ⎝

⎞ π π i ⎟ = sinh1 cosh i + cosh1 sinh i 2 2 ⎠

From equations (4.4-13) and (4.4-14) we have:

⎛π !cosh ⎜ ⎝2

⎞ π i ⎟ = cos = 0 2 ⎠

⎛π !sinh ⎜ ⎝2

⎞ π i ⎟ = i sin = i 2 ⎠

Therefore:

)2 = cos2 i z − i 2 sin2 i z

From equation (4.3-17) we have: !cosh 2 z − sinh 2 z = cos 2 i z + sin 2 i z = 1

Example 4.4-4 Show that

d tanh z = sech2z . dz

Solution: From equations (4.4-7) and (4.4-8) we have:

d d sinh z cosh z sinh 2 z cosh 2 z − sinh 2 z ! tanh z = = − = dz dz cosh z cosh z cosh 2 z cosh 2 z Using equation (4.4-21) we have: 201

d 1 2 ! tanh z = = sech z! dz cosh 2 z

! sin z = sin 2 x + sinh 2 y

where cosh z ≠ 0

Therefore:

(

)

Show that sinh z1 + z2 = sinh z1 cosh z2 + cosh z1 sinh z2 .

2

2

! sin z + cos z = 1+ sinh 2 y + sinh 2 y As shown in Example 4.3-19, for a real number y we have:

From equation (4.4-18) we have:

)

(

)

(

!sinh z1 + z2 = − i sin ⎡⎣ i z1 + z2 ⎤⎦ = − i sin i z1 + i z2

)

!cosh 2 y − sinh 2 y = 1 and so cosh 2 y = 1+ sinh 2 y . Therefore:

From equation (4.3-19) we have:

(

2

or

Solution:

(

2

! sin z + cos z = sin 2x + sinh 2 y + cos 2x + sinh 2 y

Example 4.4-5

)

( )

( )

( ) ( )

!sinh z1 + z2 = − i sin i z1 cos i z2 − i cos i z1 sin i z2

2

2

! sin z + cos z = sinh 2 y + cosh 2 y

From equation (4.4-18) we have:

)

! !

cosh z = cos y cosh x + i sin y sinh x !

(4.4-35)

Example 4.4-6

!

sinh z = cos y sinh x + i sin y cosh x !

(4.4-36)

(

!sinh z1 + z2 = sinh z1 cosh z2 + cosh z1 sinh z2

2

We also have:

2

Show that sin z + cos z = sinh 2 y + cosh 2 y .

Example 4.4-7

Solution:

Show that cosh z = cos y cosh x + i sin y sinh x .

From equations (4.3-36) and (4.3-37) we have: ! cos z = cos 2 x + sinh 2 y

Solution: Equation (4.4-2) can be written in the form: 202

e x+i y + e− x−i y !cosh z = 2

Example 4.4-8 Show that:

We then have: !cosh z =

(

)

(

e x cos y + i sin y + e− x cos y − i sin y

)

2

or −x

−x

x

2

2

4.4.6! MODULUS OF COMPLEX HYPERBOLIC FUNCTIONS ! ! !

cosh z = cosh x − sin y ! sinh z = cosh z =

sinh 2x + sin 2 y ! sinh 2x + cos 2 y !

sinh z =

sinh 2x + sin 2 y

! cosh z =

cos 2 y cosh 2 x + sin 2 y sinh 2 x

! sinh z =

cos 2 y sinh 2x + sin2 y cosh 2x

Using the identity: !cosh 2 x − sinh2x = 1

The modulus of cosh z and sinh z are given by: 2

!

and sinh z are given by:

!cosh z = cos y cosh x + i sin y sinh x

2

sinh 2x + cos 2 y

Using equations (4.4-35) and (4.4-36), the modulus of cosh z

Therefore:

!

cosh z =

Solution:

e +e ) e −e ) ( ( !cosh z = cos y + i sin y x

!

we have: (4.4-37)

(

cos 2 y cosh 2 x + sin 2 y cosh2 x − 1

! sinh z =

cos 2 y sinh 2x + sin2 y sinh2x + 1

(4.4-38) (4.4-39)

)

! cosh z =

(

)

or ! cosh z =

( cos y + sin y ) cosh x − sin y 2

2

2

2

203

! sinh z =

( cos y + sin y ) sinh x + sin y 2

2

2

both cosh z and sinh z must also be periodic with a period of

2

2 k π i . We will then have:

Therefore the modulus of cosh z and sinh z are given by: 2

!

2

! cosh z = cosh x − sin y ! sinh z = !

! !

sinh 2x + sin 2 y

z+2 k π i ) − z+2 k π i ) e( +e ( e z + e− z cosh z + 2 k π i = = = cosh z ! 2 2 (4.4-44)

(

z+2 k π i ) − z+2 k π i ) e( −e ( e z − e− z ! sinh z + 2 k π i = = = sinh z ! 2 2 ! ! (4.4-45)

(

From equations (4.4-37) and (4.4-38) we have the two

inequalities:

)

)

and so:

!

cosh z ≤ cosh x !

(4.4-40)

!

cosh z = cosh z + 2 k π i !

)

k = 0, ± 1, ± 2,!!

(4.4-46)

!

sinh z ≥ sinh x !

(4.4-41)

!

sinh z = sinh z + 2 k π i !

(

k = 0, ± 1, ± 2,!!

(4.4-47)

4.4.7! PERIODICITY OF THE COMPLEX HYPERBOLIC FUNCTIONS !

From Proposition 4.2-1 we know that the periodicity of the

complex exponential function is 2 k π i : !

e z = e z+i 2 k π !

(4.4-42)

−z

terms of e and e : !

e z + e− z ! cosh z = 2

e z − e− z ! sinh z = 2

(4.4-43)

)

Therefore both complex hyperbolic secant and complex

hyperbolic cosecant are also periodic with a period of 2 k π i . We note that both cosh x and sinh x , where x is a real variable, are not periodic. !

where k is an integer. Since cosh z and sinh z are defined in z

!

(

For the complex hyperbolic tangent we can use equation

(4.4-43) to write: !

sinh z e z − e− z e2z − 1 ! tanh z = = = cosh z e z + e− z e2z + 1

(4.4-48)

Therefore the complex hyperbolic tangent has a periodicity that is half that of the complex hyperbolic sine and cosine. The 204

complex hyperbolic tangent and hyperbolic cotangent are then periodic with a period of i π : !

(

)

tanh z = tanh z + i k π !

k = 0, ± 1, ± 2,!!

(4.4-49)

4.4.8! ZEROS OF THE COMPLEX HYPERBOLIC FUNCTIONS !

Since cosh 2 x ≥ 1 and sin 2 y ≤ 1 , we see from equation

(4.4-37) that cosh z = 0 if and only if sin y = 1 and cosh x = 1 .

(

)

For cosh z = 0 , we have x = 0 and y = 2 k + 1 π 2 = π 2 + k π

(

)

where k = 0, ± 1, ± 2, ! , and so z = π 2 + k π i . !

Similarly, since sinh 2 x ≥ 0 and sin 2 y ≥ 0 , we see from

equation (4.4-38) that we will have sinh z = 0 if and only if

(

sin y = 0 and sinh x = e x − e− x

)

2 = 0 . For sinh z = 0 we have

x = 0 and y = k π where k = 0, ± 1, ± 2, ! , and so z = k π i !

We see then that all the zeros of cosh z and of sinh z are

!

cosh z = 0 !

⎛π ⎞ when z = ⎜ + k π ⎟ i ! ⎝2 ⎠

(4.4-51)

!

tanh z = 0 !

when z = k π i !

(4.4-52)

!

sech z = 0 !

⎛π ⎞ when z = ⎜ + k π ⎟ i ! ⎝2 ⎠

(4.4-53)

!

csch z = 0 !

when z = k π i !

(4.4-54)

!

coth z = 0 !

when z = k π i !

(4.4-55)

4.5!

COMPLEX LOGARITHMIC FUNCTION

!

as the inverse to the real exponential function e x so that x = e y where the real variable x > 0 . We then have:

pure imaginary. The complex hyperbolic tangent, tanh z , has

!

the same zeros as sinh z since tanh z = sinh z cosh z . Similarly, the

and

complex sech z has the same zeros as the complex cosh z , the

The real natural logarithmic function y = log e x is defined

ey = e

log e x

( )

= x!

log e e x = x !

(4.5-1)

complex csch z has the same zeros as the complex sinh z , and

!

the complex coth z has the same zeros as the complex sinh z .

Both log e x and e x are one-to-one functions, and so the inverses

! !

In summary, we have for k = 0, ± 1, ± 2,! :

sinh z = 0 !

when z = k π i !

(4.5-2)

exist. (4.4-50) 205

where r = z . From equations (4.5-5) and (4.5-8) we have:

4.5.1! !

DEFINITION OF THE COMPLEX LOGARITHMIC FUNCTION

The complex logarithmic function w = ln z cannot be

defined, however, as simply the inverse of the complex z

exponential function e , since the complex exponential function is multivalued, as given in equation (4.2-32): z

e =e

!

z+i 2 k π

k = 0, ± 1, ± 2,!!

!

eu = r !

!

or since u and r are both real numbers: !

u = log e r !

!

z ≠0!

(4.5-3)

(4.5-4)

We can write:

!

z = e w = eu+i v = eu ei v !

(4.5-5)

z= z e

i arg ( z )

!

(4.5-6)

Since: !

()

arg z = θ + 2 k π !

k = 0, ± 1, ± 2, ! ! (4.5-7)

we can write: !

i θ +2 k π ) z= z e( = r eiθ + i 2 k π !

k = 0, ± 1, ± 2, ! ! (4.5-10)

function w = u + i v = ln z :

(

or

k = 0, ± 1, ± 2, ! ! (4.5-11)

()

()

w = ln z = log e z + i arg z = log e r + i arg z !

!

(4.5-12)

()

In each branch of w = ln z we will have arg z = θ continuous.

! !

PERIODICITY OF THE COMPLEX LOGARITHMIC FUNCTION

Unlike the real exponential function:

x = ey !

(4.5-13)

which has only one logarithmic solution: !

k = 0, ± 1, ± 2, ! ! (4.5-8)

)

w = ln z = log e r + i θ + 2 k π !

!

4.5.2!

or in exponential form: !

v = θ + 2kπ !

where log e r is the real natural logarithm of the real variable

that are solutions of the equation:

z = ew !

k = 0, ± 1, ± 2, ! ! (4.5-9)

r > 0 . We then have the multivalued complex logarithmic

For any given value of z there are infinitely many values of w !

v = θ + 2kπ !

y = log e x !

(4.5-14)

the complex exponential function: 206

z = ew !

!

(4.5-15)

Solution:

has an infinity of logarithmic solutions as given by equation

We have from equations (4.5-12) and (4.5-11):

(4.5-11): !

(

)

w = ln z = log e r + i θ + 2 k π !

()

k = 0, ± 1, ± 2, ! ! (4.5-16)

!r = i = 1 !

resulting from its relation to the complex exponential function which has that periodicity. Since there are an infinity of

The complex logarithmic function w = ln z does not return

contour, and so the origin is a branch point for w = ln z .

k = 0, ± 1, ± 2,!

(

)

4k +1 π ⎛π ⎞ !w = ln i = i ⎜ + 2 k π ⎟ = i ! 2 ⎝2 ⎠

Example 4.5-1 !

k = 0, ± 1, ± 2,!

Real logarithms of negative numbers do not exist, but it is

possible to determine complex logarithms of negative numbers.

Solution: We have from equation (4.5-11) since r is real:

=e

k = 0, ± 1, ± 2,!

Since log e1 = 0 , all solutions to e w = i are:

to its initial value after encircling the origin with a closed

!e

π + 2 kπ ! 2

⎛π ⎞ !w = ln i = log e1+ i ⎜ + 2 k π ⎟ ! ⎝2 ⎠

complex logarithm.

ln z

()

arg i =

Therefore:

logarithmic solutions, there exist an infinity of branches for the

Show that eln z = r eiθ = z .

)

where:

The periodicity of the complex logarithmic function is 2 π i

!

(

w = ln z = log e r + i arg z = log e r + i θ + 2 k π

log e r+i (θ +2 k π )

iθ

= re e

i 2kπ

Example 4.5-2 Determine all values of w = ln i .

iθ

= re = z!

k = 0, ± 1, ± 2,!

The complex logarithm is defined for all nonzero numbers. For any negative number the complex logarithm is not real (see Example 4.5-6). Example 4.5-3 Determine all values of ln z when z = − 7 . 207

Example 4.5-5

Solution:

Determine all values of ln z when z = −1.

We have: !r = z = − 7 = 7

Solution:

( )

(

)

!arg −7 = π + 2 k π = 2 k + 1 π !

k = 0, ± 1, ± 2,!

!r = z = − 1 = 1

From equation (4.5-11) we have:

( )

(

)

!ln −7 = log e 7 + i 2 k + 1 π !

We have:

k = 0, ± 1, ± 2,!

( )

!arg −1 = π + 2 k π !

k = 0, ± 1, ± 2,!

From equation (4.5-11) we have:

( )

(

)

!ln −1 = log e1+ i π + 2 k π !

Example 4.5-4 Determine all values of ln z when z = 1.

k = 0, ± 1, ± 2,!

and since log e1 = 0 we have:

( ) (

) (

)

!ln −1 = i π + 2 k π = 2 k + 1 π i !

Solution:

k = 0, ± 1, ± 2,!

We have: Example 4.5-6

!r = z = 1 = 1

()

!arg 1 = 0 + 2 k π = 2 k π !

k = 0, ± 1, ± 2,!

From equation (4.5-11) we have:

(

)

!ln z = log e1+ i 2 k π !

k = 0, ± 1, ± 2,!

and since log e1 = 0 we have: !ln1 = 2 k π i !

Show that the logarithm of a negative real number is never real. Solution: From equation (4.5-12) we have:

()

!ln z = log e r + i arg z

k = 0, ± 1, ± 2,!

Since z is real and negative, we have: 208

()

!arg z = π + 2 k π !

k = 0, ± 1, ± 2,!

and so:

(

)

!ln z = log e r + i π 1+ 2 k !

k = 0, ± 1, ± 2,!

which is never a real number.

!

!

Determine all values of ln z when z = 1+ i .

We have: !r = z = 1+ 1 = 2

)

π + 2kπ ! 4

We also have:

)

!ln 1+ i = log e

arg z = Arg z + 2 k π !

by restricting the value of arg z to − π < θ ≤ π (see Section

k = 0, ± 1, ± 2,!

From equation (4.5-11) we have:

⎛π ⎞ 2 + i ⎜ + 2kπ ⎟ ! ⎝4 ⎠

(4.5-17)

will be holomorphic, and the domain will be a branch of ln z .

!

(

()

w = ln z = log e r + i arg z !

k = 0, ± 1, ± 2, ! ! (4.5-18) () () where Arg ( z ) is the principal value of the argument obtained

Solution:

!arg 1+ i =

As long as arg z varies continuously in a domain, the

complex logarithmic function given in equation (4.5-12):

Example 4.5-7

(

4.5.3! PRINCIPAL VALUE OF THE COMPLEX LOGARITHMIC FUNCTION

1.10-3). For the complex logarithm the domain corresponding to this principal value of the argument is known as the principal branch of the complex logarithmic function.

k = 0, ± 1, ± 2,!

or

! !

We can write:

()

Ln z = log e r + i Arg z !

()

− π < Arg z ≤ π !

(4.5-19)

where Ln z is defined to be the principal value of the complex

(

)

!ln 1+ i =

⎛π ⎞ 1 log e 2 + i ⎜ + 2 k π ⎟ ! 2 ⎝4 ⎠

k = 0, ± 1, ± 2,!

()

logarithmic function. Since Arg z is discontinuous along the negative real axis and since the logarithmic function is not defined at the origin, the negative real axis is a branch cut and

209

the origin is a branch point for the complex logarithm. We know that log e r is not continuous at the origin. ! !

Determine the principal value of w when e w = i .

Using z = r eiθ we can also write:

Ln z = log e r + i θ !

−π < θ ≤ π !

(4.5-20)

Since Ln z corresponds to only one of the branches of ln z , the function Ln z is single-valued. ! !

The k branches of ln z are given by:

wk = ln z = Ln z + i 2 k π !

k = 0, ± 1, ± 2, ! ! (4.5-21)

Solution: From Example 4.5-2 we have:

(

()

! wk = ln z = log e r + i Arg z + i 2 k π !

π 2

k = 0, ± 1, ± 2, ! ! (4.5-22)

If the complex exponential function is restricted to a

Example 4.5-9 Determine the value of Ln 5 .

w = Ln z is a single-valued inverse of z = e w . To show this, we

Solution:

can use equation (4.5-19) to write: Ln z log r i Arg z e ( ) = e e e ( )!

()

− π < Arg z ≤ π !

We have: (4.5-23)

Therefore: !

eLn z = r e

()

!Arg 5 = 0 From equation (4.5-19):

iArg ( z )

= z!

()

− π < Arg z ≤ π !

(4.5-24)

Since z = e w we then have: !

k = 0, ± 1, ± 2,!

Therefore the principal value of e w = i is:

domain in which its argument has the principal value, then

!

)

4k + 1 π ⎛π ⎞ !w = ln i = i ⎜ + 2 k π ⎟ = i ! 2 ⎝2 ⎠

!w = Ln i= i

or

!

Example 4.5-8

w = Ln z !

()

!Ln 5 = log e 5+ i Arg 5 Therefore:

(4.5-25)

!Ln 5 = log e 5 210

!ln z a = Ln z a + i 2 k π = a Ln z + i 2 k π ≠ a ln z !

Example 4.5-10

( )

Determine the value of Ln − e .

4.5.4!

Solution: !

We have:

ALGEBRAIC PROPERTIES OF THE COMPLEX LOGARITHMIC FUNCTION

If z1 and z2 are two complex numbers, using equation

(4.5-12) we can write:

!r = − e = e

!

From equation (4.5-19):

( )

k = 0, ± 1, ± 2,!

( )

( )

ln z1 + ln z2 = log e z1 + i arg z1 + log e z2 + i arg z2 !

(4.5-26)

where from equation (4.5-7):

( )

!Ln − e = log e e + i Arg − e = log e e + i π Therefore:

( )

!Ln − e = 1+ i π

( )

k1 = 0, ± 1, ± 2, ! !

(4.5-27)

( )

k2 = 0, ± 1, ± 2, ! !

(4.5-28)

!

arg z1 = θ1 + 2 k1 π !

!

arg z2 = θ 2 + 2 k2 π !

and so equation (4.5-26) can be written:

(

Example 4.5-11

!

Show that a ln z ≠ ln z a where a is a real number.

!

Solution:

where k = k1 + k2 . Since log e z1

From equation (4.5-21) we have: !ln z = Ln z + i 2 k π ! Therefore:

)

or

k = 0, ± 1, ± 2, ! !

(4.5-29)

and log e z2

are natural

k = 0, ± 1, ± 2,!

(

ln z1 + ln z2 = log e z1 z2 + i θ1 + θ 2 + 2 k π

! !

(

!

)

logarithms of real variables, we have:

k = 0, ± 1, ± 2,!

!a ln z = a Ln z + i 2 k π = a Ln z + i 2 a k π !

ln z1 + ln z2 = log e z1 + log e z2 + i θ1 + θ 2 + 2 k π

!

)

k = 0, ± 1, ± 2, ! !

(4.5-30)

If z1 and z2 are both real numbers, then k = 0 . We can write equation (4.5-30) as: 211

!

(

)

ln z1 z2 = ln z1 + ln z2 !

(4.5-31)

only if the branches of ln z1 and ln z2 are appropriately chosen. ! ! !

Using z = r eiθ we have from equation (4.5-20):

Ln z1 = log e r1 + i θ1 !

− π < θ1 ≤ π !

Ln z2 = log e r2 + i θ 2 !

− π < θ2 ≤ π !

(4.5-32) (4.5-33)

iθ

z1 z2 = r1 e 1 r2 e

iθ 2

i θ +θ = r1 r2 e ( 1 2 ) !

(

)

( ) (

ln z1 z2 = Ln r1 r2 + i θ1 + θ 2 + 2 k π

(4.5-34)

! ! !

(4.5-35)

Following similar procedures we obtain:

⎛z ⎞ ln ⎜ 1 ⎟ = ln r1 − ln r2 + i θ1 − θ 2 + 2 k π ⎝ z2 ⎠ k = 0, ± 1, ± 2, ! !

(

⎛z ⎞ ln ⎜ 1 ⎟ = ln r1 − ln r2 ! ⎝ z2 ⎠

⎡(1 n ) log e r+i (θ +2 k π ) 1 n ln z !e( ) = e ⎣

k = 0, ± 1, ± 2,!

)

n ⎤⎦

!

k = 0, ± 1, ± 2,!

and so: 1 n ) ln z

= r1 n e (

i θ +2 k π ) n

k = 0, ±1, ± 2,!

!

The only distinct values will result from k = 0, 1, 2, !, n − 1 . Therefore we have: !e (

1 n ) ln z

= r1 n e (

i θ +2 k π ) n

k = 0, 1, 2, !, n − 1

!

From equation (1.14-12) we have: (4.5-36)

We can write equation (4.5-36) as: !

1 n ⎡ log r+i θ +2 k π ) ⎤⎦ 1 n ln z !e ( ) = e ( ) ⎣ e ( !

!e (

)

k = 0, ± 1, ± 2, ! !

! ! !

From equation (4.5-11) we have:

or

Therefore: !

1 n ln z Show that z1 n = e( ) .

Solution:

and !

Example 4.5-12

!z1 n = r 1 n e i (

θ +2 k π ) n

!

k = 0, 1, 2, !, n − 1

and so: (4.5-37)

1 n ln z !z1 n = e( )

only if the branches of ln z1 and ln z2 are appropriately chosen. 212

4.5.5! !

We are thereby considering the principal branch, Ln z , of the

DERIVATIVE OF THE COMPLEX LOGARITHMIC FUNCTION

complex logarithmic function. Both r and θ are continuous

The complex logarithmic function is multivalued, and so

the value of the derivative of the function at a point z will be unique only when a branch is specified. We can see this from the equation: !

( ) = lim ln ( z + Δz ) − ln z ! Δ z →0

negative real axis. A discontinuity of 2 π exists across the negative real axis. Therefore Ln z is continuous in the domain defined by r ≠ 0 and − π < arg z ≤ π , and is discontinuous at

r = 0 and across arg z = π .

d ln z dz

everywhere in the z-plane except at the origin and along the

(4.5-38)

Δz

!

The complex logarithm then has a branch cut along the

negative real axis. The point z = 0 is a branch point for the

For this derivative to exist, the two logarithmic functions in the

complex logarithm. Therefore the branch cut and branch point

numerator of this equation must tend to the same limit. In other

for the principal branch of the complex logarithm are the same

words, these two functions must be single-valued in the

as shown in Figure 4.1-2 for the complex square root function.

neighborhood of the point z . Only single-valued functions can have derivatives. By selecting a single branch for the

Proposition 4.5-1:

logarithmic function, we can ensure that it is single-valued.

For z = r eiθ a complex logarithm Ln z is holomorphic when

!

− π < θ ≤ π and r > 0 .

!

We then have:

( ) = 1!

d ln z dz

(4.5-39)

z

!

z

In any given branch, ln z is the unique inverse of e . !

We will now consider the derivative of the complex

z = r eiθ !

!

From equation (4.5-20) we have:

w = u + i v = Ln z = l og e r + i θ !

− π < θ ≤ π ! (4.5-41)

Therefore:

logarithmic function where: !

Proof:

−π < θ ≤ π !

(4.5-40)

!

u = log e r !

v =θ !

(4.5-42) 213

and so:

(

)

(

!

)

!

∂u ∂ log e r 1 = = ! ∂r ∂r r

∂u ∂ log e r = = 0! ∂θ ∂θ

(4.5-43)

!

∂v ∂θ = = 0! ∂r ∂r

∂v ∂θ = = 1! ∂θ ∂θ

(4.5-44)

Show that

1 ∂u ∂v =0=− ! r ∂θ ∂r

!

dz

⎢ ⎢⎣

∂r ⎥⎦

∂r

Let!

!

(

dz

) = e− iθ = r

1 1 ! = iθ z re

w = ln g !

()

g= f z

We then have:

() ()

⎥ (4.5-46) ∂r ⎥⎦

dw dw dg 1 dg f ′ z ! = = = dz dg dz g dz f z

or

d Ln z

(

( ))

Solution:

) = e− iθ ⎡ ∂u + i ∂v ⎤ = e− iθ ⎡ ∂( loge r ) + i ∂θ ⎤ ! ⎢ ∂r ⎣

() ()

f′ z d ln f z = dz f z

(4.5-45)

the derivative given in equation (3.5-8):

(

(4.5-48)

Example 4.5-13

The derivative of Ln z can be calculated using the polar form of

d Ln z

( ( ))

as shown in Example 4.5-13.

(3.5-11) are satisfied:

∂u 1 1 ∂v ! = = ∂r r r ∂θ

() ()

f′ z d ln f z = ! dz f z

!

We see that the Cauchy-Riemann equations given in equation

!

The derivative of the complex logarithm of a function is:

(4.5-47)

4.5.6! MAPPING PROPERTIES OF THE COMPLEX LOGARITHMIC FUNCTION

where − π < θ ≤ π and r > 0 . Since Ln z satisfies the Cauchy-

!

Riemann equations and since Ln z is continuous in the domain

logarithmic function w = Ln z = l og e r + i θ

defined by r ≠ 0 and − π < arg z ≤ π , the complex logarithm Ln z

Since the complex logarithmic function w = Ln z is the inverse

is holomorphic in this domain.

of z = e w , mapping of w = Ln z (Figure 4.5-1) is the opposite of

■

We will now consider the mapping of the complex where − π < θ ≤ π .

214

()

mapping z = e w (Figure 4.2-2). The domain of f z = Ln z is the

4.6!

COMPLEX POWER FUNCTION

set of all nonzero complex numbers z , and the range of

!

w = u + i v = Ln z is the horizontal strip −π < v < π .

given by:

Raising a complex number z to an integer power n is

w = zn!

!

(4.6-1)

as discussed in Section 1.13.3. We can rewrite this equation in the form: !

( )

w = z n = eln z

n

= en ln z !

z ≠0!

(4.6-2)

From equation (4.5-21) we have: ! w = z n = en Ln z + i 2n k π = en Ln z e i 2n k π !

k = 0, ± 1, ± 2, ! ! (4.6-3)

and so using equation (1.13-12): !

w = z n = en Ln z !

(4.6-4)

Therefore the integer power function w = z n is single-valued, Figure 4.5-1! Mapping of w = Ln z where u = ln r = log e r . !

and so an inverse exists. We will now consider the process of raising a complex number to a complex power.

To visualize the mapping of the complex logarithmic

plane. It is mapped into the w-plane as a vertical line u = ln r .

4.6.1! DEFINITION OF THE COMPLEX POWER FUNCTION

We next consider a ray Arg z = θ in the z-plane. It is mapped

!

into the w-plane as a horizontal line v = θ where − π < θ ≤ π

complex power. We then have the complex power function:

function, we will first consider a circle of radius z = r in the z-

()

(see Figure 4.5-1).

!

We can generalize equation (4.6-2) for the case of a

( )

w = z α = eln z

α

α ⎡⎣ log e r + i arg ( z )⎤⎦ !

= eα ln z = e

(4.6-5) 215

where α is a complex constant. From equation (4.5-16) we also

(

have: α ⎡⎣ log e r + i (θ +2 k π ) ⎤⎦

!

w = zα = e

!

!

= r α ei α θ eα i 2 k π !

k = 0, ± 1, ± 2, ! !

(4.6-6)

Since α is not an integer, the complex power function

!

w = z α is multivalued and will have an infinity of different values. The particular branch of the logarithm employed determines the branch of the complex power function. The point z = 0 is a branch point for the complex power function,

()

Example 4.6-1

(

)

k = 0, ± 1, ± 2,!

Example 4.6-2 Determine the value of w = i i . Solution: From equation (4.6-5) we can write: !w = i i = ei ln i From Example 4.5-2 we have:

and a discontinuity exists across the negative real axis

arg z = π for the complex power function (see Section 4.5.3).

)

!w = cos 2 2 k π + isin 2 2 k π !

!ln i = i

( 4k + 1) π !

k = 0, ± 1, ± 2,!

2

Therefore: − 4k+1 π !w = i i = ei ln i = e ( )

Determine the value of w = 1 2 . Solution:

2

− 1 =e (

2+ 2 k ) π

!

k = 0, ± 1, ± 2,!

and so the value of w = i i is real.

From equation (4.6-6) we can write for θ = 0 : !w = 1

2

0

e e

2 i2kπ

!

k = 0, ± 1, ± 2,!

or

( )i

Determine the value of w = −1

and so: !w = e 2

Example 4.6-3

2 ikπ

!

k = 0, ± 1, ± 2,!

Solution: From equation (4.6-5) we can write:

216

i i ln −1 !w = −1 = e ( )

α ⎡⎣ log e r +i Arg ( z ) ⎤⎦

( )

! w = z α = eα Ln z = e

From equation (4.5-16) we have:

( )

(

) (

)

k = 0, ± 1, ± 2,!

!w = −1 = e

i ln ( −1)

=e

− (π +2 k π )

( )

=e

− (1+ 2 k ) π

!

k = 0, ± 1, ± 2,!

value.

Because of the relation of the complex power function to

the complex logarithmic function as given in equation (4.6-5), the principal branch of the two functions is the same. We can write equation (4.6-5) in the form: α ⎡⎣ Ln z+ i 2 k π ⎤⎦

!

w = z α = eα ln z = e

!

!

α ⎡⎣ log e r +i Arg ( z ) ⎤⎦ α i 2 k π

=e

e

−i ⎡⎣ log e1+i π 2 ⎤⎦

( )−i eπ 2

= 1

and so the principal value of w = i − i is: !w = i − i = e π

2

This can be written as: − −1

= eπ

2

This is another example of how an imaginary number

e

k = 0, ± 1, ± 2, ! !

i α Arg ( z ) α i 2 k π

!w = i − i = e−i Ln i = e

! −1

(4.6-7)

and so:

w = zα = r α e

Determine the principal value of w = i − i .

From equation (4.6-9) we have:

4.6.2! PRINCIPAL VALUE OF THE COMPLEX POWER FUNCTION

!

z ≠ 0 ! (4.6-9)

Solution:

i

and so the value of w = −1 is real.

!

!

Example 4.6-4

Therefore:

( )

i α Arg ( z )

The principal value of the complex power function is a unique

!ln −1 = log1+ i π + 2 k π = i π + 2 k π !

i

= rα e

raised to an imaginary power can be real. This particular example is often mentioned in the literature of mathematics (see Nahin, 1998).

!

k = 0, ± 1, ± 2, ! !

()

(4.6-8)

Restricting the value of ln z to the principal value, we have

Example 4.6-5

the principal value of the complex power function w :

Determine the principal value of w = z 1 4 . 217

Solution:

!

From equation (4.6-9) we can write:

(1 4) ⎡⎣ loge r + i Arg ( z ) ⎤⎦ !w = z1 4 = e

!w = z1 4 = z

1 4

e

( i Arg ( z ) 4 )

()

Since Arg 1 = 0 , from Example 4.6-5 we have:

()

()

⎡ Arg 1 Arg 1 ⎤ !w = 11 4 = ⎢ cos + i sin ⎥ =1 4 4 ⎢⎣ ⎥⎦

ALGEBRAIC PROPERTIES OF THE COMPLEX POWER FUNCTION

From equation (4.6-7) we can write: α

α2

α1 ln z α 2 ln z

=e

Therefore we have:

α2

=z

α1 +α 2

!

(4.6-12)

z z

α1

α2

α − α Ln z 2 π i k α − k α = e( 1 2 ) e ( 1 1 2 2 ) !

(4.6-13)

!

e

z z

α1

α2

=z

α1− α 2

!

(4.6-14)

Equations (4.6-12) and (4.6-14) may not hold for principal values of the complex power function. Also the relations: !

(z ) α1

α2

=z

α1 α 2

!

(4.6-15)

and

(see Example 1.14-4).

z 1z

α

(4.6-11)

and so:

Solution:

!

α + α Ln z 2 π i k α + k α = e( 1 2 ) e ( 1 1 2 2 ) !

z 1z

!

Determine the principal value of w = 11 4 .

!

α2

By similar arguments, we have:

Example 4.6-6

4.6.3!

α

and so: !

or

z 1z

α Ln z + 2 π ik1 ) α 2 ( Ln z + 2 π ik 2 ) ! (4.6-10) = e 1( e

!

( z1 z2 )α = z1α z2α !

(4.6-16)

are not generally true. ! !

If α1 = 0 and α = α 2 in equation (4.6-14), we have:

1 = z −α ! α z

(4.6-17)

We also have: 218

!

zα = z α !

( ) ( )

dw d i ! = z = i i i−1 = i i dz dz

(4.6-18)

From Example 4.6-4 we then obtain:

4.6.4! DERIVATIVE OF THE COMPLEX POWER FUNCTION !

( )

dw d i ! = z = e− π dz dz

Since the complex logarithm is multivalued, the complex

2

power function is not continuous in the complex plane and so principal branch, however, the complex power function is

4.6.5! MAPPING PROPERTIES OF THE COMPLEX POWER FUNCTION

continuous, and the complex logarithm is holomorphic.

!

!

y≥0

is not holomorphic across the complex plane. Within the

The derivative of the complex power function in the

principal branch is:

( )

(

)

dw d α d α Ln z α α ! = z = e = eα Ln z = z α = α z α −1 ! dz dz dz z z

(4.6-19)

We will consider the mapping of the upper half-plane by the complex power function

w = z α π , where

0 < α < 2 π . Letting z = r eiθ we have: !

( )

w = r eiθ

α π

i αθ π ) ! = rα π e (

(4.6-20)

i αθ π ) Therefore z = r eiθ in the z-plane becomes w = r α π e ( in the

Example 4.6-7

w-plane. The modulus r of z is scaled to r α π . The point z = 0

Determine the derivative of the principal value of w = z i when z = i .

is mapped into the point w = 0 . With this transformation the set

Solution:

!

From equation (4.6-19) we have:

( )

dw d i ! = z = i z i−1 dz dz At z = i this becomes:

of rays emanating from z = 0 specified by:

()

arg z = θ !

0 ≤θ ≤ π !

(4.6-21)

are mapped into an image set of rays emanating from w = 0 in the w-plane: !

( )

arg w =

αθ ! π

0≤

αθ ≤α ! π

(4.6-22) 219

4.7.1! INVERSE COMPLEX SINE AND COSECANT FUNCTIONS ! !

The inverse complex sine function is designated by:

w = sin −1 z !

(4.7-1)

To determine the inverse complex sine we use equation (4.3-2): !

e i w − e− i w ! z = sin w = 2i

(4.7-2)

Multiplying by 2 i e i w , this equation can be written as: !

e2 i w − 2 i z e i w − 1 = 0 !

(4.7-3)

which is a quadratic equation in the variable ei w . Using the Figure 4.6-1! Mapping of the upper half-plane y ≥ 0 by the complex power function w = z α π . Since z is rotated by a factor of α π radians, the image of

!

(

)

quadratic formula to solve this equation, we have: !

(

iw

e = i z + 1− z

2

)

1 2

!

(4.7-4)

ray L1 with θ = π is rotated to α π π = α . The image of ray L2

where the complex square root is multivalued. Taking the

with θ = 0 is not rotated. The upper half-plane is then mapped

logarithm of equation (4.7-4), we obtain the inverse complex

onto the wedge image shown in Figure 4.6-1.

sine:

4.7!

INVERSE COMPLEX TRIGONOMETRIC FUNCTIONS

!

(

⎡ w = sin z = − i ln ⎢ i z + 1− z 2 ⎣ −1

)

1 2⎤

⎥⎦ !

(4.7-5)

We will now consider the inverse complex trigonometric

In this equation the complex square root and the complex

functions. These functions can all be expressed in terms of the

logarithm are multivalued, and so they are restricted to certain

complex logarithm, and so these functions are all multivalued.

branches. The points z = ±1 are branch points.

!

220

!

Similarly we can obtain the inverse complex cosecant:

!

⎡ i + z2 − 1 ⎛ ⎞ 1 ⎢ csc −1 z = sin −1 ⎜ ⎟ = − i ln ⎢ z ⎝ z⎠ ⎢⎣

(

)

1 2

⎤ ⎥ ⎥! ⎥⎦

Example 4.7-2 Show that w = sin −1 z is defined for all z .

z ≠ 0 ! (4.7-6)

Solution:

Since the expressions for the inverse sine and cosecant

From equation (4.7-5) we see that w = sin −1 z is defined for

include the complex logarithmic function, the inverse complex

all z except i z = − 1− z 2 or − z 2 = 1− z 2 . But this never

sine and cosecant are multivalued functions.

occurs since 1 ≠ 0 . Therefore w = sin −1 z is defined for all z .

!

Example 4.7-1

4.7.2!

Determine all solutions of w = sin −1 2 . Solution:

!

Since z > 1 we know that w cannot be real. From equation (4.7-5) we have:

(

(

!w = sin −1 z = − i ln 2 i + 1− 4

)1 2 ) = − i ln ⎡⎢⎣( 2 +

)

3 i ⎤⎥ ⎦

We then have from equation (4.5-11):

(

)

⎡ ⎤ ⎛π⎞ !w = − i ⎢log e 2 + 3 + i ⎜ ⎟ + i 2 k π ⎥ ! ⎝ 2⎠ ⎣ ⎦

!

INVERSE COMPLEX COSINE AND SECANT FUNCTIONS

The inverse complex cosine function is designated by:

w = cos −1 z !

(4.7-7)

Using equation (4.3-2) we have: !

e i w + e− i w ! z = cos w = 2

(4.7-8)

This equation can be written as:

k = 0, ± 1, ± 2, !

!

e2i w − 2 z e i w + 1 = 0 !

(4.7-9)

Using the quadratic formula to solve this equation, we have:

or !w =

(

)

π − i log e 2 + 3 + 2 k π ! 2

k = 0, ± 1, ± 2, !

!

iw

(

2

e = z + z −1

)

1 2

!

(4.7-10)

221

where the complex square root is multivalued. Taking the !

logarithm, we obtain the inverse complex cosine:

(

⎡ w = cos −1 z = − i ln ⎢ z + z 2 − 1 ⎣

!

)

1 2⎤

⎥⎦ !

(4.7-11)

This equation can be written as: !

In this equation the complex square root and the complex logarithm are multivalued, and so they are restricted to certain

sin w 1 ei w − e−i w ei w − e−i w z = tan w = = = − i i w −i w ! (4.7-14) cos w i ei w + e−i w e +e

z ei w + z e−i w = − i ei w + i e−i w !

(4.7-15)

( i + z ) ei w = ( i − z ) e−i w !

(4.7-16)

or

branches. The points z = ±1 are branch points.

!

!

Therefore:

!

!

Similarly we can obtain the inverse complex secant:

(

⎡ + 1− z 2 ⎛ ⎞ 1 ⎢1 −1 −1 sec z = cos ⎜ ⎟ = − i ln ⎢ z ⎝ z⎠ ⎢⎣

)

1 2

⎤ ⎥! ⎥ ⎥⎦

z ≠ 0 ! (4.7-12)

!

e2 i w =

i − z 1+ i z ! = i + z 1− i z

Taking the complex logarithm:

Since the expressions for the inverse cosine and secant

include the complex logarithmic function, the inverse complex

!

2i w = ln

i− z ! i+ z

cosine and secant are multivalued functions.

we obtain the inverse complex tangent:

4.7.3!

!

!

INVERSE COMPLEX TANGENT AND COTANGENT FUNCTIONS

The inverse complex tangent function is designated by:

!

w = tan −1 z !

(4.7-13)

w = tan −1 z =

1 i − z i 1+ z i 1− i z ! ln = ln = ln 2i i + z 2 1− z 2 1+ i z

(4.7-18)

(4.7-19)

Similarly we can obtain the inverse complex cotangent: !

To determine the inverse complex tangent we use equation (4.3-3):

(4.7-17)

!

1 1 z+i i z−i ! w = cot −1 z = tan −1 = ln = ln z 2i z − i 2 z + i

(4.7-20)

Since the expressions for the inverse complex tangent and

inverse complex cotangent include the complex logarithmic 222

function, the inverse complex tangent and inverse complex

Example 4.7-4

cotangent are multivalued functions. The points z = ± i are

(

branch points.

Solution:

Example 4.7-3

Using equation (4.7-20):

Determine all solutions of sin z = cos z .

(

)

!w = cot −1 2 + i =

Solution: Since sin z = cos z we have:

!w =

Therefore using equation (4.7-19):

()

i 2+i−i i 2 ln = ln 2 2 + i + i 2 2 + 2i

or

sin z ! = tan z = 1 cos z

!z = tan −1 1 =

)

Determine all solutions of w = cot −1 2 + i .

i 1 i i i ln = ln 1− ln 1+ i = − ln 1+ i 2 1+ i 2 2 2

(

)

(

)

From equation (4.5-11) we then have:

i 1− i i ln = ln −i 2 1+ i 2

( )

!w = −

or

⎞ i 1 ⎛π ln 2 + ⎜ + 2 k π ⎟ ! 2 2⎝4 ⎠

k = 0, ± 1, ± 2, !

Therefore:

!z =

(

( ) ) = 2i ⎛⎜⎝ − i2π + i 2 k π ⎞⎟⎠ !

i log e 1+ i arg −i 2

k = 0, ± 1, ± 2, !

!w =

π i − ln 2 + k π ! 8 4

k = 0, ± 1, ± 2, !

and so:

π !z = + k π ! 4

k = 0, ± 1, ± 2, !

Compare with Example 4.3-4.

4.7.4! BRANCH POINTS OF INVERSE COMPLEX TANGENT FUNCTIONS !

Branch points of the multivalued inverse complex tangent

function w = tan −1 z given in equation (4.7-19) are z = ±i . Two possible branch cuts corresponding to these branch points are 223

shown in Figure 4.7-1. Either of these paths will prevent a path

inverse complex trigonometric functions can be calculated for

in any branch from encircling a branch point.

that domain. !

To determine the derivative of the inverse complex sine

function, we can write: !

z = sin w !

(4.7-21)

We then have: !

1 = cos w

dw ! dz

(4.7-22)

or !

dw d 1 1 ! = sin −1 z = = 2 dz dz cos w 1− sin w

(4.7-23)

From equation (4.7-21) we have: Figure 4.7-1!

4.7.5! !

Possible branch cuts for w = tan −1 z .

DERIVATIVES OF INVERSE COMPLEX TRIGONOMETRIC FUNCTIONS

The inverse trigonometric functions are multivalued

functions. To obtain a branch of the inverse sine or cosine function, a branch of the square root function and a branch of the complex logarithm must be chosen. To obtain a branch of the inverse tangent function, a branch of the complex logarithm must be chosen. Once a domain is chosen, the derivatives of the

! !

d 1 ! sin −1 z = 2 dz 1− z

(4.7-24)

To determine the derivative of the inverse complex cosine

function, we can write: !

z = cos w !

(4.7-25)

We then have: !

1 = − sin w

dw ! dz

(4.7-26)

or 224

!

dw d 1 1 = cos −1 z = − =− ! 2 dz dz sin w 1− cos w

(4.7-27)

!

d 1 sec −1 z = ! 2 dz z z −1

(4.7-34)

!

d 1 ! cot −1 z = − dz 1+ z 2

(4.7-35)

From equation (4.7-25) we have: ! !

d 1 ! cos −1 z = − 2 dz 1− z

(4.7-28)

To determine the derivative of the inverse complex

4.8!

tangent function, we can write: !

z = tan w !

(4.7-29)

We then have: !

1 = sec 2 w

dw ! dz

(4.7-30)

dw d 1 1 ! = tan −1z = = 2 2 dz dz sec w 1+ tan w

(4.7-31)

From equation (4.7-29) we have: ! ! !

We will now consider the inverse complex hyperbolic

functions. These functions can all be expressed in terms of the complex logarithm, and so these functions are all multivalued.

or !

!

INVERSE COMPLEX HYPERBOLIC FUNCTIONS

d 1 ! tan −1z = dz 1+ z 2

! !

The inverse complex hyperbolic sine is designated by:

w = sinh −1z !

(4.8-1)

To determine the inverse complex hyperbolic sine we use (4.7-32)

equation (4.4-2): !

Similarly, we can obtain the derivatives:

d 1 ! csc −1 z = − 2 dz z z −1

4.8.1! INVERSE COMPLEX HYPERBOLIC SINE AND COSECANT

(4.7-33)

e w − e− w ! z = sinh w = 2

(4.8-2)

This equation can be written as: !

e2 w − 2 z e w − 1 = 0 !

(4.8-3) 225

This is a quadratic equation in the variable e w . Using the

Example 4.8-1

quadratic formula to solve this equation, we have: !

(

2 z + 4z 2 + 4 e = = z + z2 + 1 2 w

)

1 2

!

Determine all solutions of w = sinh −1 i . (4.8-4)

Solution: From equation (4.8-5) we have:

where the complex square root is multivalued. Taking the

(

⎛ !w = sinh −1 i = ln ⎜ i + i 2 + 1 ⎝

logarithm of this equation, we obtain the inverse complex hyperbolic sine:

(

⎛ w = sinh z = ln ⎜ z + z 2 + 1 ⎝ −1

!

)

1 2⎞

⎟⎠ !

)

1 2⎞

⎟⎠ = ln i

From Example 4.5-2 we have:

(4.8-5)

(

)

4k +1 π ⎛π ⎞ !w = i ⎜ + 2 k π ⎟ = i ! 2 ⎝2 ⎠

In this equation the complex square root and the complex

k = 0, ± 1, ± 2,!

logarithm are multivalued, and so they are restricted to certain branches. !

Similarly, we have the inverse complex hyperbolic

cosecant: !

!

(

⎡ + 1+ z 2 ⎛ ⎞ 1 ⎢1 −1 −1 cosech z = sinh ⎜ ⎟ = ln ⎢ z ⎝ z⎠ ⎢⎣

)

1 2

4.8.2! !

⎤ ⎥! ⎥ ⎥⎦

z ≠ 0 ! (4.8-6)

Since the expressions for the inverse hyperbolic sine and

cosecant include both the square root function and the complex

!

INVERSE COMPLEX HYPERBOLIC COSINE AND SECANT

The inverse complex hyperbolic cosine is designated by:

w = cosh −1z !

(4.8-7)

To determine the inverse complex hyperbolic cosine we use equation (4.4-2): !

logarithmic function, the inverse complex sine and cosecant are

e w + e− w ! z = cosh w = 2

multivalued functions. The points z = − ∞ and z = ±1 are branch

This equation can be written as:

points of the inverse hyperbolic sine and cosecant functions.

!

e2 w − 2 z e w + 1 = 0 !

(4.8-8)

(4.8-9) 226

Using the quadratic formula to solve this equation, we have: !

2 z + 4 z2 − 4 e = = z + z2 − 1 2

(

w

)

1 2

!

(4.8-10)

4.8.3! INVERSE COMPLEX HYPERBOLIC TANGENT AND COTANGENT !

The inverse complex hyperbolic tangent is designated by:

w = tanh −1z !

where the complex square root is multivalued. Taking the

!

logarithm of this equation, we obtain the inverse complex

To determine the inverse tangent we use equation (4.4-3):

hyperbolic cosine: !

(

⎛ w = cosh z = ln ⎜ z + z 2 − 1 ⎝ −1

)

!

1 2⎞

⎟⎠ !

(4.8-11)

logarithm are multivalued, and so they are restricted to certain

!

branches.

or

!

Similarly, we have the inverse complex hyperbolic secant:

!

!

⎡ + z2 − 1 1 ⎛ 1⎞ ⎢1 sech −1z = cosh −1 ⎜ ⎟ = ln ⎢ z ⎝ z⎠ ⎢⎣

!

)

sinh w e w − e− w ! z = tanh w = = cosh w e w + e− w

(4.8-14)

Rewriting equation (4.8-14):

In this equation the complex square root and the complex

(

(4.8-13)

2

⎤ ⎥! ⎥ ⎥⎦

z ≠ 0 ! (4.8-12)

Since the expressions for the inverse hyperbolic cosine and

z e w + z e− w = e w − e− w !

(4.8-15)

(1− z ) ew = (1+ z ) e− w !

(4.8-16)

Therefore: !

e2 w =

1+ z ! 1− z

Taking the complex logarithm:

secant include both the square root function and the complex logarithmic function, the inverse complex cosine and secant are multivalued functions. The points z = − ∞ and z = ±1 are branch points of the inverse hyperbolic cosine and secant functions.

(4.8-17)

!

2 w = ln

1+ z ! 1− z

(4.8-18)

we obtain inverse complex hyperbolic tangent: !

w = tanh −1 z =

1 1+ z ! ln 2 1− z

(4.8-19) 227

and the inverse complex hyperbolic cotangent: ! !

⎛ 1 ⎞ 1 ⎡ z + 1⎤ coth −1 z = tanh −1 ⎜ ⎟ = ln ⎢ ! ⎝ z ⎠ 2 ⎣ z − 1 ⎥⎦

!

z ≠ ± i ! (4.8-20)

Since the expression for the inverse hyperbolic tangent

are branch points of the inverse hyperbolic tangent.

dw ! dz

(4.8-22)

or !

includes the complex logarithmic function, the inverse hyperbolic tangent is a multivalued function. The points z = ±1

1 = cosh w

dw d 1 1 ! = sinh −1 z = = 2 dz dz cosh w sinh w + 1

From equation (4.8-21) we then have: !

d sinh −1 z = dz

1 2

z +1

!

4.8.4! DERIVATIVES OF INVERSE COMPLEX HYPERBOLIC FUNCTIONS

!

!

hyperbolic cosine function, we can write:

The inverse hyperbolic trigonometric functions are multi-

valued functions. To obtain a branch of the inverse hyperbolic sine or cosine function, a branch of the square root function and a branch of the complex logarithm must be chosen. To obtain a branch of the inverse hyperbolic tangent function, a branch of

!

functions can be calculated for that domain. !

To determine the derivative of the inverse complex

hyperbolic sine function, we can write: !

z = sinh w !

Therefore:

(4.8-24)

To determine the derivative of the inverse complex

z = cosh w !

(4.8-25)

Therefore: !

the complex logarithm must be chosen. Once a domain is chosen, the derivatives of the inverse complex trigonometric

(4.8-23)

1 = sinh w

dw ! dz

(4.8-26)

or !

dw d 1 1 ! = cosh −1z = = 2 dz dz sinh w cosh w − 1

(4.8-27)

From equation (4.8-25) we then have: (4.8-21) !

d cosh −1z = dz

1 2

z −1

!

(4.8-28) 228

!

To determine the derivative of the inverse complex

complex base α . From equations (4.9-1) and (4.9-2) we have:

hyperbolic tangent function, we can write:

z = tanh w !

!

(4.8-29)

Therefore:

dw ! dz

(4.8-30)

dw d 1 1 ! = tanh −1z = = 2 2 dz dz sech w 1− tanh w

!

(4.9-3)

(4.8-31)

d 1 ! tanh −1z = dz 1− z 2

(4.8-32)

(4.9-4)

The natural logarithm of equation (4.9-1) is:

w ln α = ln z !

!

From equation (4.8-29) we then have: !

lnα z

w = lnα α w !

! !

or !

z =α

! and

1 = sech2 w

!

where w is given by the complex logarithm of z taken to a

z ≠0!

(4.9-5)

and so the complex-based logarithmic function is given by:

ln z ! ln α

!

w = lnα z =

Letting

i θ +2 k π ) z = re (

z ≠0!

and α = ρ e i (φ +2 mπ ) ,

(4.9-6) equation

(4.9-6)

becomes:

4.9! ! !

COMPLEX-BASE LOGARITHMIC FUNCTION If α is a complex constant, and we have:

α = z! w

z ≠0!

! (4.9-1)

then we can determine w using the complex-base logarithmic function: !

w = lnα z !

z ≠0!

w = lnα z =

!

( )! ln e ρ + i (φ + 2 m π ) ln e r + i θ + 2 k π

z ≠0!

(4.9-7)

The principal value of w = lnα z is defined to be:

!

w = Lnα z =

Ln z ! Ln α

z ≠0!

(4.9-8)

The derivative of w = logα z can be obtained from equation (4.9-2)

(4.9-6): 229

!

(

d lnα z dz

)=

1 ! z ln α

z ≠0!

(4.9-9)

230

Chapter 5 Complex Integration

!∫

C

()

f z dz = 0

231

!

In this chapter we will show that integration of a complex

function can be performed in the complex plane. We will define

Δxi . Letting xi be some arbitrary point within the subinterval of length Δxi , we can then write: n

curves in the complex plane that can serve as integration paths, and we will present contour integrals. We will prove the

∑ f ( x ) Δx !

!

i

(5.1-2)

i

i =1

Cauchy-Goursat theorem for both simply and multiply

as an approximation for the integral in equation (5.1-1). The

connected domains.

sum in equation (5.1-2) is known as a Riemann sum.

5.1!

INTEGRATION OF REAL-VALUED FUNCTIONS

!

We now let the number of subintervals increase so that

n → ∞ , and each Δxi → 0 . The sum in equation (5.1-2) then

We will begin our discussion of complex integrals with a

effectively becomes a sum over all points xi in the interval from

review of integrals of real-valued functions. We note first that

point x0 to point xn . This makes it possible to define the

continuous functions can always be integrated.

definite integral of a real-valued function f x as the limit (if

!

5.1.1! !

DEFINITE INTEGRALS ALONG THE X-AXIS

The definite integral of a continuous real-valued

()

it exists) of the sum in equation (5.1-2):

∫

!

()

function f x of a single real variable x has the form: !

∫

xn x0

()

f x dx !

! (5.1-1)

where x0 and xn are the endpoints of some interval along the xaxis. The definite integral in equation (5.1-1) is defined in terms

xn x0

()

n

f x dx = lim

n→ ∞

∑ f ( x ) Δx ! i

i

(5.1-3)

i =1

()

For a piecewise continuous real-valued function f x to

be integrable, the limit of the sum in equation (5.1-3) must exist independently both of how the subintervals Δxi are chosen and of the selection of xi .

of a limiting process involving subintervals along the x-axis. We

5.1.2!

LINE INTEGRALS IN THE CARTESIAN PLANE

can divide the integration interval between points x0 and xn

!

along the x-axis into n subintervals, each of arbitrary length

of two real variables x and y has the form:

( )

The integral of a continuous real-valued function f x, y

232

!

∫

C

( )

f x, y ds !

(5.1-4)

n

!

arc length =

∫ ds = lim ∑ Δs ! C

n→∞

where integration is performed along some path C in the Cartesian plane. The integral in equation (5.1-4) is defined in

!

i

(5.1-7)

i =1

If the integrals in equations (5.1-6) and (5.1-7) are over the

(

)

terms of a limiting process involving n chords of arbitrary

same path C , then when f xi , yi = 1 , we see that the two

length Δsi inscribed on the smooth curve of the path C (see

integrals are equal. Therefore clearly the function f xi , yi

(

)

Figure 5.1-1). Letting xi , yi be some arbitrary point within the

(

)

is

not the equation of the path C .

chord of length Δsi , we can then write: n

!

∑ f ( x , y ) Δs ! i

i

(5.1-5)

i

i =1

as an approximation for the integral in equation (5.1-4). !

We now let the number of chords increase indefinitely so

that n → ∞ and each Δsi → 0 . The sum in equation (5.1-5) then

(

effectively becomes a sum over all points xi , yi

)

on the curve

C . This makes it possible to define the definite integral of a

( )

real-valued function f x, y as the limit (if it exists) of the sum in equation (5.1-5): !

∫

C

(

)

n

f xi , yi ds = lim

n→∞

∑ f ( x , y ) Δs ! i

i

i

(5.1-6)

i =1

Equation (5.1-6) is a known as a real line integral. The arc length of the path C is given by:

Figure 5.1-1!

Curve C together with approximating chords Δsi with i = 1, 2, 3, 4, 5 . 233

Along any path C that is a smooth curve, the line integral

!

in equation (5.1-4) can be represented parametrically so that

()

()

5.2.1!

x = x t and y = y t , where t is some real-valued parameter. Between two fixed endpoints t = a and t = b along the path

!

!

∫

( )

f x, y ds =

C

∫

( ( ) ( ))

f x t ,y t

a

()

()

2

2

⎡⎣ x ′ t ⎤⎦ + ⎡⎣ y ′ t ⎤⎦ dt ! (5.1-8)

ds =

∫

!

zn z0

()

f z dz !

(5.2-1)

where z0 and zn are fixed two points in the z-plane, and where

where !

The definite integral of a continuous complex function

f z of a complex variable z has the form:

a ≤ t ≤ b we then have: b

()

DEFINITE INTEGRALS IN THE COMPLEX PLANE

()

2

()

2

⎡⎣ x ′ t ⎤⎦ + ⎡⎣ y ′ t ⎤⎦ dt =

2

()

f z is a continuous function along some continuous path from

2

⎛ dx ⎞ ⎛ dy ⎞ ⎜⎝ dt ⎟⎠ + ⎜⎝ dt ⎟⎠ dt !

(5.1-9)

the initial point z0 to the terminal point zn . Infinitely many possible paths can exist between the points z0 and zn .

The derivatives in equation (5.1-9) exist and are continuous since the curve is smooth. If the path C is a piecewise-smooth

5.2.2!

curve so that it consists of a number of connected smooth arcs, then each arc requires its own parameterization.

!

CONTOUR INTEGRALS IN THE COMPLEX PLANE

A piecewise-smooth path C in the complex plane is

known as a contour, and consists of a connected set of points

5.2! !

INTEGRATION OF COMPLEX FUNCTIONS The integral of a continuous function of a complex

( x, y ) . Therefore the integral of a complex function along such a

path is referred to as a contour integral, path integral, or complex line integral.

()

The integral of a complex function f z

variable is not restricted to integration over an interval of the x-

!

of a single

axis, but can be evaluated over some smooth path in the

complex variable z along a piecewise-smooth path C can be

complex plane. The integral of a complex function is similar to

written as:

a line integral of a real function in the Cartesian plane. 234

!

∫

()

f z dz !

C

(5.2-2)

()

selection of zi . The integral is dependent, however, on both the

()

real and imaginary parts of the complex function f z , and can

where f z is a continuous function defined on the specified

also be dependent on the path C .

path C . The integral in equation (5.2-2) is defined in terms of a

!

limiting process involving n chords of arbitrary arc length Δ zi

!

along the path C . Letting zi be a point within the chord Δ zi of the path C , we can then write the Riemann sum: n

!

∑

( )

f zi Δzi !

( )

(5.2-5)

( )

()

where u x, y and v x, y are continuous functions since f z is continuous (see Proposition 2.4-2). The contour integral of a

(5.2-3)

i =1

as an approximation for the integral in equation (5.2-2). !

() w = f ( z ) = u ( x, y ) + i v ( x, y ) !

The complex function f z can be written in the form:

()

continuous complex function f z along a path C then takes the form:

∫

!

C

We now let the number of chords increase indefinitely so

()

f z dz =

∫ (u ( x, y ) + i v ( x, y )) ( dx + i dy )!

(5.2-6)

C

that n → ∞ and each Δ zi → 0 . The sum in equation (5.2-3) then

where we have used dz = dx + i dy . Therefore the contour

effectively becomes a sum over all points zi on the curve C .

integral of a continuous complex function can be defined as:

This makes it possible to define the contour integral of a

()

continuous complex function f z as the limit (if it exists) of the sum in equation (5.2-3): !

∫

C

!

()

!

∫

C

()

f z dz =

!

n→ ∞

∑ f (z ) Δ z ! i

C

(5.2-7)

C

n

f z dz = lim

∫ (u ( x, y ) dx − v ( x, y ) dy ) + i ( v ( x, y ) dx + u ( x, y ) dy ) ! ∫

(5.2-4)

i

or

i =1

()

For a continuous complex function f z to be integrable

along the path C , the limit in equation (5.2-4) must exist independently of how the chords Δ zi are chosen and of the

!

∫

C

!

()

f z dz =

∫ u ( x, y ) dx − ∫ v ( x, y ) dy + i v ( x, y ) dx + i u ( x, y ) dy ! (5.2-8) ∫ ∫ C

C

C

C

235

!

When a contour integral is expressed in the form of

equation (5.2-8), the real and imaginary parts of the integral are real line integrals of real-valued functions. This means that the evaluation of a contour integral of a complex function can consist of the evaluation of real line integrals of real-valued continuous functions. These real integrals will exist. Example 5.2-1 Evaluate

∫

C2

z 2 dz where C2 is the line shown in Figure 5.2-1.

Solution: We have: !

()

(

f z = z2 = x + i y

)2 = x 2 − y 2 + i 2 x y = u + i v

so that u = x 2 − y 2 and v = 2 x y . Using equation (5.2-8): !

∫

z 2 dz =

C2

∫ (x C2

2

)

− y 2 dx −

∫

2x y dy

∫

2x y dx + i

+i

!

Figure 5.2-1!

C2

C2

∫ (x

2

C2

For path C2 we have y = 0 and dy = 0 . We then have: !

∫

C2

z 2 dz =

∫

−1

1

x3 2 x dx = 3

−1

1

⎛ 1 1⎞ 2 = ⎜− − ⎟ = − 3 ⎝ 3 3⎠

)

− y 2 dy

Two integration paths between the same points.

5.2.3! CONTOUR DEFINITIONS ! !

Contours can be defined as follows: 1.! A contour is a piecewise-smooth curve in the complex plane (see Section 2.2.8), and so by definition, each segment of a contour will be continuous and will have 236

!

some continuous complex-valued function z t , where t is a

Integration can then be performed along the contour.

real-valued parameter that increases monotonically as the path

2.! A simple contour is a simple piecewise-smooth curve

3.! A closed contour is a contour in the complex plane whose initial and terminal points are the same point.

!

4.! A simple closed contour is a simple contour in the complex plane whose initial and terminal points are the same point.

!

5.! An open contour is any contour in the complex plane

curves. All piecewise-smooth curves are rectifiable.

!

()

a ≤ t ≤ b!

()

()

and y t

(5.2-9)

are real continuous

single-valued functions of the real parameter t : !

()

()

x=x t !

y= y t !

a ≤ t ≤ b!

(5.2-10)

by:

A contour or path C in the complex plane will consist of a

connected set of points constituting one or more path segments (see Section 2.2.8). Each path segment of a contour is a smooth

( )

curve defined in terms of a continuous function z x, y

( )

z x, y

!

() ()

()

() ()

z a = x a +iy a !

()

z b = x b +iy b !

(5.2-11)

By parameterizing an integration path in the complex

plane, integration of a complex function along the contour can

PARAMETRIZATION OF A CONTOUR IN THE COMPLEX PLANE

representing a point set. The contour

() ()

z t = x t +iy t !

where the path functions x t

!

Contours that are finite in length are called rectifiable

5.2.4!

!

We then have:

and where the endpoints of the range of contour C are given

that is not closed. !

segment is traversed from t = a to t = b . !

in the complex plane. !

()

continuous derivatives at all points along the segment.

can be

parameterized so that each path segment is defined in terms of

be greatly simplified, and a direction provided for integration along the path. The contour then acquires an orientation. !

()

()

The derivative z ′ t of z t with respect to t is continuous

within each path segment and is given by:

() () () (5.2-12) where x ′ ( t ) and y ′ ( t ) are continuous within each path segment of the parametric contour. The derivative z ′ ( t ) may not be !

z ′ t = x′ t + i y′ t !

continuous at the endpoints of a path segment. 237

Example 5.2-2

5.2.5!

PARAMETRIZATION OF CONTOUR INTEGRALS

Parametrize the contour shown in Figure 5.2-2.

We can use the following parameterizations: !

CR :!

!

x-axis:!

()

If a complex function w = f z

!

Solution:

it

z = Re ! z=t!

0≤t ≤π −R ≤ t ≤ R

is continuous, it can be

parameterized as follows:

() ()

()

( ( )) !

w t =u t +iv t = f z t

!

()

()

(5.2-13)

where u t and v t are continuous real-valued functions on a

()

contour C . The contour integral of f z is then given by: !

∫

C

()

f z dz =

∫

b

a

b ( ) f ( z ( t )) dt = f ( z ( t )) z ′ ( t ) dt ! ∫ dt

dz t

a

(5.2-14)

where a ≤ t ≤ b defines the contour C . We can also write the

() ()

()

( ( ))

contour integral of w t = u t + i v t = f z t

∫

!

C

()

f z dz =

b

b

as:

b

∫ w(t ) dt = ∫ u (t ) dt + i ∫ v (t ) dt ! a

a

(5.2-15)

a

or !

∫

C

Figure 5.2-2!

Closed contour C consisting of two path segments: a semicircle CR and a real axis interval − R, R in the z-plane. Note the directions on the contour provided by the parametrization.

(

)

()

f z dz =

∫ (u ( x (t ), y (t )) + i v ( x (t ), y (t ))) ( x′ (t ) + i y′ (t )) dt b

a

!

(5.2-16)

and so we have:

238

!

∫

()

f z dz =

C

+i

!

∫ (u ( x (t ), y (t )) x′ (t ) − v ( x (t ), y (t )) y′ (t )) dt b

Example 5.2-4

a

∫ (v ( x (t ), y (t )) x′ (t ) + u ( x (t ), y (t )) y′ (t )) dt !

Evaluate

b

a

C1

(5.2-17)

5.2-1.

where the integrals on the right side of this equation are real.

Solution:

Equation (5.2-17) provides the definition of a contour integral

()

( ( ))

with a parameterized complex function w t = f z t

Parameterizing the path C1 :

. Contour

integration produces a complex number, and so a contour

!

integral is a complex number.

∫

C

!

We can use the following parameterization:

C :!

z = eit !

!

∫

C

∫

2π

0

1 it i e dt = i it e

0 ≤θ ≤ π

!

∫

2π

0

dt = 2 π i

∫

π

e

2 iθ

iθ

i e dθ =

0

∫

π

i e3iθ dθ

0

⎛ e3iθ ⎞ 2 z dz = ⎜ i ⎟ ⎝ 3i ⎠ C1

∫

π

0

e i 3π 1 cos3π + i sin3π 1 = − = − 3 3 3 3

We then have:

0 ≤ t ≤ 2π !

where dz = i eit dt . Therefore:

1 dz = z

dz = i eiθ dθ !

r = 1!

and so: !

Solution:

∫

2

z dz =

C1

1 dz where C is a circle centered at the origin z

and having a radius of 1.

!

z = r eiθ ! Therefore:

Example 5.2-3 Evaluate

∫

z 2 dz where C1 is the semicircle shown in Figure

∫

1 1 2 z 2 dz = − − = − 3 3 3 C1

We see that using this path of integration we obtain the same result as was obtained using a completely different path between the same two points (see Example 5.2-1). Path 239

independence for integrals of complex functions is generally Example 5.2-6

not true, however. !

Evaluate

The integral of a complex function in the complex plane

C2

is generally dependent on the path of integration chosen. It is often the case, therefore, that different results will be obtained

Solution:

for an integral when different paths are chosen between the

We have:

initial and terminal points in the complex plane (see Examples !

5.2-5 and 5.2-6.

∫

C1

∫

z dz where C1 is the semicircle in Figure 5.2-1.

!

Solution:

( x − i y ) ( dx + i dy )

C2

∫

z dz =

∫

−1

1

x2 x dx = 2

−1

1

⎛ 1 1⎞ = ⎜ − ⎟ =0 ⎝ 2 2⎠

different result from that obtained using the path in

z = eiθ !

z = e−iθ !

dz = i eiθ dθ !

0 ≤θ ≤ π

Example 5.2-5, although the initial and terminal points of the two paths are the same.

Therefore:

∫

z dz =

C1

π

∫e

−iθ

iθ

i e dθ =

0

∫

C1

z dz = i θ

π

∫ i dθ 0

and so: !

∫

We see that using this path of integration we obtain a

Let:

!

z dz where C2 is the line shown in Figure 5.2-1.

For path C2 we have y = 0 and dy = 0 . We then have:

C

!

z dz =

C2

Example 5.2-5 Evaluate

∫

5.2.6! !

π 0

=πi

REPARAMETRIZATION FOR CONTOUR INTEGRALS

While the value of a contour integral can be dependent

upon the contour, it will not be dependent upon the particular parameterization chosen to represent the contour. This makes it 240

possible to select the most convenient parametrization for

valued line integrals in this equation will apply to the complex

evaluating a given integral.

line integral.

!

For example, without changing the value of the integral in

equation (5.2-14), we can change from a parameter t to a

5.2.8!

parameter s where t = h s . We then have dt = h′ s ds where

!

h′ s > 0 :

consisting of m smooth curves joined end-to-end), then, the

()

()

b

!

b

()

∫ f ( z (t )) z′ (t ) dt = ∫ f ( z (s)) = ∫ f ( z (s)) z′ (s) ds ! a

a

INTEGRAL OVER A SIMPLE CONTOUR

If C is a simple contour (a simple piecewise-smooth curve

contour is given by:

dz dh ds dt dh ds dt

C = C1 ∪ C2 ∪ C3 ∪!∪ Cm !

!

(5.2-19)

b

!

a

and so the contour integral can be decomposed into a finite

(5.2-18)

number m of separate integrals (one integral over each smooth

Equation (5.2-18) is an equality, being simply a change in the variable of an integral. !

The mapping function employed for a path to change

curve) which are summed together: !

∫

from one parameterization to another must be piecewise-

()

f z dz =

C

∫

C1

()

f z dz +

∫

C2

()

f z dz +!+

∫

()

f z dz

Cm

smooth, and so continuously differentiable. It must also be a

! !

one-to-one mapping so that it is invertible.

This follows from the properties of real line integrals when the

(5.2-20)

complex integrals are written in the form of equation (5.2-7) or

5.2.7! !

EXISTENCE OF A CONTOUR INTEGRAL

(5.2-17).

If the curve C is a simple contour, and if the function

() ()

()

( ( )) is continuous on C , the line integrals

w t =u t +iv t = f z t

()

of the continuous real-valued functions u t

()

and v t

in

5.2.9! PROPERTIES OF CONTOUR INTEGRALS !

Properties of contour integrals can be obtained from

equation (5.2-15) will exist. Therefore the complex line integral

properties of complex numbers and from the definition of the

in equation (5.2-15) will exist, and the properties of the real-

Riemann integral of a complex function. If f z and g z are

()

()

241

continuous complex functions at all points of a simple contour

C , and if k is a complex constant, then we have the following

()

b

∫ k f ( z (t )) z′ (t ) dt = k ∫ f ( z (t )) z′ (t ) dt !

!

a

()

properties of contour integrals of f z and g z : !

∫ k f ( z ) dz = k ∫ C

!

()

∫ ⎡⎣ f ( z ) + g ( z )⎤⎦ dz = ∫ C

a

b

f z dz !

C

b

(5.2-21)

()

f z dz +

C

∫ g ( z ) dz ! C

a

∫ f ( z (t )) z′ (t ) dt = − ∫ f ( z (t )) z′ (t ) dt !

!

a

b

b

(5.2-22)

!

(5.2-27)

(5.2-28)

b

∫ ⎡⎣ f ( z (t )) + g( z (t ))⎤⎦ z′ (t ) dt = ∫ f ( z (t )) z′ (t ) dt + g ( z ( t )) z ′ ( t ) dt ! (5.2-29) ∫ a

a

b

!

∫

()

f z dz =

C = C1 ∪ C2

∫

()

f z dz +

C1

∫

C2

()

f z dz !

(5.2-23)

!

a

c

!

∫

()

f z dz =

C

!

Re

∫

C

∫

C

()

f z dz =

()

f z dz !

!

C

∫ f ( z (t )) z′ (t ) dt = ∫ f ( z (t )) z′ (t ) dt + ∫ f ( z (t )) z′ (t ) dt ! a

(5.2-24)

a

c

!

∫ Re f ( z ) dz !

b

b

(5.2-25) b

Im

∫

C

()

f z dz =

∫ Im f ( z ) dz ! C

a

(5.2-26)

b

!

Re

Properties of integrals of complex functions can be written

in parametric form where t is a real variable:

b

!

Im

a

a

(5.2-32)

b

∫ f ( z (t )) z′ (t ) dt = ∫ Im f ( z (t )) z′ (t ) dt ! a

(5.2-31)

b

∫ f ( z (t )) z′ (t ) dt = ∫ Re f ( z (t )) z′ (t ) dt ! a

5.2.10! PROPERTIES OF CONTOUR INTEGRALS IN PARAMETRIC FORM !

b

∫ f ( z (t )) z′ (t ) dt = ∫ f ( z (t )) z′ (t ) dt !

! !

(5.2-30)

a

(5.2-33)

242

Solution:

Example 5.2-7 Evaluate

i

We have two path segments: C = C1 ∪ C2 which is a straight

1

line from z = 1 to z = 1+ i followed by a straight line from

∫ z dz along the path shown in Figure 5.2-3 using

z = 1+ i to z = i .

both of the following parameterizations.

C1 :!

z = 1+ i t !

dz = i dt !

0 ≤ t ≤1

C2 :!

z = t + i!

dz = dt !

1≥ t ≥ 0

For the first parameterization:

C1 :!

z = 1+ i log e t !

C2 :!

t z = +i! e

1

1

0

0

∫ z dz = ∫ (1+ it ) i dt + ∫ (t + i) dt

!

and

i dz = dt ! t 1 dz = dt ! e

i

1

and so:

1≤ t ≤ e e≥t ≥0

1

0

⎛ ⎛ t2 ⎞ t2 ⎞ 1 1 z dz = ⎜ it − ⎟ + ⎜ + it ⎟ = i − − − i = −1 2⎠ 2 2 ⎝ ⎝2 ⎠ 1 1 0

∫

!

i

For the second parameterization:

∫

!

i

z dz =

1

∫

e

1

i 1+ i log e t dt + t

(

)

∫

0

e

⎛t ⎞ 1 ⎜⎝ e + i ⎟⎠ e dt

and so: !

(

⎛ log e t z dz = ⎜ i log e t − 2 ⎜⎝ 1

∫

i

)

2

e

0 ⎞ ⎛ t2 ⎞ t 1 1 ⎟ + ⎜ 2 + i ⎟ = i − − − i = −1 e⎠ 2 2 ⎟⎠ ⎝ 2e e 1

We see that the parameterization can be changed without Figure 5.2-3! Integration path C = C1 ∪ C2 for Example 5.2-7.

changing the value of the integral. 243

5.3! !

CONTOUR INTEGRATION Contour integration of a complex function can depend

contour is considered to be negatively oriented. Unless otherwise stated, integration is always taken in a positive direction around a closed contour.

upon properties of the function in the region over which the

!

integration is being performed. Contour integration can also

positive direction is:

depend upon the direction of integration along a closed

For a closed contour C , the notation for integration in the

!∫

()

f z dz !

contour.

!

5.3.1! DIRECTION OF INTEGRATION ALONG A CLOSED CONTOUR

and the notation for integration in the negative direction is:

!

The direction of integration of complex functions along a

C

!

!∫ f ( z ) dz ! C

(5.3-1)

(5.3-2)

simple closed contour in the complex plane is important. The positive direction of integration is taken to be that direction for which the interior of a region bounded by a closed contour lies to the left as the contour is being traversed. This is illustrated in Figure 5.3-1 for two closed contours bounding an annular domain. Note that the positive direction of integration is in opposite directions for the inner and outer contours about an annular region. !

It is common practice in complex analysis to consider the

direction of integration as a property of the contour itself. For example, if the direction of integration on a closed contour is positive, the contour is considered to be positively oriented. If the direction of integration on a closed contour is negative, the

Figure 5.3-1!

Annular domain enclosed by circular contours. The positive direction of integration is indicated by arrows on the two contours. 244

()

If f z is an integrable complex function over a simple closed

()

contour C in the positive direction, then f z is integrable over the contour C in the negative direction, and we have:

!∫

!

()

f z dz = −

C

!∫

C

∫

!

Proposition 5.3-1, Direction of Integration:

()

f z dz !

(5.3-3)

− C1

()

f z dz =

a

∫ f ( z ( a + b − t )) z′ (a + b − t ) (−1) dt ! b

(5.3-6)

Changing parameterization to τ = a + b −t and dτ = − dt , we can rewrite equation (5.3-6) as:

∫

!

− C1

()

f z dz =

a

∫ f ( z (τ )) ( z′ (τ ) ) dτ ! b

(5.3-7)

and so: Proof: !

We can parameterize the simple closed contour C

∫

!

− C1

traversed in the positive and negative directions with the

()

f z dz = −

b

∫ f ( z (τ )) z′ (τ ) dτ ! a

(5.3-8)

Since t and τ are just two different parameterizations of

parameter t . We will use the notation C for the positive

!

direction and − C for the negative direction.

the same integral, we therefore have from equations (5.3-4) and

!

From equation (5.2-14) we have for integration along a

segment C1 of the closed contour C from a point t = a to a

(5.3-8):

∫

!

− C1

point t = b on C :

∫

!

()

f z dz =

C1

!

∫ f ( z (t )) z′ (t ) dt !

(5.3-4)

a

()

(

)

()

!∫

!

C

Changing direction of integration along C1 , we have for

z t → z a + b−t !

∫

()

f z dz !

C1

(5.3-9)

This equation can then be written for the closed contour C :

b

integration from a point t = b to a point t = a : !

()

f z dz = −

(

)

z′ t → − z′ a + b − t !

where the contour is now designated − C1 :

!∫

C

()

f z dz !

(5.3-10)

■

! (5.3-5)

()

f z dz = −

Taking the two closed contours in Figure 5.3-1 to be

centered at the origin, the positive orientation of the contours can be represented parametrically as: 245

! !

()

(

)

z1 t = r1 eit = r1 cost + isint !

(5.3-11)

i 0+2 π −t ) z2 0 + 2 π − t = r2 e ( = r2 e−it = r2 cost − isint ! (5.3-12)

(

)

(

()

where 0 ≤ t ≤ 2 π and where z1 t

)

()

and z2 t are the outer and

inner contours, respectively.

5.3.2! !

()

interval x1 ≤ x ≤ x2 , and if we have:

()

()

()

only on the function, but also on the integration path chosen. This is because antiderivatives do not exist for most complex functions. The fundamental theorem of integral calculus does not extend, therefore, to the integrals of all continuous complex happen to have an antiderivative does the fundamental

If f x is a continuous real-valued function defined in an

f x = F′ x =

Most integrals of complex functions will be dependent not

functions. Only for those continuous complex functions that

ANTIDERIVATIVES

!

!

theorem of integral calculus apply. Proposition 5.3-2, Fundamental Theorem of Calculus for

( )!

Complex Functions:

dF x

(5.3-13)

dx

()

If a complex function f z is holomorphic in a simply

()

then F x is called an antiderivative, primitive function, or

connected domain D , and if f z has an antiderivative

indefinite integral of f x in that interval. Every continuous

F z which is holomorphic in D so that:

()

()

real-valued function of a real variable has an antiderivative. We can write: !

∫

()

x2 x1

()

f x dx =

∫

x2 x1

( ) dx = F ( x ) − F ( x ) !

dF x dx

2

1

∫

!

of integration. This equation relating the antiderivative and the

z2 z1

()

f z dz =

∫

and

integral of a function is known as the fundamental theorem of integral calculus.

( )!

dF z

(5.3-15)

dz

then in the domain D :

(5.3-14)

where the integral depends only on the endpoints of the path

()

f z = F′ z =

!

!

∫

()

f z dz =

∫

z2 z1

( ) dz = F ( z ) − F ( z ) !

dF z dz

2

( ) dz = F ( z ) + c !

dF z dz

1

(5.3-16)

(5.3-17) 246

where c is a complex constant. Proof: !

()

()

Since f z has an antiderivative F z , we can write:

∫

!

z2 z1

()

f z dz =

∫

( ) dz !

z2

dF z dz

z1

(5.3-18)

Let z be parameterized using a real-valued parameter t :

() ()

()

z t = x t +iy t !

!

(5.3-19)

()

We will consider the smooth path a ≤ t ≤ b , where z1 = z a and

()

z2 = z b . We can then write equation (5.3-18) as: !

∫

z2 z1

()

f z dz =

∫

b

a

( ( )) dz dt ! dt

or

2

dz

z1

1

(5.3-23)

similarly, we have:

∫

!

()

f z dz =

∫

( ) dz = F ( z ) + c !

dF z

!

(5.3-24)

dz

where c is a complex constant.

■

The fundamental theorem of calculus basically states that

()

()

()

if the integral of f z is F z , then the derivative of F z is

()

f z . Integration and differentiation are inverse operations.

()

(5.3-20)

some constant. Proposition 5.3-3:

∫

z2 z1

()

f z dz =

∫

b

a

( ( ))

d ⎡ F z t ⎤⎦ dt ! ⎣ dt

∫

z2 z1

()

()

If a complex function f z is continuous in a simply connected

(5.3-21)

()

()

domain D , and if f z has an antiderivative F z that is

()

holomorphic in D , then f z is holomorphic in D .

Therefore we have: !

∫

( ) dz = F ( z ) − F ( z ) !

dF z

antiderivatives of a complex function f z can differ only by

or !

z1

()

f z dz =

z2

While it is true that antiderivatives are not unique, any two

dF z t dt

∫

!

z2

( ( )) ( ( ))

f z dz = F z b − F z a !

(5.3-22)

Proof: !

()

()

Since f z is continuous and has an antiderivative F z

()

()

()

in D , we have f z = F ′ z . Since F z is holomorphic in D ,

()

then all higher order derivatives of F z

will exist (see 247

Proposition 6.2-1), and so from equation (5.3-15) we see that

()

f z must also be holomorphic in D .

■

()

If F z and G z are distinct antiderivatives of a complex

()

()

() simply connected domain D , then F ( z ) and G ( z ) differ by

∫

!

function f z , and if F z and G z are holomorphic in a some constant c .

(5.3-26)

dz

z2 z1

()

f z dz =

∫

z2 z1

( ) dz = F ( z ) − F ( z ) !

dF z

2

dz

()

and so we see that if f z

Let H z = F z − G z where H z will be holomorphic in

()

From Proposition 3.7-4 we then have H z = c in D , and so

()

()

F z and G z differ by some constant c .

■

Proposition 5.3-5, Independent of Path:

()

The contour integral of a function f z that is holomorphic in a simply connected domain D will be independent of path in D if

()

()

f z has an antiderivative F z which is holomorphic in D . Proof:

(5.3-27)

1

is holomorphic and has an

antiderivative that is holomorphic in a simply connected

()

() () () () D since F ( z ) and G ( z ) are holomorphic in D . We can write: dH ( z ) dF ( z ) dG ( z ) ! = − = f (z) − f (z) = 0! (5.3-25) dz dz dz

!

( )!

dF z

domain D , then the integral of f z

Proof: !

()

From Proposition 5.3-2 we can write:

Proposition 5.3-4:

()

()

f z = F′ z =

!

depends only on the

endpoints of the path in D , and not upon the path.

■

Example 5.3-1

()

Show that the integral of f z = z 2 between two points in the complex plane is independent of path. Solution:

()

The function f z = z 2 is continuous and is an entire

()

function. It has an antiderivative of F z = z 3 3 everywhere

()

in the complex plane. Therefore f z = z 2 is independent of the path of integration in the complex plane. This is why the same result is obtained for an integral of this function for the two different integration paths shown in Examples 5.2-1 and 5.2-4.

We have: 248

The contour C can be expressed using a parameter θ given

Proposition 5.3-6:

()

If a complex function f z

()

by z θ = z0 + r ei θ where 0 ≤ θ ≤ 2 π . The radius r is some

is holomorphic in a simply

()

real number greater than zero. We then have:

()

connected domain D , and if f z has an antiderivative F z

which is holomorphic in D , then along every simple closed

()

z θ − z0 = r e !

!

contour C in D we must have:

!∫

!

()

f z dz = 0 !

C

(5.3-28) !

∫

()

f z dz =

z1

∫

( ) dz = F

dF z

z2

dz

z1

( z2 ) − F ( z1 ) !

(5.3-29)

∫

!

!∫

1 dz = i z − z0

()

f z dz = 0 !

z1

()

0

()

dz θ 1 dθ = z θ − z0 dθ

∫

2π

0

i r eiθ dθ r eiθ

!

C

∫

2π

dθ = 2 π i

0

()

Note that since f z is not holomorphic at the point z0 ,

For closed contours, z1 = z2 , and so: z1

∫

2π

or

From Proposition 5.3-2 we have:

!

!∫

1 dz = z − z0

C

z2

dθ

and so:

Proof: !

( ) = i r eiθ

dz θ

iθ

Proposition 5.3-6 does not apply, and the value of the (5.3-30)

()

integral of f z over a closed circle C centered at point z0 is not zero. Also note that this integral is independent of the

■

radius r of the circular contour C .

Example 5.3-2 Evaluate

!∫

C

()

f z dz =

!∫

C

1 dz where C is a closed circle z − z0

Example 5.3-3 Determine:

centered at point z0 . Solution:

1.!

∫ sin z dz 249

2.!

∫

e z dz

5.3.3!

Solution:

!

1.! From equation (4.3-8) we have:

(

d cos z

!

dz

! !

integrals.

()

f z = sin z

()

is F z = − cos z .

Proposition 5.3-7, Contour Integral Inequality:

()

Therefore from equation (5.3-17) we have:

If C is a simple contour and f z is a complex function

∫ sin z dz = − cos z + c

continuous over C , then:

∫

!

C

where c is a complex constant.

( )=e

d ez

!

!

()

f z

C

dz !

(5.3-31)

The result of integrating a complex function along a

dz

!

∫

contour can be represented as a complex number r eiθ :

z

The antiderivative of

()

f z dz ≤

Proof:

2.! From equation (4.2-9) we have: !

We will next consider an inequality for contour integrals

that is very useful in proving many propositions involving such

) = − sin z

The antiderivative of

CONTOUR INTEGRAL MODULI UPPER BOUNDS

()

f z =e

z

()

is then F z = e z .

!

C

Therefore from equation (5.3-17) we have:

and so:

∫

!

e z dz = e z + c

where c is a complex constant.

∫

()

f z dz = r eiθ !

r = e−iθ

∫

C

()

f z dz =

(5.3-32)

∫

C

()

e−iθ f z dz !

(5.3-33)

Parameterizing the contour C : 250

b

∫ e f ( z (t )) z′ (t ) dt !

r=

!

From equations (5.3-32) and (5.3-38) we can write:

−iθ

(5.3-34)

a

where a ≤ t ≤ b defines the contour C . This equation can be written in the form: ! r = Re

b

∫

e

−iθ

a

C

( ( )) z′ (t ) dt + i Im ∫

b

f z t

( ( )) z′ (t ) dt

e−iθ f z t

a

∫

( ( )) z′ (t ) dt = ∫

e−i θ f z t

a

a

( ( )) z′ (t )⎤⎦ dt

!

r≤

∫

e

−iθ

a

( ( )) z′ (t ) dt = ∫

f z t

e

−iθ

()

( ( )) z′ (t ) dt

e−iθ = 1 !

(5.3-38)

∫ f ( z (t )) z′ (t ) dt = ∫ a

C

()

f z

dz !

(5.3-42)

∫

(5.3-43)

(5.3-39)

()

f z dz ≤ M L !

Proof: !

r≤

be a

f z ≤ M!

C

and so: b

()

then: !

From equations (1.11-9) and (1.11-10) we have:

!

()

!

(5.3-37)

e iθ = 1 !

(5.3-41)

number M along C :

! !

!

C

dz !

continuous complex function on C . If f z is bounded by a real

f z t

a

()

f z

Let C be a simple contour of length L , and let f z

and with the inequality in equation (1.8-2): b

∫

Proposition 5.3-8, ML-Inequality or Darboux Inequality:

(5.3-36)

b

(5.3-40)

■

Re ⎡⎣ e−i θ f z t

! !

()

f z dz ≤

C

Equating the real parts of this equation: b

∫

!

(5.3-35)

! r = Re

()

f z dz = r eiθ = r eiθ = r !

Therefore equations (5.3-40) and (5.3-39) give us:

! !

b

∫

!

!

From Proposition 5.3-7 we can write:

∫

C

()

f z dz ≤

∫

C

()

f z

dz ≤ M

∫

C

dz !

(5.3-44) 251

In parametric form as given in equations (5.2-14) and (5.2-18): !

∫

()

f z dz ≤ M

C

∫

b

a

()

dz t

dt = M

dt

!

C

b

∫ ds ! a

Example 5.3-5

ds is differential arc length. We then have:

∫

Determine the upper bound of

∫

C

()

f z dz ≤ M L !

C

π 2

(5.3-45)

where t = a and t = b are the endpoints of the contour C , and

!

∫

e z dz ≤ e3

1 dz when C is the 2 z −1

circle of radius z = 2 .

(5.3-46)

which is the upper bound for moduli of contour integrals.

Solution:

This equation is known as the ML-inequality or the Darboux

The length L along the circle is 4 π . We also have:

inequality.

■

!

Example 5.3-4

and so:

∫

Determine the upper bound of

e z dz when C is a

C

quarter circle of radius z = 3 .

!

1 1 ≤ z2 − 1 3 Therefore from Proposition 5.3-8:

Solution: The length L along the circular arc is 1 4 of 2 π equals π 2 .

(

!

∫

C

We also have: !

z2 − 1 ≥ z2 − 1 = 3

)

(

1 4π dz ≤ 3 z2 − 1

)

e z = e x sin y + i cos y = e x sin y + i cos y = e x

()

Since x = Re z ≤ 3 along C , we have M = e3 . Therefore from Proposition 5.3-8: 252

5.4! !

CAUCHY-GOURSAT THEOREM FOR SIMPLY CONNECTED DOMAINS We will now consider the integration of a complex

()

function f z

that is holomorphic in a simply connected

holomorphic function around any closed contour within any simply connected domain is zero if no discontinuities (singularities) exist for the function within or on the contour.

5.4.1!

CAUCHY’S INTEGRAL THEOREM

domain D . We will integrate in the positive direction along a

!

simple closed contour C within the domain D such as shown

only to domains having continuous first order derivatives.

The following proof of Cauchy’s integral theorem applies

in Figure 5.4-1. Proposition 5.4-1, Cauchy’s Integral Theorem:

()

If f z is a holomorphic complex function with continuous first order derivatives in a simply connected domain D , then along every simple closed contour C that lies within D we must have:

!∫

!

()

f z dz = 0 !

C

(5.4-1)

Proof: ! !

We first recall Green’s theorem in the plane:

!∫

C

Figure 5.4-1!

!

A simple closed contour C in a simply connected domain D .

We will present both Cauchy’s theorem and the Cauchy-

Goursat theorem which each state that the integral of any

( )

( )

P x, y dx + Q x, y dy =

( )

( )

⎛ ∂Q x, y ∂P x, y ⎞ ⎜ ∂x − ∂ y ⎟ dx dy A ⎝ ⎠

∫∫

!!

(5.4-2)

where C is a contour enclosing an area A lying within a simply

( )

( )

connected domain D , and P x, y and Q x, y are real-valued 253

()

functions having continuous first-order partial derivatives in

This equation applies if f z is holomorphic with continuous

D.

first order derivatives in a simply connected domain D , and C

! !

is a simple closed contour C that lies within D .

From equation (5.2-7) we have:

!∫

C

()

f z dz =

!∫ (u ( x, y ) dx − v ( x, y ) dy ) +i !∫ (v ( x, y ) dx + u ( x, y ) dy ) !

!

C

!

(5.4-3)

C

!

() ( )

( )

Since f z = u x, y + i v x, y

has continuous first order

( )

derivatives in the simply connected domain D , then u x, y and

( )

v x, y also have continuous first order derivatives in D . We can therefore use Green’s theorem in the plane and equation (5.4-3) to write: !

!∫

C

!

()

f z dz =

⎛ ∂v ∂u ⎞ ⎜ − − ∂ y ⎟⎠ dx dy + i A ⎝ ∂x

∫∫

⎛ ∂u ∂v ⎞ − ⎟ dx dy ⎜ ∂y⎠ A ⎝ ∂x

∫∫

!

(5.4-4)

()

Since f z

∂u ∂v ! = ∂x ∂ y

!∫

C

()

possible to identify the function f z as having a holomorphic antiderivative, and so this proposition is more general than Proposition 5.3-6.

5.4.2! !

CAUCHY-GOURSAT THEOREM

Equation (5.4-6) was first derived by Louis-Augustin

Cauchy in 1825. Cauchy’s integral theorem was later expanded by Edouard Goursat who showed that continuity of first-order

() f (z)

derivatives of f z is not required for the theorem to be valid. The function

need only be holomorphic in a simply

connected domain. Proposition 5.4-2, Cauchy-Goursat Theorem:

∂v ∂u ! =− ∂x ∂y

If

(5.4-5)

()

f z dz = 0 !

()

f z

is a holomorphic complex function in a simply

connected domain D , then for every simple closed contour C that lies within D we must have:

and so equation (5.4-4) becomes: !

Cauchy’s integral theorem is valid even when it is not

is holomorphic on the domain D , the Cauchy-

Riemann equations apply: !

■

(5.4-6)

!

!∫

C

()

f z dz = 0 !

(5.4-7) 254

Proof: !

To prove this theorem we will choose the closed contour

!

C

C to be the perimeter of a large triangle Δ 0 that lies within D

()

as shown in Figure 5.4-2. The function f z

will then be

!

()

f z dz ≤

!∫

C1

()

f z dz +

!∫

C2

!∫

()

f z dz + +

!∫

()

f z dz

C3

C4

holomorphic on C since C lies within D . !

!∫

()

f z dz ! (5.4-9)

We will first construct four similar nested triangles by

connecting the midpoints of the sides of the triangle Δ 0 (see Figure 5.4-2). We will choose the closed contours C1 , C2 , C3 , and C4 to be the perimeters around these four nested triangles. Each of these triangular contours is chosen to have a positive

()

orientation. We will integrate f z

along each of the closed

contours C1 , C2 , C3 , and C4 , and sum the result. !

Due to internal cancellations, all contributions to the sum

from integration along the interior contour C2 are completely cancelled by paths of integration traversed in opposite directions along the other contours, as shown by the arrows in Figure 5.4-2. The result equals integration over the exterior contour C , and so we can write:

!∫

!

C

!

()

f z dz =

!∫

C1

()

f z dz +

!∫

C2

()

f z dz +

!∫

C3

!

()

f z dz +

!∫

C4

()

f z dz (5.4-8)

Figure 5.4-2! Triangular contour C (in red) is the perimeter of ! triangle Δ 0 which contains four nested triangles, ! all within a simply connected domain D .

From equation (5.4-8) and the triangle inequality we have: 255

!

We will now assume that all the contours are labeled

according to decreasing absolute value of the corresponding integrals in equation (5.4-9), so that: !

!∫

C1

()

f z dz ≥

!∫

()

f z dz ≥

C2

!∫

()

f z dz ≥

C3

!!

!∫

C4

()

f z dz ! (5.4-10)

so that we can write: !

!∫

()

f z dz ≤ 4

C

!∫

C1

()

f z dz !

(5.4-11)

We will define: !

M=

!∫

()

f z dz !

C

(5.4-12) Figure 5.4-3! Triangular contour C1 (in solid red) within a ! simply connected domain D .

From equation (5.4-11) we then have: !

!∫

C1

()

f z dz ≥

M ! 4

(5.4-13)

!

The contours around each of these nested triangles have

been labelled C2 , C3 , C4 , and C5 according to decreasing

We now will designate the triangle enclosed by the

absolute value of the corresponding integrals. We then form a

contour C1 to be Δ1 (see Figure 5.4-2). We next subdivide the

sum of these integrals as in equation (5.4-8). From the triangle

!

triangle Δ1 by going through the same process of constructing four similar nested triangles (see Figure 5.4-3).

inequality we can write: !

!∫

C1

()

f z dz ≤ 4

!∫

C2

()

f z dz !

(5.4-14) 256

We then can write:

Therefore using equation (5.4-13) we have:

!∫

!

C2

!

1 f z dz ≥ 4

()

!∫

C1

M f z dz ≥ 2 ! 4

()

(5.4-15)

!

iterations we obtain:

!∫

!

Cn

!

()

every one of the nested triangles Δ 0 ⊃ Δ1 ⊃ Δ 2 ⊃ Δ 3 ⊃! ⊃ Δ n .

()

is holomorphic inside and along the closed

contour C , and since all the contours Cn are inside C , the

()

complex function f z must be holomorphic at point z0 , and so

( )

f ′ z0 exists. For any real ε > 0 a real number δ > 0 must then exist such that: !

()

( )− f′

f z − f z0 z − z0

0 < z − z0 ≤

(5.4-16)

Each triangle contains a closed and bounded set of points.

Since f z

( ) ( z − z0 )

(

)

< ε z − z0 !

(5.4-18)

Triangle Δ n has perimeter Ln , and so we have:

!

There will be a unique limit point z0 that belongs to each and !

( )

when 0 < z − z0 < δ .

We can continue this subdivision process until after n

M f z dz ≥ n ! 4

()

f z − f z0 − f ′ z0

!

Ln ! 2

(5.4-19)

since the separation of two points within a triangle cannot exceed half the perimeter of the triangle. !

Let L be the perimeter of the triangular contour C . From

the construction of the triangles we then have: ! L1 =

L ! 2

L2 =

L1 2

=

L 22

!

L3 =

L2 2

=

L 2

! 3

!

!!

Ln =

L 2n

(5.4-20)

If we choose n large enough, we have from equations (5.4-20), (5.4-19) and (5.4-17):

( z0 )

< ε ! when 0 < z − z0 < δ ! (5.4-17)

!

0 < z − z0 ≤

L 2 n+1

0 can be arbitrarily small, but M ≥ 0 has a definite fixed value, we can therefore conclude that M = 0 . From equation (5.4-12) we then have:

or !

0≤ M ≤

!

!∫

Cn

( ) !∫

()

f z dz − f z0

Cn

( ) !∫ ( z − z0 ) dz

dz − f ′ z0

C

Cn

≤

!

!∫

Cn

!∫

!

ε L dz ! (5.4-24) n 22

()

f z dz = 0 !

(5.4-28)

()

if f z is holomorphic in a simply connected domain D , and C is any simple closed contour in D .

■

The last two integrals on the left side of this inequality are

!

equal to zero since they are integrals over a closed path of

valid for any simple closed curve. This is not surprising since a

functions that have antiderivatives: z is the antiderivative of 1

simple curve can be approximated by a polygon, which can, in

and

( z − z0 )2

2 is the antiderivative of z − z0 . We then have

using Ln = L 2 n : !

!∫

Cn

ε L ε L2 ! f ( z ) dz ≤ Ln ≤ n n 22 24

M ≤ 4n

turn, be divided into some finite number of triangles. Proposition 5.4-3:

!∫

Cn

()

f z dz ≤ 4 n

()

If f z is an entire function, then along every simple closed

(5.4-25)

contour C in the complex plane we must have:

and so from equation (5.4-16): !

The Cauchy-Goursat theorem can similarly be shown to be

!

ε L ε 2 = L ! 2 4n 2

!∫

C

2

(5.4-26)

()

f z dz = 0 !

(5.4-29)

258

!

Proof: !

r n+1 i (n+1) θ z dz = e n +1

!∫

n

C

Since all entire functions are holomorphic over the entire

2π 0

r n+1 ⎡ i (n+1) 2π ⎤ r n+1 ⎡⎣1− 1⎤⎦ = 0 = e −1 = ⎣ ⎦ n +1 n +1

complex plane, Proposition 5.4-3 follows from the Cauchy-

Since z n is an entire function and the unit circle is a closed

Goursat theorem.

contour, this result is consistent with Proposition 5.4-2.

!

■

From the Cauchy-Goursat theorem we can conclude that if

Example 5.4-2

()

for a complex function f z we have:

!∫

!

Evaluate

()

f z dz ≠ 0 !

C

!∫

e z dz where C is any simple closed positively

C

(5.4-30)

oriented contour in the complex plane.

()

in any simple closed contour C , then f z is not holomorphic

Solution:

in the region enclosed by the contour.

Since e z is an entire function (see Section 4.2.2), from Proposition 5.4-3 we have for any simple closed contour in

Example 5.4-1 Evaluate

!∫

the complex plane:

n

z dz where C is the circle z = r , and show

C

that the result is consistent with the Cauchy-Goursat

!

∫

e z dz = 0

C

theorem (Proposition 5.4-2).

!

Solution:

Example 5.4-3

Using z = r ei θ , we have z n = r n ei n θ and dz = r i ei θ dθ .

Evaluate

!∫

C

z n dz =

∫

2π

0

r nei n θ r i ei θ dθ = i r n+1

∫

and so with equation 1.13-12 we have:

2π

0

!∫

z dz where C is the circle z = 1 .

C

i n+1 θ e ( ) dθ

Solution: Using z = ei θ , we have z = e− i θ and dz = i ei θ dθ . Therefore: 259

!

!∫

z dz =

C

∫

2π

e

−iθ

iθ

i e dθ =

0

∫

2π

i dθ = 2 π i

!∫ f ( z ) dz = 0 !

!

0

for every simple closed contour C within D , then

Although this integral is not equal to zero, it is nevertheless

()

consistent with the Cauchy-Goursat theorem since f z = z is not a holomorphic function (see Example 3.4-4).

(5.4-31)

C

∫ f ( z ) dz !

!

(5.4-32)

is independent of path within D . Example 5.4-4 Evaluate

!∫

C

1 dz where C is the circle z = 1 . z2 + 4

Solution:

!

Let C1 and C2 be any two contours within a domain D ,

and let both contours have the same initial point z1 and the same terminal point z2 in D (see solid red line and solid blue

(

)(

)

and so the integrand

line in Figure 5.4-4). The two curves then intersect only at the

1 z 2 + 4 is not holomorphic at points z = 2 i and z = − 2 i .

initial point z1 and the terminal point z2 . If we change the

Since both of these points are outside the unit circle z = 1 ,

direction of integration on contour C2 so that we integrate from

We have

(

)

z 2 + 4 = z − 2i z + 2i

we have from the Cauchy-Goursat theorem: !

Proof:

!∫

C

(

z2 to z1 , then the combined contour C = C1 ∪ − C2

!

PATH INDEPENDENCE

Proposition 5.4-4, Independent of Path:

!∫

()

()

f z dz =

∫

()

f z dz +

C1

∫

()

f z dz = 0 !

− C2

(5.4-33)

Changing the direction of integration on contour C2 : !

If a complex function f z is continuous in a simply connected domain D , and if

becomes a

closed contour. We then have using equation (5.4-31):

1 dz = 0 z2 + 4

C

5.4.3!

)

⎡ f z dz + ⎢ − C1 ⎢⎣

∫

()

⎤ f z dz ⎥ = 0 ! C2 ⎥⎦

∫

()

(5.4-34)

or 260

!

∫

C1

()

f z dz =

∫

C2

()

f z dz !

(5.4-35)

Therefore the theorem is proved for any two paths C1 and C2 that intersect only at their endpoints.

example if contours C1 and C2 in Figure 5.4-4 extend to point

z3 , then the above procedure can be applied to the paths between points z2 and z3 , and the result combined with the result for points z1 to z2 .

■

Proposition 5.4-5, Antiderivative Theorem:

()

If a complex function f z is continuous in a simply connected domain D , and if

!∫ f ( z ) dz = 0 !

!

(5.4-36)

C

()

for every simple closed contour C within D , then f z has an

()

antiderivative F z which is holomorphic in D so that:

( )= f

dF z

!

dz

(z) !

(5.4-37)

Proof:

()

Since it is given that f z

!

is continuous in a simply

connected domain D , we know from Proposition 5.4-4 that Figure 5.4-4! Two integration paths in a domain D . !

If two contours intersect at a finite number of points in

addition to the endpoints, the above proof can be applied between all consecutive pairs of intersecting points. For

()

f z will be independent of the path between any two points z0 and z1 in D , where the path lies within D . We then can write: !

(

)

G z0 , z1 =

∫

z1 z0

()

f s ds !

(5.4-38) 261

(

)

where G z0 , z1 is some function that depends only upon the endpoints z0 and z1 . We will now show that: !

(

)

( )

( )

G z0 , z1 = F z1 − F z0 !

( )

(5.4-39)

( )

()

where F z0 and F z1 are antiderivatives of f z . !

We will begin by fixing the point z0 and letting the point

z1 = z be any point in D . The integral in equation (5.4-38) then

(

)

()

can be written as a function G z0 , z1 = F z of only z : !

()

F z =

∫

z

z0

()

f s ds !

(5.4-40)

!

∫

()

f s ds +

z0

∫

z

()

f s ds +

z +Δ z

∫

z0

()

f s ds = 0 !

z

(5.4-41)

where the path consists of the segments from z0 to the point segment from z to z0 , forming a closed contour. The line

! lim

neighborhood of z . We then have: !

∫

z0

()

f s ds −

∫

z0

()

f s ds =

Using equation (5.4-40) we obtain:

z +Δ z

z

()

f s ds

z

!

(5.4-43)

(

( ) = dF ( z ) = lim

)

Δz

1 Δz →0 Δ z

dz

∫

z +Δ z

()

f s ds ! (5.4-44)

z

()

We can now add and subtract f z from equation (5.4-44):

!

( )−f

dF z

!

dz

!

( )−f

dF z dz

1 z = lim Δ z → 0 Δz

()

∫

z+Δ z z

()

()

⎡⎣ f s − f z ⎤⎦ ds !

(5.4-45)

()

f s ds !

(5.4-42)

(z)

1 = lim Δz →0 Δ z

(z)

1 ≤ lim Δz →0 Δ z

∫

z +Δ z z

()

()

()

()

⎡⎣ f s − f z ⎤⎦ ds ! (5.4-46)

or !

( )−f

dF z

!

∫

∫

F z + Δz − F z

Δz →0

segment Δz is arbitrarily small so that the point z + Δ z is in a z

()

Dividing by Δz ≠ 0 and taking the limit as Δ z → 0 , we have:

z + Δ z and then from z + Δ z to z combined with a return path

z +Δ z

)

and so:

From equation (5.4-36) we can write: z +Δ z

(

F z + Δz − F z =

!

z +Δ z

dz

∫

z +Δ z z

f s − f z ds !

(5.4-47)

()

Since f z is holomorphic, it is continuous. For any real

number ε > 0 there must then exist a real number δ > 0 such that: !

()

()

f s − f z 0 can be an arbitrarily small number, we have: !

Proof:

()

(z) !

(5.4-55)

()

where f z and F z are holomorphic in a simply connected domain D , then from Proposition 5.3-6 we must have:

!∫

!

C

()

f z dz = 0 !

(5.4-56)

■

for every simple closed contour C within D . 263

5.5!

Proposition 5.4-7, Independent of Path: If

()

f z

is a holomorphic complex function in a simply !

connected domain D , then: (5.4-57)

exists within the domain (see Figure 5.5-1). By circling any hole in a domain with a simple closed contour, the Cauchy-Goursat

is independent of path within D . Proof:

The Cauchy-Goursat theorem also applies to multiply

connected domains. A domain is multiply connected if any hole

∫ f ( z ) dz !

!

CAUCHY-GOURSAT THEOREM FOR MULTIPLY CONNECTED DOMAINS

theorem can be applied.

()

Since f z is holomorphic in a simply connected domain

!

D , then from the Cauchy-Goursat theorem (Proposition 5.4-2) we must have:

!∫

!

C

()

f z dz = 0 !

(5.4-58)

for every simple closed contour C within D . Moreover, since

()

()

f z is holomorphic in D , f z must be continuous in D (see Proposition 3.7-1). Therefore from Proposition 5.4-4 we know that: !

∫ f ( z ) dz !

is independent of path within D .

(5.4-59) ■

Figure 5.5-1! Multiply connected domain D with inner boundary encircled by contour C2 . Arrows show direction of integration used to prove the deformation theorem. 264

and only an infinitesimal distance apart. The location of the

()

Proposition 5.5-1, Deformation Theorem:

crosscut C3 can be anywhere as long as the function f z is

Let C1 and C2 be two simple closed contours within a domain

continuous across the cut. The integrals in opposite directions

D , and let C2 be interior to C1 . If f z is holomorphic at all

along C3 in equation (5.5-2) will then cancel.

()

points within or on C1 , and outside or on C2 , then:

!∫

!

C1

()

f z dz =

!∫

()

f z dz !

C2

(5.5-1)

Proof: Examples of the simple closed contours C1 and C2 are

!

illustrated in Figure 5.5-1. By introducing a crosscut C3 between

C1 and C2 , the multiply connected domain D is converted into a simply connected domain between C1 and C2 as shown in Figure 5.5-1. We can then use the Cauchy-Goursat theorem to write: !

!∫

()

f z dz +

C1

∫

()

f z dz +

C3 +

!∫

C2

()

f z dz −

∫

()

f z dz = 0 ! (5.5-2)

C3 −

Figure 5.5-2! Arrows show direction of integration. Reversing the direction of integration on contour C2 using

where the direction of integration is shown by arrows in Figure

!

5.5-1. The direction of integration around the closed contours

Proposition 5.3-1, we can write equation (5.5-2) as:

C1 and C2 is chosen so that the interior of the annulus between C1 and C2 always lies to the left as the contours are traversed. !

Along the crosscut C3 integration is performed in opposite

directions on parallel paths C3 + and C3 − that are of equal length

!

!∫

C1

()

f z dz −

!∫

C2

()

f z dz = 0 !

(5.5-3)

or 265

!

!∫

()

f z dz =

C1

!∫

C2

()

f z dz !

(5.5-4)

!∫

!

C

()

f z dz =

n

∑ !∫ k =1

Ck

()

f z dz = 0 !

(5.5-5)

where the direction of integration along contours C1 and C2 is now the same (see Figure 5.5-2 as compared to Figure 5.5-1). !

■

From equation (5.5-4) we see that integration of a complex

()

function f z along some simple closed contour C1 within a domain D can be equivalently replaced by integration along any other simple closed contour C2 within D as long as the

Proof: !

Follows using crosscuts to convert a multiply connected

domain D into a simply connected domain as in the proof for Proposition 5.5-1 (see Figures 5.5-3 and 5.5-4). The direction of integration is positive for all the integrals in equation (5.5-5).

■

()

function f z is holomorphic in the closed region between the two contours. This replacement process is known as the principle of deformation of contour or the principle of deformation of path. The contours C1 and C2 must be such that C2 can be continuously deformed into C1 without intersecting C1 . The contour C2 can then be contracted or expanded. Proposition 5.5-2, Cauchy-Goursat Theorem for Multiply !

Connected Domains: Let C be a simple closed contour within a domain D , and let

Ck where k = 1, 2, !, n be disjoint simple closed contours

()

interior to the contour C . If f z is holomorphic at all points

Figure 5.5-3!

Multiply connected domain D .

within or on C , and outside or on each Ck , then: 266

!∫

!

C

Proof: !

1 dz = 2 π i ! z − z0

(

(5.5-6)

)

Since 1 z − z0 is holomorphic over the complete complex

plane except at point z = z0 , we can use the principle of deformation of contours to change the contour. Let C1 be interior to the contour C , and let C1 be a circle of small radius r centered at point z0 :

z − z0 = r !

!

(5.5-7)

From Proposition 5.5-1 we then have: Figure 5.5-4!

!

Multiply connected domain D with integration directions reversed from those in Figure 5.5-3.

()

We see that if a complex function f z

is holomorphic

()

!∫

!

C

dz = z − z0

z − z0 = r ei θ !

!

the positive direction along all closed contours within D will

We then have:

have a value of zero. !

!∫

C

Proposition 5.5-3:

a point inside the contour C , then:

C1

(5.5-8)

For the contour C1 we can write:

within a multiply connected domain D , the integral of f z in

If C is a simple closed contour within a domain D , and if z0 is

!∫

dz ! z − z0

dz = z − z0

∫

2π

0

(see Example 5.3-2). !

dz = r i eiθ dθ !

1 −i θ e r i ei θ dθ = r

∫

2π

i dθ = 2 π i !

(5.5-9)

(5.5-10)

0

■

We can generalize Proposition 5.5-3 as follows: 267

C

If C is a simple closed contour within a domain D , and if z = z0 is a point inside the contour C , then:

!∫

!

C

1

( z − z0 )

!∫

!

Proposition 5.5-4:

n

⎧⎪ 2 π i n = 1 dz = ⎨ ! 0 n ≠ 1 ⎩⎪

where n is an integer.

!

!∫

C

!

( z − z0 )

n

dz = 2 π i !

n = 1!

We next examine the case for n ≠ 1 . Since 1

(5.5-12) !

!∫ ( z − z ) ∫ dz

( z − z0 )n

is

contours (Proposition 5.5-1) to change the contour. Let C1 be a contour interior to the contour C , and let C1 be a circle of small radius r centered at point z0 :

We then have:

)

n

!

(5.5-14)

dz = r i eiθ dθ !

(5.5-15)

n

=

2π

0

0

1 −i n θ i iθ e r i e d θ = rn r n−1

∫

2π

1− n i θ e( ) dθ !

0

(5.5-16)

!∫

C

point z = z0 , we can use the principle of deformation of

z − z0 = r !

(

z − z0

and so:

holomorphic over the complete complex plane except at the

!

C1

! !

For n = 1 we have from Proposition 5.5-3:

1

!∫

z − z0 = r ei θ !

C

!

n

dz

We then have: !

Proof:

)

z − z0

=

For the contour C1 we can write: !

(5.5-11)

(

dz

dz

( z − z0 )

n

=

1−n iθ i e( )

(

)

r n−1 1− n i

θ = 2π

= θ=0

e ( (1− n) 1

r

n−1

i 2 (1−n ) π

!!

)

−1 ! (5.5-17)

or !

!∫

C

dz

( z − z0 )

n

=

e ( (1− n) 1

r

n−1

i 2π

)

e−i 2 n π − 1 !

(5.5-18)

From equation (1.13-12) we have: (5.5-13)

!

e i 2 n π = 1!

n = 0, ± 1, ± 2, !!

(5.5-19)

and so:

268

!∫

!

C

(

1 z − z0

)

n

dz = 0 !

From Proposition 5.5-3 we have:

n ≠ 1!

(5.5-20) !

C

■

!

!∫

The value of the integral in equation (5.5-14) does not

1 2π i 2π i dz = − =0 4i 4i z2 + 4

Also see Example 5.4-4.

depend on the radius of the circular contour C1 (see Example 5.3-2). Example 5.5-1 Evaluate

!∫

C

1 dz where C is the circle z = 4 . 2 z +4

Solution: Using partial fractions we have: !

1 1 1 1 1 1 = = − 4i z − 2i 4i z + 2i z − 2i z + 2i z2 + 4

(

)(

)

The points z = 2 i and z = − 2 i lie inside the circle z = 4 . Therefore the integrand is not holomorphic at these points. Enclosing the point z = 2 i with contour C1 ; z − 2 i = 1 and the point z = − 2 i with contour C2 : z + 2 i = 1 , (see Figure 5.5-5), we can use Proposition 5.5-2 to write: !

!∫

C

1 1 dz = 4i z2 + 4

!∫

1 1 dz − 4i C1 z − 2 i

!∫

C2

1 dz z+2i

Unit circle contours around points z = ± 2 i .

Figure 5.5-5!

Example 5.5-2 Evaluate

!∫

C

4z + 2 dz where C is the circle z − 4 = 2 . z2 − 4 z + 3 269

(

as we will now show. The integer n C, z0

Using partial fraction expansion we have:

winding number of C around z0 or the index of C around z0 .

!

!∫

C

!∫

C

1 dz + 7 z −1

!∫

C

!

!

is not holomorphic at the point z = 3 . From the CauchyGoursat theorem (Proposition 5.4-2) and from Proposition

C

5.6! !

)

!

If C is a closed contour within a domain D , and if z = z0 is

(

the contour C completely circles z0 is given by:

(

)

!∫

C

1 2π i

2π

∫

0

r i ei θ 1 d θ = 2π r ei θ

∫

2π

dθ = 1!

(5.6-4)

0

1 2π i

∫

2 nπ

0

r i ei θ 1 d θ = 2π r ei θ

∫

2 nπ

0

(

)

dθ = n C, z0 !

(5.6-5)

as the contour C circles the point z0 a total of n C, z0

a point inside the contour C , then the number of times n C, z0

!

(5.6-3)

(

WINDING NUMBER

1 dz ! z − z0

dz = r i eiθ dθ !

(5.6-4) becomes:

4z + 2 dz = − 3 0 + 7 2 π i = 14 π i z2 − 4 z + 3

1 n C, z0 = 2π i

z − z0 = r ei θ !

If we have 0 ≤ θ ≤ 2 n π where n is an integer, then equation

5.5-3 we have:

() (

(5.6-2)

When the contour C circles around z0 once, we can write:

z = 3 lies inside the circle z − 4 = 2 . Therefore the integrand

!∫

1 dz = 2 π i ! z − z0

Letting:

1 dz z−3

The point z = 1 lies outside the circle z − 4 = 2 , and the point

!

!∫

C

and so from Proposition 5.5-2 we have:

4z + 2 dz = − 3 z2 − 4 z + 3

is known as the

From Proposition 5.5-3 we have:

4z + 2 −3 7 = + z2 − 4 z + 3 z − 1 z − 3

!

!

)

Solution:

)

)

complete times. This provides the rationale for the name

(

)

winding number. The sign of n C, z0 indicates the direction of integration around the contour (positive for counterclockwise). Examples of winding numbers are shown in Figure 5.6-1. For a

(5.6-1)

discussion of the winding number around the origin, see Proposition 9.4-2. 270

Figure 5.6-1!

(

)

Winding numbers n C, z0 for contours around a point z = z0 .The contour with n = 2 is one continuous curve.

271

Chapter 6 Cauchy’s Integral Formula

1 f z0 = 2π i

( )

!∫

C

()

f z

z − z0

dz

272

!

In this chapter we will present a number of theorems that

determine the value of a holomorphic function at any given

1 f z0 = 2π i

( )

!

point inside a closed contour using only the values of the

!∫

C

( ) dz !

f z

z − z0

(6.1-1)

function on the contour. Included in these discussions will be

Proof:

the important integral formulas of Cauchy. We will also show

!

that the real and imaginary parts of a holomorphic function

connected domain D . Let C be any simple closed contour

satisfy the two-dimensional Laplace's equation.

within D . Also let C1 be a circle of small radius r centered at

6.1! !

CAUCHY’S INTEGRAL FORMULA FOR HOLOMORPHIC FUNCTIONS

()

We are given that f z

is holomorphic in a simply

point z = z0 such that C1 is interior to C .(see Figure 6.1-1).

Cauchy’s integral formula states that if a complex

()

function f z is holomorphic at all points inside and on a

()

simple closed contour C , then the value of f z at any point

z = z0 inside the contour C is completely determined by just

()

the values of f z on this contour. This is a special property of holomorphic functions that provides a method of computing their value at any point within the contour C . There is no equivalent property for real-valued functions of a real variable. Proposition 6.1-1, Cauchy’s Integral Formula:

()

Let f z

be a holomorphic complex function in a simply

connected domain D . If C is a simple closed contour that lies within D , and if z = z0 is a fixed point inside C , then:

Figure 6.1-1! A point z = z0 encircled by closed contours C and C1 within a domain D . 273

Along either of the contours C or C1 , we will have

!

z − z0 ≠ 0 since z will be points on the contour while z = z0 is a

( ) ( z − z0 ) is therefore

point inside the contour. The function f z

1 ! 2π i

!∫

C

( ) dz −

f z

z − z0

( )

f z0

holomorphic on both contours.

or by Proposition 5.3-7:

!

1 ! 2π i

From the deformation theorem (Proposition 5.5-1) we can

write:

!∫

!

C

( ) dz =

f z

z − z0

( ) dz !

f z

!∫

C1

(6.1-2)

z − z0

!

!∫

C

( ) dz − f

f z

z − z0

( z0 )

1 = 2π i

≤

!∫

C1

()

z − z0

()

!∫

C1

( ) dz

f z − f z0

( )

1 f z − f z0 2π z − z0

! (6.1-6)

dz ! (6.1-7)

We will now show that the integral on the right side of

() f ( z ) is

equation (6.1-7) must be zero. The complex function f z is

( )

Adding and subtracting the constant f z0 in the numerator of

holomorphic in the simply connected domain D , and so

the integrand of the integral on the right, we have:

continuous in D (see Proposition 3.7-1). We can let ε equal the

!∫

!

C

( ) dz =

f z

z − z0

!∫

( ) ( ()

( ))

f z0 + f z − f z0 z − z0

C1

()

dz !

(6.1-3)

!∫

C

( ) dz = f

f z

z − z0

( z0 ) !∫

C1

()

( ) dz !

f z − f z0

!∫

dz + z − z0

z − z0

C1

(6.1-4)

ε = max

(6.1-8)

!∫

C

( ) dz = 2 π i f

f z

z − z0

We can now write:

( z0 ) + !∫

C1

!

!∫

C1

()

( )

1 f z − f z0 2π z − z0

dz ≤

ε 2π r

!∫

dz !

(6.1-9)

C1

or

From Proposition 5.5-3 we then have: !

!

on the circle C1 :

where ε is a function of the radius r = z − z0 . We then have:

or !

( ) f ( z ) − f ( z0 ) !

maximum value of f z − f z0

()

( ) dz !

f z − f z0 z − z0

(6.1-5)

!

!∫

()

z − z0

C1

! !

( )

f z − f z0

dz ≤

ε 2π r = ε ! 2π r

(6.1-10)

()

Since f z is continuous in D, we have:

lim ε = 0 !

r→0

(6.1-11) 274

By decreasing the arbitrary radius r of the circular contour C1 , the value of the integral in equation (6.1-10) will change since ε

Example 6.1-1

changes. The value of the integral on the left side of equation

Evaluate the integral

z0 do not change. To avoid an inconsistency, we then must have:

()

!∫

!

( )

f z − f z0 z − z0

C1

dz = 0 !

Solution: Letting

(6.1-12)

!∫

C

( ) dz = 2 π i f ( z ) !

z − z0

0

or

( )

!

()

is entire, and so is

(6.1-13) with the point z0 = 2 inside the contour C , we have

f z

1 f z0 = 2π i

()

f z = z , we see that f z

holomorphic on and inside the contour C . Using equation

( )

!∫

C

()

f z

z − z0

dz !

() f ( z) dz = 2 π i f

f z0 = f 2 = 2 :

We can now rewrite equation (6.1-7) as: !

C

z = 4.

(6.1-7) is constant, however, since the contour C and the point

!∫

z dz where C is the circle z−2

(6.1-13)

!

!∫

C

z−2

( z0 ) = 2 π i f ( 2) = 2 π i ( 2) = 4 π i

Example 6.1-2 (6.1-14)

Evaluate the integral

C

z = 4.

where z are points on the contour C and z0 is a point inside

!∫

ez dz where C is the circle z−3

the contour C . Equation (6.1-14) is known as Cauchy’s integral

Solution:

formula.

The function f z = e z is entire, and so is holomorphic on

!

()

■

and inside the contour C . Using equation (6.1-13) with the

Cauchy’s integral formula whereby a holomorphic

point z0 = 3 inside the contour C , we have:

()

function f z can be determined for all points within a region using only information obtained from a contour enclosing the region is the result of the smoothness of holomorphic functions.

!

!∫

C

ez dz = 2 π i f 3 = 2 π i e3 z−3

()

275

Example 6.1-5

Example 6.1-3 Evaluate the integral

!∫

C

z = 2.

ez dz where C is the circle z−3

Evaluate the integral

Solution:

()

This integral can be rewritten as:

The function f z = e z is entire, and so is holomorphic on and inside the contour C . The point z0 = 3 lies outside the contour C . Therefore we can use the Cauchy-Goursat

!

C

ez dz = 0 z−3

z − i = 1.

z2 dz = 3z + i

!∫

C

z2 3 i z+ 3

dz

()

The function f z = z 2 3 is entire, and so is holomorphic on and inside the contour C . Using equation (6.1-13) with the point z0 = − i 3 inside the contour C , we have:

Example 6.1-4 Evaluate the integral

!∫

C

theorem (Proposition 5.4-2) to obtain:

!∫

!∫

C

z = 2.

Solution:

!

z2 dz where C is the circle 3z + i

!∫

C

z+3 dz where C is the circle z−3

!

!∫

C

z2 dz = 3z + i

!∫

C

z2 3

⎛ i⎞ ⎧ −1 ⎫ 2π i dz = 2 π i f ⎜ − ⎟ = 2 π i ⎨ 3 ⎬ = − i 27 ⎝ 3⎠ ⎩3 ⎭ z+ 3

Solution: The point z0 = 3 lies outside the contour C . Therefore we can use the Cauchy-Goursat theorem (Proposition 5.4-2) to obtain: !

!∫

C

z+3 dz = 0 z−3

Example 6.1-6 Evaluate the integral

z − 3 = 3.

!∫

C

e z − 2sin z dz where C is the circle z −π

Solution: 276

The function

!

()

f z = e z − 2 sin z

is entire, and so is

holomorphic on and inside the contour C . Using equation

Example 6.1-8

(6.1-13) with the point z0 = π inside the contour C , we have:

Evaluate the integral

!∫

C

e z − 2 sin z dz = 2 π i f π = 2 π i eπ z −π

!∫

C

z − 1 = 5.

(

ez

)

z z−3

() (

z0 = 4 is not. The function f z = cos π z

dz where C is the circle

and inside the contour C , and so is holomorphic on and

!

()

The function f z = e z is entire, and so is holomorphic on

to change the integral into two integrals:

(

)

!∫

C

ez 1 dz + z 3

!∫

C

ez dz z −3

!

!∫

C

(

)

2 π i e0 2 π i e3 2 π i 3 dz = − + = e −1 3 3 3 z z−3

(

)

)(

)

!∫

(

cos π z z − 4 z−2

C

!

) dz

!∫

C

(

cos π z cos 2π dz = 2 π i = −π i z −2 z −4 2 −4

)(

)

(

Example 6.1-9

We then have:

ez

(

cos π z dz = z −2 z −4

We then have:

inside the contour C . Therefore we can use Cauchy’s integral formula. First we will use partial fraction expansion

!∫

C

and inside the contour C . The points z0 = 0 and z0 = 3 are

ez 1 dz = − 3 z z −3

) ( z − 4) is entire on

inside C . Therefore we can use Cauchy’s integral formula:

Solution:

C

)

The point z0 = 2 is inside the contour C , while the point

Evaluate the integral

!∫

)(

Solution:

Example 6.1-7

!

(

C

circle z = 3 .

( )

!∫

cos π z dz where C is the z −2 z −4

Evaluate the integral 1 circle z = . 2

!∫

C

)

ez dz where C is the 2 3z − 4 z + 1

Solution: Factoring the denominator of the integrand, we get: 277

!∫

C

ez dz = 3z 2 − 4 z + 1

!∫ ( C

ez

1 dz = 3 3z − 1 z − 1

)(

)

!∫

C

ez dz ⎛ 1⎞ ⎜⎝ z − 3 ⎟⎠ z − 1

(

)

The point z0 = 1 3 is inside the contour C while the point

()

z0 = 1 is not. The function f z = e z

( z − 1)

is entire on and

inside the contour C , and so is holomorphic on and inside the C . Therefore we can use Cauchy’s integral formula: !

( z − 1) dz = 2 π i f ⎛ 1 ⎞

!∫

ez 1 dz = 3 3z 2 − 4 z + 1

!∫

ez 2 e1 3 dz = π i = − π i e1 3 2 3 1 3z − 4 z + 1 −1 3

C

!

e

z

C

!∫

C

z −1 3

3

⎜⎝ 3 ⎟⎠

()

(

)

f z = f z0 + r ei θ !

!

(6.1-17)

We can use Cauchy’s integral formula:

1 f z0 = 2π i

( )

!

( )

f z0 =

!

centered at point z = z0 and

contained within D , then: 2π

∫ ( 0

)

f z0 + r ei θ dθ !

(6.1-15)

Proof: !

so that:

( ) dz !

f z

!∫

(6.1-18)

z − z0

C

1 2π i

∫

2π

(

f z0 + r ei θ re

0

iθ

) ir e

iθ

dθ !

(6.1-19)

or

() if Dr ( z0 ) is a disk of radius r ( )

(6.1-16)

of constant radius r centered at z0 . We then have:

If f z is a holomorphic complex function on a domain D , and

!

dz = i r ei θ dθ !

where we will integrate around the contour C that is the circle

Proposition 6.1-2:

1 f z0 = 2π

z − z0 = r ei θ !

!

For points z on the boundary of a disk centered at point

z = z0 and having radius r we have:

1 f z0 = 2π

( )

!

2π

∫ ( 0

)

f z0 + r ei θ dθ !

(6.1-20)

■

Proposition 6.1-3:

() ()

() a domain D , and if Dr ( z0 ) is a disk of radius r

If w = f z = u z + i v z is a holomorphic complex function on centered at

point z = z0 and contained within D , then: 278

1 u z0 = 2π

( )

!

( )

v z0 =

!

1 2π

2π

∫ u ( z + r e ) dθ ! iθ

0

2π

∫ ( 0

(6.1-21)

0

)

v z0 + r ei θ dθ !

within D , and if z = z0 is a fixed point that lies on the contour

(6.1-22)

C , then:

!∫

!

■

C

Proposition 6.1-4:

()

be a holomorphic complex function in a simply

connected domain D . If C is a simple closed contour that lies

Follows from Proposition 6.1-2.

Let f z

()

Let f z

Proof: !

Proposition 6.1-5:

()

f z

z − z0

( )

dz = π i f z0 !

(6.1-24)

Proof:

be a holomorphic complex function in a simply

!

Let C be any simple closed contour that lies within D .

connected domain D . If C is a simple closed contour that lies

Also let C1 be a circle of small radius r centered at point z0 (see

within D , and if z = z0 is a fixed point inside the contour C ,

Figure 6.1-2). The circle C1 intersects the contour C at points z1

and if n loops are made around the contour C , then:

(

n C, z0

!

(

1 f z0 = 2π i

) ( )

!∫

C

()

f z

z − z0

and z2 as shown in Figure 6.1-2. The circle C1 is given by: !

dz !

(6.1-23)

z − z0 = r ei θ !

dz = i r ei θ dθ !

(6.1-25)

so that on this circle:

)

where n C, z0 is the winding number of C around z0 .

!

()

(

)

f z = f z0 + r ei θ !

(6.1-26)

We will now define a new contour C2 consisting of

Proof:

!

!

contour C from points z1 to z2 , and the lower half of the

Follows from the definition of a winding number given in

equation (5.6-1) and from Proposition 6.1.1.

■

circular contour C1 from points z2 to z1 ( C2 is the red contour in Figure 6.1-2). Note that the point z0 is not within or on the contour C2 . 279

( ) dz =

f z

!∫

!

C2

!

∫

z − z0

z2

( ) dz +

f z

z − z0

z1

−π

∫ ( 0

)

f z0 + r ei θ i dθ = 0

!

(6.1-28)

As r → 0 we will have:

∫

!

( ) dz →

f z

z2 z1

z − z0

( ) dz !

f z

!∫

(6.1-29)

z − z0

C

and −π

∫ (

!

0

Therefore: Figure 6.1-2! A point z0 on a closed contour C and within a contour C1 . Contours C and C1 are both within the domain D .

!∫

!

C

)

( )

f z0 + r ei θ i dθ →− π i f z0 !

()

f z

z − z0

(6.1-30)

( )

dz = π i f z0 !

(6.1-31)

■

Proposition 6.1-6: !

From the Cauchy-Goursat theorem (Proposition 5.4-2) we

()

Let f z

can write: !

!∫

C2

! or

!

()

f z

z − z0

dz =

∫

z2 z1

( ) dz +

f z

z − z0

∫

−π

0

(

f z0 + r e r ei θ

iθ

) ir e

be a holomorphic complex function in a simply

connected domain D . If C1 and C2 are two concentric circles iθ

within D , and if z = z0 is a fixed point between the two circles,

dθ = 0

(6.1-27)

then: !

( )

f z0 =

1 2π i

!∫

C1

()

f z

z − z0

dz −

1 2π i

!∫

C2

()

f z

z − z0

dz !

(6.1-32) 280

We then have:

Proof: !

The two concentric circles C1 and C2 together with a

crosscut C3 connecting the two circles are shown in Figure

1 f z0 = 2π i

( )

!

6.1-3. Let the contour C be defined as C = C1 ∪ C2 ∪ C3 .

!∫

C1

1 − 2π i

!

()

f z

1 dz + z − z0 2π i

!∫

!∫

()

f z

()

f z

C3

1 dz − 2π i C 2 z − z0

z − z0

!∫

C3

dz

()

f z

z − z0

dz !

(6.1-34)

The two integrals along the contour C3 cancel and so equation (6.1-34) becomes:

1 f z0 = 2π i

( )

!

!∫

C1

()

f z

1 dz − z − z0 2π i

!∫

C2

()

f z

z − z0

dz !

(6.1-35)

■

Proposition 6.1-7: Let

()

f z

()

and g z

be two complex functions that are

holomorphic within and on a simple closed contour C . If the

() () f ( z ) equals the value of g ( z ) everywhere within C .

value of f z on C equals the value of g z on C , then the value of Figure 6.1-3! !

Contour C = C1 ∪ C2 ∪ C3 within a domain D .

Using Cauchy’s integral formula (Proposition 6.1-1) we

can write: !

1 f z0 = 2π i

( )

!∫

C

()

f z

z − z0

Proof: !

Follows from the Cauchy integral formula given in

Proposition 6.1-1.

dz !

■

(6.1-33) 281

domain D , and if z0 is any given point inside the contour C ,

Proposition 6.1-8:

()

then f z has derivatives of all orders at the point z0 given by:

()

!If a complex function f z is holomorphic on a domain D , and if C is a simple closed contour C that lies within D , then:

!∫

!

C

⎧ 0 ⎪ f z ⎪ dz = ⎨ 2 π i f z0 z − z0 ⎪ ⎪⎩ π i f z0

()

( ) ( )

z0

outside C

z0

inside C ! (6.1-36)

z0

on C

where n ∈! .

!

From Cauchy’s integral formula given in equation (6.1-1)

we have:

Follows from Propositions 5.4-2, 6.1-1, and 6.1-5.

■

1 f z0 = 2π i

( )

!

6.2! !

(6.2-1)

Proof:

Proof: !

! !

( ) dz ! !∫C ( z − z0 )n+1 f z

n! n f ( ) ( z0 ) = 2π i

CAUCHY’S INTEGRAL FORMULA FOR DERIVATIVES Holomorphic functions possess derivatives of all orders.

They differ in this regard from real-valued functions that often have derivatives only up to a certain order. If a complex

()

function f z

()

has a derivative at a point z0 , then f z has

derivatives of all orders at z0 , and these derivatives can be expressed in integral form using an integral formula of Cauchy. Proposition 6.2-1, Cauchy’s Integral Formula for Derivatives:

()

Let f z

be a holomorphic complex function in a simply

connected domain D . If C is a simple closed contour within the

!∫

C

()

f z

z − z0

dz !

(6.2-2)

( ) dz !

(6.2-3)

and so:

1 d f ′ z0 = 2 π i dz0

( )

! !

f z

!∫

C

z − z0

()

Since f z is holomorphic on the closed contour C , we

can exchange the order of the differentiation and integration operations in equation (6.2-3) as shown in Appendix C. We now differentiate the Cauchy integral formula under the integral sign with respect to z0 and obtain:

1 ! f ′ z0 = 2π i

( )

!∫

C

()

d f z 1 dz = dz0 z − z0 2π i

(

)

( ) dz ! !∫C ( z − z0 )2 f z

(6.2-4) 282

Differentiating again with respect to z0 , we have: !

2 2 f ( ) z0 = 2π i

( )

( ) dz ! !∫C ( z − z0 )3 f z

Example 6.2-1 sin 2 z Evaluate dz where C is the circle z = 2 . 4 z C

!∫

(6.2-5)

Solution:

2 where the bracketed exponent on f ( ) ( z0 ) indicates the order of

()

The function f z = sin 2 z is entire and so is holomorphic on

the derivative. ! !

and inside the contour C . We will use Cauchy’s integral

With repeated differentiation we see that:

n! n f ( ) ( z0 ) = 2π i

( ) dz ! !∫C ( z − z0 )n+1 f z

formula for derivatives as given in equation (6.2-7): (6.2-6)

!

where n ∈! . This is Cauchy’s integral formula for derivatives.

are holomorphic. !

■

For n = 0 equation (6.2-6) becomes Cauchy’s integral

formula. It is possible to determine any order derivative of a function which is holomorphic in a simply connected domain using Cauchy’s integral formula for derivatives. We can rewrite equation (6.2-6) in a form useful for evaluating integrals: !

() !∫C ( z − z0 )n f z

dz =

2π i

( n − 1) !

n−1 f ( ) z0 !

( )

dz =

(

2π i

)

n −1 !

n−1 f ( ) z0

( )

where n = 4 . The point z0 = 0 is inside the contour C :

Derivatives of all orders exist for complex functions at any point in a simply connected domain in which the functions

() !∫C ( z − z0 )n f z

!

!∫

C

sin 2 z 2π i dz = sin′′′ 2 z 3! z4

z=0

=−

2π i 8 8 cos 2 0 = − π i 3! 3

()

Example 6.2-2 Evaluate

!∫

C

sin 2 z

( z − π i)

2

dz where C is the circle z− π i = 4 .

Solution:

()

The function f z = sin 2 z is entire and so is holomorphic on and inside the contour C . We will use Cauchy’s integral

(6.2-7)

formula for derivatives as given in equation (6.2-7) where

n = 2 . The point z0 = π i is inside the contour C . 283

!∫

!

sin 2z

C

( z − π i)

2

dz = 2 π i

d sin 2 z dz

= 4π i cos 2 π i

z =π i

Example 6.2-4 Evaluate

or using equation (4.3-56):

!∫

!

sin 2z

C

( z − π i)

2

!∫

(

cos z

z z −π

C

)

2

dz where C is the circle z = 5 .

Solution:

dz = 4 π i cosh 2 π

()

The function f z = cos z is entire and so is holomorphic on and inside the contour C . First we will use partial fraction

Example 6.2-3 Evaluate

!∫

C

expansion to change the integral into three integrals:

ez

2

( z − i)

3

dz where C is the circle z = 4 .

C

Solution:

()

The function f z = e

z

2

is entire and so is holomorphic on

!∫

C

( z − i)

3

dz =

( )

2 π i z 2 ′′ e 2!

=πi z=i

(

2 d 2 z ez dz

)

z=i

!

!∫

C

( z − i)

3

(

2

2

)

z=i

(

C

(

z −π 1 π

!∫

)

C

dz cos z

( z −π )

2

dz

z0 = π are inside the contour C .

!∫

C

dz = π i 2 e z + 4 z 2 e z

C

!∫

cos z

+

!

e

!∫

cos z 1 dz − 2 z π

formula for derivatives where n = 2 . The points z0 = 0 and

or z2

)

1 π2

integral formula and the third integral using his integral

n = 3. The point z0 = i is inside the contour C . e

(

z z −π

2

dz =

We then evaluate the first two integrals using Cauchy’s

formula for derivatives as given in equation (6.2-7) where

!

cos z

!

and inside the contour C . We will use Cauchy’s integral

z2

!∫

!

)

= π i 2 e−1 − 4 e−1 = −

2π i e

(

cos z

z z −π

)

2

dz =

( ) − 2 π i cos (π ) + 2 π i cos′ z z = π

2 π i cos 0

π2

π

π2

or !

!∫

C

cos z

(

z z −π

)2

dz =

2π i

π2

+

2π i

π2

−

( ) = 4i

2 π i sin π

π

π

284

Proof: Example 6.2-5

!∫

Evaluate

!

ln z

C

( z − 2)

3

From Cauchy’s integral formula (Proposition 6.1-1) we

have:

dz where C is the circle z − 2 = 1 .

Solution:

1 n f ( ) z0 = 2π i

( )

!

()

The function f z = ln z is holomorphic on and inside the

6.2-1) we have

!∫

ln z

( z − 2)

3

!∫

C

2π i dz = ln z ′′ 2!

( )

d ⎛ 1⎞ =πi ⎜ ⎟ dz ⎝ z ⎠ z=2

⎛ 1⎞ = π i⎜− 2⎟ ⎝ z ⎠ z=2

ln z

( z − 2)3

dz = −

z=2

πi 4

C

6.3!

( z ) dz = n!

z − z0

where n ∈! .

(6.2-10)

( ) dz = n!

n f() z

z − z0

( ) dz ! !∫C ( z − z0 )n+1 f z

(6.2-11)

■

simple closed contour C , then we have:

!∫

!∫

C

()

f

( ) dz ! !∫C ( z − z0 )n+1 f z

Therefore: !

If a complex function f z is holomorphic inside and on a

( n)

n! n f ( ) ( z0 ) = 2π i

!

Proposition 6.2-2:

!

(6.2-9)

derivatives as given in equation (6.2-7) where n = 3 . The

or

!

z − z0

C

From Cauchy’s integral formula for derivatives (Proposition

C

!

!∫

contour C . We will use Cauchy’s integral formula for point z0 = 2 is inside the contour C . !

( ) dz !

f ( n) z

( ) dz ! !∫C ( z − z0 )n+1 f z

HIGHER ORDER DERIVATIVES OF A HOLOMORPHIC FUNCTION

Proposition 6.3-1, Higher Order Derivatives of a Holomorphic Function Exist and are Holomorphic:

(6.2-8)

() ()

If a complex function f z is holomorphic at a point z = z0 , then its derivatives f ( n) z of all orders n exist and are holomorphic at the point z0 . 285

Proof: !

n z0 , then its derivatives f ( ) z

()

()

If f z is holomorphic at a point z = z0 , then there must be

()

a neighborhood of z0 in which f z is holomorphic and where

()

f ′ z exists. Let Q be the set of all points z ∈! such that: z − z0 < δ !

!

holomorphic at point z0 . !

(6.3-1)

()

From Proposition 6.3.1 we can conclude that if a function

orders will exist and be holomorphic in D . Proposition 6.3-2:

holomorphic at all the points z satisfying equation (6.3-1).

() ( )

a point z = z0 in a domain D , then its real and imaginary parts

disk centered at z0 having a radius less than δ . Now let C be a

( )

including the point z0 . Therefore from Cauchy’s integral

Proof:

theorem for derivatives (Proposition 6.2-1), we can write:

!

( )

( )

2 2π i

!∫

C

()

f z

( z − z0 )

and so f ′′ z0 exists inside C . !

( )

3

dz !

(6.3-2)

()

and so holomorphic at point z0 . We can similarly show that

( )

f ′′′ z0 exists and so f ′′ z0 is holomorphic at point z0 . !

This logic can be continued to higher order derivatives to

()

If the complex function:

() ( )

( )

f z = u x, y + i v x, y !

!

(6.3-3)

is holomorphic at point z = z0 , then from Proposition 6.3-1 we

Since f ′′ z0 exists inside C , f ′ z must be differentiable

( )

have continuous partial derivatives of all

orders at the point z0 .

will then be holomorphic on and inside C

f ′′ z0 =

( )

u x, y and v x, y

circular contour with radius δ 2 and center at point z0 . The

!

( )

If f z = u x, y + i v x, y is a holomorphic complex function at

These points lie inside a circle of radius r = δ , and so form a

()

■

f z is holomorphic in a domain D , then its derivatives of all

where δ is any given positive real number such that f z is

function f z

()

of all orders n exist and are

show that if f z is a complex function holomorphic at a point

()

know that f ′′ z

()

exists and is holomorphic at point z0 .

Therefore f ′ z must be continuous at point z0 . The first order

()

partial derivatives of f z

must then also be continuous at

point z0 . !

Similar arguments can be made for all higher order

()

()

derivatives of f z . Therefore if f z is holomorphic at a point 286

()

z0 , the real and imaginary parts of f z

partial derivatives of all orders at the point z0 .

have continuous ■

() F′(z) = f (z) !

antiderivative F z which is holomorphic in D so that:

() F ( z ) holomorphic in a simply connected domain D such that: dF ( z ) f (z) = F′(z) = ! (6.3-4) dz

If f z is a continuous complex function with an antiderivative

()

!

()

must also be holomorphic in D . Therefore if:

!∫

!

C

()

f z dz = 0 !

(6.3-7)

()

for every simple closed contour C within D , f z will be then

Proof:

()

Since F z

is holomorphic in the domain D , we know

()

from Proposition 6.3-1 that F z

()

()

then has derivatives of all

()

orders. Therefore F ′′ z = f ′ z must exist in D , and so f z is also holomorphic D .

()

■

Morera’s theorem can be considered a converse of the

Cauchy-Goursat theorem.

()

f z dz = 0 !

!

CAUCHY’S INEQUALITY

connected domain containing a point z = z0 , then all derivatives of f z

(6.3-5)

()

If a holomorphic function f z is bounded in a simply

()

domain D , and if:

C

!

6.4!

If f z is a continuous complex function in a simply connected

!∫

holomorphic in D .

■

Proposition 6.3-4, Morera’s Theorem:

!

(6.3-6)

for all points z in D . From Proposition 6.3-3 we see that f z

then f z is holomorphic in the domain D :

!

()

From Proposition 5.4-5 we know that f z will have an

!

Proposition 6.3-3:

!

Proof:

are also bounded at the point z0 , as specified by a

relation known as Cauchy’s inequality.

()

for every simple closed contour C within D , then f z is holomorphic in D . 287

()

Let f z

be a holomorphic complex function on a simply

connected domain D , and let C be a circular contour within D of radius r = z − z0

centered at z = z0 . If for some M > 0 ,

()

where M is a function of r , we have f z ≤ M for all points on C , then:

( )

!

(6.4-1)

where n ∈! .

(6.4-4)

n! M n f ( ) z0 ≤ n ! r

(6.4-5)

or

( )

!

This inequality is known as Cauchy’s inequality or Cauchy’s estimate.

n! M n f ( ) z0 ≤ n ! r

!

n! M n f ( ) z0 ≤ 2π r ! 2 π r n+1

( )

!

Proposition 6.4-1, Cauchy’s Inequality:

6.5! !

■

LIOUVILLE’S THEOREM Liouville’s theorem states that the only bounded entire

functions are constant functions. If a bounded complex function Proof:

is holomorphic, it must have a singularity somewhere in the

!

complex plane unless it is a constant function. No function can

From Cauchy’s integral formula for derivatives given in

Proposition 6.2-1, we have: !

n! n f ( ) ( z0 ) = 2π i

therefore be holomorphic and finite everywhere, except a

( ) dz ! !∫C ( z − z0 )n+1 f z

(6.4-2)

constant. Proposition 6.5-1, Liouville’s Theorem:

()

For z on C , and with f z ≤ M we have: !

n f ( ) z0 ≤

( )

n! 2π i

!∫

C

M n!M dz = r n+1 2 π r n+1

()

If an entire function f z is bounded, it must be a constant over

!∫

dz !

C

From the ML inequality (Proposition 5.3-8) we obtain:

the complex plane.

(6.4-3)

Proof: !

()

We will suppose that f z

is an entire function that is

bounded. There must then exist a real constant M > 0 such that 288

()

f z ≤ M for all values of z in the complex plane. From Cauchy’s inequality (Proposition 6.4-1) with n = 1 we have for

Proposition 6.5-2:

()

the derivative of f z at any given point z0 :

( )

f ′ z0 ≤

!

A complex function bounded at infinity and not constant must

M ! r

have at least one singularity in the finite complex plane.

(6.5-1)

Proof:

where r = z − z0 is the radius of a circular contour centered at

!

point z = z0 , and where M is independent of r since it applies

Follows from Proposition 6.5-1.

■

Example 6.5-1

to the whole complex plane. We can then let r be arbitrarily

()

large for the same value of M . From equation (6.5-1) we see

Show that if f z is an entire function having a real part

that f ′ z0

can be arbitrarily small, and so we must have:

that is nonpositive, then f z must be a constant function.

( )

(6.5-2)

Solution:

()

()

( )

!

f ′ z0 = 0 !

Therefore f ′ z = 0 . Since z0 is an arbitrary point, and f z is

()

an entire function, we see that we must have f ′ z = 0 for all

()

()

Let f z = u + i v . Since u ≤ 0 we can write: !

f z e ( ) = eu+iv = eu eiv = eu ≤ 1

points in the complex plane. From Proposition 3.7-4 we know

f z Therefore we see that e ( ) is bounded. From Liouville’s

that since f z is holomorphic with f ′ z = 0 , then f z must

theorem (Proposition 6.5-1) we know that f z must then be

be a constant function over the complex plane.

a constant function over the entire complex plane.

()

!

()

()

■

All entire complex functions that are not constant, such as

sin z and cos z , are unbounded (see Section 4.3.5). This is in sharp contrast to differentiable real functions that are bounded, but are not constant, such as sin x and cos x .

()

Example 6.5-2

()

()

Show that if f z is an entire function such that f z → ∞

()

as z → ∞ , then f z

must have at least one zero in the

complex plane. 289

Solution:

!

()

!

( )

For points z of a disk Dr z0 centered at point z = z0 and

Seeking a contradiction, we will assume f z ≠ 0 in the

having boundary radius r we will let:

complex plane. Then the function:

!

()

g z =

1

()

dz = i r eiθ dθ !

(6.6-1)

so that:

f z

()

is also an entire function, and we have g z → 0 as z → ∞ .

()

We see that g z is bounded over the complex plane. From Liouville’s theorem (Proposition 6.5-1) we know that

()

z − z0 = r ei θ !

g z must then be a constant function over the complex

()

!

()

(

)

f z = f z0 + r eiθ !

(6.6-2)

We can use Cauchy’s integral formula: !

1 f z0 = 2π i

( )

( ) dz !

f z

!∫

(6.6-3)

z − z0

C

plane. Therefore f z must also be a constant function over

where we will integrate around the boundary contour C of the

the complex plane. This is a contradiction to our definition

disk Dr z0 . We then have:

()

()

of f z , however, and so f z must have at least one zero in the complex plane.

6.6!

GAUSS’ MEAN-VALUE THEOREM

Proposition 6.6-1, Gauss’s Mean Value Theorem:

()

( ) is equal to f ( z ) around the boundary of any disk Dr ( z0 )

If f z is holomorphic on a domain D , then f z0 the mean value of

of radius r centered at point z = z0 and contained within D . Proof:

( )

!

1 f z0 = 2π i

( )

∫

2π

(

f z0 + r ei θ re

0

iθ

) ir e

iθ

dθ !

(6.6-4)

or !

1 f z0 = 2π

( )

2π

∫ ( 0

)

f z0 + r eiθ dθ !

(6.6-5)

and so: !

1 f z0 = 2π

( )

∫

2π

0

()

f z dθ !

(6.6-6)

290

() f (z)

Therefore the mean value of a holomorphic function f z on any circle C within a domain is equal to the value of

()

!

at

the center of the circle. The function f z is then said to have the mean value property.

1 2π

■

Equating the real parts of equation (6.6-6), we have:

1 u x0 , y0 = 2π

(

!

)

∫

2π

0

( )

u x, y dθ !

(

!

)

∫

2π

!

0

( )

v x, y dθ !

!

Example 6.6-1

defined by

!

!

Solution: iθ

The contour C is defined by z = 2 i + e . The mean value of

() 1 2π

!

∫ (

or

0

f 2i + e

0

i 2θ

) dθ 2π

)

1 dθ = 2π

i 2θ 4i ei θ e ⎤ ⎡ 2 + ⎢ 4i θ + ⎥ i 2i ⎣ ⎦0

∫ f (2i + e )

1 dθ = 2π

⎡ i e4 π i i ⎤ 2π i e − 8 π + 4 − − 4 + ⎢ ⎥ 2 2⎦ ⎣

1 2π

∫ (

1 2π

2π

f 2i + ei θ

0

2π

iθ

0

1 2π

∫ ( 2π

)

f 2i + ei θ dθ =

0

1 2π

(

) (

)

⎡ ⎤ i 4π i 2π i ⎢− 8 π + 4 e − 1 − 2 e − 1 ⎥ ⎣ ⎦

e 2π i = 1 !

!

e 4π i = 1

iθ

)

1 dθ = 2π

∫ (2i + e ) dθ 2π

0

iθ 2

1 2π

∫ ( 2π

0

)

f 2i + eiθ dθ =

1 ⎡ − 8 π ⎤⎦ = − 4 2π ⎣

()

This is obviously the value of f z = z 2 at point z = 2 i

f z = z on the contour is given by: 2π

+ 4i ei θ + e

and so:

z − 2 i = 1.

2

2

From equation (1.13-12) we have:

Find the value of f z = z 2 at point z = 2 i by determining the mean value of

∫ (4i 2π

Therefore:

(6.6-8)

() f ( z ) on the circle contour C

0

)

1 dθ = 2π

or

(6.6-7)

and equating the imaginary parts of equation (6.6-6), we have:

1 v x0 , y0 = 2π

f 2i + e

iθ

and so: !

!

∫ ( 2π

6.7! !

MODULUS THEOREMS We will now present several propositions which are

known as modulus theorems. 291

Proposition 6.7-1, Modulus Theorem:

()

( )

() ( ) f ( z ) is constant on Dδ ( z0 ) .

If f z is holomorphic on a disk Dδ z0 , and if f z ≤ f z0

( )

for all z within Dδ z0 , then Proof: !

()

Since f z

( )

()

is holomorphic within Dδ z0 , then f z

( )

is

continuous within Dδ z0 . Let 0 < r < δ , and let C be the circle defined by: !

z − z0 = r ei θ !

!

(6.7-1)

so that on the boundary: ! !

()

(

!

)

(6.7-2)

From Gauss’ mean-value theorem (Proposition 6.6-1) as

!

( )

∫ f ( z +re ) 0

iθ

0

1 dθ = 2π

∫

2π

0

( )

()

f z dθ !

(6.7-3)

contour C . We then have: !

( )

f z0

(6.7-5)

occurs at point z0 , we have:

(

()

f z = f z0 + r ei θ

!

) ≤ f (z ) !

(6.7-6)

0

Therefore:

1 ! 2π

2π

∫ f (z +re )

1 dθ ≤ 2π

iθ

0

0

∫

2π

0

( )

( )

f z0 dθ = f z0 ! (6.7-7)

must apply, and so:

∫ ( f ( z + r e ) − f ( z ) ) dθ = 0 ! 2π

!

iθ

0

0

(6.7-8)

()

equation (6.7-8): !

(

f z0 + r ei θ

) = f (z ) ! 0

0 ≤ θ ≤ 2π !

(6.7-9)

or for all z within C :

∫ ( 0

) dθ !

Given the continuity of f z within C , we then obtain from

()

2π

()

0

and so f z0 is equal to the mean value of f z over the closed

1 = 2π

0

(

f z0 + r ei θ

Since we are assuming that the maximum value of the

modulus f z

expressed in equations (6.6-5) and (6.6-6), we have: 2π

∫

2π

From equations (6.7-5) and (6.7-7), we see that the equal signs

f z = f z0 + r ei θ !

1 f z0 = 2π

( )

f z0

!

1 ≤ 2π

)

f z0 + r ei θ dθ !

(6.7-4)

!

()

( )

f z = f z0 !

(6.7-10)

Therefore from Proposition 5.3-7: 292

()

Therefore f z is constant for all z within C . Since r is

!

() ( ) f ( z ) is constant on Dδ ( z0 ) . ■

arbitrary within the range 0 < r < δ , we have f z = f z0

( )

all z within Dδ z0 , and so

Solution:

for

From equation (4.3-37) we have:

Proposition 6.7-2, Maximum Modulus Theorem:

()

sin z = sin 2 x + sinh 2 y

!

Since the maximum of sin 2 x occurs at x = π 2 and the

is a holomorphic and non-constant complex function

maximum of sinh 2 y within R occurs at y = 1 , we have the

within a closed bounded region R , and if f z is continuous on

maximum of f z occurs only at the point π 2,1 which is

R , then the maximum value of

on the boundary of R .

If f z

()

f z

()

()

must occur on the

(

)

boundary of R . Proposition 6.7-3, Minimum Modulus Theorem: Proof:

()

Since f z

!

()

If f z ≠ 0 is a holomorphic and non-constant complex function

is holomorphic and non-constant within the

closed bounded region R , then from Proposition 6.7-1 we know that

()

f z

()

()

must occur somewhere because f z

is

non-constant, we can conclude that the maximum of the

()

modulus f z

must occur on the boundary of R .

■

Example 6.7-1 Verify the maximum modulus theorem (Proposition 6.7-2)

()

R , then the minimum value of

cannot have a maximum within R . Since the

maximum of f z

for f z = sin z in a closed rectangular region R defined by

0 ≤ x ≤ π and 0 ≤ y ≤ 1 .

()

within a closed bounded region R , and if f z is continuous on

()

f z

must occur on the

boundary of R . Proof:

()

Since f z

!

is holomorphic and non-constant within the

()

()

closed bounded region R , and since f z ≠ 0 , then 1 f z

must be holomorphic within R . From Proposition 6.7-2 we know that 1

()

f z

cannot have a maximum within R , and

since the maximum of 1

()

()

f z

must occur somewhere because

f z is non-constant, we can conclude that the minimum of the 293

()

modulus f z

()

f z

must occur on the boundary of R .

6.8! !

cannot occur in C . Therefore the minimum of

!

()

!

()

Pn z zn

polynomial having complex coefficients into a product of linear factors.

and

Proposition 6.8-1, Fundamental Theorem of Algebra:

!

()

Every non-constant polynomial Pn z :

()

2

Pn z = a0 + a1 z + a2 z +!+ an−1 z

n−1

n

+ an z !

(6.8-1)

of finite degree n ≥ 1 , and having coefficients a0 , a1 , !, an−1 , an

()

are not necessarily distinct). Therefore Pn z can be completely

() (

) ( z − z2 ) ( z − z3 ) ! ( z − zn ) !

(6.8-2)

where the zeros zi can be complex numbers, and where c is a complex constant.

z

n

()

() a0

=

zn

+

a0

≥

z

a1

n

z n−1

+

+

a1 z

n−1

a2

+

z n−2

+

a3 z n−3

a2 z

n−2

+

+!+ an !

a3 z

n−3

+!+ an !

(6.8-3)

(6.8-4)

()

z n → an

as z → ∞ , we see that Pn z

z n is

bounded everywhere in the complex plane. For some δ > 0 we then have: !

factored so that:

Pn z = c z − z1

()

Pn z

Since Pn z

that are complex constants with an ≠ 0 , will have n roots (which

Proof:

is an entire function.

plane. Dividing Pn z by z n we have:

We will now show that it is possible to factor any

!

()

Therefore f z will be holomorphic everywhere in the complex

FUNDAMENTAL THEOREM OF ALGEBRA

!

()

all values of z so that f z = 1 Pn z

■

()

Seeking a contradiction, we will assume that Pn z ≠ 0 for

()

Pn z z

n

≥

an 2

!

z ≥δ !

(6.8-5)

z ≥δ !

(6.8-6)

and so: !

()

f z =

1

()

Pn z

≤

2 an z

n

!

()

Therefore the entire function f z is bounded everywhere in the complex plane. 294

!

()

From Liouville’s theorem (Proposition 6.5-1) we see that

()

f z is then a constant, and so Pn z must be a constant. This is

Proposition 6.8-2:

clearly a contradiction since n ≥ 1 . Therefore we must have

Complex roots of polynomial equations having real coefficients

Pn z = 0 for some value of z .

only occur in conjugate pairs.

()

!

()

At least one root z = z1 of Pn z must then exist. Factoring

out this root, we have: !

() (

)

Proof: !

()

Pn z = z − z1 Qn−1 z !

(6.8-7)

()

()

If z = z1 is a complex root of the polynomial Pn z , we will

have:

( )

Pn z1 = a0 + a1 z1 + a2 z12 + a3 z13 +!+ an z1n = 0 ! (6.8-10)

where Qn−1 z is a polynomial of degree n − 1 .

!

!

Taking the complex conjugate of both sides of this equation:

Following the same argument as above, we find that at

()

least one root z = z2 of Qn−1 z

must exist. Factoring out this

root, we have:

( ) ( )( ) ( ) where Rn−2 ( z ) is a polynomial of degree n − 2 . !

!

Pn z = z − z1 z − z2 Rn−2 z !

(6.8-8)

This procedure can be continued n times until the

()

(

Pn z = c z − z1

) ( z − z2 ) ( z − z3 ) ! ( z − zn ) !

(6.8-9)

where the zeros zi can be complex numbers. We have then shown that any polynomial of degree n ≥ 1 has n roots (which

a0 + a1 z1 + a2 z12 + a3 z13 +!+ an z1n = 0 = 0 !

(6.8-11)

a0 + a1 z1 + a2 z12 + a3 z13 +!+ an z1n = 0 !

(6.8-12)

or !

remaining polynomial is a complex constant c . We then have: !

!

Since the coefficients are real, we then have: !

a0 + a1 z1 + a2 z12 + a3 z13 +!+ an z1n = 0 !

(6.8-13)

or !

( )

( )

Pn z1 = Pn z1 = 0 = 0 !

(6.8-14)

()

are not necessarily distinct). This theorem is known as the

Therefore the complex conjugate z1 is also a root of Pn z . We

fundamental theorem of algebra.

can conclude that non-real roots of polynomial equations

■

having real coefficients only occur in conjugate pairs.

■

295

Example 6.8-1

Proposition 6.9-1, Harmonic Function Theorem:

Determine the roots of the polynomial: z 3 − 2 z 2 + z − 2 = 0 .

is holomorphic on a simply

()

Solution:

connected domain D if and only if f z is harmonic on D . The

We have a polynomial with real coefficients. Factoring it we

imaginary part v x, y of f z is then the harmonic conjugate

have:

(

)(

) (

)(

)(

)

!

z3 − 2 z2 + z − 2 = z2 + 1 z − 2 = z + i z − i z − 2 = 0

!

Therefore the complex roots occur as a conjugate pair.

6.9! !

()

A complex function w = f z

HARMONIC FUNCTIONS A real-valued function that satisfies the two-dimensional

Laplace’s equation is called a harmonic function. A complex

()

( ) () of the real part u ( x, y ) of f ( z ) . Proof:

() ( ) ( ) imaginary part v ( x, y ) of f ( z ) will be the harmonic conjugate of the real part u ( x, y ) of f ( z ) on D . If z is any interior point of !

If w = f z = u x, y + i v x, y is harmonic on D , then the

D , then u and v will have continuous first and second order !

( )

( )

part v x, y

each satisfy the two-dimensional Laplace's

( ) called the harmonic conjugate of the real part u ( x, y ) . Note that

equation at any point in D . The imaginary part v x, y is then

()

We will now define F z as:

function w = f z = u + i v holomorphic on a domain D is called a harmonic function if its real part u x, y and its imaginary

()

derivatives in D at point z since f z is harmonic on D .

!

()

( )

( )

∂u ! ∂x

( )

F z = U x, y + iV x, y =

∂u ∂u −i ! ∂x ∂y

(6.9-1)

where: !

harmonic conjugates are real variables, and should not be

U x, y =

( )

( )

V x, y = −

∂u ! ∂y

(6.9-2)

( )

confused with complex conjugates which are complex

Since u x, y and v x, y are harmonic and have continuous first

variables.

and second order derivatives on D , we have: !

∂U ∂2u ∂2u ∂V ! = 2 =− 2 = ∂x ∂x ∂y ∂y

(6.9-3) 296

Taking the derivative with respect to x :

and

∂U ∂ 2u ∂ 2u ∂V ! = = =− ∂ y ∂y ∂x ∂x ∂y ∂x

! !

(6.9-4)

()

The Cauchy-Riemann equations then hold for F z , and

()

!

()

!

!

()

()

(6.9-5)

()

and so f ′ z must exist since F z

()

exists. Therefore f z is

differentiable on D . Moreover all higher order derivatives of

()

()

F z must exist and so all higher order derivatives of f z

()

must also exist on D . Therefore f z is holomorphic on D . !

Conversely, if a complex function:

!

w = f z = u x, y + i v x, y !

() ( )

( )

any interior point of D , then the real part u x, y and imaginary

()

part v x, y of f z will have continuous partial derivatives of

()

all orders at point z (see Proposition 6.3-2). Moreover, if f z is

!

∂u ∂v ! = ∂x ∂ y

(6.9-7)

∂ 2u ∂2v ! =− 2 ∂y ∂x ∂y

(6.9-9)

Since continuous partial derivatives of all orders exist at

these equations are therefore independent of the order of differentiation, and so we can write:

∂ 2u ∂2 v ∂2 v ∂ 2u = = =− 2! ∂x 2 ∂x ∂y ∂ y ∂x ∂y

(6.9-10)

∂ 2u ∂ 2 u + 2 = 0! 2 ∂x ∂y

(6.9-11)

or !

Similarly we have: !

∂u ∂v =− ! ∂y ∂x

∂ 2u ∂2v ! = ∂y ∂x ∂ y 2

equations (6.9-8) and (6.9-9). The mixed partial derivatives in

holomorphic at z , the Cauchy-Riemann equations must hold at this point:

(6.9-8)

point z , the second partial derivatives are continuous in

(6.9-6)

is holomorphic on a simply connected domain D , and if z is

( )

!

!

( )

∂2u ∂2 v =− 2 ! ∂x ∂y ∂x

Taking the derivative with respect to y:

so F z is holomorphic on D . From equation (3.4-24) we have:

∂u ∂u f′ z = −i =F z ! ∂x ∂y

∂ 2u ∂2 v = ! ∂x 2 ∂x ∂y

∂2 v ∂2u ∂2u ∂2 v ! − 2= = = ∂x ∂ y ∂ y ∂x ∂y 2 ∂x

(6.9-12)

or 297

∂2 v ∂2 v + = 0! ∂x 2 ∂y 2

!

(6.9-13)

Proposition 6.9-3, Mean Value of a Harmonic Function:

()

Equations (6.9-11) and (6.9-13) are both two-dimensional

If a function f z is harmonic on a simply connected domain

Laplace equations. Therefore u x, y and v x, y are harmonic

D , and if z = z0 is a point within D , then f z0 is equal to the

functions, and so f z must be a harmonic function. Moreover

mean value of f z around the boundary of any disk Dr z0 of

!

( )

( )

() the imaginary part v ( x, y ) is then the harmonic conjugate of the real part u ( x, y ) of f ( z ) . ■ !

Therefore the real and imaginary parts of a holomorphic

()

function f z

will always each be solutions of the two-

dimensional Laplace’s equation. This makes holomorphic functions very important in the solution of two-dimensional potential problems in physics and engineering.

()

( )

radius r centered at point z0 and contained within D . Proof: !

Follows from Propositions 6.9-1 and 6.6-1. We have from

equation (6.6-5): !

1 f z0 = 2π

( )

2π

∫ ( 0

)

f z0 + r eiθ dθ !

(6.9-14)

() f (z )

Therefore the mean value of a harmonic function f z on any

Proposition 6.9-2, Harmonic Functions are Infinitely Differentiable:

()

circle C within a domain is equal to the value center of the circle.

A harmonic function f z holomorphic on a simply connected

!

domain D is infinitely differentiable.

harmonic function.

0

at the

■

Proposition 6.9-3 is sometimes used as the definition of a

Example 6.9-1

Proof: !

Follows from Propositions 6.9-1 and 6.3-2.

!

Proposition 6.9-2 is simply confirmation that harmonic

functions are smooth.

( )

■

()

Show that f z = e z is a harmonic function for z ∈! . Solution: We have: 298

()

Solution:

! f z = u + i v = e z = e x+i y = e x cos y + i e x sin y

We have:

()

The function f z = e z is an entire function. We have: !u = e cos y !

v = e sin y

∂u ! = 2x! ∂x

∂u ! = e x cos y ! ∂x

∂u = − e x sin y ∂y

∂2 u ! 2 = 2! ∂x

∂v ! = e x sin y ! ∂x

∂v = e x cos y ∂y

We then have:

∂2 u ! 2 = e x cos y ! ∂x

∂2 u x = − e cos y ∂ y2

x

2

∂ v ! 2 = e x sin y ! ∂x

x

( )

and so the real part u x, y = x 2 + y 2 is not a solution of the

2

∂ v = − e x sin y 2 ∂y

two-dimensional Laplace’s equation. By definition of a

( )

complex harmonic function then, u x, y = x 2 + y 2 cannot be the real part of any harmonic function.

∂2 v ∂2 v + =0 ∂x 2 ∂y 2

!

()

( )

( )

If u x, y and v x, y are conjugate harmonic functions that

each satisfy Laplace’s equation, they are both potential

and so f z = e z is a harmonic function for z ∈! .

functions. Associated with these potential functions are the equipotential curves:

Example 6.9-2

( )

2

Show that any function having the real part u x, y = x + y is not holomorphic.

∂2 u =2 ∂ y2

∂ 2u ∂ 2u ! 2 + 2 =4 ∂x ∂y

Therefore we have the Laplace’s equations:

∂ 2u ∂ 2 u ! 2 + 2 = 0! ∂x ∂y

∂u =2y ∂y

2

!

( )

u x, y = ck !

( )

v x, y = dk !

(6.9-15)

where for any given set of curves k , the equipotential curves are orthogonal. 299

Proposition 6.9-4, Orthogonal Equipotential Curves:

( )

( )

If u x, y and v x, y are conjugate harmonic functions, then the set of equipotential curves

( )

u x, y = ck !

!

( )

v x, y = dk !

( )

From equation (6.9-16) the differential of u x, y is:

∂u ∂u du = dx + dy = 0 ! ∂x ∂y

!

= k

∂u ∂ y ! ∂u ∂x k

(6.9-22)

dy dx

=− k

∂u ∂x ! ∂u ∂ y k

(6.9-18)

( )

∂v ∂v dx + dy = 0 ! ∂x ∂y

(6.9-19)

( )

dy dx

=− k

∂v ∂x ! ∂v ∂ y k

From the Cauchy-Riemann equations we have:

(6.9-23)

( )

Therefore the slope of the curve u x, y = ck given in

equation (6.9-18) is orthogonal to the slope of the curve

From equation (6.9-16) the differential of v x, y is:

dv =

⎛ ∂u ∂x ⎞ ⎛ ∂u ∂ y ⎞ ⎜ − ∂u ∂ y ⎟ ⎜ ∂u ∂x ⎟ = −1 ! ⎝ k⎠ ⎝ k⎠

! !

We then have for the slope of the curve v x, y = dk : !

dy dx

!

(6.9-17)

( )

!

( )

and so slope of the curve v x, y = dk as given in equation

(6.9-16)

We then have for the slope of the curve u x, y = ck : !

(6.9-21)

From equations (6.9-18) and (6.9-22) we then have:

Proof:

!

∂u ∂v =− ! ∂y ∂x

(6.9-20) becomes:

for any given k are orthogonal, where ck and dk are constants.

!

∂u ∂v = ! ∂x ∂ y

!

( )

v x, y = dk given in equation (6.9-22) at any point where the two curves intersect. !

■

( )

If we are given a real-valued function u x, y

that is

harmonic in a simply connected domain D , then it is always

( )

( )

possible to find a harmonic conjugate v x, y to u x, y . Proposition 6.9-5, Harmonic Conjugate:

(6.9-20)

( )

If a real-valued function u x, y

is harmonic in a simply

connected domain D , then there exists a real-valued harmonic 300

( ) ( ) f ( z ) = u + i v is holomorphic on D .

conjugate function v x, y of u x, y , defined up to an arbitrary constant, such that Proof: !

( )

!

Since u x, y is harmonic, continuous partial derivatives of

( )

u x, y

( )

exist. If a harmonic conjugate function v x, y

( )

( )

Since u x, y is harmonic, we then obtain:

exists,

dv =

!

∂v ∂v dx + dy ! ∂x ∂y

( )

equation (6.9-26) can be integrated to obtain v x, y : !

(6.9-24)

!

!

∂u ∂v =− ! ∂y ∂x

dv = −

∂u ∂u dx + dy ! ∂y ∂x

(6.9-26)

dv = P dx + Q dy !

(6.9-27)

we have: !

∂P ∂ ⎛ ∂u ⎞ ∂2 u = − =− 2! ∂ y ∂ y ⎜⎝ ∂ y ⎟⎠ ∂y

( )

We can take derivatives of v x, y

( )

(6.9-30) to determine if it is a

∂v ∂ = ∂x ∂x

!

∂v ∂ = ∂y ∂y

∫

⎛ ∂u ∂u ⎞ ∂u ! − dx + dy = − ⎜⎝ ∂ y ∂x ⎟⎠ ∂y

(6.9-31)

∫

⎛ ∂u ∂u ⎞ ∂u ! − dx + dy = ⎜⎝ ∂ y ∂x ⎟⎠ ∂x

(6.9-32)

Equations (6.9-31) and (6.9-32) are just the Cauchy-Riemann

( ) We see that v ( x, y ) is must then be a harmonic conjugate u ( x, y ) . Therefore, if a real-valued function u ( x, y ) is

equations. Therefore v x, y is a holomorphic function.

Writing this equation in the form: !

∫

⎛ ∂u ∂u ⎞ − dx + dy + c ! ⎜⎝ ∂ y ∂x ⎟⎠

! (6.9-25)

equation (6.9-24) becomes: !

v=

harmonic conjugate of u x, y :

Using the Cauchy-Riemann equations:

∂u ∂v ! = ∂x ∂ y

(6.9-29)

Therefore dv = P dx + Q dy is an exact differential, and so

continuous partial derivatives of v x, y must also exist. We can then write:

∂P ∂Q = ! ∂ y ∂x

! of

harmonic in a simply connected domain D , then a real-valued

∂Q ∂ ⎛ ∂u ⎞ ∂2 u ! (6.9-28) = = ∂x ∂x ⎜⎝ ∂x ⎟⎠ ∂x 2

( )

()

harmonic conjugate function v x, y exists so that f z = u + i v is holomorphic on D .

■

301

!

Every harmonic function in a simply connected domain is

the real part of some holomorphic function of a complex variable. If the domain of the function is not simply connected, however, a single-valued harmonic conjugate may not exist.

( )

If a real-valued function u x, y

is harmonic in a simply

( )

u x, y differ only by a constant.

( )

( )

u x, y exist. If a harmonic conjugate function v x, y exists, it must satisfy the Cauchy-Riemann equations. Let v1 and v2 be two harmonic conjugates of u . We must then have: !

!

)

(6.9-33)

(6.9-34)

( )

Therefore any two harmonic conjugates of u x, y ■

∂u ∂u dx + dy = 2 y dx + 2x dy = 2 d x y ∂y ∂x

( )

Integrating: !v = 2 x y + c

( )

( )

v x, y for the harmonic function u x, y = x 2 − y 2 is:

( )

)

only by a constant.

!dv = −

!v x, y = 2 x y + c

∂v ∂u ∂u v1 − v2 = − + = 0! ∂x ∂y ∂y

(

∂u ∂v =−2y=− ∂y ∂x

where c is a real constant. Therefore the harmonic conjugate

∂ ∂u ∂u v1 − v2 = − = 0! ∂y ∂x ∂x

(

∂u ∂v ! = 2x = ! ∂x ∂y

From equation (6.9-26) we have:

Since u x, y is harmonic, continuous partial derivatives of

( )

( )

function u x, y = x 2 − y 2 .

for the harmonic

From the Cauchy-Riemann equations we have:

connected domain D , then any two harmonic conjugates of

!

( )

Determine a harmonic conjugate v x, y

Solution:

Proposition 6.9-6:

Proof:

Example 6.9-3

can differ

() ! f ( z ) = x 2 − y 2 + i (2 x y + c )

and so a function f z having harmonic parts is then:

Testing this function, we find: 302

∂u ! = 2x! ∂x

∂u =−2y ∂y

∂2 u ! 2 = 2! ∂x

∂2 u = −2 ∂ y2

∂v ! = 2 y! ∂x

∂v = − 2x ∂y

2

∂ v ! 2 = 0! ∂x

!dv = − or

(

!dv = d − e x cos y

2

!v = − e x cos y + c

2

∂ v =0 ∂ y2

∂u ∂u ! 2 + 2 = 0! ∂x ∂y

2

)

Integrating:

where c is a real constant. Therefore we have a function having harmonic parts:

We see that we obtain Laplace’s equation: 2

∂u ∂u dx + dy = − e x cos y dx + e x siny dy ∂y ∂x

(

()

)

! f z = e x sin y + − e x cos y + c i

2

∂v ∂v + 2 =0 2 ∂x ∂y

Example 6.9-5 Example 6.9-4

( )

Determine a harmonic conjugate v x, y

( )

for the harmonic

function u x, y = e x sin y .

( )

for the harmonic

function v x, y = 2 x y . Solution:

Solution: From the Cauchy-Riemann equations we have:

∂u ∂v ! = e x sin y = ! ∂x ∂y

( )

Determine a harmonic conjugate u x, y

∂u ∂v = e x cos y = − ∂y ∂x

From equation (6.9-26) we have:

()

We are looking for a holomorphic function f z = u + i v where v = 2 x y . Since we know how to find a harmonic

( )

() step is to change the function so that v ( x, y ) = 2 x y the real part u ( x, y ) of some function.

conjugate given the real part u x, y of f z = u + i v , the first becomes

303

()

This is the same function found in Example 6.9-3.

We can do this by multiplying f z by i :

()

!i f z = iu − v = − v + iu

Example 6.9-6

Letting U = −v and V = u , we have:

(

()

!i f z = U + iV = − 2 x y + iV

real part?

from the Cauchy-Riemann equations we have:

∂U ∂V ! ! = −2 y = ∂x ∂y

Solution:

∂U ∂V = − 2x = − ∂y ∂x

We have:

From equation (6.9-26) we have: !dV = −

(

∂U ∂U dx + dy = 2 x dx − 2 y dy = d x 2 − y 2 ∂y ∂x

)

Integrating:

∂u =1 ∂y

∂2 u ! 2 = 2! ∂x

∂2 u = 0! ∂ y2

∂2 u ∂2 u + ≠0 ∂x 2 ∂ y 2

( )

(

(

()

!i f z = U + iV = − 2 x y + i x 2 − y 2 + c

)

Multiplying by i :

(

!− f z = − 2 x y i − x 2 − y 2 + c

Example 6.9-7 If φ is a two-dimensional harmonic function, write Laplace’s

)

equation for φ using Wirtinger derivatives.

( )

A function having the harmonic part v x, y = 2 x y is then: ! f z = x2 − y2 + 2 x y i

)

not exist since u x, y = x2 + y is not harmonic.

where c is a real constant. Therefore:

()

∂u ! = 2x! ∂x

Therefore a harmonic function having u x, y = x 2 + y does

!V = x 2 − y 2 + c

()

)

Does a harmonic function exist having u x, y = x2 + y as the

Solution: We are given: 304

()

∂2φ ∂2φ ! 2 + 2 =0 ∂x ∂y

!f ′ z =

()

Therefore:

We can write this equation in the form:

()

! f′ z =

⎛ ∂ ∂ ⎞⎛ ∂ ∂⎞ !⎜ +i ⎟ ⎜ −i ⎟φ =0 ∂ y ⎠ ⎝ ∂x ∂y⎠ ⎝ ∂x

1⎛ ∂ ∂⎞ − f z 2 ⎜⎝ ∂x ∂ y ⎟⎠

()

and so:

Using the Wirtinger derivatives given in equations (3.6-4)

()

and (3.6-5):

! f′ z

∂ 1⎛ ∂ ∂⎞ ! = ⎜ −i ⎟ ! ∂z 2 ⎝ ∂x ∂y⎠

1⎛ ∂ ∂⎞ − f z 2 ⎜⎝ ∂x ∂ y ⎟⎠

∂ 1⎛ ∂ ∂⎞ = ⎜ +i ⎟ ∂z 2 ⎝ ∂x ∂y⎠

2

1 ⎛ ∂2 ∂2 ⎞ = ⎜ 2 + 2⎟ f z 4 ⎝ ∂x ∂y ⎠

()

or

⎛ ∂2 ∂2 ⎞ !⎜ 2 + 2 ⎟ f z ∂y ⎠ ⎝ ∂x

()

we then have:

∂2φ ! =0 ∂z ∂z

2

2

()

=4 f′ z

2

Proposition 6.9-7:

()

If a complex function f z

Example 6.9-8

()

2

()

=4 f′ z

be represented using Wirtinger derivatives:

2

Solution:

()

connected domain D , then the Laplacian operator on f z can

()

If f z is an holomorphic function, show that:

⎛ ∂2 ∂2 ⎞ !⎜ 2 + 2 ⎟ f z ∂y ⎠ ⎝ ∂x

is holomorphic on a simply

⎛ ∂2 ∂2 ⎞ ∂ ∂ ⎡ ⎤ ⎡⎣ f z ⎤⎦ ! + f z = 4 ⎜ 2 2⎟ ⎣ ⎦ ∂z ∂z ∂y ⎠ ⎝ ∂x

()

! Proof:

()

()

(6.9-35)

Using the Wirtinger derivative from equation (3.6-4), we

!

have:

D , continuous partial derivatives of f z exist in D . From the

Since f z is holomorphic on a simply connected domain

()

305

Wirtinger derivatives given in equations (3.6-4) and (3.6-5) we have:

∂ 1⎛ ∂ ∂⎞ = ⎜ −i ⎟ ! ∂z 2 ⎝ ∂x ∂y⎠

!

(6.9-36)

(6.9-37)

()

()

(6.9-38)

or

()

()

(6.9-39)

Therefore:

⎛ ∂ ∂ ⎞ ∂ ∂ ⎡ ⎤ ⎡⎣ f z ⎤⎦ ! + f z = 4 ⎜ 2 2⎟ ⎣ ⎦ ∂z ∂z ∂y ⎠ ⎝ ∂x 2

2

()

∂ ∂z

()

(6.9-41)

is holomorphic on the simply connected

()

()

1⎛ ∂ ∂⎞ 1⎛ ∂ ∂⎞ ! − i + i 2 ⎜⎝ ∂x ∂ y ⎟⎠ 2 ⎜⎝ ∂x ∂ y ⎟⎠

(6.9-42)

2

1 ⎛ ∂2 ∂2 ⎞ = ⎜ 2 + 2 ⎟ ⎡⎣ f z ⎤⎦ ! 4 ⎝ ∂x ∂y ⎠

(6.9-43)

()

From equation (6.9-40) we also have:

⎛ ∂2 ∂2 ⎞ ∂ ∂ ⎡ ⎤ ⎡⎣ f z ⎤⎦ ! + f z = 4 ⎜ 2 2⎟ ⎣ ⎦ ∂z ∂z ∂y ⎠ ⎝ ∂x

()

()

(6.9-44)

Therefore:

()

f′ z

! Proposition 6.9-8:

=

()

f′ z

! (6.9-40)

2

and so:

■

()

∂ ∂ ⎡⎣ f z ⎤⎦ ! ∂z ∂z

From equations (3.6-4) and (3.6-5) we have:

!

∂ ∂ 1 ⎛ ∂2 ∂2 ⎞ ⎡ f z ⎤⎦ = ⎜ 2 + 2 ⎟ ⎡⎣ f z ⎤⎦ ! ∂z ∂z ⎣ 4 ⎝ ∂x ∂y ⎠

!

()

Since f z

!

∂ ∂ 1⎛ ∂ ∂⎞⎛ ∂ ∂⎞ ⎡⎣ f z ⎤⎦ = ⎜ −i ⎟ ⎜ + i ⎟ ⎡⎣ f z ⎤⎦ ! ∂z ∂z 4 ⎝ ∂x ∂ y ⎠ ⎝ ∂x ∂y⎠

!

Proof:

=

domain D , continuous partial derivatives of f z exist in D .

We can write: !

f′ z

!

!

∂ 1⎛ ∂ ∂⎞ = ⎜ +i ⎟! ∂z 2 ⎝ ∂x ∂y⎠

!

()

2

2

=

∂ ∂ ⎡⎣ f z ⎤⎦ ! ∂z ∂z

()

(6.9-45)

■

If a complex function f z is holomorphic on a simply connected domain D , then we have: 306

Proposition 6.9-9:

()

If a complex function f z is holomorphic and non-constant in

( )

a simply connected domain D , then the real part u x, y

()

of

f z cannot have a maximum or minimum value within D . Proof:

f z We have f z = u x, y + i v x, y . Let w = e ( ) . The function f z w = e ( ) will be holomorphic in D . From the maximum and

!

() ( )

( )

minimum modulus theorems (Propositions 6.7-2 and 6.7-3), we f z know that e ( ) cannot have a maximum or minimum value within D . We can write: !

f z e ( ) = eu+i v = eu !

(6.9-46)

Therefore eu cannot then have a maximum or minimum value

( )

in D . We see then that u x, y minimum value in D .

cannot have a maximum or

■

307

Chapter 7 Complex Sequences

()

()

lim fn z = f z

n →∞

308

!

In this chapter we will define and review the properties of

Example 7.1-1

infinite sequences of complex numbers. We will present

Determine the first few terms of the following complex

required criteria for a complex sequence to converge to a limit,

sequences:

including the important Cauchy’s convergence criteria. We will

⎧1+ n i ⎫ 1.! ⎨ n ⎬ ⎩ 2 ⎭

discuss sequences of complex functions and the concept of uniform convergence of sequences of complex functions.

7.1! !

⎧⎪ i n + n2 i ⎫⎪ 2.! ⎨ ⎬ n ⎩⎪ ⎭⎪

SEQUENCES OF COMPLEX NUMBERS A sequence of complex numbers

{ zn }

is obtained if an

Solution:

infinite set of complex numbers zn can be put in one-to-one correspondence with the set of positive integers, and if the

1.!

⎧1+ n i ⎫ 1+ i 1+ 2i 1+ 3i 1+ 4i 1+ 5i 1+ 6i , , , , , ,! ⎨ n ⎬= 2 3 4 5 6 2 2 2 2 2 2 2 ⎩ ⎭

2.!

⎧⎪ i n + n2 i ⎫⎪ −1+ 4i 8i 1+ 16i 26i −1+ 36i , , , , ,! ⎨ ⎬ = 2i, n 2 3 4 5 6 ⎩⎪ ⎭⎪

complex numbers zn have a definite order of arrangement:

{ zn } = z1 , z2 , z3 , ! !

!

(7.1-1)

A complex sequence then forms a countably infinite set of complex numbers. When we refer to a sequence, it will be understood to mean an infinite complex sequence unless otherwise stated. !

Each individual zn is called a term or element of the

complex sequence

{ zn } , and

zn is referred to as the nth term.

7.1.1! !

CONVERGENCE OF COMPLEX SEQUENCES

If we have:

! where

lim z = z0 !

(7.1-2)

n→∞ n

z0

is finite, then the sequence

{ zn }

is called a

The terms of a sequence need not be distinct, and the index of

convergent sequence, and z0 is the limit to which the sequence

the initial term need not be n = 1 (it can be any nonnegative

converges. Equation (7.1-2) can also be written zn → z0 as

integer).

n → ∞ . If a sequence has a limit, it can have only one limit (see 309

zn − z0

Proposition 7.1-2). If a sequence has a limit equal to zero, the sequence is called a null sequence. !

successive term of the sequence:

A necessary and sufficient condition for a complex

{ }

sequence zn to converge to a limit z0 is that, for any number

ε > 0 , there exists an integer N > 0 such that: zn − z0 < ε !

!

becomes smaller than any arbitrary ε > 0 . Each

when n > N !

!

!, zn−2 − z0 , zn−1 − z0 , zn − z0 , zn+1 − z0 , ! ! (7.1-4)

will be smaller than the previous term. If ε is very small, N (7.1-3)

may have to be very large if the sequence converges slowly.

where N is generally a function of ε . We can then write

()

N = N ε . The rate at which a sequence

{ zn }

converges to a

limit z0 is determined by N , since N specifies the number of terms n necessary to satisfy equation (7.1-3). !

When equation (7.1-3) holds, all points zn where n > N

will lie within the circle zn − z0 < ε of radius ε centered at the point z = z0 . If a sequence converges to a limit z0 , therefore, an infinity of terms zn of the sequence

{ zn }

will lie within an ε

neighborhood of z0 . That is, all but a finite number of the zn

( )

will cluster within the open disk Dε z0

as shown in Figure

7.1-1. !

The point z = z0 will be a limit point or an accumulation

point for the sequence (see Section 2.2.5.4). The limit point z0 need not belong to the sequence. As n increases with n > N , the included points zn will lie ever more closely to the limit point

{ }

Figure 7.1-1! Complex sequence zn converging to a limit z0 . Infinitely many points of the sequence inside the circle are not shown.

z0 . As n becomes larger than N , therefore, the difference 310

!

The convergence of a sequence is then determined not by

!

A sequence will also diverge if the lim zn does not n→∞

the first terms of the sequence, but by the behavior of the tail of

converge to any limit including infinity (the sequence

the sequence. This means that a finite number of the first terms

oscillates). A divergent sequence that oscillates can have several

of a sequence can be omitted without changing the convergence

limit points, although it does not converge to any one of them

properties of the sequence. That is, beginning terms in a

(see Example 7.1-4).

sequence do not change the nature of the convergence of the

Example 7.1-2

sequence.

Determine convergence criteria for the following sequence:

Proposition 7.1-1, Convergent Sequence Requirement:

{ } converges to a limit z0

A complex sequence zn

if and only if

!

we have: (7.1-5)

n→ ∞

Proof: !

!

Follows from the definition of sequence convergence

{ }

{ }

the sequence zn to converge, the criteria are:

If a complex sequence zn does not converge, it is called a

lim z = ∞ !

n→∞ n

⎧ i ⎫ lim ⎨5 + 2 ⎬ = 5 n →∞ ⎩ n ⎭ and so this sequence converges as n → ∞ to the limit 5. For

■

divergent sequence. A sequence will diverge if we have: !

⎭

We have: !

given in equation (7.1-3).

⎩

Solution:

lim zn − z0 = 0 !

!

{ zn } = ⎧⎨5 + ni2 ⎫⎬

!

R > 0 no matter how large, there exists an N such that zn > R

when n > N

Therefore:

(7.1-6)

so that no finite limit exists. Equation (7.1-6) will hold if for any

zn − 5 < ε !

!

⎛ i ⎞ i 1 zn − z0 = ⎜ 5 + 2 ⎟ − 5 = 2 = 2 < ε ⎝ n ⎠ n n

for all n ≥ N . 311

!

n> N =

L1− L2 ≤ L1− zn + zn − L2 = zn − L1 + zn − L2 !

!

and so for convergence we need:

1

Therefore:

ε

L1− L2
0 we can find integers N1 > 0 and N2 > 0 such that: !

ε zn − L1 < ! 2

for all n > N1 !

(7.1-7)

and !

zn − L2
N !

(7.1-11)

assumption was wrong, therefore. A convergent complex

{ zn }

must have a unique limit. A convergent

sequence can then have only one limit. !

A complex sequence

{ zn }

■

is said to be bounded if there

exists a real number M > 0 such that:

zn ≤ M !

!

for all n > 0 !

(7.1-12)

All points zn of the sequence will then be contained within a disk of radius M centered at the origin. Example 7.1-3

ε ! 2

for all n > N2 !

(7.1-8)

)

L1− L2 = L1− zn + zn − L2 !

Are the following sequences bounded?

⎧ i ⎫ 1.! ⎨ 2 ⎬ ⎩n ⎭

Let N = max N1 , N2 . We can then write: !

ε ε + =ε! 2 2

Since ε > 0 can be arbitrarily small, we must have L1 = L2 .Our sequence

Proof: !

(7.1-10)

(7.1-9)

2.!

{e } i nθ

312

⎧⎪ ⎛ 1 ⎞ n ⎫⎪ 3.! ⎨ ⎜ ⎟⎠ ⎬ 1+ i ⎝ ⎪⎩ ⎪⎭ 4.!

Proof:

{ n + in} 2

lim z = z0 !

!

i ≤ 1! 2 n

for all n > 0 !

ei nθ = 1 !

zn − z0 < ε !

!

zn = zn − z0 + z0 ≤ zn − z0 + z0 !

!

zn < ε + z0 !

!

The sequence is bounded since we have:

⎛ 1 ⎞ ⎜⎝ 1+ i ⎟⎠

n

⎛ 1 ⎞ =⎜ ⎟ = 1+ i ⎝ ⎠

!

n + in =

4

2

(7.1-15)

for all n > N !

(7.1-16)

for all n > 0 !

(7.1-17)

We will then have:

1

( 2)

n

≤

1 2

!

for all n > 0

2

n + n = n n +1 !

zn ≤ M !

!

where M is given by:

4.! The sequence is not bounded since we have: 2

(7.1-14)

or using equation (7.1-14):

for all n > 0

! n

for all n > N !

We can write:

2.! The sequence is bounded since we have:

3.!

(7.1-13)

n→∞ n

For any real number ε > 0 , we then have for some N > 0 :

1.! The sequence is bounded since we have:

!

{ zn }

converges:

Solution:

!

Let z0 be the limit to which the complex sequence

!

for all n > 0

M = max ⎡⎣ z1 , z2 , z3 , !, zn , ε + z0 ⎤⎦ !

!

(7.1-18)

{ }

Therefore if a complex sequence zn converges, it is bounded. ■

Proposition 7.1-3, Convergent Sequences are Bounded:

{ }

If a complex sequence zn converges, then it is bounded.

!

A complex sequence that is not bounded is considered

unbounded. If a complex sequence is unbounded, then no limit 313

exists for the sequence, and so the sequence diverges. A sequence can, however, be bounded, but still not converge (see

Proposition 7.1-4, Monotonic Sequence Convergence:

Example 7.1-4), and so the converse of Proposition 7.1-3 is not

A monotonic real-valued sequence

true.

only if it is bounded.

Example 7.1-4

Proof:

{ }

Does the sequence i n converge and is it bounded?

!

!

{ an }

is convergent if and

is a monotonic real-valued sequence that is

bounded, then by the least upper bound principle (see Section

Solution:

2.2.11) there will exist some number M such that an ≤ M for all

We have:

n > 0 . There will then also exist a term aN of the sequence such

{ i } = i, − 1, − i, 1, !

that:

n

and these four terms keep repeating. Therefore the sequence

!

aN > M − ε !

(7.1-19)

has four limit points that it oscillates between, and so it

where ε > 0 can be arbitrarily small. Equation (7.1-19) follows

never reaches a unique limit. The sequence therefore

from the fact that M is the least upper bound (not M − ε ). All

diverges. Since we have i n = 1 for all n , the sequence i n

terms in the monotonic increasing real sequence an

{ }

is bounded by 1 although it does not converge. !

If

{ an }

A real-valued sequence

{ an }

{ an }

is defined as monotonic

is defined as monotonic decreasing if an ≥ an+1

for all integers n > 0 .

will then be bounded by: !

increasing if an ≤ an+1 for all n > 0 . Similarly, a real-valued sequence

{ } after aN

M − ε < an ≤ M !

for all n > N !

(7.1-20)

an − M < ε !

for all n > N !

(7.1-21)

or ! and so: !

lim a = M !

n→∞ n

(7.1-22) 314

{ }

Therefore a limit L exists for the real sequence an , and this limit is L = M :

a subsequence of the sequence z1 , z2 , z3 , z4 , z5 , z6 , z7 , z8 , z9 , !

lim a = L !

!

(7.1-23)

n→∞ n

( n = 1, 2, 3, 4, ! ).

{ }

and so if an is bounded, it is convergent.

{ }

Conversely, if an

!

sequence z2 , z4 , z6 , z8 , ! ( nk = 2, 4, 6, 8, ! , and k = 1, 2, 3, 4, ! ) is

Proposition 7.1-5, Subsequences Converge to the Same Limit

is a monotonic increasing real-valued

as the Sequence:

sequence that is convergent, then by Proposition 7.1-3 as

If a complex sequence

applied to complex numbers having a zero imaginary part, we

{ }

see that an must be bounded.

7.1.2!

infinite subsequences

■

SUBSEQUENCES

{ }

{ }

{ } occur within { z } in the same order,

{ }

k

of zn if all terms of zn

n

k

but not necessarily consecutively. The k th term of subsequence

{ z } is a term of the sequence { z } that we will designate the n term. The running index of a subsequence { z } is the k nk

n

th k

nk

index. We will always have nk ≥ n . !

{ } is then an infinite sequence formed

The subsequence zn

k

by deleting terms of

{ }

{ zn }

without changing the order of any

{ }

terms in zn . Gaps can be present in the subsequence zn compared to the sequence

converges to a limit z0 , then all

{ z } of { z } also converge to the limit nk

n

z0 .

A portion of a sequence zn is called a subsequence zn

!

{ zn }

k

as

Proof: !

{ }

Since the complex sequence zn

converges to a limit z0 ,

for any real number ε > 0 we must have for some N > 0 : !

zn − z0 < ε !

for all n > N !

(7.1-24)

{ } since n ≥ n , we also have:

For the subsequence zn

k

k

!

zn − z0 < ε !

for all nk > N !

k

and so the subsequence

{ }

limit z0 as does zn .

(7.1-25)

{ z } of { z } converges to the same nk

n

■

{ zn } . For example, we see that the 315

!

{ }

There will be at most only N terms of the subsequence

zn

Proposition 7.1-7:

for which:

k

zn − z0 ≥ ε !

!

nk ≤ N !

k

{ } of { z } that converges to z .

(7.1-26)

{ } satisfying equation

but that leaves an infinity of terms of zn

k

(7.1-25).

{ }

A point z0 is a limit point of an infinite bounded sequence zn if there is a subsequence zn Proof: !

7.1.3!

BOLZANO-WEIERSTRASS THEOREM

Proposition 7.1-6, Bolzano-Weierstrass Theorem for Complex

Every bounded sequence of complex numbers has a convergent subsequence.

If

{ zn }

is an infinite bounded sequence, then from the

Bolzano-Weierstrass theorem (Proposition 2.9-1), we know that

{ zn } .

This limit point will be the

{ }

convergence point for some subsequence zn . k

■

0

neighborhood of z0 will contain all but a finite number of the points of

{ z }. nk

Therefore every neighborhood of z0 will

{ }

limit point of the sequence zn .

!

a limit point z0 exists for

n

k

7.1.4!

Proof:

0

{ } of { z } converges to z , then every

If a subsequence zn

contain an infinite number of points of

Sequences:

!

n

k

{ zn } , and so must be a

■

CAUCHY’S CONVERGENCE CRITERIA

{ }

If a complex sequence zn converges to a limit, then for a

real number ε > 0 there must exist some N ∈! such that: !

zn − zk < ε !

for all n > k > N !

(7.1-27)

or !

zn + p− zn < ε !

for all n > N and p = 1, 2, ! ! (7.1-28)

The condition for convergence given in equation (7.1-27) is known as Cauchy’s convergence criteria. Basically, this criteria states that, if a sequence converges, then all but a finite number 316

of points of the sequence will be very close to each other. If a

Let

sequence satisfies Cauchy’s convergence criteria, it is known as

!

(

M = max zm

a Cauchy sequence. !

Cauchy’s convergence criteria can be written in terms of

the distance between two points in the sequence using equation (1.7-4):

(

)

d zn , zk < ε !

!

for all n > k > N !

zn − zk → 0 !

for all large enough n and k !

m = 1, 2, 3, !, k !

(7.1-33)

for all n > k !

(7.1-34)

so that:

zn < M !

!

and so the Cauchy sequence (7.1-29)

{ zn }

is bounded. From

Propositions 2.9-1 and 7.1-6 we know that

{ zn }

will then

possess a limit point z0 , and that some infinite subsequence

Cauchy’s convergence criteria is equivalent to: !

)+ε !

(7.1-30)

Proposition 7.1-8, Cauchy’s Convergence Criteria:

{ } converges if and only if it is a Cauchy

A complex sequence zn sequence.

{z } nk

of

{ zn }

will converge to z0 . Since

{ zn }

is a Cauchy

sequence, for any ε > 0 there will be an integer N1 such that:

zn − z0
N1 !

(7.1-35)

We now need to show that the limit point z0 for the

{ }

{ }

subsequence zn . is also the limit point for the sequence zn . k

Proof: !

{ }

If a complex sequence zn is a Cauchy sequence, then for

n > k we have for ε > 0 : !

We can write:

zn − zk < ε !

! for all n > k > N !

(7.1-31)

zn − z0 = zn − zn + zn − z0 ≤ zn − zn + zn − z0 ! k

zn = zn − zk + zk ≤ zk + zn − zk < zk + ε !

k

(7.1-36)

{ }

! (7.1-32)

k

Since zn is a Cauchy sequence we have:

We also can write: !

k

zn − zn < k

ε ! 2

(7.1-37)

317

From equations (7.1-37) and (7.1-35) we obtain: !

ε zn − z0 < ! 2

which is Cauchy’s convergence criteria. We see then that a complex sequence

for all n > k !

and so the Cauchy sequence

{ }

(7.1-38)

{ zn } converges to the same limit

k

■

Cauchy’s convergence criteria shows that if the difference

Conversely, if a complex sequence

{ zn }

sequence will converge. This criteria for convergence has the converges to a

limit z0 , we can select an N ∈! such that for any real number

ε > 0 we have: !

!

will converge if and only if it is a

between terms of a sequence becomes small enough, then the

z0 as the subsequence zn . !

Cauchy sequence.

{ zn }

advantage of not requiring that the limit of a sequence first be determined in order to prove that the sequence converges. Rather this criteria relies only on the convergence of sequence

ε zn − z0 < ! 2

for all n > N !

(7.1-39)

terms with respect to each other, and so the only information needed to test the sequence for convergence is the sequence

{ zn }

Since the terms zk in equation (7.1-31) constitute a subsequence

terms themselves. For a complex sequence

of

Cauchy convergence criteria, we have for the distance between

{ zn } , from Proposition 7.1-5 we see that { zk } the same limit as { zn } . Therefore we also have: !

ε zk − z0 < ! 2

converges to

for all n > k > N !

points zn and zk : n,k → ∞

(7.1-40)

)

(7.1-43)

Proposition 7.1-9:

(

) (

zn − zk = zn − z0 − zk − z0

)

≤ zn − z0 + zk − z0 !

zn − zk
k > N !

(7.1-42)

n,k → ∞

zn − zk = 0 !

(7.1-44)

Proof: !

Follows from Proposition 7.1-8.

■

318

We have: Proposition 7.1-10, Cauchy Sequence Converges:

!

zn − zk =

Every complex Cauchy sequence converges. and so:

Proof: !

1+ n 1+ k 1 1 i− i ≤ − n k n k

Follows from Proposition 7.1-8.

■

!

Proposition 7.1-11, Unique Limit of Cauchy Sequence:

lim zn − zk ≤ lim

n→ ∞ k→∞

n→ ∞ k→∞

1 1 − =0 n k

{ }

We see that zn is a Cauchy sequence. Convergence can be

Every complex Cauchy sequence has a unique limit.

shown directly: Proof: !

Follows from Propositions 7.1-10 and 7.1-2.

■

Proposition 7.1-12, Cauchy Sequence Bounded:

!

n→ ∞

SEQUENCE CONVERGENCE PROPERTIES

{ } can be

The terms of a sequence of complex numbers zn

written in the form:

Proof: Follows from Propositions 7.1-10 and 7.1-3.

■

{ }

Determine if the sequence zn is Cauchy and if it converges

1+ n i. n

!

zn = xn + i yn !

n = 1, 2, 3, ! !

(7.1-45)

We see then that each sequence of complex numbers has

Example 7.1-5

when zn =

1+ n 1 i = lim + i = i n→ ∞ n n→ ∞ n

lim zn = lim

7.1.5!

Every complex Cauchy sequence is bounded.

!

!

associated with it two corresponding sequences of real

{ } and { yn } .

numbers xn

Solution: 319

!

Proposition 7.1-13:

{

A complex sequence zn = xn + i yn

} converges to a limit

z0 = x0 + i y0 so that: (7.1-46)

n→∞ n

{ }

if and only if xn

converges to a limit x0 and

{ yn } converges

to a limit y0 so that:

lim xn = x0 !

!

(7.1-47)

n→∞

lim yn = y0 !

!

(7.1-48)

n→∞

!

If equations (7.1-47) and (7.1-48) hold, then we must have

for any real number ε > 0 some integer N1 > 0 such that: !

xn − x0
N1 !

(7.1-49)

!

yn − y0
N2 !

)

Letting N = max N1 , N2 , we have:

(7.1-50)

yn − y0
N !

(7.1-52)

(7.1-53)

(

) (

)

(

) (

)

(7.1-54)

zn − z0 ≤ xn − x0 + i yn − y0 !

(7.1-55)

zn − z0 ≤ xn − x0 + yn − y0 !

(7.1-56)

or

zn − z0 = xn − x0 + i yn − y0 !

We then have:

or !

Using equations (7.1-51) and (7.1-52) we obtain:

zn − z0
0 such that:

(7.1-51)

zn − z0 = xn + i yn − x0 + i y0 !

!

! Proof:

for all n > N !

We can write:

!

and

ε ! 2

and !

lim z = z0 !

!

xn − x0
N !

{ zn }

(7.1-57)

converges and we have

lim z = z0 if equations (7.1-47) and (7.1-48) hold.

n→∞ n

320

!

Conversely, if a sequence

{ zn }

converges to a limit

!

The limit of a complex sequence is therefore equivalent to

z0 = x0 + i y0 so that lim zn = z0 , then we must have for any real

the limit of two real sequences. This makes it possible to apply

number ε > 0 :

much of the theory of real sequences to complex sequences.

n→∞

(

) (

zn − z0 = xn + i yn − x0 + i y0

!

) N !

(7.1-58)

(

) (

zn − z0 = xn − x0 + i yn − y0

) N !

(7.1-59)

(

) (

)=

(

) (

)=

xn − x0 ≤ xn − x0 + i yn − y0 yn − y0 ≤ xn − x0 + i yn − y0

!

xn − x0 < ε ! yn − y0 < ε !

!

zn − z0 ! zn − z0 !

(7.1-60)

lim x = x0 !

n→∞ n

lim yn = y0 !

!

n→∞

■

)

n→ ∞

n→ ∞

(7.1-66)

(7.1-61)

Does the following sequence converge? !

for all n > N ! for all n > N !

{ zn } = ⎧⎨ 23nn −+ ii ⎫⎬ ⎩

⎭

Solution: (7.1-62) (7.1-63)

We have: !

and so we have: !

n→ ∞

Example 7.1-6

Therefore from equation (7.1-59): !

n→ ∞ n

provided the limits exist.

We can write: !

(

lim z = lim xn + i yn = lim xn + i lim yn !

!

or !

From Proposition 7.1-13 we have:

2 n + i 2 n + i 3n + i 6 n2 − 1 5n zn = = = 2 +i 2 3n − i 3n − i 3n + i 9 n + 1 9 n + 1 and so:

(7.1-64) ! (7.1-65)

1 n2 ! Re zn = 1 9+ 2 n

( )

6−

( )

Im zn =

5 n 9+

1 n2

We see that: 321

!

( )

x0 = lim Re zn n→ ∞

2 = ! 3

{ } then { zn } converges to 2

( )

!

y0 = lim Im zn = 0 n→ ∞

{ } converges to 0 ,

3.

!

Example 7.1-7

!

{ zn } = ⎧⎨( −1)n + n5i2 ⎫⎬ ⎩

⎭

Solution:

Solution:

We see that:

We see that the real part of each term of this sequence

!

1 x0 = lim Re zn = lim 2 = 0 n→ ∞ n→ ∞ n

!

y0 = lim Im zn = lim 2 i = 2i

oscillates between +1 and −1 and so does not converge to a

( )

n→ ∞

( )

unique value. Therefore the sequence does not converge. Proposition 7.1-14:

n→ ∞

{ }

If a sequence of complex numbers zn converges to a limit z0 :

Therefore: !

ε

Does the following sequence converge?

following sequence converge?

{ zn }

1

n>

Example 7.1-8

For any positive number ε , for what value of n does the

!

1 +2i−2i 0 :

zn − 0 = zn < ε !

!

for all n > N !

(7.1-81)

Therefore:

zn − 0 < ε !

!

lim z = z0 !

! for all n > N !

(7.1-82)

and so:

{

}

and ζ n − zn converges to a limit 0:

(

)

lim ζ n − zn = 0 !

n→ ∞

(7.1-83)

n→ ∞

(7.1-87)

n→ ∞ n

!

lim zn = 0 !

!

{ } and {ζ n } are two sequences of complex numbers, and if { zn } converges to a limit z0 :

If zn

(7.1-88)

{ }

then ζ n converges to a limit z0 : Conversely, if we have:

lim z = 0 !

!

Proof:

then from Proposition 7.1-13 we will have:

lim xn = 0 !

n→∞

lim yn = 0 !

n→∞

(7.1-85)

!

By the definition of convergence given in equation (7.1-3)

we must have for any real number ε > 0 some integer N1 > 0 such that:

and so:

lim zn = 0 !

!

(7.1-89)

n→ ∞

(7.1-84)

n→ ∞ n

!

lim ζ n = z0 !

!

n→ ∞

(7.1-86)

!

■

zn − z0
N1 !

(7.1-90)

and some integer N2 > 0 such that: !

ε ζ n − zn < ! 2

for all n > N2 !

(7.1-91) 324

(

)

Letting N = max N1 , N2 , we can write:

ζ n − z0 = ζ n − zn + zn − z0 !

!

!

for all n > N !

zero, these terms can be omitted without changing the limit of (7.1-92)

or

ζ n − z0 ≤ ζ n − zn + zn − z0 !

!

for all n > N ! (7.1-93)

Therefore:

ζ n − z0
0 such that:

ε ε + =ε! 2 2

for all n > N ! (7.1-94)

and so we can conclude that:

!

z0

zn − z0
N !

(7.1-98)

z0 − zn − z0 !

(7.1-99)

We can write:

lim ζ n = z0 !

!

{ }

If some finite number of terms of the sequence zn equal

(7.1-95)

n→ ∞

!

■

(

zn = z0 − z0 − zn

)≥

and so: Proposition 7.1-18: If a sequence of complex numbers

{ zn }

converges to a limit

z0 ≠ 0 : (7.1-96)

n→ ∞ n

{

!

}

then the sequence 1 zn converges to a limit 1 z0 : !

lim 1 zn = 1 z0 !

n→ ∞

zn ≥ z0 −

2

=

z0 2

> 0!

(7.1-100)

We can write:

lim z = z0 !

!

!

z0

(7.1-97)

z −z 1 1 − = 0 n ! zn z0 zn z0

(7.1-101)

If we now choose ε > 0 for some integer N1 > N > 0 such that: 2

Proof:

!

zn − z0
N1 !

(7.1-102) 325

then we have:

Proposition 7.2-1: 2

z0 ε 1 1 2 − < ! zn z0 zn z0

!

for all n > N1 !

{ ( ) } and { b ( z )} are two convergent sequences of complex numbers, then { a ( z ) + b ( z ) } is also a convergent sequence of If an z

(7.1-103)

n

2 z0

2

=ε !

(7.1-104)

lim 1 zn = 1 z0 !

n→∞

(7.1-105)

■

7.2! !

COMPLEX SEQUENCE OPERATIONS Convergent complex sequences can be added, subtracted,

Let

()

()

{ a ( z )} n

{ b ( z )}

converge to L1 and

n

(7.2-1)

converge to L2 .

ε > 0 there must then exist some N1 ∈! so that: ε an z − L1 < ! 2

()

!

sequences involve a limiting process, proofs of the propositions for these arithmetic operations are very similar to those given

for all n > N1 !

(7.2-2)

for all n > N2 !

(7.2-3)

and some N2 ∈! so that:

()

bn z − L2
N : 326

ε ε ⎡⎣ an z ± bn z ⎤⎦ − ⎡⎣ L1 ± L2 ⎤⎦ < + = ε ! 2 2

()

!

()

for n > N ! (7.2-5)

and since ε > 0 can be arbitrarily small:

()

!

()

lim ⎡ a z ± bn z ⎤⎦ = L1 ± L2 ! n→ ∞ ⎣ n

!

(7.2-6)

n

converges. Moreover we

n

have:

()

()

()

()

lim ⎡ a z ± bn z ⎤⎦ = lim an z ± lim bn z ! n→ ∞ ⎣ n n→∞ n→∞

!

{ a ( z ) } converge to L and { b ( z )} converge to L . { a ( z ) } and { b ( z )} converge, they are bounded (see

Let

Since

n

1

n

n

2

n

Proposition 7.1-3) so there exists some positive number M such

{ a ( z ) + b ( z )}

and so the sequence

Proof:

(7.2-7)

that: !

()

an z ≤ M !

()

bn z ≤ M !

for all n ∈! !

(7.2-9)

and the limits will then also be bounded:

■

!

Proposition 7.2-2:

L1 ≤ M !

L2 ≤ M !

for all n ∈! !

(7.2-10)

From the definition of a sequence limit, for any real number The sum or difference of two null sequences is a null sequence. Proof: !

!

Follows from Proposition 7.2-1.

■

()

an z − L1 ≤

()

bn z − L2 ≤

{ ( ) } and { b ( z )} are two convergent sequences of complex numbers, then { a ( z ) b ( z )} is also a convergent sequence of

!

complex numbers with:

! an z bn z − L1 L2 =

If an z

n

n

() ()

n

()

()

lim ⎡⎣ an z bn z ⎤⎦ = lim an z lim bn z ! n→ ∞ n→ ∞

n→ ∞

ε ! 2M

for all n > N1 !

(7.2-11)

and some N2 ∈! so that:

Proposition 7.2-3:

!

ε > 0 , there must then exist some N1 ∈! so that:

(7.2-8)

ε ! 2M

for all n > N2 !

(7.2-12)

We can write:

() ()

!

() ()

()

()

an z bn z − an z L2 + an z L2 − L1 L2 !

(7.2-13) 327

or

Proposition 7.2-4:

() ()

{ ( ) } and { b ( z )} are two convergent sequences of complex numbers, then { a ( z ) b ( z ) } is also a convergent sequence of

an z bn z − L1 L2 =

!

If an z

() ()

()

an z ⎡⎣ bn z − L2 ⎤⎦ + L2 ⎡⎣ an z − L1 ⎤⎦ !

!

(7.2-14)

n

()

{ ( )} is not zero. We will then have:

limit of bn z

() ()

! an z bn z − L1 L2 ≤

( ) bn ( z ) − L2

an z

()

+ L2 an z − L1 !

(7.2-15)

ε ε ! an ( z ) bn ( z ) − L1 L2 < M +M 2M 2M

!

(

! (7.2-16)

)

Let N = max N1 , N2 . We then have:

() ()

an z bn z − L1 L2
N ! (7.2-17)

()

lim b z !

n→ ∞ n

(7.2-20)

n→ ∞

(7.2-18)

{ a ( z ) b ( z )} n

n

converges. Moreover we

have:

() ()

()

()

lim ⎡⎣ an z bn z ⎤⎦ = lim an z lim bn z ! n→ ∞ n→ ∞

n→ ∞

n

n

1

n

2

n

such that:

()

an z ≤ M !

()

bn z ≤ M !

for all n ∈! ! (7.2-21)

and the limits will then also be bounded:

() ()

and so the sequence

{ a ( z ) } converge to L and { b ( z )} converge to L . { a ( z ) } and { b ( z )} converge, they are bounded (see

Let

Since

!

lim ⎡⎣ an z bn z ⎤⎦ = L1 L2 !

!

■

n→ ∞ ⎣ n

Proposition 7.1-3), so there exists some positive number M

and since ε can be arbitrarily small:

!

lim an ( z ) ( ) bn ( z ) ⎤⎦ = n→ ∞

lim ⎡ a z

! Proof:

or

!

n

complex numbers providing that bn z ≠ 0 for all n > 0 and the

Using the triangle inequality, we have:

!

n

!

L1 ≤ M !

L2 ≤ M !

for all n ∈! ! (7.2-22)

From the definition of a sequence limit, for any real number

ε > 0 , there must then exist some N1 ∈! so that: (7.2-19)

!

()

an z − L1 < ε
N1 !

(7.2-23) 328

( ) L1 − bn ( z ) L2

and some N2 ∈! so that: !

L2

()

bn z − L2 < ε
N2 !

!

2

(7.2-24)

(

( ) L1 − bn ( z ) L2

an z

=

()

()

an z L2 − L1 bn z

()

L2 bn z

!

() ()

(7.2-25)

( )2

Adding and subtracting an z bn z in the numerator and L2

( ) L1 − bn ( z ) L2

an z

=

()

() () 2 ⎡⎣ L2 ⎤⎦ + L2 ⎡⎣ bn ( z ) − L2 ⎤⎦ ()

an z ⎡⎣ L2 − bn z ⎤⎦ + bn z ⎡⎣ an z − L1⎤⎦

( ) L1 − bn ( z ) L2

!

( ) L1 − bn ( z ) L2

or

(7.2-29)

2

≤

( ) bn ( z ) − L2

an z

L2

2

( ) an ( z ) − L1 ! bn ( z ) − L2

+ bn z

− L2


N !

() ()

⎡ an z ⎤ L1 ! lim ⎢ ⎥= n→ ∞ b z L ⎢⎣ n ⎥⎦ 2

!(7.2-26)

and so the sequence

(7.2-30)

()

(7.2-31)

{ a (z) b (z) } n

n

converges providing that

bn z ≠ 0 . Moreover we have:

(7.2-27)

lim an ( z ) ( ) bn ( z ) ⎤⎦ = n→ ∞

lim ⎡ a z

!

n→ ∞ ⎣ n

()

lim b z !

n→ ∞ n

(7.2-32)

■

Proposition 7.2-5:

Therefore: !

L2 − L2

L2

and since ε can be arbitrarily small: !

and so from the triangle inequality:

an z

2

ε!

)

an z

!

in the denominator, we have: !

2M


N !

a region R if and only if for every sequence

() ()

an z bn z < M

!

ε =ε! M

for all n > N !

{ ( ) ( ) } is a null sequence.

and so an z bn z

7.3! !

(7.2-35)

SEQUENCES OF COMPLEX FUNCTIONS A sequence of complex functions

()

when a complex function fn z

converges to z0 , the sequence f zn

{ f ( z )} n

is obtained

that is defined and single-

0

Proof: ! If f z is continuous at z0 , we have for any real number ε > 0:

()

()

( )

f z − f z0 < ε !

!

■

when z − z0 < δ !

{ }

zn − z0 < δ !

!

for all n > N !

one correspondence with the set of positive integers. For any

!

( )

!

{ f ( z )} = f ( z ) , f ( z ) , f ( z ) , !, f ( z ) ! n

1

2

3

n

( )

f zn − f z0 < ε !

given point z in the region R , the sequence of complex n

(7.3-3)

()

Since f z is continuous at z0 , we then obtain:

{ f ( z )} becomes a sequence of complex numbers:

(7.3-2)

If zn converges to z0 , we also have:

valued in a region R of the complex plane can be put in one-to-

functions

in R that

{ ( )} converges to f ( z ) .

(7.2-34)

where ε → 0 as n → ∞ . Therefore we have:

{ zn }

for all n > N !

Because ε is arbitrary, the sequence

( )

{ f ( z )} n

(7.3-4) converges to

f z0 .

(7.3-1) 330

()

Conversely, if f z is not continuous at z = z0 , then for

!

( ( )) will not contain the

some real number ε > 0 the disk Dε f z0

()

image created by f z

of any of the neighborhoods of z0 .

{ }

Therefore a sequence of points zn can exist in R such that:

( )

where δ = 1 n !

zn ∈Dδ z0 !

!

(7.3-5)

( ( ))!

( )

f zn ∉Dε f z0

{ f ( z )} converges to some function f ( z ) : n

()

( )

where f z − f z0 < ε ! (7.3-6)

n→ ∞ n

(7.3-9)

then the sequence of complex functions

{ f ( z )} n

is called a

pointwise convergent sequence and is said to converge functions

{ f ( z )} defines a function f ( z ) in R . The value of the

()

n

limit f z depends of the specific point z . The pointwise limit

()

for all n ∈! !

(7.3-7)

{ zn } converge to z0 and yet not have f ( z0 ) , unless f ( z ) is continuous at z0 . ■

Proposition 7.3-2, Cauchy’s Convergence Criteria for Complex

we can have a sequence

{ f ( z )} converge to n

Functions: A sequence of complex functions

()

{ f ( z )} to converge to a limit

f z at a point z is that, for any ε > 0 , there exists N ∈! such

()

()

fn z − f z < ε !

( )

()

()

fn z − fk z < ε !

for all n > k > N !

(7.3-10)

(7.3-8)

The number N can depend on both ε and z , and so we write

N = N ε, z .

such that: !

for all n > N !

n

Cauchy. For any real number ε > 0 , there then exists N ∈!

n

that:

{ f ( z )} converges pointwise

on a domain D if and only if in D the sequence is pointwise

A necessary and sufficient condition for a sequence of

single-valued complex functions

!

()

f z need not be continuous in R . zn − z0 < δ = 1 n !

!

()

lim f z = f z !

!

pointwise to some f z within R . The sequence of complex

Since !

sequence

()

and !

If at each point z within the region of convergence the

!

Proof: !

Follows from Proposition 7.1-8.

■

331

Proposition 7.3-3:

Proposition 7.3-4:

If a sequence of single-valued complex functions continuous at a point z = z0 , then the sequence

{ f ( z )} is

If a sequence of continuous single-valued complex functions

n

{ f ( z ) } is also

{ f ( z )} converges pointwise at a point z = z within its region of convergence R , then { f ( z )} converges to f (z0): n

n

continuous at point z0 .

!

n

Since

{ f ( z )} is continuous at a point z = z , we have for n

0

any real number ε > 0 :

()

( )

fn z − f z0 < ε !

!

when zn − z0 < δ !

(7.3-11)

We also have:

()

( )

()

( )

≤ fn z − f z0 !

(7.3-12)

and so

()

( )

fn z − f z0

{ f ( z )} is a continuous sequence within its region of n

()

( )

fn z − f z0 < ε !

when zn − z0 < δ !

(7.3-15)

Since ε can be arbitrarily small, δ can approach zero. Therefore greater than

some N :

0 : !

fn z − f z0

!

n→∞ n

Proof: !

()

lim f z = f z0 !

!

Proof:

0

0

■

(7.3-13)

!

zn − z0 < δ !

when n > N !

(7.3-16)

We can then rewrite equation (7.3-15) as: !

()

( )

fn z − f z0 < ε !

when n > N !

(7.3-17)

or since ε can be arbitrarily small, we see that

{ f ( z )} n

converges to: 332

()

( )

lim f z = f z0 !

!

n→∞ n

(7.3-18)

from its pointwise limit at that same point). A sequence that

()

converges pointwise to a limit function f z within R need not

■

()

converge uniformly, however, to a limit function f z within

7.4!

UNIFORM CONVERGENCE OF SEQUENCES OF COMPLEX FUNCTIONS If for the same value of N ∈! , a sequence of complex

!

functions

{ f ( z )} n

()

converges to f z

R. Proposition 7.4-1, Uniformly Convergent Sequence:

for all points z in its

A sequence of single-valued complex functions

region of convergence R when n > N , then the sequence is said

()

()

()

fn z − f z < ε !

!

() N = N ( ε , z ) . A sequence of complex functions { fn ( z )} that is uniformly convergent to f ( z ) will converge at approximately the same rate to f ( z ) for all points z within its region of convergence R , and f ( z ) will be continuous within R (see uniform convergence, therefore, we have N = N ε , but not

Proposition 7.4-2). !

()

f z

()

for any real number ε > 0 there exists N = N ε ∈! such that:

()

()

pointwise to f z for all points z within R (the uniform limit of the sequence at any particular point z in R cannot differ

()

fn z − fk z < ε ! for all z ∈R when n, k > N ! (7.4-2)

! Proof: !

If

{ f ( z )} converges uniformly to f ( z ) within its region of n

convergence R , then for any real number ε > 0 we have for some N ∈! : !

ε fn z − f z < ! 2

!

If

()

A sequence that converges uniformly to a limit function within its region of convergence R also converges

()

if and only if the sequence is uniformly Cauchy within R so that

for all z ∈R when n > N ! (7.4-1)

The number N can then depend on ε , but not z . For

!

n

converges uniformly within its region of convergence R to f z

to converge uniformly to f z , and we have for any real number ε > 0 :

{ f ( z )}

()

{ f (z)} k

for all z ∈R when n > N !

is a subsequence of

7.1-5 we know that

{ f (z)} k

(7.4-3)

{ f ( z )} , from Proposition n

converges to the same limit as

{ f ( z )} . Therefore we also have: n

333

ε fk z − f z < ! 2

()

!

()

for all z ∈R when k > N !

(7.4-4)

We can write:

( ) ( fn ( z ) − f ( z ) ) − ( fk ( z ) − f ( z ) ) !

()

fn z − fk z =

!

()

()

()

(7.4-5)

()

()

()

()

()

! !

(7.4-6)

ε ε + =ε 2 2 for all z ∈R when n, k > N !

()

(7.4-9)

{ f ( z )} n

{ f ( z )} is uniformly n

■

Proposition 7.4-2:

n

{ f ( z )} is uniformly Cauchy n

within its region of convergence R , then for any real number

ε > 0 we have for some N ∈! such that n, k > N : ε fn ( z ) − f k ( z ) < ! 2

()

(7.4-7)

n

Conversely, if the sequence

within

its region of convergence R , then f z is continuous within R .

Therefore if a sequence of single-valued complex functions

!

ε N , and so

or by the triangle inequality: !

()

lim fn z − fk z = fn z − f z ≤

!

for all z ∈R when n, k > N ! (7.4-8)

Holding n constant but with n > N , and letting k vary, we can

Proof: !

Since

{ f ( z )} n

()

converges uniformly to f z

within its

region of convergence R , for any real number ε > 0 we can choose n so that for some N : !

ε fn z − f z < ! 3

()

()

for all z ∈R and n > N ! (7.4-10)

where N generally depends upon ε . !

Since

{ f ( z )} converges to f ( z ) at every point z within its n

region of convergence R , if z0 is some given point within R , we must have:

write: 334

( )

!

( )

fn z0 − f z0
N !

()

Since we are given that each term fn z

(7.4-11) is continuous

within R , for the same ε > 0 there must exist a δ > 0 such that:

()

! !

( )

fn z − fn z0

!


0 : !

lim

!

We are given that:

within its region of

convergence R , then for any contour C within R we have the

convergence R , and if within R the terms of the sequence are all

Proof:

()

converges uniformly to f z

n

within its region of

We then have on C : ! !

()

()

lim f z = f z !

n→ ∞ n

(7.4-33)

()

Since the functions fn z are all continuous within R , they

()

are integrable. From Proposition 7.4-2 we know that f z is 337

()

continuous within R , and so f z is also integrable. We then have:

∫

!

∫

()

lim f z dz =

n C n→ ∞

()

(7.4-34)

We can write: !

∫

C

()

∫

()

f z dz =

C

∫

C

()

()

⎡⎣ fn z − f z ⎤⎦ dz ! (7.4-35)

∫

()

fn z dz −

C

∫

()

f z dz ≤

C

∫

C

∫

C

()

fn z dz =

∫

()

lim f z dz =

n C n→ ∞

∫

()

f z dz !

C

(7.4-40)

Equation (7.4-40) is not necessarily true if

()

{ f ( z )} does not n

converge uniformly to f z within R . Proposition 7.4-6:

or using Proposition 5.3-7: !

n→ ∞

■

!

fn z dz −

lim

!

f z dz !

C

Using equation (7.4-34) we have finally:

()

()

fn z − f z

If a sequence of holomorphic functions

()

dz ! (7.4-36)

{ f ( z )} n

converges

()

uniformly to f z in a simply connected domain D , then f z is holomorphic in D .

From the ML inequality (Proposition 5.3-8) and equation Proof:

(7.4-31) we then have: !

∫

()

fn z dz −

C

∫

C

ε f ( z ) dz < L ! for all n > N ! (7.4-37) L

Since the sequence of functions

∫

C

()

∫

()

f z dz < ε !

C

for all n > N !

(7.4-38)

single-valued

()

obtain:

Proposition 7.4-5:

lim

∫

C

()

fn z dz =

∫

C

()

f z dz !

(7.4-39)

{ f ( z )} n

converges

()

Proposition 7.4-2 we know that f z will be continuous in D . !

n→ ∞

functions

uniformly to f z in a simply connected domain D , then by

Therefore since ε > 0 becomes arbitrarily small as n → ∞ , we

!

n

in the domain D , we know from Proposition 3.7-1 that they will continuous

fn z dz −

{ f ( z )} are holomorphic

be continuous single-valued functions. Since the sequence of

and so: !

!

!

For any simple closed contour C within D , we have from

lim

n→∞

!∫

C

()

fn z dz =

!∫

C

()

f z dz !

(7.4-41) 338

()

Since the functions fn z are all holomorphic in D , we have

!

from the Cauchy-Goursat theorem (Proposition 5.4-2):

if z = z0 is any given point inside the contour C , then from

!∫

!

C

()

fn z dz = 0 !

Proposition 6.2-1 we know that derivatives of all orders of

for each n > 0 !

!∫

C

()

f z dz = 0 !

(7.4-43)

()

that f z is holomorphic in D .

■

{ f ( z )} is a sequence of continuous complex functions that are differentiable, and if { f ( z )} converges uniformly to f ( z ) in a domain D , it need not be true that { f ′ ( z )} converges If

n

n

()

( ) dz ! !∫C ( z − z0 )k+1 fn z

(7.4-44)

( z − z0 )k+1 is continuous on C , and { fn ( z )} uniformly to f ( z ) in D , we have on C : fn ( z ) f (z) ! ! lim = k+1 k+1 n→ ∞ ( z − z0 ) ( z − z0)

()

{ f ( z )} be a sequence of holomorphic functions in a simply connected domain D . If { f ( z )} converges uniformly to f ( z )

k! k lim f n( ) z0 = lim n→ ∞ n→ ∞ 2 π i

( )

!∫

C

{

( ) } converges uniformly to

k in D , then f n( ) z

k f ( ) z in D .

()

(7.4-45)

( ) dz ( z − z0 )k+1 fn z

( ) dz ! !∫C n→ ∞ ( z − z0 )k+1

(7.4-46)

() (k ) z ! dz = f ( 0) k+1 ( z − z0 )

(7.4-47)

k! 2π i

=

!

n

n

converges

Using Proposition 7.4-5 we have: !

Proposition 7.4-7:

Proof:

k! k f n( ) ( z0 ) = 2π i

!

n

to f ′ z , or that f ′ z even exists.

Let

fn z in D exist at point z0 , and are given by:

Since 1

From Morera’s theorem (Proposition 6.3-4) we can conclude

!

()

(7.4-42)

Therefore, for any simple closed contour C within D : !

If C is a simple closed contour within the domain D , and

lim

fn z

and so: !

k! k lim f n( ) ( z0 ) = n→ ∞ 2π i

!∫

C

f z

339

Since z0 is any point in D , we see that k uniformly to f ( ) z in D . ■

()

{ f ( ) ( z ) } converges k

n

340

Chapter 8 Complex Series

∑

f (z) =

n≥0

n) ( f ( z0 )

n!

( z − z0 )

n

341

!

In this chapter we will define and review the properties of

infinite series of complex numbers. We will present several tests

{ } is known as the k th

terms of an infinite complex sequence zn partial sum:

for the convergence and divergence of infinite complex series. We will show that every holomorphic function can be

!

k

∑z = z + z + z + ! + z !

()

Sk z =

n

1

2

3

k

(8.1-2)

n =1

represented as a complex power series, and we will determine the region of convergence of such series. We will describe

As additional terms of the sequence are summed, new partial

Taylor series, Maclaurin series, and Laurent series.

sums are formed:

8.1! !

SERIES OF COMPLEX NUMBERS

!

1

()

S1 z =

{ zn } . Complex series therefore require the

n

2

()

!

S2 z =

An infinite complex series is defined as the sum of the terms zn

!

!

of an infinite complex sequence zn :

!

existence of a sequence and the algebraic operation of addition.

{ }

∞

!

∑ z = z + z + z + !! n

1

2

∑z = z + z ! n

1

2

n =1

(8.1-3)

3

()

S3 z =

∑z = z + z + z n

1

2

3

n =1

(8.1-1)

3

n =1

!

1

n =1

A complex series is defined as the sum of the terms zn of a

complex sequence

∑z = z

Since it is not possible to add an infinity of terms, it is not

possible to directly obtain the sum of an infinite complex sequence. Instead the sum of an infinite complex sequence must be determined in the form of a limit. To obtain a limit that can represent the sum of an infinite complex sequence, a

()

!

! !

()

k

Sk z =

∑z = z + z + z + ! + z n

1

2

3

k

n =1

()

and so Sk z

represents the summation of the terms of the

{ }

infinite complex sequence zn truncated at the k th term.

sequence of partial sums is used. The sum Sk z of the first k 342

{ zn } , the set of partial sums forms a complex sequence of partial sums { Sk ( z ) } : !

For any given complex sequence

{S ( z ) } = S ( z ) , S ( z ) , S ( z ) ,!, S ( z ) !

!

k

1

2

3

1

1

2

1

2

3

k

the infinite series given in equation (8.1-1) is said to converge to a sum (see Section 8.2). For the purpose of determining a limit,

!

{ zn }

can be

{ ( )} :

recovered from the complex sequence of partial sums Sk z

()

z1 = S1 z !

! !

()

!

= 1+ z + z 2 + ! + z n +!!

()

Sk z =

() (

(

) )

− z1 + z2 + z3 + z4 +!+ zk−1 !

(8.1-7)

k

∑ n=0

(8.1-9)

1− z k+1 ! z = 1− z n

(8.1-10)

Another form of the geometric series is: ∞

∑

a z n = a + a z + a z 2 + ! + a z n +!!

(8.1-11)

n=0

where a is a complex constant. This geometric series has the k th partial sum: !

()

Sk z =

(

a 1− z k+1 1− z

)!

(8.1-12)

If we replace z with − z in equation (8.1-9) we have: ∞

and so: !

n

(8.1-6)

Sk z − Sk−1 z = z1 + z2 + z3 + z4 +!+ zk−1 + zk

!

∑z

which has the k th partial sum (see Example 8.1-1):

!

Individual terms of a complex sequence

A very important series is the complex geometric series:

n=0

therefore, an infinite series is represented by a sequence of partial sums.

GEOMETRIC SERIES ∞

!

(8.1-5) {S ( z ) } = z , ( z + z ) , ( z + z + z ) , !! If the sequence of partial sums { S ( z ) } has a finite limit, then k

!

(8.1-4)

k

or !

8.1.1!

()

()

zk = Sk z − Sk−1 z !

!

k > 1!

(8.1-8)

n 2 3 −z = 1− z + z − z +! ! ( ) ∑

(8.1-13)

n=0

which is also a geometric series. 343

!

given in equations (8.1-9) and (8.1-13) in the forms: !

1 = 1+ z + z 2 + z 3 +! ! 1− z

z < 1!

z < 1!

()

Sk z =

∑ n=0

Show that we can write for the geometric series:

()

1− z k+1 z = 1− z

Solution: From equations (8.1-9) and (8.1-10) we have the geometric series expansion:

k

∑

!

Forming the product of Sk z and z :

()

z Sk z = z + z 2 + z 3 + ! + z k+1

k−1

1− z k = 1− z

()

2

k−1

1 zk = − 1− z 1− z

or

z n = 1+ z + z 2 + ! + z k

()

()

2

Sk −1 z = 1+ z + z + ! + z

!

n=0

!

1 zk 2 k−1 = 1+ z + z + ! + z + 1− z 1− z

!

n

We have:

Sk z =

()

Example 8.1-2

Solution:

!

1− z k+1 Sk z = 1− z

(8.1-15)

Show that the k th partial sum of a complex geometric series is given by: !

()

and so:

(8.1-14)

Example 8.1-1

k

()

!

1 = 1− z + z 2 − z 3 +! ! 1+ z

()

Sk z − z Sk z = 1− z k+1

!

and, replacing z with −z : !

()

Subtracting z Sk z from Sk z :

Finally we note that we can write the two geometric series

Sk −1 z = 1+ z + z + ! + z Therefore:

!

1 zk 2 k−1 = 1+ z + z + ! + z + 1− z 1− z 344

where for every real number ε > 0 :

8.1.2! !

OTHER SPECIAL SERIES

An important series of real numbers is the harmonic

!

series: ∞

∑

!

n =1

1 1 1 1 1 = 1+ + + + +! ! n 2 3 4 5

()

()

S z − Sk z < ε !

!

for all k > N !

(8.2-2)

The number N ∈! depends in general on both ε and z .

{ ( ) } converges

The rate at which a sequence of partial sums Sk z (8.1-16)

()

to a limit S z

is determined by N , since N specifies the

()

number of terms necessary in the partial sum Sk z to satisfy

()

which diverges very slowly. This series is useful for comparison

equation (8.2-2). The limit S z is recognized as the series sum.

testing of certain complex series.

Convergence of a series is then defined as the convergence of a

!

sequence of partial sums.

Related to the harmonic series are the p-series: ∞

∑

!

n =1

1 ! np

(8.1-17)

!

The properties of an infinite complex series are defined in

terms of properties of its sequence of partial sums. If a sequence

{ ( ) } of a series does not converge, the series

When p = 1 we have the divergent harmonic series. When p > 1

of partial sums Sk z

we have a real number p-series that converges.

is called a divergent series. A series will diverge either because

8.2! !

CONVERGENT AND DIVERGENT COMPLEX SERIES A series is said to converge to a limit when its sequence of

{ ( ) } has a limit S ( z ) that is finite such that:

partial sums Sk z !

()

()

k

S z = lim Sk z = lim k→∞

k→∞

∑ n =1

zn !

()

the lim Sk z k→∞

does not exist (the sequence of partial sums

()

oscillate), or because lim Sk z = ∞ . A sequence of partial sums k→∞

that oscillates can have several limit points, although it does not converge to any one of them. ∞

!

A complex series

∑z

n

is said to be:

n =1

(8.2-1)

k

!

!

Convergent if lim

k→∞

∑z

n

converges.

n =1

345

k

!

!

Absolutely convergent if lim

k→∞

∑z

n

converges.

Proposition 8.2-1, Series Convergence Requirement:

n =1

!

!

Conditionally convergent if lim

k→∞

k

!

but lim

!

k→∞

∑z

n

∑

!

!

Divergent if lim

k→∞

∑

()

sum S z , it is necessary that the k th term zk of the sequence

zn converges,

{ zk } have the limit zero:

n =1

does not converge.

lim z = 0 !

! Proof:

zn does not converge.

!

n =1

From the definition of an absolutely convergent series, we

see that it is the moduli of the terms that are added to form the partial sums. Since partial sums of a real sequence

{ z } will n

constitute a sequence of positive numbers, this sequence of partial sums must either converge absolutely to a limit or If a series of complex numbers converges absolutely, then

the series will also converge (see Proposition 8.3-4). Also if a

!

Proposition 8.2-10). This is not true for conditionally convergent series. Rearranging the terms of a conditionally convergent series can produce different sums.

()

()

()

k

Sk z =

∑z !

()

k−1

Sk−1 z =

n

n=2

∑z ! n

(8.2-4)

n=2

()

Since the series converges to S z , we must have: !

()

()

lim S z = S z !

k→∞ k

lim S

k → ∞ k−1

(z) = S (z)!

(8.2-5)

From Proposition 2.3-7 we have: !

complex series is absolutely convergent, the terms of the series can be rearranged without affecting the value of the sum (see

(8.2-3)

k→∞

We will let Sk z and Sk−1 z equal the partial sums:

increase without bound. !

lim zk = 0 !

k→∞ k

n =1

k

!

{ ( ) } to converge to a

For a complex sequence of partial sums Sk z

k

( ()

S ( z ) − lim S k−1 ( z ) ! ( ) ) = klim →∞ k k→∞

lim Sk z − Sk −1 z

k→∞

(8.2-6)

and so: !

( ()

( ) ) = S (z) − S (z) = 0 !

lim Sk z − Sk −1 z

k→∞

(8.2-7)

From equation (8.1-8) have: 346

()

()

Sk z − Sk−1 z = zk !

!

k > 1!

(8.2-8)

We then obtain:

lim z = 0 !

!

lim zk = 0 !

k→∞ k

k→∞

(8.2-9)

()

The converse of Proposition 8.2-1 is not true, however. It is

possible to have lim zk = 0 , and yet for the sequence of partial k→∞

sums not to converge. Therefore lim zk = 0 is a necessary but

()

S z =

!

{ ( )}.

Show that the complex geometric series converges if z < 1

k→∞

1 1 − lim z k+1 1− z 1− z k → ∞ k→∞

!

sequence of partial sums Sk z Example 8.2-1

⎡ 1 k+1 ⎤ 1 − lim ⎢ z ⎥ k → ∞ 1− z k → ∞ 1− z ⎣ ⎦

If z < 1 then lim z k+1 = lim z

k→∞

not sufficient condition for convergence of the complex

()

or

Proposition 8.2-1 provides a test for the divergence of a

and diverges if z > 1 .

()

S z = lim Sk z = lim

!

series. If equation (8.2-3) does not hold, then the series diverges. !

()

Therefore:

■

!

⎡ 1 1 k+1 ⎤ S z = lim Sk z = lim ⎢ − z ⎥ k→∞ k → ∞ 1− z 1− z ⎣ ⎦

!

()

S z =

k→∞

= 0 and we have

1 , and the complex geometric series converges. 1− z

If z > 1 then lim z k+1 = lim z k→∞

!

k+1

k+1

= ∞ , and the complex

k→∞

geometric series diverges.

()

(

)

We see that the series S z = lim 1+ z + z 2 + ! + z k only k→∞

()

converges when z < 1 , while the function S z = 1 1− z is Solution:

valid for all values of z except z = 1.

From equation (8.1-10) we have: !

1− z k+1 1 1 k+1 Sk z = = − z 1− z 1− z 1− z

()

Example 8.2-2 Does the following complex series converge?

and so: 347

∞

∑

!

k =1

()

Rk z = zk +1 + zk +2 + zk +3 + !!

!

k 3k + 5

(8.2-12)

Proposition 8.2-2, Series Remainder Theorem:

{ ( ) } will converge to a

Solution:

A complex sequence of partial sums Sk z

We have:

sum S z

()

k = lim k → ∞ 3k + 5 k→∞

lim zk = lim

!

k→∞

1 3+

5 k

=

1 3

∞

that the complex series

∑ k =1

k does not converge. 3k + 5

() () () (8.2-10) where Rk ( z ) is defined as the k th remainder of the series or the S z = Sk z + Rk z !

remainder after k terms. We then have: !

()

()

()

Rk z = S z − Sk z !

Since the k

th

()

lim Rk z = 0 !

!

Let !

of the series

converges so that:

Therefore lim zk ≠ 0 and so from Proposition 8.2-1 we see k→∞

()

if and only if the remainder Rk z

k→∞

(8.2-13)

Proof: ! !

If we have:

()

lim Rk z = 0 !

k→∞

(8.2-14)

then for any for every real number ε > 0 , there must exist an N such that: !

()

Rk z < ε !

for all k > N !

(8.2-15)

for all k > N !

(8.2-16)

From equation (8.2-11) we have: (8.2-11)

!

remainder of the series is what remains of the

series after removing the partial sum of the first k terms, we

or

have:

!

()

()

S z − Sk z < ε !

()

()

lim S z = S z !

k→∞ k

(8.2-17) 348

{ ( ) } converges

and so the complex sequence of partial sums Sk z

()

!

Follows from Proposition 8.2-2.

■

to the sum S z . !

Proposition 8.2-4:

Conversely, if the complex sequence of partial sums

{ S ( z ) } converges to the sum S ( z ) , we must have:

If a complex sequence of partial sums

()

k

()

()

lim S z = S z !

!

k→∞ k

()

k

converges to a

sum S z , then the sequence of partial sums is bounded. (8.2-18) Proof:

We can then write:

()

S z − lim Sk z = 0 !

!

{ S ( z )}

k→∞

(8.2-19)

or

!

Follows from Proposition 7.1-3.

!

Since the partial sums of a series form a sequence, the

■

Cauchy criterion for convergence of a sequence can be used to

( ()

( )) = 0!

lim S z − Sk z

!

k→∞

(8.2-20)

Proposition 8.2-5, Cauchy’s Series Convergence Criteria:

From equation (8.2-11) we must then have:

()

lim Rk z = 0 !

!

k→∞

determine if a series converges to a limit sum.

A complex series converges if and only if, for any real number

(8.2-21)

ε > 0 there exists N ∈! such that:

■

Proposition 8.2-3:

{ ( ) } and the remainders { R ( z ) } will

For any k the complex sequence of partial sums Sk z corresponding complex sequence of converge or diverge together. Proof:

()

k

()

Sn z − Sk z

!

k > N !

(8.2-22)

Proof: !

From Proposition 7.1-8 we know that a complex sequence

converges if and only if the sequence is Cauchy. Considering

()

the partial sums Sn z

()

and Sk z

as terms in a complex

sequence, based on the Cauchy criteria for sequences we see 349

that equation (8.2-22) provides the necessary and sufficient

()

requirement for a sequence of partial sums of a series to converge. !

()

Sn z − Sk z = zk+1 + zk+2 + zk+3 +!+ zn !

!

(8.2-25)

or

■

The partial sums of a series must then cluster as the series

()

()

Sn z − Sk z =

!

j

!

n>k > N !

(8.2-26)

and so:

series first be determined in order to prove that the series converges. We can restate Cauchy’s convergence criteria for

∑z

j = k+1

progresses if the series is to converge. Cauchy’s convergence criteria has the advantage of not requiring that the limit of a

n

n

∑z

!

j

k > N !

(8.2-27)

j = k+1

series in the following way: ■

Proposition 8.2-6, Cauchy’s Series Convergence Criteria: A complex series converges if and only if, for any real number

ε > 0 there exists N ∈! such that:

∑z

j

k > N !

From Cauchy’s convergence criteria for series we see that

the necessary and sufficient condition for the convergence of an infinite complex series is that, for any real number ε > 0 , there exists N ∈! such that the remainder of the series beyond N is

n

!

!

(8.2-23)

j = k+1

always smaller than ε . This requires that terms of the series approach zero as a limit (see Proposition 8.2-1).

Proof:

Example 8.2-3

!

Show that the harmonic series diverges.

From Cauchy’s convergence criteria (Proposition 8.2-5) we

have: !

()

()

Sn z − Sk z < ε !

n>k > N!

(8.2-24)

Solution: The harmonic series is:

We can write: 350

∞

!

∑ k =1

1 1 1 1 1 = 1+ + + + +! k 2 3 4 5

n , we have:

∑

1 1 1 1 1 1 = + +!+ >n = k n + 1 n + 2 2 n 2 n 2 k = n+1 Letting ε = 1 2 , we see that we will have:

!

∑

∑z

j

k > N!

(8.2-28)

If we now eliminate a finite number of terms from the

(

we delete, then letting N1 = max k1 , N

)

we can write for the

modified series: !

2n

!

()

beginning of the series so that k1 is the index of the largest term

2n

1 >ε! k k = n+1

()

Sn z − Sk z =

j = k+1

If we consider any n adjacent terms of the series beginning at

!

!

n

()

n

()

Sn z − Sk z =

∑z

j

k > N1 !

(8.2-29)

j = k+1

when n > N

The beginning terms in the sum do not change the nature of the

for all N ∈! . The harmonic series does not satisfy Cauchy’s

convergence of the series. If the original series converges, the

convergence criteria, and so the harmonic series diverges.

modified series will also converge. The sum for the two series will not be the same, however.

Proposition 8.2-7, Omit beginning Terms of a Series:

■

Proposition 8.2-8:

An infinite complex series that is convergent will remain

{ ( ) } will converge to a sum

A sequence of partial sums Sk z

convergent if a finite number of terms are omitted from the

()

()

()

S z = Re S z + i Im S z if and only if the sequences of partial

beginning of the series.

sums: Proof: !

Since the series is convergent, from Cauchy’s criteria for

convergence (Proposition 8.2-5) for any real number ε > 0 there must exist N ∈! such that:

k

!

∑ Re ( S ( z )) ! k

n =1

k

∑ Im ( S ( z )) ! k

(8.2-30)

n =1

converge. 351

!

Proof: !

We will consider the partial sums:

()

()

∑ Re ( S ( z )) !

k

S z = lim

k

Re Sk z =

!

!

From Proposition 8.2-8 we have:

k→∞

k

∑ z = lim ∑( x + i y ) n

k→∞

n =1

n

n =1

(8.2-31)

n

k

= lim

!

n =1

k→∞

()

n

n =1

k→∞

n

(8.2-36)

n =1

Therefore certain properties of real series also apply to complex

k

Im S k z =

k

∑ x + i lim ∑ y !

and !

n

∑ Im ( S ( z )) !

(8.2-32)

n

series.

n =1

()

()

()

If the partial sum Sk z = Re Sk z + i Im Sk z

exists, we have

Proposition 8.2-9, Addition and Subtraction of Series:

{ }

from equation (8.2-1):

()

!

()

S z = lim Sk z !

!

k→∞

(8.2-33)

obtained if and only if for the sequences of partial sums

{Re S ( z ) } and {Im S ( z )} we have:

!

∞

! !

k→∞

()

()

! (8.2-34)

∑b !

n

n

(8.2-37)

n =1

are convergent infinite series, then:

∑( a ± b ) ! n

(8.2-38)

n

n =1

is also a convergent series and

and !

∑a ! ∞

k

lim Re Sk z = Re S z !

∞

n =1

From Proposition 7.1-13 we know that equation (8.2-33) is

k

{ }

If an and bn are two sequences of complex numbers, and if:

()

()

lim Im Sk z = Im S z !

k→∞

(8.2-35)

Therefore this proposition and its converse must be true.

■

∞

!

∑( n =1

⎡ ∞ ⎤ ⎡ ∞ ⎤ an ± bn = ⎢ an ⎥ ± ⎢ bn ⎥ ! ⎢ ⎥ ⎢ ⎥ ⎣ n =1 ⎦ ⎣ n =1 ⎦

)

∑

∑

(8.2-39)

352

Proof: !

From associative and commutative properties of addition

Proposition 8.2-10, Rearrangement of Series:

{ } { } { bn } is any rearrangement of { an } , and if:

and subtraction, we can write:

⎡ k ⎤ ⎡ k ⎤ ⎢ an ⎥ ± ⎢ bn ⎥ = ⎢ ⎥ ⎢ ⎥ ⎣ n =1 ⎦ ⎣ n =1 ⎦

∑

!

∑

If an and bn are two sequences of complex numbers where k

∑( a ± b ) ! n

(8.2-40)

n

n =1

∞

∑a

!

n

Taking the limit, we have from Proposition 2.3-7:

⎡ k ⎤ ⎡ k ⎤ lim ⎢ an ⎥ ± lim ⎢ bn ⎥ = lim k→∞ ⎢ ⎥ k→∞ ⎢ ⎥ k→∞ ⎣ n =1 ⎦ ⎣ n =1 ⎦

∑

!

∑

k

∑( an ± bn ) !

is an absolutely convergent infinite series, then the two series

(8.2-41)

∞

∑a

n =1

n =1

⎡ ∞ ⎤ ⎡ ∞ ⎤ an ± bn = ⎢ an ⎥ ± ⎢ bn ⎥ ! ⎢ ⎥ ⎢ ⎥ ⎣ n =1 ⎦ ⎣ n =1 ⎦

∑a

n

∑

)

∞

∑b

n

n =1

and

n =1

∑b

n

are two convergent infinite series of ∞

n =1

convergent infinite series.

∑( a ± b ) n

n

■

∞

A complex series ∞

complex series

∑a

n

n =1

∑

bn is defined as a rearrangement of a

n =1

if a one-to-one correspondence exists

between indices n and m so that an = bm .

(8.2-44)

n

n =1

n =1

Proof: !

is also a

n =1

∞

n

(8.2-42)

∞

converge to the same sum:

∑ a =∑ b !

!

∑

complex numbers therefore, then

!

and

n ∞

∑( If

∞

n =1

∞

!

(8.2-43)

n =1

and so: !

!

!

We will define the partial sums:

()

S ka z =

k

∑a

n

n =1

!

()

S kb z =

k

∑b

n

!

(8.2-45)

n =1

()

where we are given that the sequence of partial sums S ka z

converges absolutely as k → ∞ . Both series in equation (8.2-45) must then converge absolutely since they contain the same positive terms (just in a different order): 353

()

()

()

S a z = lim S ka z !

!

k→∞

()

S b z = lim S kb z ! k→∞

(8.2-46)

We then have for any given value of k :

(z) ≤ S (z)! (z) ≤ S (z)! (8.2-47) since S a ( z ) and S b ( z ) are composed of all terms in the series. S ka

!

b

S kb

a

8.3.1!

DIVERGENCE TEST

Proposition 8.3-1, nth Term Test for Divergence: ∞

A complex series

When k → ∞ , equation (8.2-47) becomes:

()

()

()

Sa z ≤ Sb z !

!

()

()

Sb z ≤ S a z !

()

Sa z = Sb z !

(8.2-48)

(8.2-49)

n

n =1

Follows from Proposition 8.2-1.

8.3.2!

n

■

COMPARISON TEST

If zn ≤ Mn where Mn are nonnegative real numbers, and where

(8.2-50)

n =1

the series:

■

∞

!

An infinite complex series that is only conditionally

!

convergent may change the value of its sum when a change is made to the order arrangement of its terms.

8.3! !

n→ ∞

Proposition 8.3-2, Comparison Test:

∞

∑ a =∑ b !

!

diverges if lim zn ≠ 0 .

Proof: !

or ∞

n

n =1

and so both series converge to the same sum: !

∑z

TESTS FOR CONVERGENCE AND DIVERGENCE OF A COMPLEX SERIES Convergence and divergence tests for complex series are

generally similar to those for real series.

∑M ! n

(8.3-1)

n =1

!

converges in ! , then the series: ∞

!

∑z ! n

(8.3-2)

n =1

will converge absolutely in ! . 354

is a p-series with p = 3 as given in equation (8.1-17). This

Proof: !

∞

Since

∑

series is known to converge. Since:

Mn converges, it satisfies Cauchy’s convergence

n =1

!

criteria given in Proposition 8.2-6: n

∑

!

Mj < ε !

n>k > N !

by the comparison test we see that the series:

(8.3-3)

j = k+1

∞

!

Since zn ≤ Mn , we have: n

j

j

n>k > N!

(8.3-4)

j = k+1

Example 8.3-2

∞

Therefore the series

∑z

n

Does the following complex series converge?

also satisfies Cauchy’s convergence

n =1

criteria, and so the series is absolutely convergent.

■

∞

!

Example 8.3-1

∑

i i i = i + + +! 3 3 3 n 2 3 n =1

!

∑ n =1

1 1 1 = 1+ + 3 +! 3 3 n 2 3

2

3

The series: ∞

! ∞

n

⎛ 1+ i ⎞ 1+ i ⎛ 1+ i ⎞ ⎛ 1+ i ⎞ = + + +! ⎜⎝ 3 ⎟⎠ 3 ⎜⎝ 3 ⎟⎠ ⎜⎝ 3 ⎟⎠

Solution:

∞

Solution:

∑ n =1

Does the following complex series converge? !

i i i = i + + +! 3 3 3 n 2 3

is absolutely convergent.

n

j = k+1

∑ n =1

∑ z ≤ ∑M k > N!

(8.3-16)

j = k+1

(8.3-13)

From the comparison test (Proposition 8.3-2) we see that:

()

∞

X x =

!

j

or

xn ≤ zn !

!

∑z

!

∑x ! n

( )

∑z

!

j

∑y ! n

(8.3-14)

From the triangle inequality given in Proposition 1.8-2 we have:

n =1

n

must both converge absolutely.

N!

(8.3-17)

n>k > N!

(8.3-18)

j = k+1

Proposition 8.3-4, An Absolutely Convergent Series is and so:

Convergent:

n

If a complex series converges absolutely, then it converges. That ∞

is, if

∑

∞

zj converges, then

j =1

∑

!

This is just Cauchy’s criterion for the convergence of

∑z

j

.

■

j =1

∞

Since

0 there must exist N ∈! such that:

!

It is often easier to first test for the convergence of a

complex series using the absolute values of the terms rather than the terms themselves, and to then use Proposition 8.3-4. The following convergence tests are useful in this regard. 357

8.3.3!

We then have:

RATIO TEST

!

zN+1 < r zN

!

zN+ 2 < r zN +1 < r 2 zN

!

zN+ 3 < r zN +2 < r 3 zN !

Proposition 8.3-5, Ratio Test: Given a series: ∞

∑z !

!

(8.3-19)

n

n =1

!

lim

!

n→ ∞

zn+1 zn

!

= q!

(8.3-20)

q < 1 , the series converges absolutely

!

q > 1 or q = ∞ , the series diverges

∞

!

q = 1, the test is inconclusive

)

∞

∑z


N so that we obtain: !

(

zN + zN+1 + zN+ 2 +! < zN 1+ r + r 2 + r 3 +! ! (8.3-23)

or

Proof: !

We then have: !

!

!

zN+ k < r zN +k−1 < r k zN

!

then if:

!

!

!

where:

(8.3-22)

zn+1 zn

∞

!

∑z

N

k =0

k

r =

zN 1− r

is the geometric series which converges for 0 < r < 1. Based

N !

(8.3-21)

upon the comparison test (Proposition 8.3-2), we see that: 358

∞

∞

∑z

!

=

N +k

k =1

∑

zn !

(8.3-26)

n = N +1

then converges absolutely for q < 1 . We also have: ∞

∑ z =∑ z +∑ z

!

n

n

n =1

n =1

n

!

lim zN+k ≠ 0 !

!

lim z ≠ 0 !

! (8.3-27)

n = N +1

From the divergence test (Proposition 8.3-1), we see that the series diverges if q > 1 .

sum of terms which are all finite. Therefore the series:

!

∞

∑

zn !

(8.3-28)

converges absolutely for q < 1 . From Proposition 8.3-4 we then know that the series also converges if q < 1 .

If q = 1 the ratio test provides no information.

Does the following complex series converge? ∞

!

∑ n! = e

z

Solution:

If q > 1 , select any r such that 1 < r < q . Given the limit in

Using the ratio test, we have:

obtain:

z n+1 zn+1 zn

>r!

for all n > N !

(8.3-29)

zN+ k > r

k

zN !

!

lim

n→ ∞

( n + 1) ! z

n

= lim

n→ ∞

z n+1 n! z

n

( n + 1) !

= lim

n→ ∞

z n +1

= 0 N so that we

!

■

Example 8.3-3

n =1

!

(8.3-32)

n→ ∞ n

where the first term on the right must converge since it a finite

!

(8.3-31)

k→∞

Therefore we will also have:

∞

N

and so:

(8.3-30)

We see by the ratio test that the series representing e z converges absolutely for all z in the complex plane. 359

i n+1

Example 8.3-4 Does the following complex series converge?

lim

!

n→ ∞

∞

!

∑z

n

2 n+1 in 2n

=

i 1 = 2 2 ∞

in We see by the ratio test that the series converges n 2 n=0 absolutely.

∑

n=0

Solution: Using the ratio test, we have:

z n+1 !

lim

n→ ∞

z

n

Example 8.3-6

= z

Does the following complex series converge? ∞

∞

We see by the ratio test that the series absolutely for all z < 1 .

∑z

n

converges

∑

!

e− n z

n=0

n=0

Solution: Example 8.3-5

Using the ratio test, we have:

Does the following complex series converge? ∞

!

∑ n=0

n

i 2n

Solution: Using the ratio test, we have:

− n+1 z e ( ) −z −x lim = e = e n→ ∞ e− n z

!

Therefore: !

For x > 0 we have e− x < 1 and the series converges.

!

For x < 0 we have e− x > 1 and the series diverges.

360

For x = 0 we have e− x = 1 , and so the ratio test is inconclusive. If x = 0 then e− n z = e−i n y and for no value of y will lim e

−i n y

n→ ∞

= 0 . We can conclude

from Proposition 8.2-1 that the series diverges.

!

If q < 1 , select any r such that q < r < 1 . Given the limit q

in equation (8.3-34), it is then possible to choose n > N so that we obtain: !

zn < r !

n

for all n > N !

(8.3-35)

for all n > N !

(8.3-36)

We then have:

8.3.4!

ROOT TEST

!

Proposition 8.3-6, Cauchy’s Root Test:

We can write: ∞

Given a series:

!

∞

∑z !

!

n

(8.3-33)

lim

n→ ∞

= r N + r N r + r N r 2 + r N r 3 + !!

n

zn = q !

!

q < 1 , the series converges absolutely

!

!

q > 1 or q = ∞ , the series diverges

!

!

q = 1, the test is inconclusive

(8.3-37)

or !

then if:

Proof:

n

∞

where:

!

∑r n= N

n=0

!

zn ≤ r n !

∑

(

n= N

(8.3-34)

)

r n = r N 1+ r + r 2 + r 3 + ! !

(8.3-38)

Since r < 1 , we can use the fact that the geometric series converges for r < 1 (see Example 8.2-1) together with the comparison test (Proposition 8.3-2) to conclude from equations (8.3-38) and (8.3-36) that the series: ∞

!

∑z ! n

(8.3-39)

n=0

361

converges absolutely if q < 1 . From Proposition 8.3-4 we then

Example 8.3-7

know that the series also converges if q < 1 .

Does the following complex series converge?

!

If q > 1 , then it is possible to choose n > N so that we

∞

!

obtain:

zn > 1!

n>N!

Solution:

(8.3-40)

Using the root test, we have:

and so: !

zn > 1n = 1 !

n>N!

(8.3-41)

!

Therefore we can never have: !

n

n=0

n

!

∑n

zn

lim z = 0 !

n

nn

n→ ∞

= lim

n→ ∞

z n

=0

We see by the root test that the series converges absolutely

(8.3-42)

n→ ∞ n

lim

zn

for all z .

and so from the divergence test (Proposition 8.3-1), we see that the series diverges.

Example 8.3-8

■

Does the following complex series converge? !

If the limit in equation (8.3-34) does not exist, a limit based

upon the concept of supremum or superior value discussed in

∞

!

Section 2.2.11 can be used: !

lim sup n zn = q !

n→ ∞

n=0

Solution:

(8.3-43)

Using the root test, we have:

The advantage of this formulation is that the limit supremum always exists (even though it may be infinity).

∑

in 2n

!

lim

n→ ∞

n

in i 1 = = 2 2 2n 362

∞

We see by the root test that the series

n=0

absolutely.

8.4! !

∑

in converges 2n

If the terms of a series are not simply constant complex

()

numbers zn , but are functions f n z of complex numbers, then the series is a complex function series. Such a function series is

()

{ f ( z )} of complex functions:

of an infinite sequence

∑

!

()

()

()

()

fn z = f0 z + f1 z + f 2 z + ! !

(8.4-1)

n=0

!

()

The sum Sk z of the first k terms of an infinite sequence

of complex functions

{ f ( z )} is known as the k n

th

partial sum

(see Section 8.1): !

()

Sk z =

k

∑

()

()

()

()

f n z = f0 z + f1 z + f 2 z + ! + f k z ! (8.4-2)

n=0

!

For any given sequence of complex functions, the set of

partial sums forms a sequence of partial sums

()

{ S ( z ) } . This k

sequence consists of all the terms Sk z as defined in equation (8.4-2):

!

1

2

(8.4-3)

k

{S ( z ) } = f ( z ) , ⎡⎣ f ( z ) + f ( z )⎤⎦ , ⎡⎣ f ( z ) + f ( z ) + f ( z )⎤⎦ , ! k

0

0

1

0

1

2

!!

(8.4-4)

8.5! !

CONVERGENCE OF COMPLEX FUNCTION SERIES

()

If a complex function f n z is defined and single-valued in

the region R , the sequence of complex functions

{ f ( z )} n

becomes a sequence of complex numbers. If the sequence of partial sums

{S ( z ) }

()

of the function f n z

k

converges for all

points in a region R , then R is called a region of convergence for the function series. !

()

0

a region R of the complex plane, then for any given point z in

n

∞

k

or

SERIES OF COMPLEX FUNCTIONS

defined as the sum of terms f n z

{S ( z ) } = S ( z ) , S ( z ) , S ( z ) ,!, S ( z ) !

!

If the convergence of a complex function series depends

upon the point z chosen, the convergence is called pointwise

()

convergence. That is, the series converges pointwise to f z if we have: !

()

()

k

f z = lim Sk z = lim k→∞

k→∞

∑ f (z)! n

(8.5-1)

n=0

such that for every real number ε > 0 : 363

()

()

Sk z − f z < ε !

!

for all k > N !

(8.5-2)

where N ∈! depends on both ε and z .

()

The limit f z is called the sum of the series at point z .

!

()

For complex function series we will use the notation S z and

()

()

()

f z interchangeably, where S z ≡ f z . If a series of complex functions does not converge, it is said to diverge.

a necessary but not sufficient condition for the convergence of a series of functions.

()

!

We will define Rk z as:

!

Rk z = f z − Sk z !

()

()

()

(8.5-5)

()

The term Rk z in equation (8.5-5) is called the remainder of the series of complex functions or the remainder after k terms. We then have:

Proposition 8.5-1:

()

()

If a series of complex functions converges to a sum f z :

()

∞

f z =

!

∑ f (z)!

(8.5-3)

n

()

Rk z < ε !

!

8.6!

()

lim f k z = 0 !

!

k→∞

{S ( z ) } k

Follows from Proposition 8.2-1.

■

(8.5-6)

for all k > N !

(8.5-7)

If for the same value of N ∈! , a sequence of partial sums

! !

()

UNIFORM CONVERGENCE OF COMPLEX FUNCTION SERIES

(8.5-4)

Proof:

()

From equation (8.5-2) we then have:

n=0

then:

()

Rk z = fk +1 z + fk +2 z + fk +3 z + !!

!

()

converges to f z

for all points z in its region of

convergence R when n ≥ N , then the sequence of partial sums

()

is said to converge uniformly to f z , and we will have for !

The converse of Proposition 8.5-1 is not true, however. It is

()

possible to have lim f k z = 0 , and yet for the sequence of k→∞

partial sums not to converge. Proposition 8.5-1 then represents

any real number ε > 0 : !

()

()

Sk z − f z < ε !

for all k > N !

(8.6-1)

364

!

For uniform convergence the number N can depend on ε ,

()

( )

but not z . We therefore have N = N ε , but not N = N ε , z . If a series converges uniformly, then term by term integration is

Keeping n fixed and letting k → ∞ , we have: !

Proposition 8.6-1, Cauchy Criteria for Uniform Convergence:

!

∑ f (z)!

(8.6-2)

n

!

()

for all z ∈R !

(8.6-6)

∑ f (z)!

(8.6-7)

n

n=0

n=0

converges uniformly and absolutely to a limit for every point z

converges uniformly and absolutely to a finite limit for every

within its region of convergence R if and only if the sequence is

point z in R .

uniformly Cauchy within R such that: n

∑

!

()

fj z

k > N !

! (8.6-3)

j = k+1

Proof:

Conversely, if a complex function series converges

()

uniformly and absolutely to a limit f z for all points z ∈R , then for any real number ε > 0 we can let: !

ε fn z − f z < ! 2

for all z ∈R and n > k > N !

(8.6-8)

!

ε fk z − f z < ! 2

for all z ∈R and k > N !

(8.6-9)

()

()

If the Cauchy criteria for uniform convergence holds, for

any real number ε > 0 we will have:

! !

()

lim f z = f z !

n→∞ n

∞

∞

!

(8.6-5)

and so the complex function series:

A complex function series:

!

()

and so letting n → ∞ :

possible (see Proposition 8.6-5).

!

()

fn z − f z < ε !

()

()

fn z − f k z =

n

∑ f (z) < ε ! j

j = k+1

for all z ∈R and n > k > N !

()

()

()

where N = N ε . From equation (8.6-4) we have:

(8.6-4) 365

n

∑ f (z) =

!

j

()

()

()

()

fn z − f z + f z − f k z !

(8.6-10)

!

j = k+1

∑ f (z) ≤

!

j

()

()

Since the series in equation (8.6-13) converges uniformly

()

to f z , for any real number ε > 0 there must then exist a real

and so by the triangle inequality: n

Proof:

()

()

fn z − f z + fk z − f z !

(8.6-11)

()

From equations (8.6-8), (8.6-9), and (8.6-11) we obtain: n

!

∑

()

fj z
k > N ! (8.6-12) 2 2

!

()

Sn z − f z
0 where N = N ε such that:

ε ! 3

for all z ∈R and n > N ! (8.6-14)

Also since the series in equation (8.6-13) converges, if

z = z0 is any given point within R we must have:

( )

( )

ε ! 3

!

Sn z0 − f z0
N ! (8.6-15)

ε > 0 there must then exist a δ > 0 such that:

()

( )

Sn z − Sn z0

!


N where

Proposition 8.6-3:

N ∈! .

If a sequence of continuous single-valued complex functions

{ f ( z )} converges uniformly to f ( z ) at every point within its n

region of convergence R , and if z = z0 is a point within R ,then

!

we have:

equation (1.8-17), we can write: ∞

!

Proof:

lim

z→z0

∞

∑ f ( z ) =∑ lim f ( z) ! n

n=0

n=0

z→z0 n

Using the generalized triangle inequality given in n

(8.6-20)

!

n

∑ f (z) ≤ ∑ f (z) ! j

j = k+1

j

(8.6-23)

j = k+1

367

()

test. It is possible for a series to converge uniformly, and yet not

Using fn z ≤ Mn we then have: n

()

have terms that satisfy the relation fn z ≤ Mn . The Weierstrass

n

M-test can be considered a generalization of the comparison

∑ f (z) ≤ ∑ M !

!

j

(8.6-24)

j

j = k+1

j = k+1

test for uniform convergence. Example 8.6-1

{ } is a sequence∞of positive constants which

Note that since Mn

are not a function of z , and since

∑M

n

Does the series:

is convergent, then for

n=0

()

any real number ε > 0 there must exist an N = N ε ∈! such

∞

!

n =1

that:

converge uniformly and absolutely for z > 1 ?

n

∑ M k > N ε !

(8.6-25)

Solution:

j = k+1

Let Mn equal:

where N is not a function of z . We therefore have: n

∑ f (z) < ε !

!

j

for all z ∈D and for n > k > N ! (8.6-26)

!

j = k+1

!

From

Cauchy’s

criteria

for

uniform

convergence

1 n2

∞

!

∞

∑ f ( z ) converges uniformly and absolutely in D .

■

While the Weierstrass M-test is a sufficient test for uniform

convergence, and therefore very important, it is not a necessary

∑ n =1

∞

Mn =

∑ n =1

1 n2

converges. We also have:

n=0

!

Mn =

The series:

(Proposition 8.6-1), we see that the series of complex functions n

∑

1 n2 z n

!

1 ≤ Mn 2 n n z 368

where N = N ε , and where L is the length of some simple

that the series:

contour C that lies within D along which we will integrate. We

∞

!

()

and so by the Weierstrass M-test (Proposition 8.6-4) we see

∑ n =1

then have:

1 n2 z n

()

If a series of continuous complex functions f k z

converges

uniformly to the sum f z in a domain D , then for any simple

()

contour C that lies within D the integral of f z along C can be determined using term-by-term integration of the series:

lim

n→ ∞

∫∑ C

k=0

()

n

fk z dz = lim

n→ ∞

∑∫ k=0

C

()

fk z dz !

Proof: !

()

(8.6-29)

(8.6-27)

()

is integrable. From Proposition 8.6-2 we

()

know that their sum f z is continuous in D , and so it also is integrable. We can therefore write:

⎡ n ⎤ ⎢ fk z − f z ⎥ dz ⎢ ⎥ C ⎣ k=0 ⎦

n

!

∫ ∑ f ( z ) dz − ∫ f ( z ) dz = ∫ ∑ C

k

C

k=0

()

()

! !

(8.6-30)

We have from Proposition 5.3-7:

()

Since the series of continuous complex functions f k z

()

for all n > N !

Since the functions fk z are all continuous in D , their

partial sum Sn z

Proposition 8.6-5, Term-by-Term Integration:

n

ε ! L

where Sn z is the nth partial sum of the complex functions. !

()

()

()

converges uniformly and absolutely for z > 1 .

!

()

Sn z − f z
0 there must exist an integer N > 0

Since the series of continuous functions converges uniformly,

such that:

we have from equations (8.6-28), (8.6-30), and (8.6-31): n

!

∑ k=0

ε fk ( z ) − f ( z ) < ! L

n

for all n > N !

(8.6-28)

!

∫∑ C

k=0

()

fk z dz −

∫

C

()

f z dz
N ! (8.6-32)

! !

As n → ∞ , then ε > 0 can be arbitrarily small, and so:

∫

!

!

()

n→ ∞

∫ ∑ f ( z ) dz ! C

Since

{ f ( z ) } is a sequence of holomorphic functions in a k

simply connected domain D , we know from Proposition 3.7-1

n

f z dz = lim

C

Proof:

(8.6-33)

k

k=0

()

that the functions fk z

also know that their sum

sum f z we have:

!

()

()

n→ ∞

∑ f (z) !

(8.6-34)

k

k=0

Therefore:

C

n

lim

n→ ∞

n

∑ f ( z ) dz = lim ∑ ∫ k

k=0

By definition of a series of complex functions converging

()

lies within the domain D : !

∫

!

n→ ∞

k=0

C

!∫ f ( z ) dz = !∫ C

()

fk z dz !

(8.6-35)

C

n

lim

n→ ∞

∑ f ( z ) dz ! k

k=0

n

!

!∫ f ( z ) dz = lim ∑ ∫ n→ ∞

C

Proposition 8.6-6:

{ ( ) } is a sequence of holomorphic functions in a simply

(8.6-37)

We then have from Proposition 8.6-5:

■

k=0

C

()

fk z dz !

(8.6-38)

()

If fk z

Since the functions fk z are all holomorphic in D , we have

connected domain D , and if the series:

from the Cauchy-Goursat theorem (Proposition 5.4-2):

∞

!

() f ( z ) must be continuous in D .

k

to a sum f z , we have for any simple closed contour C that

n

f z = lim

{ f (z)}

converges uniformly to f z in D , from Proposition 8.6-2 we

By definition of a series of complex functions converging to a

!

are continuous in D . Since

n

∑ f (z) !

(8.6-36)

k

k= 0

()

()

converges uniformly to f z in D , then f z is holomorphic

!

lim

n→ ∞

∑ !∫ k=0

C

()

fk z dz = 0 !

for each k !

(8.6-39)

Therefore, for every simple closed contour C within D :

on D . 370

!∫

!

()

f z dz = 0 !

C

(8.6-40)

!

()

()

that f z is holomorphic in D .

()

()

()

()

()

()

()

!

()

()

converges

()

multiplied by any function g z bounded in D , the resulting

(8.6-44)

From equations (8.6-42) and 8.6-44) we obtain:

()

()

()

()

g z fn+1 z + g z fn+2 z +! < M

! !

series will also converge uniformly in D .

!

ε =ε! M !

n>N (8.6-45)

for all points z in D . Since ε > 0 can be arbitrarily small, the series in equation (8.6-45) must be uniformly convergent.

()

Since g z is bounded, for some M > 0 we have:

()

g z ≤M!

for all z in D !

The series of continuous functions

(8.6-41)

()

fn z

(8.6-42)

()

converges

uniformly to the sum f z in a domain D , then the derivative

()

of f z at any point z in D can be determined by term-by-term differentiation of the series:

using equation (8.6-3):

n> N!

()

If a series of continuous complex functions fn z

converges

Proposition 8.6-1). For any real number ε > 0 we can then write

ε ! fn+1 ( z ) + fn+2 ( z ) +! < M

■

Proposition 8.6-8, Term by Term Differentiation:

uniformly in D , and so it must be a Cauchy series (see

!

()

fn+1 z + fn+2 z +!

g z fn+1 z + g z fn+2 z +! < M fn+1 z + fn+2 z +!

uniformly to the sum f z in a domain D , and if the series is

!

()

and so: !

If a series of continuous complex functions fn z

!

()

(8.6-43)

■

Proposition 8.6-7:

!

()

!

From Morera’s theorem (Proposition 6.3-4) we can conclude

Proof:

()

g z fn+1 z + g z fn+2 z +! = g z

!

k f ( ) ( z) =

∞

∑

k f n( ) z !

()

(8.6-46)

n=0

for all points z in D . We also have: Proof: 371

Let C be simple closed contour that lies within a domain

!

!

From Proposition 6.2-1 we see that each of the terms of

D , and let z be a point inside C . The series in parametric form

this equation are the derivatives of the corresponding terms in

is:

equation (8.6-47). We then have:

()

()

()

()

f t = f0 t + f1 t + f 2 t +! !

!

(8.6-47)

()

()

() () f ′ ( z ) of f ( z ) at any point z

f ′ z = f0′ z + f1′ z + f 2′ z +! !

!

(8.6-51)

where we will let the parameter t encompass the range of z on

and so the derivative

the contour C .

D can be determined by term-by-term differentiation of the

!

series fn z .

Multiplying equation (8.6-47) by the bounded function:

1

!

(

2π i t − z

)

! 2

(8.6-48)

we have:

()

!

Following a similar argument by multiplying equation

(8.6-47) by the bounded function: !

()

f t

!

(

2π i t − z

)

2

=

()

f0 t

(

2π i t − z

)

2

+

()

f1 t

(

2π i t − z

)

2

+! !

(8.6-49)

(

2π i t − z

we know that this series can be integrated term-by-term along

!

!

C

(8.6-52)

()

()

term differentiation of the series fn z :

f

!

(k )

∞

( z) = ∑ fn( k ) ( z) !

(8.6-53)

n=0

()

f t dt

(t − z )

!

()

the contour C :

!∫

)

k+1

k it is possible to show that the k th derivative f ( ) z of f z at

converge uniformly in the domain D . From Proposition 8.6-5

!

1

any point z in a domain D can also be determined by term-by-

From Proposition 8.6-7 we know that this series will also

1 2π i

in a domain

2

1 = 2π i

!∫

C

()

f0 t dt

(t − z )

2

1 + 2π i

!∫

C

■

()

f1 t dt

(t − z )

2

+" (8.6-50)

8.7! !

POWER SERIES A complex power series is an infinite series of complex

functions having the form: 372

∞

!

∑

(

an z − z0

)n = a0 + a1 ( z − z0 ) + a2 ( z − z0 )2 !!

! (8.7-1)

n=0

If the circle of convergence of a power series encompasses

the entire complex plane, then r = ∞ , and the power series represents an entire function. If the circle of convergence of a

where an , z, z0 ∈! . This series is also known as a power series

power series encompasses only the point of expansion z0 , then

in z − z0 . The point z0 is a fixed point in the complex plane, and

r = 0 by definition.

is called the center of the series or the point of expansion of the series. The coefficients an are complex constants that are

Proposition 8.7-1, Circle of Convergence of a Power Series:

generally dependent on the particular center of the series z0 . If

an = a for all n where a is a constant, then equation (8.7-1) is a geometric series. The coefficients of a power series for any

Every power series: ∞

(8.7-2)

an ! r = lim n→ ∞ an+1

!

power series itself may not be an entire function, however. !

)n

has a circle of convergence z − z0 = r with radius given by:

Each term of a power series is holomorphic over the

entire finite complex plane, and so is an entire function. The

(

an z − z0 !

n=0

given point of expansion z0 determine the series. !

∑

!

(8.7-3)

For every complex power series there is always a unique

if this limit exists. Within the circle of convergence the power

circle centered at z0 and having a radius z − z0 = r ≥ 0 such that

series will converge absolutely, and outside this circle it will

the series converges absolutely at every point inside the circle,

diverge.

and diverges at all points outside of the circle (see Proposition 8.7-1). This circle is known as the circle of convergence or disk of convergence, and its radius r is known as the radius of convergence. A power series may converge on some or none of the points that are on its circle of convergence. Each power series can have only one radius of convergence.

Proof: !

From the ratio test (Proposition 8.3-5) an infinite series will

converge absolutely if q < 1 , and diverge if q > 1 , where: !

lim

n→ ∞

( )n+1 n an ( z − z0 )

an+1 z − z0

= q!

(8.7-4) 373

A power series will then converge absolutely if: !

z − z0 lim

an+1

n→ ∞

an

< 1!

providing the limit in equation (8.7-7) exists (see Proposition 8.7-4). (8.7-5)

The limit in equation (8.7-5) depends only upon the coefficients of the power series. We see then that the power series will converge absolutely if: !

z − z0 < lim

n→ ∞

1 ≡r! an+1

(8.7-6)

an

From equation (8.7-8) we see that every power series has a

circle of convergence inside of which it will converge absolutely, and outside of which it will diverge. A power series always converges at its point of expansion. !

■

Convergence of a power series for points on its circle of

convergence must be determined individually for each point. For at least one point on the circle of congruence, the power

We have now identified the radius of convergence z − z0 = r of a power series as: !

!

series will not converge (see Proposition 8.8-7). Proposition 8.7-2, Infinite Radius of Convergence:

an ! r = lim n→ ∞ an+1

∞

(8.7-7)

If for a power series

∑

(

an z − z0

)n we have:

n=0

if the limit in this equation exists. !

Since the radius of convergence is a function only of the

coefficients of the power series, every complex power series

!

lim

n→ ∞

an

an

= 0!

(8.7-9)

converges for all z (for the entire z-plane).

power series can be found using the ratio test:

1 =q= ! r

n→ ∞

then the radius of convergence is r = ∞ , and the power series

has a circle of convergence. The radius of convergence of a

an+1

lim

!

an+1

Proof: (8.7-8)

!

Follows from equation (8.7-6). For an entire function the

circle of convergence of any point of expansion has a radius of 374

convergence equal to infinity, and the power series converges at

This is the geometric series. The radius of convergence is

every point z .

given by equation (8.7-7):

■

!

Proposition 8.7-3, Zero Radius of Convergence: ∞

If for a power series

an+1

lim

!

an

n→ ∞

∑

(

an z − z0

)n we have:

and so the geometric series converges for z − z0 < 1 and

n=0

= ∞!

diverges for z − z0 > 1 (see Example 8.2-1).

(8.7-10)

then the radius of convergence is r = 0 , and the power series

Example 8.7-2

converges only at the point of expansion z0 .

Determine the radius of convergence for the power series: ∞

Proof: !

an 1 r = lim = lim =1 n→ ∞ an+1 n→ ∞ 1

Follows from equation (8.7-6). If r = 0 the power series

!

)n

n=0

diverges for all z ≠ z0 , and the circle of convergence z − z0 = r of the power series consists of the single point z0 .

∑

(

n! z − z0

Solution:

■

The radius of convergence is given by equation (8.7-7): Example 8.7-1 Determine the radius of convergence for the power series: ∞

!

∑ (z − z )

!

an r = lim n→ ∞ an+1 We have:

n

0

n=0

Solution:

!

r = lim

n→ ∞

n!

( n + 1) !

= lim

n→ ∞

1 =0 n +1

375

and so the series will converge only if z = z0 . Therefore the series converges only at the point of expansion z0 .

Proposition 8.7-4, Radius of Convergence: The radius of convergence r of a power series:

Example 8.7-3

∞

Determine the radius of convergence for the power series:

∑ n=0

1 z − z0 n!

(

)n

r = lim

n→ ∞

1

r=

!

lim

n→ ∞

The radius of convergence is given by equation (8.7-7):

an an+1

1 n!

r = lim

n→ ∞

r = lim

!

1

( n + 1)!

we have: !

∑ n=0

zn = ez n!

n

!

an

(8.7-12)

n→ ∞

an an+1

!

(8.7-13)

if the limits exist.

= lim n + 1 = ∞ n→ ∞

and so the series converges for all values of z . When z0 = 0 ∞

(8.7-11)

and by:

We have:

!

)n

is given by:

Solution:

!

(

an z − z0 !

n=0

∞

!

∑

!

Proof: !

From the root test (Proposition 8.3-6) a power series will

converge absolutely if q < 1 , and diverge if q > 1 , where: !

lim

n→ ∞

n

1 zn = q = ! r

(8.7-14)

Using equation (8.7-11), a power series will converge absolutely if: 376

lim

!

n

n→ ∞

(

an z − z0

) ( n

) n→ ∞

= z − z0 lim

n

an < 1!

∞

(8.7-15)

1 lim

n

≡r! an

n→ ∞

(8.7-16)

!

When z = z0 , every power series: ∞

∑

!

where we have identified the radius of convergence r as:

r=

1 lim

!

n

an

n→ ∞

(8.7-17)

As shown in Proposition 8.7-1, the radius of convergence

is reduced to the single term a0 , and so will converge to this value. Therefore every power series converges for at least one

n→ ∞

an

If a power series:

an+1

∞

!

(8.7-18)

■

Proposition 8.7-6, Abel’s Theorem:

(Proposition 8.3-5):

r = lim

)n = a0 + a1 ( z − z0 ) + a2 ( z − z0 )2 ! ! (8.7-20)

n=0

for some power series can also be found using the ratio test

!

(

an z − z0

point, its center (point of expansion) z0 .

if the limit exists. !

(8.7-19)

Proof:

z − z0
0 for z = z1 , and so:

(

an z1 − z0

!

)

n

≤ M!

for all n !

(

! an z − z0

)

(

= an z1 − z0

)

z − z0 ⎞ ⎜z −z ⎟ ⎝ 1 0⎠

n⎛

n

z − z0

≤M

z1 − z0

∑

!

(

)n

an z − z0 !

(8.7-26)

n=0

diverges at a point z = z1 where z1 ≠ z0 , then it diverges for

! (8.7-23)

every point z outside the circle r = z1 − z0 . Proof:

∞

∑ a (z − z ) n

0

n

∞

0 Proposition 8.7-10, Continuity of a Function Represented by a

there must then exist a real integer N > 0 such that:

Power Series:

()

If a power series represents a function f z : ∞

()

f z =

!

∑ a (z − z ) ! n

n

0

()

(8.7-34)

Since z1 is a point inside z − z0 = r where the series

!

converges uniformly, we must also have:

()

( )

∑

(

)n

an z − z0 !

()

continuous). For the same ε > 0 there then must exist a δ > 0 (8.7-35)

such that:

n=0

()

converges uniformly to f z within its circle of convergence

z − z0 = r > 0 . We will now consider two points inside its circle of convergence: a point z1 and a general point z near z1 . !

{ ( ) } represent the sequence of partial sums of the

Let Sn z

()

power series f z at the point z inside z − z0 = r , where: !

()

Sn z =

n

∑

(

ak z − z0

)k !

(8.7-38)

z − z0 = r (the sum of a finite number of continuous functions is

From Proposition 8.7-9 we know that a power series: ∞

ε ! 3

Because Sn z is a polynomial in z , it is continuous inside

!

Proof:

( )

Sn z1 − f z1
N ! (8.7-37)

()

then f z will be continuous at all points within its circle of

f z =

ε ! 3

where N = N ε .

n=0

!

()

Sn z − f z
0 . Therefore: !

1 2

1 f ′ −1 = 2 2

( )

!

n z f ( ) z0 = e 0

( )

From equation (8.8-1) we then have the Taylor series: ∞

!

z

e =

∑ n=0

z

e0 z − z0 n!

(

)n 393

Example 8.8-4

()

f z =

!

()

()

⎛ π⎞ z − ⎜ 2 ⎟⎠ 2 n +1 ! ⎝

2 n+1

)

()

If a complex function f z can be represented as a Taylor series

radius of convergence is ∞ . We have:

f z = cos z !

( −1)n

Proposition 8.8-4:

Since the cosine function is entire (see Section 4.3.2), the

!

∑( n=0

Determine the Taylor series for f z = cos z about z0 = π 2 . Solution:

∞

()

about a point z = z0 , then f z will be divisible by z − z0 if and

( )

⎛π⎞ f ⎜ ⎟ =0 ⎝ 2⎠

()

only if f z0 = 0 so that z0 is a zero of f z . Proof:

!

! ! !

⎛π⎞ f ′ ⎜ ⎟ = −1 ⎝ 2⎠

!

()

⎛π⎞ f ′′ ⎜ ⎟ = 0 ⎝ 2⎠

!

()

⎛π⎞ f ′′′ ⎜ ⎟ = 1 ⎝ 2⎠

or

()

f ′ z = − sin z !

f ′′ z = − cos z ! f ′′′ z = sin z ! 4 f ( )( z ) = cos z !

4 ⎛π⎞ f ( )⎜ ⎟ = 0 ⎝ 2⎠

and so: !

2n ⎛ π ⎞ f ( ) ⎜ ⎟ = 0! ⎝ 2⎠

2 n+1 ⎛ π ⎞ f ( ) ⎜ ⎟ = −1 ⎝ 2⎠

( )n

From Proposition 8.8-1 we have:

()

f z =

∞

∑

n f ( ) z0

n=0

()

( )

( )

n!

f z = f z0 +

!

∞

∑ n =1

( z − z0 )n ! n f ( ) z0

( )

n!

( z − z0 )n !

(8.8-14)

(8.8-15)

Therefore z − z0 is a factor of this equation if and only if

( )

f z0 = 0 .

■

From equation (8.8-1) we then have the Taylor series: 394

Proposition 8.8-5, Uniform Convergence of Taylor Series:

Proposition 8.8-6, Taylor Series Representation of Holomorphic Functions:

If a Taylor series:

()

∞

f z =

!

∑

n f ( ) ( z0 )

n!

n=0

()

If a complex function f z

( z − z0 )n !

connected domain D , and if z = z0 is some given point in D ,

(8.8-16)

has a radius of convergence r , then the series converges uniformly on any closed disk of radius 0 < ρ < r .

is holomorphic in a simply

()

then f z can be expanded as a Taylor series centered at z0 :

()

∞

f z =

!

∑

(

an z − z0

)

n

∞

=

n=0

∑ n=0

n f ( ) z0

( )

n!

( z − z0 )n !

(8.8-18)

( )

Proof:

that converges on a disk Dr z0 having a radius r such that the

!

disk is the largest that can be contained within D .

Follows from Proposition 8.7-9 with:

an =

! ■

!

n f ( ) z0

( )!

(8.8-17)

n!

()

If a function f z can be represented by a Taylor series,

()

then f z

is holomorphic at every point inside its circle of

convergence (see Proposition 8.7-15). We will now use Cauchy’s integral formula to show the converse: every function

( )

holomorphic in a disk Dr z0

can be expanded as a Taylor

series centered at z0 that will converge in the disk. The

Proof: !

( )

Let z be an arbitrary fixed point in the disk Dr z0 , where

( )

Dr z0 is the largest disk that can be contained within D (see Figure 8.8-1). Let C be the circular contour enclosing the disk. !

We will describe C using the parameter s so that s will

always be a point on C . The circular contour C enclosing the disk is then given by r = s − z0 . !

Using Cauchy’s integral formula (Proposition 6.1-1) we

( )

following proposition is one of the most important theorems of

can write for an arbitrary fixed point z in the disk Dr z0 :

complex variables, and it has wide applications in the sciences.

!

()

f z =

1 2π i

!∫

C

( ) ds !

f s

s− z

(8.8-19)

395

!

For any fixed point z in the interior of the disk, we have:

r = s − z0 > z − z0 and so: !

z − z0 s − z0

< 1!

(8.8-22)

From Example 8.1-2 we have the geometric series expansion:

2

n−1

⎛ z − z0 ⎞ ⎜⎝ s − z ⎟⎠ 0

n

z − z0 ⎛ z − z0 ⎞ ⎛ z − z0 ⎞ 1 ! = 1+ + +!+ ⎜ ⎟ + z − z0 z − z0 s − z0 ⎜⎝ s − z0 ⎟⎠ ⎝ s − z0 ⎠ 1− 1− s − z0 s − z0 !! Figure 8.8-1!

!

( )

The maximum disk Dr z0 centered at z0 which can be contained within the domain D .

We can rewrite equation (8.8-19) subtracting and adding

z0 in the denominator of the integrand: !

()

f z =

1 2π i

() !∫C ( s − z0 ) − ( z − z0 ) ds ! f s

(8.8-20)

or

(

)n

2 n−1 z − z0 z − z0 ⎛ z − z0 ⎞ ⎛ z − z0 ⎞ 1 ! = 1+ + +!+ ⎜ ⎟ + n−1 z − z0 s − z0 ⎜⎝ s − z0 ⎟⎠ ⎝ s − z0 ⎠ s − z s − z 1− 0 s − z0 !! (8.8-24)

(

!

)(

)

Since a power series converges uniformly within its circle

of convergence (see Proposition 8.7-9), we can integrate the

or !

(8.8-23)

1 f z = 2π i

()

!∫

C

()

f s

1 ds ! z − z0 s − z0 1− s − z0

series term by term in its region of convergence (see (8.8-21)

Proposition 8.7-11). Using equation (8.8-24), we can then write equation (8.8-21) as: 396

!

1 f z = 2π i

()

()

f s

!∫

s − z0

C

ds +

z − z0 2π i

() ds 2 ( s − z0 )

!

f s

!∫

C

!

2

n z − z0 ) ( +

!

2π i

f s

( ) ds ! !∫C (s − z ) ( s − z0 )n

convergence and point z0 (see Figure 8.8-1). We then have:

ds

f s

(8.8-25)

From equations (6.2-2), (6,2-4), (6.2-5), and (6.2-6) we see that we then have: !

()

( )

f z = f z0 +

( )

f ′ z0 1!

+

!

( z − z0 ) +

( )

f ′′ z0 2!

( ) n−1 z − z0 ) + Rn ( z ) ! ( ( n − 1) !

) (

)

and so:

1

! !

s− z

≤

1 ! r−d

(8.8-30)

()

Since f z is holomorphic within the disk enclosed by the

contour C , we know from the maximum modulus theorem

(8.8-26)

!

()

Rn z =

( z − z0 )n 2π i

() !∫C (s − z ) ( s − z0 )n f s

!

()

Rn z

n z − z0 ) ( =

2π i

( ) ds ! !∫C (s − z ) ( s − z0)n f s

()

f z =

n−1

∑ k =0

k f ( ) ( z0 )

k!

( z − z0 )k + Rn ( z ) !

has a maximum value M on C .

We can then write using the ML-inequality (Proposition 5.3-8):

!

dn M ds ≤ 2π r 2π r − d r n

(

)

(8.8-31)

or (8.8-27) !

We can rewrite equation (8.8-26) as: !

(

s − z = s − z0 − z − z0 ≥ s − z0 − z − z0 = r − d ! (8.8-29)

()

where !

!

(Proposition 6.7-2) that f z

( z − z0 )2 +!

n−1 f ( ) z0

()

Rn z → 0 as n → ∞ . Let

d = z − z0 , the distance between any point z in the circle of

n−1 z − z f ( s) ( ) ( ) ( 0) + ∫C ( s − z0 )3 ds +"+ 2π i !∫C ( s − z0 )n 2π i !

z − z0

We will now show that

n

Mr ⎛ d⎞ ! Rn z ≤ ⎜ ⎟ r−d ⎝ r⎠

()

(8.8-32)

( d r ) < 1 and so ( d r )n → 0 as n → ∞ . Therefore we will also have Rn ( z ) → 0 as n → ∞ . Notice too that Since d < r we have

(8.8-28)

397

()

Rn z → 0 independently of z , showing again that the Taylor

expressed as power series and functions that are holomorphic

series converges uniformly in its circle of convergence.

in some simply connected domain.

!

n−1

()

f z = lim

!

n→ ∞

∑

k f ( ) ( z0 )

k!

k =0

( z − z0 )k !

into a Taylor series at any point where it is holomorphic (8.8-33)

Changing indices we have: ∞

()

f z =

!

∑ n=0

n f ( ) z0

( )

n!

()

If f z is a multivalued function, it can still be expanded

!

We can now rewrite equation (8.8-28) as:

provided that the same branch is used for the derivation of all coefficients of the series. Proposition 8.8-7, Singularity on Circle of Convergence:

( z − z0 )

n

The circle of convergence of a Taylor series representing a

!

()

(8.8-34)

()

function f z contains at least one singularity of f z on the circumference of the circle.

which is just the Taylor’s series centered at z0 that converges to

()

f z at every point z inside its circle of convergence. We see,

()

Proof:

therefore, that if a complex function f z is holomorphic in a

!

simply connected domain D , and if z = z0 is a fixed point in D ,

series centered at a point z = z0 and representing a function

()

then f z can be represented as a Taylor series centered at z0 :

()

f z =

!

∞

∑ n=0

(

an z − z0

)n

∞

=

∑ n=0

n f ( ) z0

( )

n!

( z − z0 )n !

(8.8-35)

Seeking a contradiction we will assume that, for a Taylor

()

f z , no singularity exists on the circumference of the circle of convergence z − z0 = r . This means that the Taylor series must also converge inside the circle z − z0 = r + ε , where ε > 0 is some real number. The actual circle of convergence for the

■

Taylor series is then z − z0 = r + ε , contrary to the given radius From Proposition 8.7-15 we also know that if a complex

of z − z0 = r . Therefore our assumption must be wrong, and a

function f z can be expressed as a power series, then f z is

singularity point must exist on the circumference of the circle

!

()

()

holomorphic at every point within its circle of convergence.

z − z0 = r .

■

There is then an equivalence between functions that can 398

!

The circle of convergence of a Taylor series representing

()

an analytic function f z will always extend to the nearest

!

()

f z =

1 = 1− 1+ 1− 1+! 1+ z 2

singular point of the function. Therefore the radius of convergence z − z0 = r of a Taylor series expansion centered at

Proposition 8.8-8:

z0 will equal the distance from z0 to the nearest singularity.

()

Every entire complex function f z can be represented as a

Only within this circle of convergence will the Taylor series

Taylor series.

()

expansion centered at z0 converge to f z , and so provide a

()

valid representation of f z .

Proof: !

Example 8.8-5

complex plane, we see from Proposition 8.8-6 that an entire

Explain why the power series expansion of the real function: !

()

f x =

Since an entire function is holomorphic everywhere in the

1 = 1− x 2 + x 4 − x 6 +! 2 1+ x

()

function f z can be represented as a Taylor series everywhere in the complex plane.

Proposition 8.8-9, Uniqueness of Taylor Series:

converges only if −1 < x < +1 .

( )

points in a disk Dr z0 , the coefficients an of the series:

By considering x as part of a complex number z , we have:

()

f z =

!

1 f z = = 1− z 2 + z 4 − z 6 +! 2 1+ z

()

∞

∑

(

)n

an z − z0 !

(8.8-36)

n=0

()

We see then that z = ± i are singularities for f z , and so

()

()

If a Taylor series about a point z = z0 converges to f z for all

Solution:

!

■

f z has singularities on the circle of convergence z = 1 .

()

Therefore f x converges only if −1 < x < +1 . For z = 1 the

()

( )

are unique for an expansion of f z in Dr z0 . Proof:

series diverges since it oscillates between two limits: 399

!

()

If two Taylor series about a point z = z0 converge to f z

can be differentiated term by term at every point inside its circle of convergence z − z0 = r , giving the derivative of the sum:

for all points in a circular domain D , then we have the two power series:

()

f z =

!

∞

∑a ( z − z ) !

!

()

∞

∑

(

)n

bn z − z0 !

(8.8-38)

n=0

an =

n f ( ) ( z0 )

n!

= bn !

n = 0, 1, 2, ! !

()

expansion of a function f z

(8.8-39)

are unique (see Proposition

■

Proposition 8.8-10, Term by Term Differentiation of a Taylor Series:

()

f z =

∑ n=0

■

∞

∑

n f ( ) z0

n=0

( )

n!

( z − z0 )n !

(8.8-42)

at every point inside its circle of convergence. Proof: Follows from Proposition 8.7-18.

■

Proposition 8.8-12:

() () () ()

If f z and g z are holomorphic in a domain D , then n n f z ≡ g z if and only if f ( ) z ≡ g ( ) z for any integer n .

A Taylor series: !

()

f z =

!

!

∞

Follows from Proposition 8.7-16.

Derivatives of all orders exist for a Taylor series:

Therefore an = bn for all n ≥ 0 , and so the coefficients of a Taylor 8.7-13).

(8.8-41)

Proposition 8.8-11, Derivatives of All Orders of a Taylor Series:

From Proposition 8.8-1 we have: !

)n−1 !

Proof:

and

f z =

∑

(

n an z − z0

n=1

(8.8-37)

0

n=0

!

f′ z =

!

n

n

()

∞

(

)n

an z − z0 !

(8.8-40)

()

()

Proof: 400

!

Follows from Propositions 8.8-10 and 8.8-9.

■

M

an ≤

!

rn

Proposition 8.8-13, Radius of Convergence of the Derivative of a Taylor Series:

Proof:

()

f′ z =

!

∑

(

n an z − z0

)

n−1

!

(8.8-43)

()

∞

()

f z =

!

has the same radius of convergence as the original Taylor series:

()

f z =

!

∑

∑

(

)n

an z − z0 !

(8.8-46)

n=0

n=1

∞

(8.8-45)

The Taylor series expansion for f z about a point z0 is:

!

The derivative of a Taylor series: ∞

n = 0, 1, 2, ! !

!

(

)

n

an z − z0 !

(8.8-44)

where n f ( ) z0

( )!

an =

!

n!

n=0

n = 0, 1, 2, ! !

(8.8-47)

From Cauchy’s inequality (Proposition 6.4-1) we have: Proof: !

Follows from Proposition 8.7-17.

n! M n f ( ) z0 ≤ n ! r

( )

!

■

n = 0, 1, 2, ! !

(8.8-48)

n = 0, 1, 2, ! !

(8.8-49)

Therefore:

Proposition 8.8-14, Taylor Series Inequality:

()

Let f z be holomorphic on a simply connected domain D and have a circle C of convergence z − z0 = r inside D . If for some

()

M > 0 , where M is a function of r , we have f z ≤ M for all

an ≤

!

M r

n

!

■

points inside and on C , then the coefficients of the Taylor series

()

representing f z are limited by:

401

8.9! !

Example 8.9-1

MACLAURIN SERIES

()

Determine the Maclaurin series for f z = e z .

A Taylor series expanded about a point z = z0 = 0 is known

as a Maclaurin series: !

()

Solution:

∞

f z =

∑

∞

n

an z =

n=0

∑ n=0

n f() 0

( ) zn !

n!

We have

(8.9-1) !

A function represented by a Maclaurin series will have a circle

n f ( ) z = ez !

!

()

(8.9-2)

!

a1 = f ′ 0 !

()

(8.9-3)

!

e =

a2 =

a3 =

( )!

f ′′ 0 2!

( )!

f ′′′ 0 3!

an =

n f() 0

( )!

n!

∑ n=0

n f() 0

( ) zn =

n!

∞

∑ n=0

zn z2 z3 z4 zn = 1+ z + + + +!+ n! 2! 3! 4! n!

is the Maclaurin series for e z . From Example 8.3-3 we see that this series converges absolutely for all z . Example 8.9-2

()

Determine the Maclaurin series for f z = cos z .

(8.9-4)

Solution: (8.9-5)

and !

z

The Maclaurin coefficients are given by equation (8.8-3):

a0 = f 0 !

!

∞

!

!

()

for all z . From equation (8.9-1) we then have:

of convergence centered at the origin. The power series for e z ,

sin z , and cos z are all Maclaurin series.

n f ( ) 0 =1

()

From equation (8.9-1) we have: !

()

f z = cos z =

∞

∑ n=0

(8.9-6)

n f() 0

( ) zn

n!

where: 402

!

n 2n f ( ) ( 0 ) = ( −1)

!

2 n+1 f ( ) (0) = 0

∞

Substituting 2 z for z : 2n 2 z ( ) n cos 2 z = ∑ ( −1) ( 2 n) ! n=0 ∞

∞

cos z =

∑ (−1) (2 n)!

!

z2 n

n

n=0

We then have:

so that: 2

!

4

6

z z z cos z = 1− + − +!+ −1 2! 4! 6!

2n

z 2n !

( )( ) n

2

⎛ 1 = ⎜ 1+ 2⎜ ⎝

!

()

Determine the Maclaurin series for f z = cos 2 z .

⎛ 1 2 cos z = ⎜ 1+ 1+ 2⎜ ⎝

∞

We will use the trigonometric identity:

( )

From Example 8.9-2 we can write:

∞

∑ (−1) n =1

n

22 n z 2 n ⎞ ⎟ 2 n ! ⎟⎠

( )

and so:

Solution:

⎛ 1+ cos 2 z ⎞ cos z = ⎜ ⎟ 2 ⎝ ⎠

2n ⎞ 2 z ( ) n ( −1) 2 n ! ⎟⎟ ∑ ( ) ⎠ n=0 ∞

or

Example 8.9-3

!

( )

1+ cos 2 z

!

is the Maclaurin series for cos z .

2

∑ (−1) (2 n)! z2n

n=0

Therefore: !

cos z =

!

n

!

2

cos z = 1 +

∑ (−1) ( n =1

n

2 2 n−1 2 n z 2n !

)

Example 8.9-4

()

Determine the Maclaurin series for f z =

1 . 1− z 403

Solution:

Solution:

We have: !

n f() z =

()

We will use the fact that:

n!

(1− z )n+1

and so: !

()

!

From equation (8.9-1) we then have: !

()

(

)

(

)

or

n f ( ) 0 = n!

1 f z = = 1− z

d 1 ln 1+ z = = 1− z + z 2 − z 3 +! dz 1+ z

!

∞

∑

n f() 0

( ) zn =

n!

n=0

∞

∑ n=0

Integrating term-by-term, we then have:

n! n z n!

!

=

∑

1 1 1 ln 1+ z = z − z 2 + z 3 − z 4 +!+ c 2 3 4

(

)

where c is a constant of integration. Setting z = 0 we have:

∞

!

d ⎡⎣ ln 1+ z ⎤⎦ = ⎡⎣1− z + z 2 − z 3 +!⎤⎦ dz

z n = 1+ z + z 2 + z 3 +!

!

n=0

(

()

)

z < 1 , this defines the circle of convergence of the

(

)

Therefore the Maclaurin series for f z = ln 1+ z is:

Therefore the Maclaurin series for 1 1− z is the geometric series. Since the geometric series converges only when

ln1 = 0 = c

!

1 1 1 ln 1+ z = z − z 2 + z 3 − z 4 +! = 2 3 4

(

)

∞

( −1)n z n+1

∑ n +1 n=0

Maclaurin series. Example 8.9-6 Example 8.9-5

()

(

Determine the Maclaurin series for f z = ln 1+ z

(

)

branch where ln 1+ z = 0 when z = 0 .

)

Determine the Maclaurin series for the sine integral: on the

!

()

Si z =

∫

z

0

sin s ds s 404

Solution:

and the series:

We will use the fact that: !

!

s 3 s5 s 7 sin s = s − + − +! 3! 5! 7!

For − z this series becomes:

and so: !

sin s s 2 s 4 s6 = 1− + − +! s 3! 5! 7!

()

Si z =

We then have:

∫

z

0

sin s ds = s

∫

z

0

ds −

∫

z

0

2

s ds + 3!

∫

z

0

s ds −! 5!

()

z3 z5 z7 Si z = z − + − +! = 3 3! 5 5! 7 7!

()

( ) ( )

( )

( −1) z 2 n+1 ∑ ( 2 n + 1)( 2 n + 1) ! n=0 ∞

or !

n

! !

Example 8.9-7

⎞ e z − e− z 1 ⎛ z3 z5 sinh z = = ⎜ 2 z + 2 + 2 +!⎟ 2 2⎝ 3! 5! ⎠

!

4

we have the Maclaurin series for Si z : !

z 2 z 3 z 4 z5 e = 1− z + − + − +! 2! 3! 4! 5! −z

!

Integrating term-by-term: !

z 2 z 3 z 4 z5 e = 1+ z + + + + + ! 2! 3! 4! 5! z

()

z 3 z5 sinh z = z + + +! = 3! 5!

!

!

e −e sinh z = 2

−z

!

n=0

z2 z3 z4 z e = 1+ z + + + +! = 2! 3! 4! z 3 z5 sin z = z − + −! = 3! 5!

We will use the definition: z

∑ (2 n + 1)! z 2 n+1

The Maclaurin series for some elementary functions are:

Determine the Maclaurin series for f z = sinh z . Solution:

∞

z2 z4 cos z = 1− + −! = 2! 4!

∞

∞

∑ n=0

zn ! n!

∑ (−1) ( n

n=0 ∞

z 2 n+1 ! 2n +1 !

)

∑ (−1) ( n=0

n

z < ∞ ! (8.9-7)

z2 n ! 2n !

)

z < ∞ ! (8.9-8)

z < ∞ ! (8.9-9) 405

π z < ! 2

z 3 2 z 5 17 z 7 tan z = z + + + +! ! 3 15 315

!

z 3 z5 sinh z = z + + +! = 3! 5!

!

∞

∑( n=0

(8.9-10)

!

From Example 8.3-3 we know that this series converges

z 2 n+1 ! 2n +1 !

)

absolutely. Letting z = iθ we have:

z < ∞ ! (8.9-11) !

z2 z4 cosh z = 1+ + +! = 2! 4!

!

tanh z = z −

!

(

)

! ln 1+ z = z −

3

5

∞

∑(

z2 n ! 2n !

n=0

z 0 . Solution: We can use the Maclaurin series for e z given in Example 8.9-1 to write:

Example 8.11-2

()

Determine the Laurent series for f z = e2

z

about the point

z = 0 valid for z > 0 .

!

⎛ 1 ⎞ 1 1 1 ! f z = z 3 ⎜ 1+ + + + +! ⎟ 2 3 4 z ⎝ ⎠ 2! z 3! z 4! z

()

()

Solution:

and so the Laurent series for f z = z 3 e1

We can use the Maclaurin series for e z given in Example 8.9-1:

z 2 z 3 z 4 z5 e = 1+ z + + + + +! 2! 3! 4! 5!

!

z 3 e1 z = z 3 + z 2 +

z

z >0

is:

z 1 1 1 + + + +! ! 2 2! 3! 4! z 5! z

z >0

z

!

()

and so the Laurent series for f z = e2 ! e

2 z

Example 8.11-4 z

is:

2 22 23 24 25 = 1+ + + + + +! = z 2! z 2 3! z 3 4! z 4 5! z 5

()

Therefore f z = e2

z

∞

∑ n=0

()

Determine the Laurent series for f z =

1 2n ! n! z n

z >0

is analytic everywhere except at z = 0 .

(

1

z 1− z

)

about the

point z = 0 for 0 < z < 1 . Solution: Using partial fractions we have: 420

!

()

f z =

(

1

z 1− z

)

=

1 1 + z 1− z

()

and so the Laurent series for f z =

For z < 1 we can use equation (8.1-14) to write: !

!

1 = 1+ z + z 2 + z 3 + ! 1− z

()

()

(

1

z 1− z

()

)

=

1 + 1+ z + z 2 + z 3 + ! z

!

We can use the Maclaurin series for cos z given in equation

()

()

f z =

ez

=

e

( z − 1) ( z − 1) 2

2

e z−1

We can use the Maclaurin series for e z−1 given in equation (8.9-7) to write:

Solution:

!

about the

We can write:

()

⎛ ⎞ z2 z4 z6 ⎜ 1− 2! + 4! − 6! +!⎟ ⎝ ⎠

( z − 1)

2

Solution:

cos z Determine the Laurent series for f z = 2 about the z point z = 0 .

(8.9-9) to write:

ez

point z = 1 .

Example 8.11-5

1 f z = 2 z

where 0 < z < ∞

Determine the Laurent series for f z =

0 < z < 1 is: f z =

cos z 1 1 z 2 z 4 = 2 − + − +! ! 2! 4! 6! z2 z Example 8.11-6

and so the Laurent series for f z about the point z = 0 for

!

cos z is: z2

!

()

f z =

(

e

)

z −1

2

(

) (

) (

)

2 3 4 ⎛ ⎞ z −1 z −1 z −1 ⎜ 1+ z − 1 + + + +!⎟ 2! 3! 4! ⎜⎝ ⎟⎠

(

)

()

and so the Laurent series for f z =

ez

( z − 1)

2

is:

421

(

!

(

) (

)

2 ⎛ ⎞ z − 1 z − 1 e 1 1 1 ⎜ + + + +!⎟ ! z ≠ 1 2 =e 2 + 2! 3! 4! ⎜⎝ z − 1 ⎟⎠ z −1 z −1 z

)

(

) (

)

()

(

) (

where 0 < z − 1 < ∞ .

where 0 < z − 1 < ∞ .

Example 8.11-7

Example 8.11-8

()

Determine the Laurent series for f z = the point z = 1 .

e2 z

( z − 1)3

(

) (

)

()

Determine the Laurent series for f z =

centered at

point z = 1 .

Solution:

Solution:

Let s = z − 1 so that z = s + 1 . We then have:

We begin by dividing z − 1 by z + 1 :

!

(

2 s+1 e ( ) e2 2s = = 3 e 3 3 s s z −1

e2 z

The Maclaurin series for e

e

!

2s

2s

2!

is:

3

3!

4

4!

5!

()

+!

⎡ ⎤ e2 2s 2 2 8 24 s 25 s 2 2 1 e =e ⎢ 3+ 2 + + + + +!⎥ s 3! 4! 5! s3 s s ⎣ ⎦ or

)

z −1 centered at the z +1

z −1 2 = 1− z +1 z +1

f z = 1−

!

5

Therefore we can write: !

(

We next need to obtain z − 1 in the denominator:

2s ) ( 2s ) ( 2s ) ( 2s ) ( = 1+ 2 s + + + + 2

()

f z =

!

)

)

2 4 5 ⎡ ⎤ 2 z − 1 2 z − 1 1 2 2 8 2⎢ f z =e + + + +!⎥ 3 + 2 + 4! 5! ⎢ z −1 ⎥ z − 1 3! z −1 ⎣ ⎦

2

( z − 1) + 2

= 1−

2 ⎛ z − 1⎞ 2 ⎜ 1+ 2 ⎟⎠ ⎝

= 1−

1 ⎛ z − 1⎞ 1+ ⎜⎝ 2 ⎟⎠

or !

⎛ z − 1⎞ f z = 1− ⎜ 1+ 2 ⎟⎠ ⎝

()

−1

Therefore: 422

!

2 ⎛ z − 1 ⎛ z − 1⎞ f z = 1− ⎜ 1+ +⎜ − 2 ⎜⎝ ⎝ 2 ⎟⎠

()

3 ⎞ ⎛ z − 1⎞ ⎜⎝ 2 ⎟⎠ +! ⎟⎟ ⎠

()

and so the Laurent series for f z =

Proof:

z −1 centered at the z +1

point z = 1 is: !

(

)2 − ( z − 1)3 +! 23

Laurent Series:

()

If a Laurent series about a point z0 converges to f z for all points in the annulus ρ1 < z − z0 < ρ2 , the absolute value of the coefficients an of the series:

()

f z =

∑

(

)n

an z − z0 !

(8.11-56)

n = 0, ± 1, ± 2, !!

()

where M = max f z

n = 0, ± 1, ± 2, !! (8.11-58)

()

f s

!∫

s − z0

C

n+1

n = 0, ± 1, ± 2, !! (8.11-59)

ds !

s − z0 = ρ ei θ !

!

ds = ρ i eiθ dθ = ρ dθ !

(8.11-60)

We can then write:

M an ≤ 2 π ρ n+1

!

∫

2π

0

M ρ dθ = 2π ρ n

∫

2π

dθ !

n = 0, ± 1, ± 2, !

0

!

(8.11-61)

or

can be estimated by:

M an ≤ n ! ρ

( ) ds ! !∫C (s − z0 )n+1 f s

Since s − z0 = ρ we have:

!

n=−∞

!

1 an ≤ 2π

!

Proposition 8.11-5, Cauchy’s Estimate for the Coefficients of a

!

1 an = 2π i

or

where 0 < z − 1 < ∞ .

∞

we have ρ1 < ρ < ρ2 . From equation

(8.11-41) we have: !

z −1 z −1 f z = + 2 22

()

Since ρ = z − z0

!

(8.11-57)

an ≤

!

M ! n ρ

n = 0, ± 1, ± 2, ! !

(8.11-62)

■

and ρ = z − z0 . 423

Chapter 9 Singularities and Zeros

( z→z

) ()

lim z − z0 f z = 0 0

424

!

In this chapter we will consider complex functions that are

A point z = z0 is called an isolated singular point or an

!

()

()

not analytic at certain points in the complex plane, and we will

isolated singularity of a function f z if f z is not analytic at

determine the nature of these singular points. We will also

z0 , but is analytic throughout some neighborhood of z0 . That is,

consider complex functions that are equal to zero at certain

a point z0 is an isolated singular point of f z if f z is not

points in the complex plane. We will see that power series can

analytic at z0 , but is analytic in some open punctured disk

provide important information on the analytic nature of

Dδ* z0 about z0 where a δ > 0 exists such that:

complex functions. Finally we will present theorems relating to

!

the number and location of the zeros of a complex function.

9.1!

()

()

not analytic at z0 . If a point z0 is a singularity of a function

()

( )

f z , then f z will always be discontinuous at z0 since f ′ z0 will not exist.

A point z = z0 in the complex plane is called an ordinary

()

()

point or a regular point of a function f z if f z is analytic at

()

0 < z − z0 < δ !

(9.1-1)

()

point or a singularity of a function f z if the function f z is

!

( )

f z is analytic can be made arbitrarily small. We see then that,

SINGULARITIES

()

()

and where the inner radius of the punctured disk in which

A point z = z0 in the complex plane is called a singular

!

()

()

for an isolated singular point z0 of a function f z , a δ neighborhood of z0 will always exist in which there are no

()

other singular points of f z . Since the δ neighborhood of an isolated singularity z0 will not contain any singularities of

() () Dδ* ( z0 ) by a Laurent series centered at z0 :

f z , it is possible to represent f z in the punctured disk

!

z0 . A function f z is called entire if it is analytic everywhere

()

f z =

∞

∑

(

an z − z0

)n !

(9.1-2)

n=−∞

in the complex plane, and so has no singularities in the complex

!

plane. If a function f z is multivalued, any reference to its

singular point or a nonisolated singularity of a function f z

analyticity applies only to a given single-valued branch of the

if there are other singular points of f z

function.

neighborhood of z0 ). A point that is a limit point of a sequence

()

A point z = z0 in the complex plane is called a nonisolated

()

()

near z0 (in every

of isolated singularities will then be a nonisolated singularity. 425

()

Any singularity of f z that is not an isolated singularity is called a nonisolated singularity.

∑

(

)n

an z − z0 !

(9.1-3)

()

( )

which defines f z on the punctured disk Dδ* z0 . This series is

()

An isolated singularity of a function f z

at a point

z = z0 can be classified using the Laurent series expansion for

()

f z =

!

n=0

9.1.1! ISOLATED SINGULARITIES !

()

∞

f z about z0 . This Laurent series expansion will converge in

( )

()

an open punctured disk Dδ* z0 . An isolated singularity of f z

at point z0 can be classified as either a removable singularity, a pole, or an essential singularity according to the number of

just a Taylor series that converges on the punctured disk

( ) f ( z ) at only the one point z0 so that: ! lim f ( z ) = f ( z0 ) = a0 ! z→ z

Dδ* z0 . The singularity at z0 can be removed by redefining (9.1-4)

0

()

( )

terms in the principal part of its Laurent series expansion about

becomes holomorphic on the disk Dδ z0 .

z0 .

9.1.1.2!

9.1.1.1! !

!

REMOVABLE SINGULARITY

()

If a complex function f z has an isolated singularity at a

()

point z = z0 , and if the Laurent series for f z about z0 has no principal part, then

()

f z

()

is said to have a removable

singularity at z0 . If f z has a removable singularity at z0 , then

()

( )

a finite lim f z exists, but this limit is not equal to f z0 . We z → z0

()

see, therefore, that f z is not continuous at z0 . !

()

The Laurent series expansion centered at z = z0 for f z

when it has a removable singularity at z0 is:

()

making f z then continuous at z0 . The function f z then

POLE

()

If a function f z has an isolated singularity at a point

()

z = z0 and, if the principal part of the Laurent series for f z

()

about z0 has a finite number of terms, then f z is said to have a pole at the point z0 . If a pole exists at z0 we will have: !

()

lim f z = ∞ !

z → z0

(9.1-5)

()

where the Laurent series for f z about z0 is convergent in the

( )

punctured disk Dδ* z0 . Such a punctured disk always exists for a pole. Equation (9.1-5) holds for any pole independent of the

()

manner in which z → z0 . The magnitude of the function f z

426

will increase without bound as z → z0

if the isolated 9.1.1.3!

singularity is a pole. All poles are isolated. !

If the principal part of a Laurent series about z = z0 can be

written with k > 0 :

a− k ≠ 0 !

!

a− ( k+1) = a− ( k+2) = ! = 0 !

(9.1-6)

()

()

z0 has an infinity of terms, then f z

isolated essential singularity at the point z0 . For a function goes to infinity, and its Laurent series has the form:

∑ a (z − z ) !

(9.1-7)

0

n ≥ −k

or !

()

f z =

( z − z0 )

k

+

a− ( k −1)

( z − z0 )

k−1

+!+

a−1 z − z0

(

)

+ a0 + a1 z − z0 + !

! ! (9.1-8) then z0 is said to be a pole of order k . Therefore the principal

()

part of the Laurent series expansion of f z is a polynomial of

(

degree k in z − z0 !

)−1 .

If the principal part of a Laurent series has exactly one

()

f z =

a−1 z − z0

(

)

(

a−2 z − z0

)

2

+

a−1 z − z0

(

)

(

)2

+ a0 + a1 z − z0 + a2 z − z0 +!

(9.1-10) where an infinity of the coefficients an with n < 0 are nonzero, and where the series is convergent in the punctured disk

( )

Dδ* z0 . 9.1.1.4! !

SINGULARITY CLASSIFICATION

()

As noted above, if a function f z

has an isolated

()

singularity at a point z = z0 , a Laurent series expansion of f z about z0 can be used to classify this isolated singularity:

term, then the series can be written as: !

()

! f z = !+ ! !

a− k

is said to have an

f z with an essential singularity at z0 , the order of the pole

n

n

()

z = z0 and if the principal part of Laurent series for f z about

()

()

f z =

()

If a function f z has an isolated singularity at a point

!

so that the Laurent series for f z has the form: !

ESSENTIAL SINGULARITY

(

)

2

+ a0 + a1 z − z0 + a2 z − z0 +! !

(9.1-9)

and the pole is said to be a pole of order one or a simple pole.

!

Removable singularity – Laurent series has no

!

!

!

Pole – principle part of Laurent series has a finite

!

principle part.

number of terms. 427

!

()

that f z is a Laurent series, and that z = 1 is a pole of

Essential singularity – principle part of Laurent series

!

order 4.

has an infinity of terms. 3.

Example 9.1-1

each of the points z = 0, ± π , ± 2π , ! where sin z → 0 and

Determine the nature of the following singularities:

f z =1 z

2.!

f z =

3.!

()

f z →∞.

()

1.!

() ()

f z =

1

( z − 1)

Example 9.1-2

4

()

Determine the nature of the singularity of f z =

1 sin z

sin z using z

the Laurent series. Solution:

Solution:

()

singularity at z = 0 . Using L’Hôpital’s rule we have:

z = 0 is a singularity. The point z = 0 is an isolated

()

singularity since f z is analytic everywhere else in the

()

The function f 0 is undefined at z = 0 , and so f z has a

()

1.! We see that as z → 0 then f z → ∞ , and so the point

!

complex plane. From equation (9.1-9) we also see that

()

2.

()

f z = 1 sin z has isolated singularities that are poles at

()

sin z cos z = lim =1 z→0 z z→0 1

lim f z = lim z →0

()

f z has a Laurent series with only one term in the

We see that lim f z is finite, and so we can anticipate that

principal part. Therefore the singularity is a simple

the singularity at z = 0 is removable. Using the Maclaurin

pole.

series expansion of sin z , we can write:

()

(

)4

f z = 1 z −1

has an isolated singularity at z = 1

()

where we see f z → ∞ . From equation (9.1-8) we see

z→0

!

⎞ sin z 1 ⎛ z 3 z5 z 7 f z = = ⎜ z − + − +!⎟ z z⎝ 3! 5! 7! ⎠

()

428

()

The Laurent series for f z = !

sin z then is: z

()

The Laurent series for f z =

sin z z2 z4 z6 f z = = 1− + − +! z 3! 5! 7!

()

()

!

⎧ sin z ⎪ f z = g z =⎨ z ⎪ 1 ⎩

()

()

()

1 1 z2 z4 f z = 2 − + − +! 2! 4! 6! z

()

!

We see that the principal part of the series has a leading

All the terms in the principal part of the Laurent series are sin z zero. We see then that f z = becomes an analytic z function at z = 0 if we define f 0 = 1 . Therefore z = 0 is a sin z removable singularity of the function f z = . We then z have the entire function g z :

()

cos z then is: z2

term with an exponent of − 2 , and so z = 0 is a pole of order 2.

()

Example 9.1-4

()

Determine the nature of the singularity of f z = e1 z .

if z ≠ 0

Solution:

if z = 0

From Example 8.11-2 we can obtain the Laurent series for

()

f z = e1 z : Example 9.1-3

()

Determine the nature of the singularity of f z =

cos z . 2 z

Using the Maclaurin series expansion of cos z we can write:

1 f z = 2 z

()

1 1 1 1 1 f z = 1+ + + + + +! ! z 2! z 2 3! z 3 4! z 4 5! z 5

()

z >0

We see that the principal part of the series has an infinity of nonzero terms, and so z = 0 is an essential singularity of

Solution:

!

!

⎛ ⎞ z z z ⎜ 1− 2! + 4! − 6! +!⎟ ⎝ ⎠ 2

4

6

()

()

f z = e1 z . Note also that f z = e1

z

will approach infinity

()

if z → 0 on the positive real axis, but that f z = e1

z

will

approach zero if z → 0 on the negative real axis. If z → 0 on

()

the imaginary axis, f z = e1

z

will oscillate wildly. Since 429

()

f z = e1

z

is analytic for all points z ≠ 0 , we see that z = 0 is

!

()

f z =

an isolated singularity.

1 z5

⎛ 1 1 1 ⎞ 1 1 1 1 1+ + + +! = = − ⎜⎝ ⎟⎠ z z2 z3 z 5 1− 1 z z 4 1− z

or using equation (8.1-14): Example 9.1-5

!

()

f z =−

Determine the nature of the singularity of the function:

(

)

1 2 3 1+ z + z + z +! 4 z

Therefore we have:

1 1 1 f z = 5 + 6 + 7 +! . z z z

()

!

Solution:

()

f z =−

1 1 1 1 2 − − − − 1− z − z −! 4 3 2 z z z z

()

We see that the function f z has a pole of order 4 at z = 0 .

While this series appears to have an essential singularity at

z = 0 , to be certain we must first determine where the series converges to see if z = 0 is included. From the ratio test we have:

!

lim

n→ ∞

1 zn

!

PROPERTIES OF ISOLATED SINGULARITIES

We will now consider isolated removable singularities,

poles, and isolated essential singularities.

1 z n+1

9.1.2!

=

9.1.2.1!

1 =q z

!

()

As we will now show, a function f z has a removable

singularity at a point z0 if and only if the following conditions

Therefore this series will converge only if 1 < z , and so we

apply:

cannot assume that this series has an essential singularity at

!

1.!

To classify the singularity of f z we can rewrite the series

!

2.!

using equation (8.1-14):

!

3.!

z = 0.

ISOLATED REMOVABLE SINGULARITY

()

()

f z is bounded in some deleted neighborhood of z0 .

(

) ()

lim z − z0 f z = 0 .

z → z0

()

lim f z exists.

z → z0

430

Proposition 9.1-1, Riemann’s Theorem on Isolated Removable Singularities:

() () f ( z ) is bounded in some neighborhood of z0 .

f z , then z0 is a removable singularity of f z if and only if

!

()

()

can be expanded in a Laurent series about the point z0 : !

()

f z =

∑

(

)n

an z − z0 !

(9.1-11)

n=−∞

where from equation (8.11-41): ! !

an =

1 2π i

( ) ds ! !∫C (s − z0 )n+1 f s

()

(9.1-13)

()

series expansion of f z will have no principal part, and so the singularity is removable. !

Since z = z0 is an isolated singularity of f z , then f z ∞

1 M M 2 π r = ! 2 π r n+1 rn

As r → 0 , then for n < 0 we have an → 0 . Therefore the Laurent

If a point z = z0 is an isolated singularity of an analytic function

Proof:

an ≤

!

()

Conversely, if z0 is a removable singularity of f z , it

()

follows that an analytic function h z must exist in the form of

( ) ( ) in the circle of convergence of f ( z ) . Moreover, since h ( z ) is analytic in a disk Dδ ( z0 ) , then h ( z ) must be continuous at z0 . Therefore lim h ( z ) must exist, z→ z and so f ( z ) is bounded near z0 (see Proposition 2.8-1). ■ a series such that f z = h z

0

Proposition 9.1-2, Isolated Removable Singularity:

n = 0, ± 1, ± 2, !!

If a point z = z0 is an isolated singularity of an analytic function

(9.1-12)

()

If f z is bounded near z0 , we have f z ≤ M for all

()

()

f z , then z0 is a removable singularity of f z if and only if:

(

) ()

lim z − z0 f z = 0 !

!

z → z0

(9.1-14)

points z near z0 ( 0 < z − z0 < δ ). Letting the contour C be a small circle of radius r = s − z0 centered at z0 where 0 < r < δ , we can use the ML-inequality (Proposition 5.3-8) together with equation (9.1-12) to write:

Proof: ! !

We will assume that:

(

) ()

lim z − z0 f z = 0 !

z → z0

(9.1-15) 431

where z = z0 is an isolated singularity of a complex function

()

( )

We can define:

f z that is analytic in a punctured disk Dδ* z0 . We will define

()

a new function g z :

(

()

()

) ()

( )

g ′ z0

if z ≠ z0 if z = z0

!

(9.1-16)

)n

(9.1-21)

z → z0

z − z0

() 2 g ( z ) = ( z − z0 ) h ( z ) !

( )

where h z is analytic on the disk Dδ z0 . We then have: !

Dδ*

z − z0

z → z0

∑

(

an+2 z − z0 !

n=0

( z0 ) . We then have: 2 g ( z ) − g ( z0 ) z − z0 ) f ( z ) ( ! = lim = lim

so that g z is analytic in !

h z =

!

2 ⎧ ⎪ z − z0 f z g z =⎨ 0 ⎪ ⎩

!

()

∞

(9.1-22)

()

()

We see from equations (9.1-16) and (9.1-22) that h z and f z (9.1-17)

( )

are the same function on the disk Dδ z0 except at the point z0

()

where f z has a singularity. This singularity at z0 must then be removable.

or

( )

(

) ()

g ′ z0 = lim z − z0 f z = 0 !

!

z → z0

(9.1-18)

() represent g ( z ) with a Taylor series:

( )

Therefore g z is also analytic in a disk Dδ z0 . We can then

()

g z =

!

∞

∑

(

)n

an z − z0 !

(9.1-19)

n=0

()

()

()

follows that an analytic function h z must exist in the form of

() () () in a disk Dδ ( z0 ) , then h ( z ) must be continuous at z0 . Therefore:

a series such that f z = h z . Moreover, since h z is analytic

z → z0

!!

when z = z0 , from equation (8.8-6) we have a1 = g ′ z0 = 0 . We

and so:

can now write equation (9.1-19) as:

!

()

! g z =

∞

∑ n=0

(

an+2 z − z0

)

n+2

(

= z − z0

)

2

∞

∑

(

)

n

an+2 z − z0 ! (9.1-20)

(

) ()

(

) ( ) ( ) z→ z (

lim z − z0 f z = lim z − z0 h z = h z0 lim z − z0

!

Since g z = 0 when z = z0 , we have a0 = 0 . Also since g ′ z = 0

( )

()

Conversely, if z = z0 is a removable singularity of f z , it

!

z → z0

0

)

(9.1-23)

(

) ()

lim z − z0 f z = 0 !

z → z0

(9.1-24)

■

n=0

432

and so:

Proposition 9.1-3, Isolated Removable Singularity: If a point z = z0 is an isolated singularity of an analytic function

()

()

f z , then z0 will be a removable singularity of f z if and

()

only if lim f z exists and is finite. z → z0

Proof: !

()

()

Dδ*

( z0 ) .

If the limit:

()

(9.1-25)

()

exists where L is finite, then f z is continuous in the disk

( ) 1.!

! !

()

f z is analytic and therefore continuous in the

( )

punctured disk Dδ* z0 .

! 2.!

C

z − z0

n−1

dz !

n =1, 2, 3, ! ! (9.1-27)

( )

Letting the boundary of the disk Dδ z0 be z − z0 = r , we can

bn ≤

!

1 2π

!∫

M r n−1 dz !

C

n =1, 2, 3, ! ! (9.1-28)

()

maximum modulus of f z on z − z0 = r . We then have:

bn ≤

!

Dδ z0 . We therefore have: !

!∫

()

f z

where C is the circle contour z − z0 = r , and where M is the

lim f z = L !

z → z0

1 2π

equation (9.1-27) to bound the coefficients:

Since the singularity of f z at a point z = z0 is isolated,

!

bn ≤

use the maximum modulus theorem (Proposition 6.7-2) and

f z will be analytic in a punctured disk !

!

()

()

f z is continuous at z0 since the limit L of f z exists at z0 .

1 M r n−1 2 π r = M r n ! 2π

n =1, 2, 3, ! ! (9.1-29)

Since we can let r be arbitrarily small, the coefficients bn of the principal part of the Laurent series must all be zero. The

() about z0 . Therefore f ( z ) the lim f ( z ) exists. z→ z

Laurent series for f z about z0 then becomes the Taylor series has a removable singularity at z0 if

0

!

From equation (8.11-34) we have the coefficients of the

principal part of the Laurent series about z0 : !

1 bn = 2π i

( ) dz ! !∫C ( z − z0 )− n+1 f z

n =1, 2, 3, ! ! (9.1-26)

!

()

Conversely, if f z has a removable singularity at z = z0 ,

()

then from Proposition 9.1-1 we know that f z is bounded in a neighborhood of z0 . Therefore a finite value L can be assigned 433

()

()

to lim f z . The function f z then becomes analytic in a disk z → z0

and so this limit exists. From Proposition 9.1-3 we see then that

( ) g ( z ) has a removable singularity at z0 .

()

f z

centered at z0 , and so from Proposition 3.7-1 we see that f z is

( )

continuous in the disk Dδ z0 , and we have:

()

( )

lim f z = f z0 = L !

!

z → z0

9.1.2.2! (9.1-30)

■

■

POLE

Proposition 9.1-5, Pole not Bounded:

()

A complex function f z that is analytic in a punctured disk

( )

Proposition 9.1-4, Isolated Removable Singularity:

() () g ( z ) have a zero of order n

() f (z) g (z)

If f z and g z are analytic functions, and if both f z and at a point z = z0 , then

has a removable singularity at z0 .

Dδ* z0 , and that has a pole at a point z = z0 cannot be bounded. Proof: !

Follows from Proposition 9.1-1.

■

Proposition 9.1-6: Proof: !

() (

) f0 ( z ) !

g z = z − z0

() (

) n g0 ( z ) !

()

()

f z = z − z0

! !

where f0 z

f0 !

( )

n

and g0 z

( )

Dδ* z0 has a pole of order k ∈! at a point z = z0 , then: (9.1-31) ! (9.1-32)

are both analytic at z0 , and where

( ) f ( z ) f0 ( z ) ! = g ( z ) g0 ( z )

z → z0

lim z − z0

( z→ z

)k f ( z ) ≠ 0 !

(9.1-34)

( z→ z

)k+1 f ( z ) = 0 !

(9.1-35)

0

z0 ≠ 0 and g0 z0 ≠ 0 . We then have: lim

()

If a complex function f z that is analytic in a punctured disk

We can write:

and !

lim z − z0 0

(9.1-33)

Proof:

434

()

If a complex function f z that is analytic in a punctured

!

( )

the limit of the product as z → z0 no longer diverges. When

(

disk Dδ* z0 has a pole of order k at a point z0 , then its Laurent

equations (9.1-34) and (9.1-35) both apply, z − z0

series expansion is given by equation (9.1-8):

pole of order k at z0 .

()

f z =

! ! !

(

a− k z − z0

)

k

+

(

a− ( k −1) z − z0

)

k−1

+!+

a−1 z − z0

(

)

+ a0 + a1 z − z0 + !

()

(9.1-36)

If a complex function f z that is analytic in a punctured disk

( z − z0 )k f ( z ) = a− k + a− k+1 ( z − z0 ) + a− k+ 2 ( z − z0 ) 2 +! ! (9.1-37)

( ) has a pole of order k ∈! at point z = z0 , then ( z − z0 ) k f ( z ) will not have a pole at z0 , but ( z − z0 )k−1 f ( z ) will Dδ* z0

have pole at z0 . We will also then have:

and so:

( z→ z

lim z − z0

!

0

)k f ( z ) = a− k ≠ 0 !

(9.1-38)

( z − z0 )

k+1

()

(

)

(

f z = a− k z − z0 + a− k+1 z − z0

)

2

(

+ a− k+ 2 z − z0

! !

)

3

+!

(9.1-39)

!

!

z → z0

(

(9.1-41)

Follows from equations (9.1-40), and (9.1-36). We can

!

( z→ z

lim z − z0 0

lim z − z0

■

z → z0

write:

and so: !

()

lim f z = ∞ !

! Proof:

We also have: !

has a

Proposition 9.1-7:

We then have: !

)k f ( z )

)

k+1

()

f z = 0!

(9.1-40)

(9.1-42)

and !

Proposition 9.1-6 shows if a complex function has a pole at

)k f ( z ) = a− k ≠ 0 !

(

lim z − z0

z → z0

(

)

k−1

)k f ( z )

()

f z =

a− k z − z0

+ a− ( k −1) → ∞ !

(9.1-43)

(

Therefore z − z0

multiplying the function by successive factors of z − z0 until

has an unremovable singularity in the form of a pole at z0 .

(

)

has no pole at z0 , while z − z0

)k−1 f ( z )

a point z = z0 , the order of the pole can be determined by

435

!

From equation (9.1-43) we have:

Example 9.1-7

()

lim f z = ∞ !

!

z → z0

()

(9.1-44)

Determine the nature of the singularity of f z where:

■

Example 9.1-6

()

Determine the nature of the singularity of f z =

1 . 2 z +9

()

The function f z

!

( z → 3i (

lim z − 3i

z → 3i

!

()

this: !

) ( z − 3i )1( z + 3i ) = 6i1 ≠ 0 1

) ( z − 3i )( z + 3i ) 2

( z→3

)4

lim z − 3

(

z + 2i

)

z−3

4

= 3+ 2i ≠ 0

and !

=0

( z→3

)5

lim z − 3

z + 2i

( z − 3)

4

(

)(

)

= lim z − 3 z + 2i = 0 z→3

()

From Proposition 9.1-6 we see that f z has a pole of order

and !

( z − 3)4

Solution:

has singularities at z = 3i and z = − 3i

that are simple poles. We can verify this:

lim z − 3i

z + 2i

We see that f z has a pole at z = 3 of order 4. We can verify

Solution:

!

()

f z =

!

)(

(

) ( z − 3i )( z + 3i ) = 0

lim z + 3i

z → −3i

1 =− ≠0 6i z − 3i z + 3i

(

lim z + 3i

z →−3i

1

2

)(

4 at z = 3 .

)

Proposition 9.1-8:

1

()

A complex function f z that is analytic in a punctured disk

()

( ) f ( z ) can be represented by the Laurent series:

From Proposition 9.1-6 we see that f z has simple poles at

Dδ* z0 will have a pole of order k at the point z = z0 if and only

z = 3i and z = − 3i .

if

436

() ! f (z) = ( z − z0 )k where g ( z ) is analytic at z0 , and g ( z0 ) ≠ 0 . g z

!

Proof:

(9.1-45)

()

If a complex function f z that is analytic in a punctured

disk

Dδ*

( z0 ) can be represented by the Laurent series: g (z) * ! ! for in ! f (z) = z D (9.1-46) δ ( z0 ) k ( z − z0 ) where g ( z ) is analytic at the point z0 and g ( z0 ) ≠ 0 , then we can represent g ( z ) as a Taylor series: !

()

g z =

∑b ( z − z ) n

(

)

(

) 2 +!!

b0 ≠ 0 !

(9.1-47)

)k

! or

() ( z − z0 )k

()

= f z =

∞

b (z − z ) ! ∑ (z − z ) 1

0

n

n

k

n=0

( z − z0 )

k−2

+! ! (9.1-50)

0

∑

(

)n

bn+k z − z0 !

(9.1-51)

which is the Laurent series representation of

()

()

f z

in a

punctured disk when f z has a pole of order k at point z0 . !

()

Conversely, since f z

( )

()

is analytic in a punctured disk

Dδ* z0 , if f z has a pole of order k at point z0 , the pole can be represented by the Laurent series: ∞

()

f z =

∑

(

an z − z0

)n

!

=

!

!

a− k

( z − z0 )

k

+

a− k +1

( z − z0 )

k −1

+!+

a−1 z − z0

(

)

+ a0 + a1 z − z0 +! ! (9.1-52)

or

Dividing equation (9.1-47) by z − z0 , we have:

g z

+

n=− k

(9.1-48)

(

∞

()

where !

( z − z0 ) ( z − z0 )

k−1

b2

n = −k

!

0

= b0 + b1 z − z0 + b2 z − z0

+

k

f z =

!

n

n=0

!

()

f z =

!

b1

and so:

!

∞

b0

(9.1-49)

!

()

f z =

∞

1

( z − z0 )

k

∑

(

)n

an−k z − z0 !

(9.1-53)

n=0

(

)k

where we have factored out the term 1 z − z0 , and where: 437

a− k ≠ 0 !

!

()

(9.1-54)

()

We can define a function g z represented by a Taylor series as: ∞

()

g z =

!

∑

(

)n

an−k z − z0 !

(9.1-55)

!

where

( ) (9.1-56) and so f ( z ) can be represented by the Laurent series: g (z) ! ! for z in Dδ* ( z0 ) ! f (z) = (9.1-57) k ( z − z0 ) where g ( z ) is analytic at z = z0 , and g ( z0 ) ≠ 0 . ■ g z0 = a− k ≠ 0 !

Proposition 9.1-9:

()

()

()

(

)

z2 z − 5

Proof: Using Proposition 9.1-8 we can write:

4

!

()

We will use Proposition 9.1-8. Writing f z in the form:

()

f z =

( )!

z2

() ()

respectively, at a point z = z0 , then f z g z will have a pole of

()

f z =

Solution:

!

( )

punctured disk Dδ* z0 , and that have poles of orders j and k ,

!

1

g z

()

If f z and g z are complex functions that are analytic in a

order j + k at z0 .

Determine the nature of the singularities of f z where:

f z =

() ! 1 where! g z = () ( ) z2 ( z − 5)4 we see that f ( z ) has a pole of order 4 at z = 5 , while g ( z ) is g z

f z =

analytic at z = 5 .

Example 9.1-8

!

()

analytic at z = 0 . Writing f z in the form:

n=0

!

()

we see that f z has a pole of order 2 at z = 0 , while g z is

()

where! g z =

1

( z − 5)4

!

()

g z =

α (z)

( z − z0 )

j

β (z)

!

(9.1-58)

!

(9.1-59)

( z − z0 ) where α ( z ) and β ( z ) are analytic at z0 . We then have: k

438

() ()

α (z)

f z g z =

!

From Proposition order j + k at z0 .

β (z)

=

α (z) β (z)

( z − z0 ) ( z − z0 ) ( z − z0 ) 9.1-8 we see that f ( z ) g ( z ) k

j

j +k

n where h z is analytic at z0 and h z0 ≠ 0 . Therefore f ( ) z

()

!

(9.1-60)

has a pole of

will have a pole of order k + n at z0 .

()

A meromorphic function f z is a single-valued function

!

()

that is meromorphic can have no singularities except poles.

()

( )

is analytic in a punctured disk

Proposition 9.1-11, Meromorphic Function:

Dδ* z0 and has a pole of order k at the point z = z0 , then the n nth derivative f ( ) z of f z will have a pole of order k + n at

()

()

() f (z).

If f z is a meromorphic function, then f ′ z will also be a

()

meromorphic function with the same poles as

z = z0 .

Proof:

Proof:

()

From Proposition 9.1-8 we known that the function f z

can be represented by the Laurent series:

()

f z =

()

where g z

() ! ( z − z0 )k

(9.1-61)

( )

is analytic at z0 and g z0 ≠ 0 . Taking the nth

()

() ! n f ( ) (z) = ( z − z0 )k+n

!

()

Using Proposition 9.1-8 we can represent the function

f z at each of the poles ak in the form:

() ! f (z) = ! n ( z − ak ) where the g k ( z ) are analytic at the poles ak gk z

g z

derivative of f z we have: !

■

discrete subset of points of D that are poles of f z . A function

If a complex function f z

!

()

that is analytic at all points in a domain D except possibly for a

■

Proposition 9.1-10:

!

( )

h z

(9.1-62)

(9.1-63)

k

( )

and g k ak ≠ 0 . We

then can write: !

()

f′ z =

1

( z − ak )

nk +1

(− n g ( z ) + ( z − a ) g′( z ) ) ! k

k

k

k

(9.1-64)

439

()

in the neighborhood of each pole. Therefore f ′ z has the same

()

poles as f z .

■

Proposition 9.1-13:

()

An analytic function f z can have an infinite number of poles Proposition 9.1-12:

only in an open or unbounded region.

() curve C , then the poles of f ( z ) inside C are finite in number.

If a complex function f z is analytic inside a simple closed

Proof:

()

Seeking a contradiction, we will assume that f z

!

Proof: !

has

Follows from Proposition 9.1-12.

9.1.2.3!

■

ISOLATED ESSENTIAL SINGULARITY

set of poles zn must then have a limit point z0 within the

() point z = z0 , then f ( z ) will not have a unique limiting value as z → z0 . Rather f ( z ) will oscillate wildly among an infinity of

region, and we have:

values (see Proposition 9.1-14). This is in contrast to the case

infinitely many poles zn within the region bounded by C . From the Bolzano-Weierstrass theorem given in Proposition 2.9-1, the

( )

( )

lim f zn = f z0 !

!

n→ ∞

!

If a function f z has an isolated essential singularity at a

() f ( z ) → ∞ as z → z0 .

when a function f z has a pole at z0 . As we have seen, a pole

(9.1-65)

will always have

By definition of a limit point, every neighborhood of z0

!

()

contains infinitely many poles of f z . Therefore z0 cannot be a

()

Proposition 9.1-14, Casorati-Weierstrass Theorem:

()

pole of f z , since no neighborhood of z0 exists that has no

If an analytic function f z has an isolated essential singularity

poles. The point z0 must then be a nonisolated singular point of

at a point z = z0 , then in any δ neighborhood of z0 , the function

()

()

f z . Therefore the singularity at the limit point cannot be a pole, but must be an essential singularity. !

()

We can conclude that if a complex function f z

f z will come arbitrarily close to any given complex value w0 . That is, given any complex value w0 , for every real number

is

()

ε > 0 we will have:

analytic inside a simple closed curve C , the poles of f z inside C must be finite in number.

■

!

()

f z − w0 < ε !

when 0 < z − z0 < δ !

(9.1-66) 440

!

Proof: !

Seeking a contradiction, we will assume that there exists a

δ neighborhood of z0 such that for any real number ε > 0 we have:

()

when 0 < z − z0 < δ !

(9.1-67)

()

and so the function f z cannot come arbitrarily close to the given complex value w0 in this neighborhood. !

()

Since z0 is an isolated essential singularity of a function

()

( )

f z , then f z is analytic in the punctured disk Dδ* z0 where: !

0 < z − z0 < δ !

(9.1-68)

( )

In the punctured disk Dδ* z0 we will let: !

()

g z =

1

()

f z − w0

!

()

()

g z =

1

()

f z − w0

≤

()

1 ! ε

(9.1-70)

We see then that g z is bounded and analytic in a punctured neighborhood of the point z0 .

(

)

(

g z = a0 + a1 z − z0 + a2 z − z0

)2 +!!

(9.1-71)

Therefore:

()

lim g z = a0 !

!

z → z0

(9.1-72)

( )

()

We can define g z0 = a0 so that g z is holomorphic at z0 . If a0 ≠ 0 we will then have from equations (9.1-69) and

!

(9.1-72):

()

lim f z = w0 + lim

z → z0

z → z0

1 1 = w0 + ! a0 g z

()

(9.1-73)

and so:

(

) ()

lim z − z0 f z = 0 !

!

where from equation (9.1-67) we have: !

()

represent g z using a power series:

! (9.1-69)

()

know that z0 is a removable singularity of g z , and so we can

!

f z − w0 ≥ ε !

!

From Riemann’s theorem (Proposition 9.1-1) we then

z → z0

(9.1-74)

Contrary to what we were given, z0 will then not be an

()

essential singularity of f z , but a removable singularity of

()

f z , as shown in Proposition 9.1-2. Therefore our assumption

()

is wrong, and the function f z will come arbitrarily close to any given complex value w0 . 441

()

singularity, an analytic function f z can approach any given !

If a0 = 0 we can write:

!

g z = z − z0

() (

finite number arbitrarily closely.

)k h ( z ) !

(9.1-75)

()

()

!

■

A similar but more definitive theorem was proved by

Picard. This theorem states that in every neighborhood of an

()

where h z is a holomorphic function at z0 , and h z ≠ 0 . From

isolated essential singularity of a function f z , there will exist

equation (9.1-73) we then have:

some point at which f z

!

k z − z0 ) ( ! lim ( z − z0 ) f ( z ) = lim ( z − z0 ) w0 + lim z→ z z→ z z→ z g (z) k

k

0

0

0

()

value infinitely many times, except at most a single value. (9.1-76) Proposition 9.1-15:

()

or !

If an analytic function f z

( z→ z

lim z − z0 0

)k f ( z ) = h 1z ≠ 0 ! ( 0)

(

lim z − z0

z → z0

)

k+1

()

f z = lim

z − z0

z → z0

( )

h z0

()

Proof:

=0!

(9.1-78)

! !

of order k . Therefore our

assumption is wrong, and the function

()

f z

will come

arbitrarily close to any given complex value w0 . !

()

lim f z = ∞ .

z → z0

From Proposition 9.1-6 we see then that, contrary to what we were given, z0 is a pole of f z

has an isolated singularity at a

point z = z0 , then the singularity z0 will be a pole if and only if

(9.1-77)

and !

actually assumes every complex

Since w0 can be any given complex number, we can

conclude that, in a δ neighborhood of an isolated essential

If we have:

()

lim f z = ∞ !

z → z0

(9.1-79)

then the isolated singularity z0 must be either a pole or an

()

essential singularity of f z

()

since it cannot be a removable

()

singularity of f z because the limit lim f z (see Proposition 9.1-3).

z → z0

does not exist

442

!

From

the

Casorati-Weierstrass

theorem

(Proposition

()

9.1-14), we see that z0 cannot be an essential singularity of f z

()

and so we can conclude that: z → z0

since from equation (9.1-79) we know that f z cannot come

()

Therefore z0 must be a pole. If z0 is a pole of f z , then

for any R ≥ 0 there must exist a δ > 0 such that:

()

for z − z0 < δ !

f z ≥ R!

!

Proposition 9.1-16:

()

If an analytic function f z (9.1-80)

z → z0

()

Conversely if z0 is a pole of f z , then there exists a

deleted neighborhood Dδ* z0 of z0 such that: !

() ! () ( z − z0 )k

f z =

()

where g z

g z

( )

for z in Dδ* z0 !

( )

( )

()

! is

() g ( z0 ) ≥ R > 0

for any value of the real number R . We then have in this

()

f z ≥

From Proposition 9.1-3 we see that a point z = z0 will be a

()

z → z0

()

From Proposition 9.1-15 we see that the point z0 will be a

()

()

pole of f z if and only if lim f z = ∞ . z→ z !

()

If lim f z z → z0

0

()

does not exist, then f z

can be neither a

removable singularity nor a pole, and so it must be an essential

( z − z0 )

k

!

()

has an isolated singularity at a point z0 such that lim f z does not exist will the singularity be an essential singularity.

R

()

singularity. Therefore, if and only if an analytic function f z z → z0

neighborhood: !

!

is finite.

(9.1-81)

continuous within Dδ* z0 , and since z0 is a pole of f z , there must exist a neighborhood in Dδ* z0 such that

Proof: removable singularity of f z if and only if lim f z exists and

is analytic, and where g z0 ≠ 0 . Since g z

( )

()

and only if lim f z does not exist, even as infinity.

()

( )

has an isolated singularity at a

point z = z0 , then this singularity is an essential singularity if

since f z is unbounded. This agrees with equation (9.1-79). !

(9.1-83)

■

arbitrarily close to any value when in a neighborhood of z0 . !

()

lim f z = ∞ !

!

■

(9.1-82)

443

()

9.1.3! NONISOLATED SINGULARITIES !

(

these points. All singularities of f z = 1 sin π z

)

are

isolated with the exception of z = 0 , which is non-

Branch points and all points on a branch cut are examples

isolated since every punctured disk centered at zero,

of nonisolated singular points. Even the smallest circuit around

Dδ* 0 , will contain an infinity of poles as the limit

a branch point of a function will cross a branch cut, and so every neighborhood of a branch point will contain at least one singularity other than the singularity at the branch point.

()

If a function f z has branch points, any δ neighborhood

!

()

of a branch point z0 will contain other singularities of f z . It is

()

( )

not possible then to represent f z in a punctured disk Dδ* z0

by a Laurent series centered at z0 . Some functions that have nonisolated singularities have an infinity of poles that converge to a limit point z0 .

()

z → 0 is approached (there will always be an n such that n > 1 δ as δ → 0 , and so the point z = 1 n = 0 will always be within the punctured disk 0 < z < δ ). The point z = 0 is then a nonisolated essential singularity.

()

2.! The function f z = Ln z is not analytic at any point on the negative real axis. Therefore all singularities of

()

f z = Ln z are nonisolated singularities since every δ

()

neighborhood of a singularity of f z = Ln z contains

()

Example 9.1-9

other singularities of f z = Ln z .

Determine the nature of the following singularities: 1.!

()

f z =

1

(

sin π z

Proposition 9.1-17:

)

()

If a complex function f z is analytic in a domain D except for

()

an infinite set of points z1 , z2 , z3 , ! which are singularities of

Solution:

this limit point must be a singularity of f z . Moreover this

2.!

1.!

f z = Ln z

()

()

f z , and if these points have a limit point z = z0 in D , then

(

f z = 1 sin π z

)

has singularities at the points z = 0

and z = 1 n where n = ±1, ± 2, ! since

()

singularity must be a nonisolated essential singularity.

()

f z → ∞ at 444

Proof: Since z = z0 is a limit point of an infinite set of singularities

!

()

of f z

()

in D , then z0 must also be a singularity of f z .

!

k−1 f ′ z0 = f ′′ z0 = f ′′′ z0 = ! = f ( ) z0 = 0 !

(9.2-2)

!

k f ( ) z0 ≠ 0 !

(9.2-3)

( )

( )

( )

( )

( )

Moreover, in every neighborhood of z0 there will be other

This definition of a zero of order k of a function provides a

singular points of f z , and so z0 must be a nonisolated

method of determining the multiplicity of a given root of the

singularity of f z . Therefore z0 cannot be a pole, and so must

function.

be a nonisolated essential singularity.

!

()

9.2! !

()

()

ZEROS

■

( )

()

()

Example 9.2-1

()

Determine the nature of the zero of f z = sin z at z = 0 .

is identically zero will all coefficients of the Taylor series that

()

represents f z be zero.

Solution:

DEFINITION OF A COMPLEX FUNCTION ZERO

()

A complex function f z that is not identically zero is said

( )

to have a zero (or root) at a point z0 if f z0 = 0 . A function

()

f z is said to have a zero of order k or a zero of multiplicity

()

k at a point z0 , if all derivatives of f z k including, f ( ) z equal zero at z0 :

()

!

( )

f z0 = 0 !

( )

then f z0 = 0 and f ′ z0 ≠ 0 .

f z can be expanded as a Taylor series about z0 . Only if f z

!

()

is said to have a simple zero at z0 . If z0 is a simple zero of f z

If a complex function f z is analytic at a point z0 , then

9.2.1!

()

A function f z having a zero of order k = 1 at a point z0

up to, but not

We have: ! !

() f ′ ( z ) = cos z !

() f ′ (0) = 1 ≠ 0

f z = sin z !

f 0 =0

()

Therefore the function f z = sin z has a simple zero z = 0 .

(9.2-1)

445

such that: Example 9.2-2

() (

)

!

an = 0 !

for n = 0, 1, 2,!, k − 1 !

(9.2-5)

!

an ≠ 0 !

for n = k !

(9.2-6)

2

Determine the nature of the zeros of f z = 1− z 2 . Solution:

() (

The function f z = 1− z 2 !

() (

f z = 1− z

()

2

)

2

has zeros at z = ±1 .

)! 2

(

2

Proof: !

( )

series are given by:

( )

!

( )

From equations (9.2-1), (9.2-2), (9.2-3), and (9.2-7) we see that,

f ±1 = 0

) (− 2 z ) !

!

f ′ z = 2 1− z

!

f ′′ z = 12 z 2 − 4 !

f ′ ±1 = 0

()

f ′′ ±1 = 8 ≠ 0

() (

Therefore the function f z = 1− z

2

)

2

As shown in Proposition 8.8-1 the coefficients of a Taylor

an =

n f ( ) z0

( )!

n = 0, 1, 2, ! !

n!

by definition, the coefficients of the Taylor series for a function

()

has zeros of order two

at ±1.

f z

having a zero of order k at a point z0 are given by

equations (9.2-5) and (9.2-6). In particular, the term a0 in the

( )

Taylor series will always be zero since f z0 = 0 . Proposition 9.2-1:

()

( )

A complex function f z that is analytic in a disk Dδ z0 will

()

!

()

f z = !

∑ n=0

(

an z − z0

)n = ak ( z − z0 )k + !

(

ak+1 z − z0

)k+1 + !

( )

A complex function f z that is analytic in a disk Dδ z0 will have a zero of order k at the point z = z0 if and only if its Taylor

represented by a Taylor series: ∞

■

Proposition 9.2-2:

have a zero of order k at the point z = z0 if and only if it can be

!

(9.2-7)

series can be expressed in the form: !

() (

)k ( )

f z = z − z0 g z !

(9.2-8)

(9.2-4) 446

()

where k is a unique integer, and g z is a power series analytic

( )

( )

in the disk Dδ z0 with g z0 ≠ 0 . Proof: !

()

()

Dδ z0 , and let f z be expressed in the form:

!

f z = z − z0

( ) (

f ′ z0 = ⎡ z − z0 ⎢⎣

( ) (

)k g ( k ) ( z ) + ( 1k ) k ( z − z0 )k−1 g ( k−1) ( z ) +! (

)k g ′ ( z ) + k ( z − z0 )k−1 g ( z ) ⎤⎥⎦ z = z= 0 !

(9.2-9)

( )

Dδ z0 , and (9.2-10)

) ( )( ) ( )

! (9.2-13) + k k − 1 ! 2 1 g z ⎤⎦ z = z0

and so: k f ( ) z0 = k! g z0 ≠ 0 !

( )

!

( ) ( )k g ( z ) ! where g ( z ) is a power series analytic in the disk where g ( z0 ) ≠ 0 . We then have: ! f ( z0 ) = 0 ! !

k f ( ) z0 = ⎡ z − z0 ⎢⎣

!

Let f z be a complex function that is analytic in a disk

( )

!

( )

(9.2-14)

The coefficients of the Taylor series are given by:

ak =

!

k f ( ) z0

( ) =g

k!

( z0 ) !

k = 0, 1, 2, ! !

(9.2-15)

and so the k th term is the first nonzero term of the Taylor series.

()

We see then that f z has a zero of order k at z0 . (9.2-11)

0

!

()

Conversely, if a complex function f z that is analytic in a

( )

disk Dδ z0 has a zero of order k , then from Proposition 9.2-1 ! !

( ) (

f ′′ z0 = ⎡ z − z0 ⎢⎣

)k g ′′ ( z ) + 2 k ( z − z0 )k−1 g ′ ( z ) (

)(

+ k k − 1 z − z0

)k−2 g ( z ) ⎤⎥⎦ z = z= 0 !

we see that it will have the Taylor series:

()

(9.2-12)

0

!

(

! f z = ak z − z0

!

()

)k + ak+1 ( z − z0 )k+1 + ak+2 ( z − z0 )k+2 + ! ! (9.2-16)

where ak ≠ 0 . We then have:

() (

! f z = z − z0

Continuing in this manner, we find that all derivatives of f z

evaluated at z = z0 are zero up to the k th derivative. For the k th

so that:

derivative we have:

!

)k ⎡⎣⎢ ak + ak+1 ( z − z0 ) + ak+2 ( z − z0 )2 + ! ⎤⎦⎥ ! (9.2-17)

() (

)k ( )

f z = z − z0 g z !

(9.2-18) 447

where

()

(

)

(

g z = ak + ak+1 z − z0 + ak+2 z − z0

!

)2 + ! !

(9.2-19)

()

g z =

( )

(

)n

an+k z − z0 !

(9.2-20)

unique integer.

n=0

() ( ) () in Dδ ( z0 ) . From equation (9.2-19) we also have: k f ( ) ( z0 ) ! g ( z0 ) = ak = ≠ 0! (9.2-21) k! ()

Therefore a complex function f z that is analytic in a disk

( )

Dδ z0 will have a zero of order k at the point z0 only if its Taylor series can be expressed in the form of equation (9.2-18).

!

To show that k is unique we will assume:

() (

)k ( ) (

)j ( )

f z = z − z0 g z = z − z0 h z !

( ) j−k g ( z ) = ( z − z0 ) h( z ) !

where h z0 ≠ 0 . If j > k , then dividing by !

() ()

( ) ( z − z0 ) j we have:

( z − z0 )k

(9.2-23)

Letting z → z0 we obtain g z0 = 0 , which is a contradiction. If

k > j , then dividing by

()

The function f z = z sin z has a zero at z = z0 = 0 . Determine

()

the nature of this zero and represent f z as a series. Solution:

f (0) = 0 () ! f ′ ( z ) = sin z + z cos z ! f ′ (0) = 0 ! f ′′ ( z ) = 2 cos z − z sin z ! f ′′ ( 0 ) = 2 ≠ 0 The function f ( z ) = z sin z has a zero of order k = 2 at z0 = 0 . !

(9.2-22) we have:

■

Example 9.2-3

where g z is analytic in the disk Dδ z0 since f z is analytic

!

(9.2-24)

must then have k = j and so g z = h z . Therefore k is a

∞

∑

)k− j g ( z ) !

Letting z → z0 we obtain h z0 = 0 , which is a contradiction. We

or !

() (

h z = z − z0

!

f z = z sin z !

Using the Maclaurin series expansion for sin z , we can write: !

⎛ ⎞ z 3 z5 z 7 f z = z sin z = z ⎜ z − + − +!⎟ 3! 5! 7! ⎝ ⎠

()

()

The Laurent series for f z = z sin z then is:

448

!

z 4 z 6 z8 f z = z sin z = z − + − +! 3! 5! 7!

()

!

2

zero of order k at z = z0 . Since both

Putting this equation in the form given in Proposition 9.2-2: !

() (

f z = z − z0

)k g ( z )

() (

f ′ z = z − z0

() (

)

( )

()

g z = 1−

2⎛

k − 1 at z0 .

()

2

4

■

Proposition 9.2-4:

()

Let a complex function f z

6

z z z + − +! 3! 5! 7!

()

domain D , and if f z has exactly one zero within C which lies at a point z0 , then:

()

( )

Let f z be a complex function analytic in a disk Dδ z0 and

z0 =

!

let it have a Taylor series that can be expressed in the form:

() (

)k ( )

f z = z − z0 g z !

()

(9.2-25)

( ) ( )

where g z is analytic in Dδ z0 , g z0 ≠ 0 , and k ≥ 2 is an

()

integer. The function f z then has a zero of order k at z = z0 ,

()

be holomorphic in a simply

connected domain D . If C is a simple closed contour within the

Proposition 9.2-3:

and f ′ z has a zero of order k − 1 at z0 . Proof:

(9.2-26)

()

⎞ z2 z4 z6 f z = z − 0 ⎜ 1− + − +!⎟ 3! 5! 7! ⎝ ⎠

!

)k−1 ( k g ( z ) + ( z − z0 ) g ′ ( z ) ) !

From Proposition 9.2-2 we see that f ′ z has a zero of order

where g z0 = g 0 = 1 : !

using equation (9.2-25) we can write: !

we have: !

( ) will have a f ( z ) and g ( z ) are analytic,

From Proposition 9.2-2 we know that f z

1 2π i

!∫

C

( ) dz ! f (z)

z f′ z

Proof: !

(9.2-27)

()

From Proposition 9.2-2 we know f z can be written in

the form: !

() (

) ()

f z = z − z0 g z !

(9.2-28)

()

where g z is a power series analytic in D . We can then write: 449

1 2π i

!

( ) dz = 1 ∫C 2π i ! f (z)

z f′ z

!∫

C

() ( ) () ( ) ()

!!

(9.2-29)

or

!

( ) dz = 1 ∫C 2π i ! f (z)

z f′ z

!∫

C

() (

) ()

z ⎡⎣g z + z − z0 g ′ z ⎤⎦ g z

()

( z − z0 )

!

1 2π i

dz !

(9.2-30)

Proof:

( ) dz = f (z)

z f′ z

!∫

C

() (

) ()

()

g z

1 2π i

!∫

C

( ) dz = z ! 0 f (z)

z f′ z

! (9.2-31) z = z0

() (

()

( ) ( )k β ( z ) ! α ( z ) and β ( z ) are power

(9.2-33)

g z = z − z0

! where

( )

(9.2-34) series analytic in the disk

() () (

) j+kα ( z ) β ( z ) ! α ( z ) β ( z ) is also

f z g z = z − z0

!

(9.2-35)

( )

analytic in Dδ z0 .

Equation (9.2-35) is a Taylor series having a zero of order j + k (9.2-32)

at z = z0 .

9.2.2!

Proposition 9.2-5:

■

ISOLATED ZERO

()

An isolated zero of a complex function f z is a point z0

!

()

that has no other zeros of f z near it. If z0 is an isolated zero

()

If f z and g z are complex functions that are analytic in a

( )

)j ( )

f z = z − z0 α z !

!

where the product

■

()

()

From Proposition 9.2-2 we know that f z and g z can

Dδ z0 . Therefore we have:

z ⎡⎣ g z + z − z0 g ′ z ⎤⎦

and so: !

z0 .

be represented by Taylor series in the form:

From the Cauchy integral formula (Proposition 6.1-1) we have: !

point z = z0 , then f z g z will have a zero of order j + k at

!

1 2π i

!

() ()

z ⎡⎣ g z + z − z0 g ′ z ⎤⎦ dz z − z0 g z

disk Dδ z0 and have zeros of orders j and k , respectively, at a

()

of f z , it is always possible to find a δ neighborhood of z0 (a

( )

disk Dδ z0 centered at z0 with δ > 0 ) containing no zero of

()

( )

()

f z except z0 . That is, if f z0 = 0 is an isolated zero of f z , 450

( )

there will exist a deleted disk Dδ* z0 with δ > 0 containing no

!

zero of f z .

second case applies, then the function f z has a zero of order

()

( )

()

z = z0 be a point in D . If f z0 = 0 , then either f z

()

of f z .

()

()

D , we can express f z in a Taylor series about the point z0 :

()

f z =

∞

∑

(

an z − z0

)n !

(9.2-36)

n=0

( )

If f z0 = 0 , then there are two possibilities:

!

1.!

All the coefficients an of the Taylor series are zero, and

()

so f z is identically zero in some neighborhood of z0 . 2.

All the coefficients an of the Taylor series are not zero, and so some coefficient ak is the first coefficient that is not zero.

)k ( )

(9.2-37)

()

where g z is analytic and is given by:

()

(

)

(

g z = ak + ak+1 z − z0 + ak+2 z − z0

!

Since f z is analytic in a domain D , and z0 is a point in

!

() (

f z = z − z0 g z !

!

is

identically zero in a neighborhood of z0 , or z0 is an isolated zero

!

()

Taylor series about the point z0 in the form:

Let a complex function f z be analytic in a domain D , and let

Proof:

()

k at z0 . Using Proposition 9.2-2 we can represent f z by a

Proposition 9.2-6, Isolated Zero:

()

If the first case applies, the proposition is proven. If the

( )

)2 + ! !

(9.2-38)

()

and so g z0 = ak ≠ 0 . Since g z is analytic in the domain D ,

()

g z

is continuous at z0 , and so, from the definition of

continuity, for some δ we will have: !

() ( )

g z − g z0 < ε !

0 < z − z0 < δ !

(9.2-39)

Choosing δ such that: !

ε=

1 g z0 ! 2

( )

(9.2-40)

we have: !

() ( )

g z − g z0
1.

!

– a pole of order k − j at z0 if j < k .

g z

■

()

If f z and g z are complex functions that are analytic in a respectively, at a point z0 , then f z

()! () g z (9.3-12) (1 ) We see that h1( z ) is analytic at z0 since h ( z ) is analytic at z0 . We also have h1 ( z0 ) ≠ 0 since g ( z0 ) ≠ 0 and g1 ( z0 ) ≠ 0 . Therefore from equation (9.3-11) and Proposition 9.2-2, we see that h ( z ) must have a zero of order k at z0 if f ( z ) = g ( z ) h ( z ) has a pole of order k at z0 .

( ) f (z) = ! h( z )

are

g z

h z = z − z0

h1 z =

( )

()

and h z

will have a pole at z0 of the same order as the zero of h z0 .

where: !

( )

!

have: !

Therefore if two complex functions g z

Proof: !

()

()

From Proposition 9.2-2 we know that f z and g z can

be represented by Taylor series in the form:

() (

)j ( )

(9.3-14)

() (

)k β ( z ) !

(9.3-15)

!

f z = z − z0 α z !

!

g z = z − z0

456

()

where α z

( )

()

and β z

( )

are power series analytic in the disk

( ) f (z) j−k α ( z ) = ( z − z0 ) ! g (z) β (z)

Example 9.3-1

Dδ z0 with α z0 ≠ 0 and β z0 ≠ 0 . Therefore we have: !

()

Determine the nature of the singularity of f z where: (9.3-16)

() ()

( )

where α z β z is also analytic in Dδ z0 . !

( ) g (z)

If j − k = 1 then f z

( z →z

lim z − z0

!

0

)β

(z)

has an isolated removable

()

The function f z has a pole of order 6 − 1 = 5 as given using z→0

= 0!

This limit has an indeterminate form, and so we can use

(9.3-17)

L’Hôpital’s rule: !

( ) g ( z ) has a zero of order

If j − k > 1 then f z

j − k at z0

( ) g ( z ) has a pole of order k − j at z0

If j < k then f z

Proposition 9.1-8).

(see

z→0

■

If a point z = z0 is an isolated singularity of a complex function

()

()

f z , then z0 is a pole of order k if and only if lim f z = ∞ .

Singularities of rational functions are always isolated

■

z → z0

Proof: !

Proof: Follows from Proposition 9.3-3.

cos z z → 0 6 z5

Proposition 9.3-5:

Proposition 9.3-4:

!

()

lim f z = lim

At z = 0 this function clearly has a pole of order 5.

(see Proposition 9.2-3). !

()

Proposition 9.3-3. To verify this we can determine lim f z .

(see Proposition 9.1-2). !

sin z z6

Solution:

singularity at z0 since:

α (z)

()

f z =

!

()

()

If lim f z = ∞ , we then must have f z ≠ 0 in the near z → z0

vicinity of z0 . 457

Since we are given that z0 is an isolated singularity of

!

()

! lim

f z , we can let:

()

h z =

!

z → z0

1

()

f z

!

(9.3-18)

( ) will be analytic and bounded f ( z ) → ∞ as z → z0 , we have:

where h z Since

()

h z = lim

!

z → z0

1

()

f z

( )

in a disk Dδ* z0 .

(9.3-19)

()

()

lim f z = ∞ !

z → z0

( ) has a pole of order k ∈! at point j lim ( z − z0 ) f ( z ) = ∞ where j = 0, 1, !, k − 1 . z→ z Proof:

()

In a deleted neighborhood of z0 we can represent f z , as

() ! () (9.3-20) k ( z − z0 ) where g ( z ) is analytic at z0 , and where g ( z0 ) ≠ 0 . Therefore we

a Laurent series:

have:

z = z0 , then

0

!

g z

■

Dδ* z0

()

()

f z =

(9.3-22)

()

Proposition 9.1-8 we have: !

0

If a complex function f z that is analytic in a punctured disk

()

Conversely, if z0 is a pole of order k of f z , then from

(9.3-21)

Proposition 9.3-6:

that z0 is a pole of order k of f z if lim f z = ∞ . !

0

0

!

is analytic in a disk Dδ z0 . From Proposition 9.3-1 we see then z → z0

=

(see also Propositions 9.1-7 and 9.1-15).

= 0!

()

)k 0 = = 0! g ( z0 ) lim g ( z ) z→ z

and so:

We see that z0 is a zero of some order k of h z . Therefore h z

( )

( z − z0 ) = lim f (z) z→ z g (z) 1

( z→ z

lim z − z0

k

!

()

f z =

a− k

( z − z0 )

k

+

a− ( k −1)

( z − z0 )

k−1

+!+

! !

a−1 z − z0

(

)

+ a0 + a1 z − z0 + ! (9.3-23)

For j = 0, 1, !, k − 1 we can write: 458

! lim

z → z0

where

(

) () j

z − z0

f z = lim

( z − z0 )k f ( z )

z → z0

1

( z − z0 )

k− j (

z − z0

) () k

f z ! (9.3-24)

has no pole at z0 (see Proposition 9.1-7).

Since k > j we can write:

lim

!

z → z0

( z − z0 ) j f ( z )

= ∞!

(9.3-25)

(9.3-27)

where the point z0 is a given fixed point, and where z is an h z f z = f z0 e ( ) !

!

()

( )

()

()

(9.3-28)

()

()

is analytic in a

()

simply connected domain D , and if f z ≠ 0 in D , then f z can be expressed in the form: g z f z = e ( )!

() where g ( z ) is analytic in D .

Proof:

(9.3-26)

(9.3-29)

we can write: g z f z = e ( )!

()

!

()

where g z is analytic in D .

(9.3-30) ■

Proposition 9.3-8:

()

If and only if a transcendental entire complex function f z has

()

()

()

()

no zeros can f z be expressed in the form:

Since f z is analytic in D , then g z and f ′ z are both

()

( ) f (z)

analytic in D . Moreover, since f z ≠ 0 in D , then f ′ z

!

is analytic in D . Because D is a simply connected domain, and

( ) f (z)

( ) ()

g z = ln f z = ln f z0 + h z !

!

If a transcendental complex function f z

f′ z

z0

Letting

Proposition 9.3-7:

!

∫

()

h z =

( ) dz = ln f z − ln f z = ln f ( z ) ! () ( 0) f z f (z) ( 0)

f′ z

arbitrary point. Therefore:

■

!

!

z

is holomorphic in D , from Proposition 5.4-7 we

( ) f ( z ) is path independent in D .

g z f z = e ( )!

() where g ( z ) is an entire function.

know that the integral of f ′ z

Proof:

Therefore this integral is a function of only the endpoints of a

!

path. We can define h z as:

is clearly nonzero and entire everywhere.

()

(9.3-31)

g z If g z is an entire function, then the function f z = e ( )

()

()

459

!

()

Conversely, if f z is an entire complex function having

()

no zeros, then as shown in Proposition 9.3-7, f z can be g z expressed in the form f z = e ( ) where g z will be analytic

()

()

()

()

where f z is analytic. Therefore g z is an entire function.

()

f z =

!

n

m z − z ( ) ∏ k

k

g z e ( )!

(9.3-34)

k =1

()

where g z is an entire function.

■

Proof: Proposition 9.3-9:

!

()

We can write:

If a transcendental analytic function f z has no zeros in a simply connected domain D , then it is possible to define a

( ( )) in D .

()

h z =

!

branch of ln f z

()

where h z

Proof: !

From Proposition 9.3-7 we have:

!

f z =e

g (z)

( ( ))

(9.3-32)

()

()

(9.3-33)

( ( ))

2

n

!

(9.3-35)

is an entire function without any zeros. From

g z h z = e ( )!

()

!

(9.3-36)

()

where g z is an entire function. Therefore:

ln f z = g z !

where g z is a branch of ln f z .

1

Proposition 9.3-8 we then have:

! () where g ( z ) is analytic in D . We then can write: !

() m m m z − z z − z ! z − z ( 1) ( 2 ) ( n ) f z

() (

f z = z − z1

!

)m ( z − z2 )m ! ( z − zn )m e g ( z ) ! 1

2

n

(9.3-37)

or

■

()

f z =

!

Proposition 9.3-10:

()

If a transcendental entire complex function f z has distinct

()

zeros zj of order mk , where k = 1, 2, 3,!, n , then f z can be

n

m z − z ( ) ∏ k

k

g z e ( )!

(9.3-38)

k =1

■

expressed in the form: 460

()

()

If a transcendental complex function f z has no zeros and no poles in a simply connected domain D , then it is possible to

( ( ))

define an analytic branch of f z

1 n

in D where n ∈! .

Proof: !

g z f z = e ( )!

() where g ( z ) is analytic in D . We can then define:

( f ( z ))

1 n

!

=e

⎡⎣ g ( z ) + 2 k π i ⎤⎦ n !

where k = 0, 1, 2, ! , n − 1 .

(9.3-39)

(9.3-40)

has removable singularities at all

()

9.3.2!

■

POLES AND ZEROS AT INFINITY

()

The nature of a singularity of f z at z = ∞ is the same as

( )

( )

(

as a one-to-one mapping from z = 0 to w = ∞ , and z = f 1 w

)

functions together provide mappings between the extended zplane and the extended w-plane. We have the definitions: is meromorphic in the complex

plane, then it can be expressed as the ratio of entire functions.

!

(9.3-42)

serves as a one-to-one mapping from w = 0 to z = ∞ . These two

()

()

A function f z is said to be analytic at the point z = ∞ if

( )

f 1 z is analytic at z = 0 .

( )

()

A function f 1 z is said to be analytic at z = ∞ if f z is

()

Since f z is meromorphic in the complex plane, we must

have:

() () () where we see that g ( z ) f ( z ) g z f z =h z !

!

that of w = f 1 z at z = 0 . The function w = f 1 z then serves

■

If a complex function f z

!

has its poles.

Therefore:

!

Proposition 9.3-12:

Proof:

()

function having its zero points where f z

poles, and so is an entire function h z .

From Proposition 9.3-7 we can write:

!

()

where h z is some complex function, and g z is an entire

Proposition 9.3-11:

( )! f (z) = g (z)

analytic at z = 0 .

()

A function f z is said to have a pole of order k at z = ∞

h z

(9.3-41)

( )

if f 1 z has a zero of order k at z = 0 . 461

()

A function f z is said to have a zero of order k at z = ∞

( )

if f 1 z has a pole of order k at z = 0 .

pole at the point z0 = ∞ .

() f (1 z ) has an essential singularity at z = 0 .

A function f z is said to have an essential singularity at

z = ∞ if !

()

z0 = ∞ has a finite number of terms, then f z is said to have a

The Laurent series can be used to classify the

singularities of a function at z = ∞ . The roles of the principal and analytic parts of the Laurent series are reversed in this classification process. A singularity of a function at the point

z = ∞ can be classified as either a removable singularity, a pole, or an essential singularity according to the number of

!

()

If a function f z has a singularity at the point z0 = ∞ , and

()

if the analytic part of Laurent series for f z about z0 = ∞ has

()

an infinity of terms, then f z is said to have an essential singularity at the point z0 = ∞ . Example 9.3-2

⎛ 1⎞ 1 Determine the nature of the singularity of f ⎜ ⎟ = 4 at ⎝ z⎠ z z = ∞.

terms in the analytic part of the Laurent series expansion of the

Solution:

function about z = ∞ .

⎛ 1⎞ 1 f z = z 4 has a zero of order 4 at z = 0 . Therefore f ⎜ ⎟ = 4 ⎝ z⎠ z has a pole of order 4 at z = ∞ .

!

()

If a function f z has an isolated singularity at the point

()

z0 = ∞ , and if the Laurent series for f z about z0 = ∞ has no

()

analytic part, then f z is said to have a removable singularity

()

at the point z0 = ∞ . The Laurent series for f z

about z0 = ∞

then has the form: !

()

0

f z =

∑

(

)n

an z − z0 !

(9.3-43)

n=−∞

!

()

If a function f z has an isolated singularity at the point

()

z0 = ∞ , and if the analytic part of Laurent series for f z about

()

Example 9.3-3

()

Determine the nature of the singularity of f z = e z at z = ∞ Solution:

⎛ 1⎞ f ⎜ ⎟ = e1 z has an essential singularity at z = 0 . Therefore ⎝ z⎠

()

f z = e z has an essential singularity at z = ∞ . 462

()

where M = w0 + ε > 0 . Therefore f z

Proposition 9.3-13:

extended complex plane. From Liouville’s theorem (Proposition

()

()

A complex function f z is analytic in the extended complex

6.5-1) we can conclude that f z must be a constant function

plane if and only if it is a constant function.

over the extended complex plane.

Proof: !

()

If f z is a constant function at all points in the extended

( )

complex plane, and so f 1 z

2 + z2 Determine the nature of f z = 2 at infinity. z −1 Solution:

()

()

must be analytic at z = 0 .

Therefore f z is analytic at z = ∞ .

()

We have:

2+

Conversely, if f z is analytic over the extended complex

()

()

plane, then f z is an entire function, and f z is analytic at

()

z = ∞ . Therefore f z

!

must be continuous at z = ∞ (see

Proposition 3.7-1), and so we have: !

()

lim f z = w0 !

z→∞

(9.3-44)

()

some R > 0 exists such that when z > R we have f z − w0 < ε . Therefore:

()

f z < w0 + ε !

(9.3-45)

or !

1 z2

2 z2 + 1 ⎛ 1⎞ f⎜ ⎟= = ⎝ z⎠ 1 1− z 2 −1 z2 At z = 0 we have:

where w0 is some finite limit. This requires that for any ε > 0

!

■

Example 9.3-4

complex plane, then it is an analytic function over the extended

!

is bounded over the

!

⎛ 1⎞ f ⎜ ⎟ =1 ⎝ z⎠ ⎛ 1⎞ 2 + z2 Since f ⎜ ⎟ is analytic at z = 0 , therefore f z = 2 is ⎝ z⎠ z −1

()

analytic at z = ∞ . Example 9.3-5

()

f z R!

!

(9.3-50)

n =1

■

order 3 at z = ∞ .

Proposition 9.3-15: Proposition 9.3-14:

()

If f z is a non-constant entire complex function having a pole

()

If f z is a complex function that is analytic at infinity, then:

()

f z =

!

∞

∑z

an n

n.

!

(9.3-47)

n =1

Proof:

()

where this series converges to f z in some region z > R .

!

! !

Proof:

()

of order n at infinity, then f z must be a polynomial of degree

( )

Let z = 1 w . The function f w

()

We can represent f z with a Taylor series:

()

f z = a0 + a1 z + a2 z 2 +! !

(9.3-51)

will be analytic in a

This function is entire (valid for all z ∈! ), and the series has an

neighborhood of w = 0 . We then can represent f w with the

infinite convergence radius. Since f z has a pole of order n at

Maclaurin series:

z = ∞ , then:

!

( )

f w =

( )

()

∞

∑ n =1

an w n !

w < R!

(9.3-48)

!

a a a ⎛ 1⎞ g ⎜ ⎟ = a0 + 1 + 22 + ! nn ! z z ⎝ z⎠ z

(9.3-52)

where: 464

()

must have a zero of order n at z = 0 . Therefore an ≠ 0 , and all

!

ak = 0 for k > n . We then have:

z = 1 w . We can then write:

()

f z = a0 + a1 z + a2 z 2 +!+ an z n !

!

(9.3-53)

()

and so f z is a polynomial of degree n .

■

()

If a complex function f z is rational, then it is meromorphic in

()

the extended complex plane, and f z is either analytic or has a pole at infinity.

()

is, by definition, the

quotient of two complex polynomials, and so has the form:

( ) = a0 + a1 z + a2 z + !+ an−1 z + an z ! ( ) Q z b + b z + b z 2 +!+ b z m−1 + b z m () 0 1 2 m−1 m P z

2

n−1

(

()

()

f z has a pole of order n − m at w = 0 . Therefore f z is either

()

()

are due to

zeros of Q z . Since zeros are always isolated, the singularities

Proposition 9.3-17:

()

If a complex function f z

()

!

Follows from Proposition 9.3-16.

( ) is meromorphic in the extended f ( z ) is a rational function, and can be

will be discrete, and so f z

complex plane, then

will be meromorphic in the

■

Proposition 9.3-18: If a complex function f z

complex plane.

has an essential singularity at

Proof:

will be poles in the complex plane. Moreover, the set of poles

()

■

infinity, then f z must be transcendental.

max m, n . The number of terms in a polynomial is finite. The only singularities that will exist in f z

()

We see then that if a complex function f z is rational, it is

meromorphic in the extended complex plane.

(9.3-54)

()

)

)

)

analytic or has a removable singularity at w = 0 . If n < m then

n

where an ≠ 0 and bm ≠ 0 , and where the degree of f z is the !

(

depends only upon the factor 1 w n−m . If m ≥ n then f 1 w is

!

A complex rational function f z

(

n n−1 n−2 ⎛ 1 ⎞ a0 w + a1 w + a2 w + !+ an−1 w + an 1 ! f⎜ ⎟= ⎝ w ⎠ b0 w m + b1 w m−1 + b2 w m−2 +!+ bm−1 w + bm w n−m ! ! (9.3-55)

analytic or has a pole at infinity.

Proof:

! f z =

is analytic at infinity, we will let

When z → ∞ we will have w → 0 . We then find that f 1 w

Proposition 9.3-16:

!

To determine if f z

represented in the form: 465

()

()

n

f z = c + f∞ z +

!

∑ f (z)!

(9.3-56)

k

()

()

n

()

g z = f z − f∞ z −

!

()

f∞ z

is the principal part of the Laurent series

()

() f ( z ) in the complex

representation of f z at infinity, and fk z is the principal part of the Laurent series representation of

plane for the k th pole, and c is a constant given by:

()

()

c = lim ⎡⎣ f z − f∞ z ⎤⎦ ! z→∞

!

(9.3-58)

k

k =1

k =1

!where

∑ f (z)!

is analytic in the entire extended complex plane. By Proposition

()

9.3-13, g z is then equal to some constant c , and so:

()

()

f z = c + f∞ z +

!

n

∑ f (z)!

(9.3-59)

k

k =1

(9.3-57)

()

As z → ∞ we will have f k z → 0 for all k = 1, 2, !, n .

!

Therefore: Proof:

()

Let f∞ z

!

be the principal part of the Laurent series

() f∞ ( z )

()

expansion of f z about the point z = ∞ . If f z has a pole at infinity, then

()

is a polynomial. If f z does not have a

()

pole at infinity, then f∞ z = 0 . For either case the function

()

()

f z − f∞ z is analytic at infinity. !

()

If the poles of a meromorphic function f z are at points

()

a1 , a2 , a3 , !, an in the extended complex plane, then f k z are

(

)

polynomials with powers of 1 z − ak . These polynomials are

()

the principal parts of the Laurent series expansions of f z

()

()

about the poles of f z . Therefore f z − function. We then have:

n

∑ k =1

()

f k z is an entire

()

()

c = lim ⎡⎣ f z − f∞ z ⎤⎦ ! z→∞

!

(9.3-60)

()

We can conclude that if a function f z is meromorphic in

!

()

the extended complex plane, then f z is a rational function having the form:

()

()

f z = c + f∞ z +

!

n

∑ f (z)! k

(9.3-61)

k =1

■

Proposition 9.3-19:

()

If a complex function f z is rational, then it can have only a finite number of poles in the complex plane. Proof: 466

!

From Proposition 9.3-16 we know that the rational

()

function f z is meromorphic in the extended complex plane.

()

Seeking a contradiction, we will assume that the function f z

has a countably infinite number of poles in the extended complex plane. From the Bolzano-Weierstrass theorem given in Proposition 2.9-1 we then know that these poles must have a limit point in the extended complex plane. Such a limit point

()

would be a nonisolated singularity, however, and so f z can not be meromorphic in the extended complex plane. Therefore,

()

if a complex function f z is rational, it can have only a finite number of poles in the complex plane.

9.4!

inside and on C . Taking the derivative of equation (9.4-2) and

()

then dividing by f z , we obtain: !

be a meromorphic function in a simply connected

()

!∫

C

!

∑ !∫ k=1

1

1

g′ z

(9.4-3)

C

(

1 z − ak

)

dz = 2 π i N !

(9.4-4)

where N are the number of zeros including multiplicities, and

( ) dz = N − P ! f (z)

f′ z

(9.4-1)

()

P

!

where N is the number of zeros of f z including multiplicities

()

inside C , and P is the number of poles of f z multiplicities inside C .

( ) ! − − − !− + ( z − b1 ) ( z − b2 ) ( z − bP ) g ( z ) 1

N

there are no poles or zeros of f z , then: !

( ) = 1 + 1 +!+ 1 f ( z ) ( z − a1 ) ( z − a2 ) ( z − aN )

f′ z

Using Proposition 5.5-3, we have:

domain D . If C is a simple closed contour in D along which

1 2π i

The function f z can be written in the form:

!

Proposition 9.4-1, Argument Principle:

()

() ( z − a1 ) ( z − a2 )!( z − an ) ! f (z) = g (z)! (9.4-2) z − b z − b ! z − b ( 1) ( 2 ) ( n ) where ai are the zeros of f ( z ) inside C and bi are the poles of f ( z ) inside C . The function g ( z ) is then analytic and nonzero !

■

ARGUMENT PRINCIPLE

Let f z

Proof:

including

∑ !∫ k=1

C

1

( z − bk )

dz = 2 π i P !

(9.4-5)

where P are the number of poles including multiplicities. From the Cauchy-Goursat theorem (Proposition 5.4-2) we also have: 467

!∫

!

C

( ) dz = 0 ! g (z)

g′ z

(9.4-6)

!

1 2π i

!∫

C

( ) dz = N − P = 2 − 0 = 2 f (z)

f′ z

Therefore integrating equation (9.4-3) over the contour C , we Example 9.4-2

obtain:

1 2π i

!

( ) dz = N − P ! f (z)

f′ z

!∫

C

1 Evaluate 2π i

(9.4-7)

■

!

!∫

C

( ) dz f (z)

f′ z

()

where f z =

z4

( z − i ) ( z − 3) 3

2

and

where C is the circle z = 2 .

( ) f (z)

The ratio f ′ z

()

is called the logarithmic derivative

Solution:

of f z . If the number and location of the zeros and poles of a

()

function in some domain are unknown, equation (9.4-1) can be

The function f z has 1 zero of order 4 at point z = 0 , and 1

successively applied to smaller and smaller regions within the

pole of order 3 at point z = i inside the circle z = 2 . It also

domain until all the zeros and poles are located.

has 1 pole of order 2 at point z = 3 , but this pole is outside the circle z = 2 . Therefore we have N = 4 and P = 3 . From

Example 9.4-1

!∫

1 2π i of radius r > 0 . Evaluate

C

( ) dz where f z = z 2 and C () f (z)

f′ z

the argument principle (Proposition 9.4-1) we then have: is a circle

Solution:

()

!

!

1 2π i

!∫

C

( ) dz = N − P = 4 − 3 = 1 f (z)

f′ z

()

We will now consider a function w = f z that is analytic

The function f z = z 2 has a zero of order 2 and no poles at

at each point on a simple closed contour C , and that is

point z = 0 . Therefore we have N = 2 and P = 0 . From the

meromorphic inside C . Such a function will generally not map

argument principle (Proposition 9.4-1) we then have:

the z-plane to the w-plane as a one-to-one mapping. Therefore 468

even a simple closed contour C in the z-plane may be mapped

()

by w = f z onto the w-plane as a closed contour Cw that is not simple. !

If the origin in the w-plane lies inside the contour Cw , then

a single trip around C can correspond to Cw being traversed a

( ) = d ln f z = d log f z + i d arg f z ! ( ( )) dz ( e ( ) ) dz ( ( )) f ( z ) dz

f′ z

!

We then can write:

( ) dz = 1 d log f z ∫C e ( ) 2π i ! f (z)

f′ z

!∫

1 2π i

!

number of times. The argument principle can be used to determine the number of times the image contour Cw of C

C

w

+

!

winds completely around the origin in the w-plane.

()

()

plane. Since log e f z

be an analytic function in a simply connected

domain D . If C is a simple closed contour in D along which

()

!∫

!

( ) dz = i d arg f z = n C, 0 2 π i ! !∫C ( ) ( ) f (z)

C

(

)

!∫ (

)

C

! (9.4-11)

w

is the number of times the contour Cw winds

completely around the origin in the w-plane. The number of times the branch cut for the logarithm is crossed is given by

(

)

n Cw , 0 .

()

()

On a branch of the logarithm of f z on which f z is

holomorphic, we can write:

is single-valued on the closed contour

( ) dz = 1 d arg f z = n C , 0 ∫C ( ( ) ) ( w ) 2π ! f (z)

f′ z

where n Cw , 0

C winds completely around the origin in the w-plane.

!

1 2π i

(9.4-8)

is the number of times the image contour Cw of

Proof:

Cw

! (9.4-10)

then have using equation (9.4-9): !

w

where n C, 0

!∫ d (arg f ( z ))

Cw , the first integral on the right in equation (9.4-10) is zero. We

there are no poles or zeros of f z , then:

f′ z

1 2π

where the contour Cw is the image of the contour C in the w-

Proposition 9.4-2, Winding Number Around the Origin: Let w = f z

(9.4-9)

!

The integral in equation (9.4-11) provides a measure of the

()

change in arg f z per circuit of 2 π around the origin in the w-

(

plane. For this reason n Cw , 0

)

is known as the winding

number of a contour Cw about the origin w = 0 in the w-plane 469

(see Section 5.6). The integral in equation (9.4-11) also provides

Proof:

the rationale for naming the principle the argument principle.

!

()

Letting f z = z , equation (9.4-11) becomes:

!

(

)

n Cw , 0 =

!

1 2π i

!∫

C

!

1 dz ! z

(9.4-12)

From equations (9.4-7) and (9.4-11) we also have:

(

)

n Cw , 0 = N − P !

!

!

(9.4-13)

()

We will now consider a theorem that allows us to

closed contour.

h z =

f z +g z

!

() f (z)

g z

< 1!

(9.5-4)

From equations (9.5-3) and (9.5-4) we then obtain:

Proposition 9.5-1, Rouché’s Theorem:

!

()

If f z and g z are analytic functions in a simply connected domain D , and if C is a simple closed contour in D where:

()

() ()

( ) ( ) = 1+ g ( z ) ! () (9.5-3) f (z) f (z) We see that w = h ( z ) has no zeros or poles on the contour C .

determine the number of roots a polynomial has within a given

!

(9.5-2)

From equation (9.5-2) we have:

ROUCHÉ’S THEOREM

()

()

for all points on C , we know that f z ≠ 0 and f z + g z ≠ 0

!

■

9.5!

()

f z > g z ≥ 0!

on C . Let:

which is equation (5.6-1) when z0 = 0 . !

Since we are given:

()

!

() f (z)

g z

()

= h z − 1 < 1!

(9.5-5)

()

Using w = h z , the contour C in the z-plane will be

(9.5-1)

mapped into a closed contour Cw in the w-plane since C is

for all points on C , then f z and f z + g z have the same

centered at the point w = 1 in the w-plane. This means that the

f z > g z ≥ 0!

()

() ()

number of zeros inside C (counting multiple zeros).

closed. All points in Cw will be contained in the disk w − 1 < 1 point w = 0 is not contained on this disk, and so 1 w is analytic 470

on and inside Cw . We then have from the Cauchy-Goursat

of the largest part of the original function, and the second

theorem (Proposition 5.4-2):

function g z acting as a perturbation to the main part. Since this perturbation will be smaller than the main function, the

!∫

1 dw = 0 ! Cw w

!

(9.5-6)

()

change in h z will be small. Equation (9.5-8) which measures this change over the contour results in an integer value which

()

Since h z = w , we also can write:

must be zero for a very small change.

()

dw = h′ z dz !

!

()

(9.5-7) Example 9.5-1

and so equation (9.5-6) becomes:

!∫

!

Cw

Determine the number of roots 16 z 3 − 20 z 2 + 8 z − 1 = 0 has

( ) dz = 0 ! h( z )

h′ z

inside the unit circle z = 1.

(9.5-8)

()

where the integrand is the logarithmic derivative of h z .

Solution:

!

Let f z + g z = 16 z 3 − 20 z 2 + 8 z − 1 . We must choose f z

() () () so that f ( z ) > g ( z ) on the contour z = 1 . We will let f ( z ) = 16 z 3 + 8 z so that g ( z ) = − 20 z 2 − 1. On the circle z = 1

From the argument principle (Proposition 9.4-1), we see

()

that h z has an equal number of zeros and poles inside Cw .

()

() f (z)

Since the poles of h z are the zeros of f z , and the zeros of

()

() ()

h z are the zeros of f z + g z , then

() ()

and f z + g z

we then have:

have the same number of zeros inside C (counting multiple

()

zeros). The particular zeros inside f z

() ()

ones inside f z + g z . !

may differ from the

f z = 16+ 8= 24

!

g z = − 20− 1 = 21

■

Rouché’s theorem can be used to determine the number

and location of roots for a complicated function by breaking the

()

function into two functions: the main function f z consisting

()

!

()

and so

()

()

f z > g z

() ()

on the contour. This means that

f z + g z = 16 z 3 − 20 z 2 + 8 z − 1 should have the same 471

()

number of zeros inside z = 1 as does f z = 16 z 3 + 8 z . We

()

can check this. We have for f z = 0 :

!

()

3

() ()

2

2

()

()

()

fn z − f z < M ≤ f z !

!

for all n > N ! (9.6-1)

()

()

()

()

()

f z + fn z − f z = f n z !

!

for all n > N ! (9.6-2)

have the same number of zeros inside C .

HURWITZ’S THEOREM

{ f ( z )} be a sequence of continuous single-valued complex ()

functions that converges uniformly to f z ≠ 0 at every point z inside and on a simple closed contour C . If the contour C does

() N > 0 such that every function fn ( z ) with n > N number of zeros inside C as the function f ( z ) .

not pass through any zeros of f z , then there exists an integer has the same

■

()

From Hurwitz’s theorem we see that every function fn z

!

Proposition 9.6-1, Hurwitz’s Theorem: n

There must then exist an integer N > 0 such that:

we then know that the functions f z and:

which can also be seen to have three zeros inside z = 1 .

Let

()

converges uniformly to f z ≠ 0 at every

for all points on C . From Rouché’s theorem (Proposition 9.5-1)

⎛ 1⎞ ⎛ 1⎞ 16 z − 20 z + 8 z − 1 = ⎜ z − ⎟ ⎜ z − ⎟ = 0 2⎠ ⎝ 4⎠ ⎝

9.6!

n

()

function f z + g z = 16 z 3 − 20 z 2 + 8 z − 1 = 0 can be written: !

{ f ( z )}

f z > 0 will have some minimum M > 0 on the contour C .

which can be seen to have three zeros inside z = 1. The

3

Since

point z inside and on a simple closed contour C , the function

⎛ 1⎞ f z = 16 z + 8 z = z ⎜ z 2 + ⎟ = 0 2⎠ ⎝

!

Proof:

with n > N can have no zeros in a neighborhood of any given

( )

point z0 unless z0 is a zero of f z0 .

9.7! !

OPEN MAPPING THEOREM We will now consider an open set of points S in a domain

()

D . If these points are mapped by a complex function f z onto an open set of points Ω , then the mapping is considered open.

()

That is, for every z ∈S the image Ω created by f z is open. 472

! Proposition 9.7-1, Open Mapping Theorem:

( )

() ( )

Dδ z0 , where from continuity we have f z − f z0 ≤ ε for

()

() g ( z ) = f ( z ) − f ( z0 ) = w − w0 !

If f z is a non-constant function analytic in a domain D in

some ε > 0 . We will let g z be given by:

the z-plane, then the image of the domain D produced by the

!

( )

mapping w = f D is a domain in the w-plane.

()

(9.7-2)

where g z will also be a non-constant function analytic in D .

Proof: !

() ( )

The function f z − f z0 will then be analytic in the disk

By definition a domain is an open connected set of points

()

(see Section 2.2.10). We need to show that the function w = f z

()

We can choose δ such that g z ≠ 0 on the circle C , and

()

g z = 0 at only the one point z0 within C (an isolated zero).

maps subsets of the domain D to open connected subsets in the

()

w-plane. That is, every point of D mapped by w = f z is an

( )

( )

interior point of f D , and f D is connected. !

() f (z)

Since f z is a non-constant function analytic in D , we

know that

is continuous in D (see Proposition 3.7-1).

Since any two points of D are connected, mapping of these

() the image f ( D ) must be connected. ! To show that the image f ( D ) is open, we will consider a points by w = f z will result in two connected points, and so

subset S of points in a neighborhood of D defined by: !

z − z0 ≤ δ !

!

(9.7-1)

where z0 is any point in D , and where δ > 0 . We will define C to be the circle z − z0 = δ .

Figure 9.7-1!

Mapping of an open set of points S in a domain D by a function f z onto an open set of points Ω.

()

473

!

There will be an image circular neighborhood Ω centered

( )

( )

at the point f z0 = w0 of the set of points S in the disk Dδ z0

() ( )

as is shown in Figure 9.7-1. We will let M = min f z − f z0

( )

on

the image circle f C in the w-plane. !

If w1 is an arbitrary point within the circular neighborhood

Ω , we can write:

()

( )

(9.7-3)

We have previously shown that a power series known as a

Taylor series can be used to represent any analytic function. A Taylor series representing an analytic function

()

f1 z

will

converge on the disk specified by the circle of convergence of

()

for the Taylor series to provide a valid representation of f1 z .

()

Beyond this circle the power series for f1 z is not defined. We

or

()

w0 − w1 < M ≤ f z − w0 !

!

(9.7-4)

We can write: 0

0

1

1

(9.7-5)

() as does f ( z ) − w0 .

From Rouché’s theorem we know that f z − w1 has the same number of zeros inside z − z0 = δ

() ( ) () f ( z ) − w1 has at least one zero, and so f ( z ) − w1 = 0 at

Since f z − f z0 = f z − w0 = 0 at z = z0 , we know that the

function

()

will now consider ways to construct another function f 2 z

()

which will serve to extend the power series for f1 z beyond the circle of convergence of its Taylor series.

( f (z) − w ) + (w − w ) = f (z) − w !

!

!

!

ANALYTIC CONTINUATION

the series. Only within this circle of convergence is it possible

w1 − w0 < M ≤ f z − f z0 !

!

9.8!

!

For a Taylor series whose point of expansion is z1 and for

which the nearest singularity is at a point zp , the radius of its circle of convergence C1 will be: !

zp − z1 = r !

!

(9.8-1)

(see Figure 9.8-1 and Proposition 8.8-7). This circle is also

()

least once. Therefore a point z within S must be mapped into

known as the singularity circle for the Taylor series for f1 z .

Ω as w1 . Finally, since w1 is an arbitrary point in Ω , for every

The point set inside the circle of convergence C1 will be

point w in the open disk f z − f z0 < ε a point z will exist in

designated as domain D1 (we will use the same subscript to

the open disk Dδ z0 such that w = f z . Therefore if f z is a

designate the circle of convergence and its associated domain).

non-constant function analytic in a domain D in the z-plane,

!

then the image w = f D is a domain in the w-plane.

chosen as the center of expansion for a new Taylor series. We

( )

( )

() ( )

()

()

■

Any point within the circle of convergence C1 can be 474

will now select another point z2 inside the circle of convergence

!

C1 , and we will make z2 a point of expansion for a new Taylor

at point zp , the circle of convergence C2 centered at point z2

series that represents some other function f 2 z . The function

includes points not inside the circle of convergence C1 centered

f 2 z will have a circle of convergence C2 different from C1 (see

at point z1 . The function f1 z is analytic within domain D1 ,

()

()

While neither circle C1 nor C2 has inside it the singularity

()

()

and the function f 2 z is analytic within domain D2 . Within the

Figure 9.8-1).

point set that is the intersection D1 ∩ D2 we will have

()

()

f1 z = f 2 z . !

()

The function f 2 z is known as the analytic continuation

()

or analytic extension of the function f1 z function

()

f1 z

into D2 . The

can equally be considered the analytic

()

continuation or analytic extension of the function f 2 z into D1 . !

We then have:

!

f1 z !

()

defined for all z ∈D1 .

!

f2 z !

()

defined for all z ∈D2 .

!

f1 z = f 2 z !

()

()

defined for all points D1 ∩ D2 .

()

We can define a function F z that is analytic on D1 ∪ D2 :

Figure 9.8-1!

Analytic continuation of a function from a circle of congruence C1 centered at z1 to a circle of congruence C2 centered at z2 . Both circles are in the z-plane.

()

()

for all z ∈D1 .

()

()

for all z ∈D2 .

!

F z = f1 z !

!

F z = f2 z !

!

If an analytic continuation exists it will be unique (see

Propositions 8.8-9 and 9.8-4). Although the functional forms of 475

()

the Taylor series for f1 z

()

and f 2 z

()

will be different, they

represent the same function F z in different domains of the

r2 = zp − z2 !

!

(9.8-4)

()

complex plane.

where the nearest singularity of f z

!

Therefore from Proposition 1.8-3:

We see then that a Taylor series representation of any

()

analytic function f1 z

can be continued by selecting a

!

different point of expansion within its circle of convergence

()

to form a new Taylor series representing a function f 2 z .

and

This new series can be analytically continued through any

!

()

points on the circle of convergence of the f1 z series that are not singular (unless the singularity is removable).

()

Let f z

be analytic in a domain D1 . If the radius of

()

r2 = zp − z2 ≤ zp + z2 !

(9.8-6)

r1 − r2 ≤ z1 − z2 !

z1, z2 ∈D !

(9.8-7)

z1, z2 ∈D1 !

!

r1 − r2 ≤ z1 − z2 !

z1, z2 ∈D !

(9.8-8)

We then have using Proposition 1.8-5:

z1 is r1 , and about a point z2 is r2 , then: r1 − r2 ≤ z1 − z2 !

(9.8-5)

or

convergence of the Taylor series expansion of f z about a point

!

r1 = zp − z1 ≤ zp + z1 !

We then can write: !

Proposition 9.8-1:

is at the point zp .

(9.8-2)

r1 − r2 ≤ z1 − z2 !

!

z1, z2 ∈D !

(9.8-9)

■

Proof: !

By definition of the radius of convergence, we have from

equation (9.8-1): ! and

r1 = zp − z1 !

!

The process of analytic continuation can be repeated many

times as long as only isolated singularities exist that can be avoided by new circles of congruence. This makes it possible

(9.8-3)

for an analytic function having only poles to be represented by Taylor series everywhere in the complex plane (except at the 476

poles) starting from its values in some neighborhood of a point of expansion z1 (see Figure 9.8-2).

()

We can then represent a function F z by the functions

!

()

f k z , each having its own circle of convergence Ck :

!

⎧ ⎪ ⎪ ⎪⎪ F z =⎨ ⎪ ⎪ ⎪ ⎪⎩

()

() f 2 ( z ) for z ∈D2 f3 ( z ) for z ∈D3 f1 z for z ∈D1

()

!

(9.8-10)

!

f k z for z ∈Dk

Example 9.8-1 Determine an analytic continuation for the complex

()

geometric series f1 z =

∞

∑z

k

.

k =0

Solution: Figure 9.8-2! !

()

Analytic continuation of a power series starting from a domain D1 centered at z1 in the z-plane.

()

A complete analytic function F z can then be obtained

by analytic continuation along some arc in the form of a chain of circles of congruence. The centers of expansion for this chain will be located on the arc. This curve will continue as long as the last circle of congruence is partially outside the previous

The function f1 z is the Maclaurin series representation of the function: !

()

f2 z =

1 1− z

This series only converges for z < 1 (see Example 8.2-1).

()

The circle of convergence of f1 z is z = 1.

circle of convergence, and has a center within the previous

The series representation of a function may be defined in a

circle of congruence (see dashed curve in Figure 9.8-2).

much smaller region than is the function itself. This is the 477

()

()

case for the functions f1 z and f 2 z . While the function

()

f1 z is defined only inside its circle of convergence z = 1 ,

()

the function f 2 z

()

()

() f1 ( z )

provides an analytic continuation of

()

over the entire

!

complex plane except at the singular point z = 1. If the process of analytic continuation encounters an

∑

z2

n

We then have: !

convergence of f1 z , we have f1 z = f 2 z . Therefore f 2 z

()

f z = z + z 2 + z 4 + z 8 +! =

n=0

is defined everywhere in the complex

plane except at the point z = 1. Inside the circle of

!

!

∞

( ) () f (z ) = f (z) − z − z f z2 = f z − z 4

!

! !

( )

()

n

f z2 = f z −

infinity of singularities, then these singularities can form a natural boundary, and the process of analytic continuation

2

n−1

∑

z2

k

k =0

()

We see then that if z = eiθ is a singularity of f z , then

( )

2n

cannot be extended beyond the boundary. If a circle of

z = eiθ

convergence exists for which a function is singular at every

we see that z = 1 is a natural boundary for the function

()

point on the circle, then the function cannot be analytically

f z .

continued across this line (see Example 9.8-2).

Proposition 9.8-2, Identity Theorem:

Example 9.8-2

() () () g ( z ) have the same value at each of the points zk of a line segment contained in D , then f ( z ) = g ( z ) everywhere in D . If f z and g z are analytic in a domain D , and if f z and

Show that the circle of convergence z = 1 is a natural

()

boundary for the function f z =

∞

∑

n

z 2 where n ∈! .

n=0

Solution:

()

The series f z is:

are also singularities for all n ∈! . Letting n → ∞ ,

Proof: !

We have: 478

( )

( )

f zk = g zk !

!

k = 1, 2, 3, ! , n !

(9.8-11)

Let z0 be an arbitrary point in D where:

zk ≠ z0 !

!

k = 1, 2, 3, ! , n !

(9.8-12)

()

complex function f z is defined and analytic in a domain D ,

()

() ()

() () () f ( z ) is considered an analytic continuation

real-valued function g x exists such that f z = g x on the

!

(9.8-13)

line interval, then

()

of g x onto D . Only one unique continuous analytic function

Therefore:

!

elementary functions of a real variable to elementary

and if within D there is an interval of the real axis on which a

h z = f z −g z !

( )

h zk = 0 !

!

Analytic continuation makes possible the extension of

functions of a complex variable as was done in Chapter 4. If a

and let: !

!

k = 1, 2, 3, ! , n !

()

(9.8-14)

()

can represent a set of points in a domain that includes some interval of the real axis.

The function h z is analytic at point z0 since f z and

() h ( z ) at all points in D as a Taylor series centered at point z0 : h′′ ( z0 ) 2 ! h ( z ) = h ( z0 ) + h′ ( z0 ) ( z − z0 ) + z − z0 ) +!! (9.8-15) ( 2

Example 9.8-3

For equation (9.8-14) to hold in D for the points zk , where

sin 2 z + cos 2 z = 1 along the entire x-axis, if we let:

g z are analytic at all points in D . Therefore we can represent

k = 1, 2, 3, ! , n and zk ≠ z0 , we must have all coefficients of the Taylor series in equation (9.8-15) equal to zero. Therefore we

Show that sin 2 z + cos 2 z = 1 for all z ∈! . Solution: Since sin z and cos z are entire functions, and since we have

!

( )

( )

( )

h z0 = h′ z0 = h′′ z0 = ! = 0 !

(9.8-16)

!

()

()

so f z = g z everywhere in D .

■

()

() ()

h z = f x − g x = sin 2 x + cos 2 x − 1 = 0 then from Proposition 9.8-2 we have for all z ∈! :

But then we must also have h z ≡ 0 for every point in D , and

()

()

g z =1

so that for all z = x :

must have: !

()

f z = sin 2 x + cos 2 x !

!

sin 2 z + cos 2 z = 1 479

and so from the identity theorem (Proposition 9.8-2) we must Proposition 9.8-3:

have in D2 :

()

() f (z) = 0

()

()

f3 z ≡ f 2 z !

If f z is an analytic function in a domain D , and if f z = 0

!

at all points on a line segment within D , then

which is contrary to our assumption.

(9.8-18)

everywhere in D . Proof: !

Follows from the identity theorem (Proposition 9.8-2).

■

Proposition 9.8-4, Unique Analytic Continuation: Let D1 and D2 be two intersecting domains, and let their union

() f 2 ( z ) that is analytic in

be the domain D . If a function f1 z is analytic in D1 , then there is at most only one function

()

D2

and equal to f1 z in D .

Figure 9.8-3!

Proof: !

The domains D1 , D2 , and D are illustrated in Figure 9.8-3.

Seeking a contradiction, we will assume that two different

()

()

functions f 2 z and f3 z exist that are both analytic in D2 and

()

both equal to f1 z in D . ! !

!

Intersecting domains where D = D1 ∩ D2 .

()

There is then at most only one function f 2 z

()

that is

analytic in D2 and equal to f1 z in D . If a natural boundary

()

()

exists for f1 z , however, no analytic continuation of f1 z will exist.

■

We then have everywhere in domain D :

()

()

()

()

f3 z − f 2 z = f1 z − f1 z = 0 !

(9.8-17) 480

()

Proposition 9.8-5:

of f ′ z .

()

If a function f x defined on some interval of the real axis can be analytically continued into the complex plane, then this continuation can only occur in one way.

!

!

The configuration of the two domains D1 and D2 is shown

()

()

()

and since f z is analytic everywhere in D = D1 ∩ D2 , we must

Follows from Propositions 9.8-4 and 9.8-2.

()

()

()

have f ′ z = g ′ z everywhere in D = D1 ∩ D2 . Therefore g ′ z

■

Proposition 9.8-5 shows why it was possible to extend the

elementary holomorphic functions discussed in Chapter 4 from the real axis into the complex plane.

()

is an analytic continuation of f ′ z .

■

Proposition 9.8-8, Analytic Continuation Across a Boundary:

()

()

Let f1 z be analytic in a domain D1 , and f 2 z be analytic in a domain D2 , and let the two domains be adjacent to each other

Proposition 9.8-6: The behavior of a function that is analytic in a domain D is completely determined by its behavior in a small neighborhood of an arbitrary point in that domain.

()

()

with D1 ∩ D2 = ∅ . Let f1 z = f 2 z

along a segment of the

()

common border between the two domains, and let both f1 z

()

and f 2 z be continuous on this segment. Let C1 be a closed contour within D1 that includes this common border segment, and let C2 be a closed contour within D2 that includes this

Proof: !

Proof: in Figure 9.8-3. Since f z = g z everywhere in D = D1 ∩ D2 ,

Proof: !

()

everywhere in D1 ∩ D2 . Then g ′ z is an analytic continuation

Follows from Propositions 9.8-2 through 9.8-5.

()

common border segment. Then f1 z

■

()

and f 2 z

are analytic

continuations of each other. Proposition 9.8-7:

()

()

Let f z be analytic in a domain D1 , and g z be an analytic

()

()

()

continuation of f z into a domain D2 so that f z = g z

481

!

Proof: !

The configuration of the two domains D1 and D2 is shown

in Figure 9.8-4. We have from the Cauchy-Goursat theorem (Proposition 5.4-2) along contour C1 : !

!∫

C1

()

f1 z d z = 0 !

!∫

C2

()

f2 z d z = 0 !

integration along the common border segment of D1 and D2 will cancel because of opposite directions of traverse (see Figure 9.8-4). !

(9.8-19)

Defining

!∫

()

f z dz =

C

(9.8-20)

C = C1 ∪ C2

without

the

common

border

segment, we then have: !

and along contour C2 : !

Summing the integrals in equations (9.8-19) and (9.8-20),

()

!∫

C1

()

f1 z d z +

!∫

C2

()

f2 z d z = 0 !

()

(9.8-21)

() f (z)

where f z is equal to f1 z for z in D1 and to f 2 z for z in

D2 . From Morera’s theorem (Proposition 6.3-4),

is then

holomorphic on the combined domains D1 and D2 . Therefore

()

()

f1 z and f 2 z are analytic continuations of each other across the boundary between domains D1 and D2 .

9.9! !

■

SCHWARZ’S REFLECTION PRINCIPLE If the domain of an analytic function includes an interval

of the real axis, under certain conditions it is possible to obtain an analytic continuation of the function across the real axis in the form of a reflection about the axis. This provides a method to continue a function using a symmetry process. Figure 9.8-4!

Adjacent domains D1 and D2 . 482

()

If f z is an analytic function defined in a domain D in the upper half-plane that includes an interval of the real axis, then !

( )

() (9.9-1) if and only if f ( x ) is real-valued function for each point of the f z = f z !

Proof:

!

()

We will first assume that f z is real-valued when z is on

()

the interval of the real axis within D . To determine if f z

()

( )

is

() (

)

(

)

!

f z = f x, −y = u x, −y + i v x, −y !

!

g z = U x, y + iV x, y !

!

g z = f z = u x, −y − i v x, −y !

( )

(

()

(

()

( )

) (

)

)

(

(

(

)

)

)

(

)

(

) (

)

U x, y = u x, −y !

∂u ∂v =− ! ∂y ∂x

(9.9-8)

∂U ∂u ! = ∂x ∂x

∂V ∂v =− ! ∂x ∂x

(9.9-2)

!

∂U ∂u =− ! ∂y ∂y

∂V ∂v ! = ∂y ∂y

(9.9-3)

Using equations (9.12-8), (9.12-9), and (9.12-10) we obtain:

(9.9-4)

!

(9.9-5)

Therefore U x, y and V x, y also satisfy the Cauchy-Riemann

∂U ∂V ! = ∂x ∂ y

(

)

∂U ∂V ! =− ∂y ∂x

(

)

(9.9-9)

(9.9-10)

(9.9-11)

(

equations and, since the partial derivatives of U x, y

We then have from equations (9.12-4) and (9.12-5): !

∂u ∂v ! = ∂x ∂ y

From equations (9.8-6), (9.8-7), and (9.8-8) we have:

write:

f z = u x, y + i v x, y !

()

The Cauchy-Riemann equations apply to f z :

!

analytic throughout D , we will let g z = f z . We then can

!

(

V x, y = − v x, −y !

must be continuous in D . !

real axis contained in D .

!

) ( ) (9.9-7) ! Since f ( z ) = u ( x, y ) + i v ( x, y ) is an analytic function, the partial derivatives of u ( x, y ) and v ( x, y ) are continuous in the domain D . Therefore the partial derivatives of u ( x, −y ) and − v ( x, −y ) are also continuous in D , and so U ( x, y ) and V ( x, y ) !

Proposition 9.9-1, Schwarz Reflection Principle:

)

and

(9.9-6) 483

(

)

()

( )

V x, y are continuous in D , we can conclude that g z = f z

()

!

be real-valued for each point of the real axis contained in D .

On the segment of the real axis within D :

() ( )

( )

f x = u x, 0 + i v x, 0 !

()

(9.9-12)

( )

v x, 0 = 0 !

!

(9.9-13)

( )

()

()

( )

f x = f x = u x, 0 !

! !

( )

()

the real axis, then f x

()

(9.9-15)

()

f z = f z !

)

(

(9.9-16)

) (

)

(

)

u x, −y + i v x, −y = u x, −y − i v x, y !

( )

( ) ( )

( )

u x, 0 + i v x, 0 = u x, 0 − i v x, 0 !

Follows from Proposition 9.9-1.

■

Example 9.9-1

()

Show that f z = 2 z + 1 = 2 z + 1 . (9.9-17)

Solution:

()

and so on the real axis: !

Proof: !

Conversely if we have:

(

can be continued analytically to the

entire complex plane.

then using equations (9.9-2) and (9.9-3), we have: !

()

including the real axis, and if f x is a real-valued function on

(9.9-14)

f z = f z !

( )

in

If f z is an analytic function defined on the upper half-plane

From Proposition 9.8-5 we then have: !

()

and f z

Proposition 9.9-2:

From equations (9.9-12) and (9.9-13) we obtain: !

()

From Proposition 9.8-8 we see that f z

■

Proposition 9.9-1 are analytic continuations of each other.

Since f z is real-valued when z is real, we have: !

()

f z must be real-valued when z is real. Therefore f z must

is analytic in D . !

( )

For this equation to be true we must have v x, 0 = 0 , and so

Since 2 x + 1 is real, we see that f z = 2 z + 1 is real when z is (9.9-18)

real. According to the Schwarz reflection principle we then have 2 z + 1 = 2 z + 1 . We can check this: 484

!

(

)

(

)

2 z +1= 2 x − i y +1= 2x +1 − i y

Example 9.9-3

and !

(

)

(

)

(

()

)

2 z +1= 2 x + i y +1= 2x +1 + i y = 2x +1 − i y

We have:

2 z +1= 2 z +1 ! Example 9.9-2

()

Show that f z = sin z = sin z . Solution: This is shown in equation (4.3-58). Using equation (4.3-2) we

()

Solution:

Therefore: !

( )

If f z = 2 z + i , does f z = f z ?

( )

()

f z = 2 z + i! Therefore

()

( )

f z =2z+i=2z −i

()

f z ≠ f z . This is to be expected since

f x = 2 x + i is not real, and so does not satisfy the criteria necessary for the Schwarz reflection principle to be applicable.

can write sin z in the form: !

ei z − e− i z sin z = 2i When z is real we then have:

!

ei x − e− i x = sin x 2i as given in equation (4.3-1). Therefore sin z is real-valued when z is real.

485

Chapter 10 Residues

!∫

C

()

( () )

f s ds = 2 π i Res f z , z0

486

!

In this chapter we will define the residue from the Laurent

series, and we will present methods for obtaining the residue. We will derive Cauchy’s residue theorem, and we will demonstrate how it can be used to evaluate definite and improper real-valued integrals. We will discuss the summation of some infinite series using residues. Finally we will consider Mittag-Leffler’s expansion theorem.

analytic function plays a key role both in evaluating integrals of the function and in determining the location of zeros and poles of the function. This coefficient is known as the residue of the !

()

If f z

is a complex function analytic in an annular

()

domain D centered at an isolated singularity z = z0 , then f z

can be represented by the Laurent series about z0 as given in equation (8.11-40): !

()

f z =

∞

∑ a (z − z ) ! n

n

1 2π i

( ) dz ! !∫C ( z − z0 )n+1 f z

0

!

n=−∞

The coefficients an are specified by equation (8.11-41):

()

Integrating the Laurent expansion of f z term-by-term

over the contour C , we can write:

!∫

∞

()

f z dz =

∑ !∫ an

( z − z0 )n dz

!

(10.1-3)

⎧⎪ 2 π i for n = −1 ! dz = ⎨ ⎪⎩ 0 for n ≠ −1

(10.1-4)

n=−∞

C

Using Proposition 5.5-4 we obtain: !

!∫ ( z − z ) 0

C

n

We therefore have: !

!∫

C

()

f z dz = 2 π i a−1 !

(10.1-5)

or !

(10.1-1)

n = 0, ± 1, + 2, + 3, ! ! (10.1-2)

contour centered at z0 and lying within D .

C

One of the coefficients in a Laurent series expansion of an

function.

an =

where C is any positively oriented simple closed circular

!

10.1! RESIDUES !

!

a−1 =

1 2π i

!∫

C

()

f z dz !

(10.1-6)

()

Obviously if a−1 is known for the function f z , then the

()

integral of the function f z over the closed contour C can be obtained immediately from equation (10.1-5). 487

!

The term with coefficient a−1 in the Laurent series

()

expansion of f z about a singularity at a point z0 is the only

!

()

!

()

()

will exist even if z0 is an isolated essential singularity of f z . Since a function will not have a representation as a Laurent series at a branch point, the residue is not defined at such a point. !

()

We will designate the residue of a function f z

at a

singularity z0 by: !

()

a−1 = Res ⎡⎣ f z , z0 ⎤⎦ !

(10.1-7)

()

a−1 = Res f z ! z = z0

(10.1-8)

!∫

C

!

()

()

f z dz = 2 π i Res ⎡⎣ f z , z0 ⎤⎦ !

)

()

(

)

(10.1-10)

()

(10.1-11)

10.1.1! RESIDUES IN THE COMPLEX PLANE !

()

If the Laurent series expansion for an analytic function

f z about a singularity z0 can be found, then the coefficient a−1 of the Laurent series can be determined by inspection, and

()

the residue of the function f z at the singularity z0 obtained. It is often necessary to calculate only a few terms of a Laurent series expansion to determine the term with coefficient a−1 . Nevertheless, finding the Laurent series expansion of a function ways of determining the single coefficient a−1 without first having to find other terms of the Laurent series expansion. Example 10.1-1

(10.1-9)

If the contour C in equation (10.1-3) has a winding

(

(

()

Using equation (10.1-7), equation (10.1-5) becomes: !

f z dz = 2 π i n C, z0 Res ⎡⎣ f z , z0 ⎤⎦ !

()

f z can be quite difficult. Fortunately there are some other

Another notation in common use for residues is: !

!∫

C

this reason a−1 is called the residue of the function f z at the singularity z0 . The residue of a function f z at a singularity z0

f z dz = 2 π i n C, z0 a−1 !

C

nonzero term left following term-by-term integration of the Laurent series over a simple closed contour C encircling z0 . For

!∫

)

number n C, z0 > 1 about the point z0 , then it is necessary to rewrite equations (10.1-5) and (10.1-9) using equation (5.6-1):

5 z2 − z + 1 Determine the residue of f z = . 3 z

()

Solution:

()

The Laurent series for f z can be determined directly: 488

1 1 5 − + z3 z2 z

()

f z =

!

found from knowledge of the other series expansions.

()

We see that a−1 = 5 , and that f z has a singularity at z = 0 .

Example 10.1-3

Therefore: !

()

In some cases the Laurent series for a function f z can be

ez Determine the residue of f z = 3 . z

()

()

Res ⎡⎣ f z , 0 ⎤⎦ = 5

Solution:

()

Example 10.1-2

The function f z has a pole of order 3 at z = 0 . Using the

()

Determine the residues of f z =

(

1

)

z z −1

Maclaurin series expansion of e z , we can write:

. !

Solution:

()

(

1

)

z z −1

=

−1 1 + z z −1

The Laurent series for f z then is: !

()

We see that f z has simple poles at z = 0 and z = 1 . Both

are: !

()

()

!

1 and we have: 2

1 Res ⎡⎣ f z , 0 ⎤⎦ = 2

()

Res ⎡⎣ f z , 0 ⎤⎦ = −1 Example 10.1-4

and !

ez 1 1 1 1 1 z z 2 f z = 3= 3+ 2+ + + + +! 2! z 3! 4! 5! z z z Therefore a−1 =

terms on the right side of this equation are Laurent series that each have only the a−1 term. Therefore the two residues

()

⎛ ⎞ z 2 z 3 z 4 z5 ⎜ 1+ z + 2! + 3! + 4! + 5! +!⎟ ⎝ ⎠

()

Using partial fractions we can write:

f z =

ez 1 f z = 3= 3 z z

()

Res ⎡⎣ f z ,1⎤⎦ =1

()

Find the residue of f z =

sin z inside the circle z = π 2 . z2 489

Solution:

()

The function f z has a pole of order 2-1=1 at z = 0 as given

!

()

f z = cot z =

cos z = sin z

using Proposition 10.3xxx-3. Using the Maclaurin series expansion of sin z , we can write: !

sin z 1 f z = 2 = 2 z z

()

⎛ ⎞ z 3 z5 ⎜ z − 3! + 5! −!⎟ ⎝ ⎠

!

()

1 −! z

()

Res ⎡⎣ f z , 0 ⎤⎦ = 1 Example 10.1-6

sin z 1 z z 2 f z = 2 = − + −! z 3! 5! z

()

()

Determine the residue of f z = e 1

z2

.

Solution:

Therefore a−1 = 1 and we have: !

z 3 z5 z − + +! 3! 5!

=

Therefore a−1 = 1 and we have:

The Laurent series for f z then is: !

z2 z4 1− + +! 2 3!

()

The function f z has a pole at z = 0 . Using the Maclaurin

()

Res ⎡⎣ f z , 0 ⎤⎦ = 1

series expansion of e z , we can write the Laurent series:

Example 10.1-5

!

()

Find the residue of f z = cot z inside the circle z = π 2 .

()

f z = e1

z2

= 1+

1 1 1 1 1 + + +! z 2 2! z 4 3! z 6

Since there are an infinity of terms in the principal part of

()

Solution:

the Laurent series expansion of f z

The function:

function f z

()

f z = cot z =

()

cos z sin z

has a pole at z = 0 inside the circle z = π 2 . Using the Maclaurin series for sin z and cos z , we can write:

about z = 0 , the

has an essential singularity at z = 0 (see

()

Example 10.1-4). Since f z has no 1 z term in the Laurent series expansion, we have a−1 = 0 , and so: !

()

Res ⎡⎣ f z , 0 ⎤⎦ = 0 490

Example 10.1-7

sin z z2 z4 z6 f z = = 1− + − +! z 3! 5! 7!

()

!

()

Find the residue of f z = z cos

1 inside the circle z = π 2 . z

All the terms in the principal part of the Laurent series are

()

define f 0 = 1. Therefore z = 0 must be a removable

Using the Maclaurin series expansion of cos z , we can write

()

singularity of f z . We also see that a−1 = 0 , and so:

the Laurent series:

1 1 1 z cos = z − + −! = z 2! z 4! z 3

∞

∑ (−1) n=0

n

1

( 2 n) ! z 2 n−1

Since there are an infinity of terms in the principal part of

()

the Laurent series expansion of f z

()

()

Res ⎡⎣ f z , 0 ⎤⎦ = 0

!

Proposition 10.1-1:

()

about z = 0 , the

If a complex function f z has a removable singularity at a point z = z0 , then:

function f z has an essential singularity at z = 0 . From the

1 z term in the Laurent series expansion, we have: !

()

zero. We see then that f z will be analytic at z = 0 if we

Solution:

1 Res ⎡⎣ f z , 0 ⎤⎦ = 2

()

()

Res ⎡⎣ f z , z0 ⎤⎦ = 0 !

!

(10.1-12)

Proof:

()

Since z0 is a removable singularity of f z , the Laurent

!

()

series for f z is:

Example 10.1-8

sin z Find the residue of f z = inside the circle z = π 2 . z

()

Solution: From Example 9.1-2 we have the Laurent series expansion:

()

(

)

(

f z = a0 + a1 z − z0 + a2 z − z0

!

)2 +!!

(10.1-13)

and so no coefficient a−1 exists. Therefore:

()

Res ⎡⎣ f z , z0 ⎤⎦ = 0 !

!

(10.1-14)

■

491

()

where P z is a polynomial of degree ≤ 2 , and a , b , and c are

Proposition 10.1-2:

()

distinct complex numbers, then:

If a complex function f z has a simple pole at a point z = z0 , then:

()

(

) ()

Res ⎡⎣ f z , z0 ⎤⎦ = lim ⎡⎣ z − z0 f z ⎤⎦ ! z → z0

! Proof: !

()

()

f z =

!

z − z0

(

)

(

+ a0 + a1 z − z0 + a2 z − z0

)

+! ! (10.1-16)

( ) ( z − z0 ) f ( z ) = a−1 + ( z − z0 ) ⎡⎣ a0 + a1 ( z − z0 ) +! ⎤⎦ !

()

(10.1-17)

Taking the limit as z → z0 , we obtain a−1 :

()

(

) ()

Res ⎡⎣ f z , z0 ⎤⎦ = lim ⎡⎣ z − z0 f z ⎤⎦ = a−1 ! z → z0

!

!

() )( )

(10.1-22)

P c

(

Representing the terms in equation (10.1-19) as a Laurent

series about the point z = a , we have:

(

(10.1-18)

A z− a

)

!

(10.1-23)

which is a single term Laurent series. We also have:

Proposition 10.1-3:

!

()

If a complex function f z has the form:

!!

(

! C = Res ⎡⎣ f z , c ⎤⎦ = c−a c−b

!

!

■

!

()

(

Proof:

Multiplying by z − z0 we have: !

(10.1-21)

P b ! B = Res ⎡⎣ f z , b ⎤⎦ = b− a b−c

f z is: 2

() )( )

!

()

Since z0 is a simple pole of f z , the Laurent series for

a−1

(10.1-20)

P a A = Res ⎡⎣ f z , a ⎤⎦ = ! a−b a−c

(10.1-15)

()

() )( )

!

()

f z =

()

P z

=

A

+

B

+

C

( z − a ) ( z − b) ( z − c ) ( z − a ) ( z − b) ( z − c )

!

(

B

−B

∞

∑(

1 B = =− b − a 1− z − a z− b n=0 b − a b− a

) (

)

)

n ! (10.1-24) z − a ( ) n+1

which is a Taylor series analytic at the point z = a , and:

(10.1-19) 492

!

(

C

−C

∞

∑(

1 C = =− c − a 1− z − a z− c n=0 c − a c−a

) (

)

( z − a) ! n

)

n+1

Example 10.1-9 (10.1-25)

5z 2 Determine the residue of f z = . z−i

()

Solution:

which is a Taylor series analytic at the point z = a . !

Using

equations

(10.1-23),

(10.1-24),

and

()

The function f z has a simple pole at z = i . From

(10.1-25),

equation (10.1-19) becomes:

()

f z =

!

A

( z − a)

∞

−

∑ n=0

Proposition 10.1-2 we have:

⎡ B ⎢ ⎢ b− a ⎣

(

⎤ ⎥ z − a n ! (10.1-26) + n+1 n+1 ⎥ c−a ⎦ C

)

(

)

(

)

!

and so using Proposition 10.1-2:

() )( ) (

()

(

!!

() )( )

()

() )( )

(

( ) ⎡ ⎤ ! C = Res ⎣ f ( z ) , c ⎦ = ( c − a ) ( c − b) ■

(

1

)

z z −1

.

Solution:

()

The function f z has simple poles at z = 0 and z = 1 . From Proposition 10.1-2 we have:

(10.1-28) !

P c

!

)

()

Using a similar derivation for the other singularities, we obtain: !

(

Determine the residues of f z =

(10.1-27)

P b ⎡ ⎤ ! B = Res ⎣ f z , b ⎦ = b− a b−c

()

Example 10.1-10

P z P a A = Res ⎡⎣ f z , a ⎤⎦ = lim = z→a z − b z − c a−b a−c

!

⎡ 5z 2 ⎤ Res ⎡⎣ f z , i ⎤⎦ = lim ⎢ z − i = 5i 2 = −5 ⎥ z→i z −i⎦ ⎣

(10.1-29)

⎡ 1 ⎤ 1 Res ⎡⎣ f z , 0 ⎤⎦ = lim ⎢ z = −1 ⎥= z→0 −1 z z − 1 ⎢⎣ ⎥⎦

()

(

)

and !

⎡ 1 ⎤ 1 Res ⎡⎣ f z ,1⎤⎦ = lim ⎢ z − 1 ⎥ = =1 z →1 z z − 1 ⎢⎣ ⎥⎦ 1

()

(

)

(

)

493

()

()

!

Using Proposition 10.1-3 we can write f z in the form of

has a simple zero at z0 so that z0 is a simple pole of f z , we

partial fractions:

then have:

()

()

Res ⎡⎣ f z , 0 ⎤⎦ Res ⎡⎣ f z ,1⎤⎦ −1 1 f z = = + = + z z z −1 z −1 z z −1

()

(

1

)

()

!

From Proposition 10.2-2 we know that since z0 is a simple

!

Example 10.1-2.

()

pole of f z ,we can write:

Example 10.1-11

()

()

(

Res ⎡⎣ f z , z0 ⎤⎦ = lim ⎡⎣ z − z0 z → z0

!

ez Determine the residues of f z = . z −1

() ()

⎡ p z ⎤ f z ⎤⎦ = lim ⎢ z − z0 ⎥ z → z0 q z ⎥⎦ ⎢⎣

) ()

(

!!

)

(10.1-31)

( )

We also have q z0 = 0 . Therefore we can write:

Solution:

()

()

Proposition 10.1-2:

()

(

)

() ()

()

() () ( )

()

(10.1-33)

and so equation (10.1-33) becomes:

Let f z = p z q z where the complex functions p z and

( )

)

p z ⎡ ⎤ ! Res ⎣ f z , z0 ⎦ = lim z → z0 ⎡ ⎤ q z − q z0 ⎢ ⎥ z − z ⎢⎣ ⎥⎦ 0

!

Proposition 10.1-4:

()

(

or

⎡ e1 z ⎤ Res ⎡⎣ f z ,1⎤⎦ = lim ⎢ z − 1 = lim e1 z = e ⎥ z →1 z − 1 ⎦ z →1 ⎣

() () ( )

⎡ ⎤ p z Res ⎡⎣ f z , z0 ⎤⎦ = lim ⎢ z − z0 ⎥ ! (10.1-32) z → z0 − q q z z0 ⎥⎦ ⎢⎣

!

The function f z has a simple pole at z = 1 . We have from

()

(10.1-30)

Proof:

These results can be compared with those obtained from

!

( ) ( )

p z0 ⎡ ⎤ Res ⎣ f z , z0 ⎦ = ! q′ z0

()

q z are both analytic at a point z = z0 . If p z0 ≠ 0 and q z

( ) ( )

p z0 ⎡ ⎤ ! Res ⎣ f z , z0 ⎦ = q′ z0

()

!

(10.1-34)

■

494

Therefore:

Proposition 10.1-5:

()

()

()

Let f z = 1 q z where the complex function q z is analytic

()

at a point z = z0 . If q z has a simple zero at z0 so that z0 is a

()

!

simple pole of f z , we then have:

()

Res ⎡⎣ f z , z0 ⎤⎦ =

!

1

( )

q′ z0

ez Res ⎡⎣ f z , 1⎤⎦ = 2

()

= z =1

e 2

and

!

(10.1-35)

!

ez Res ⎡⎣ f z , − 1⎤⎦ = 2

()

e−1 1 = = 2 2e

z = −1

Proof: !

Follows from Proposition 10.1-4.

■

Example 10.1-13

z2 Determine the residues of f z = 4 . z −1

()

Example 10.1-12

z ez Determine the residues of f z = 2 . z −1

()

Solution:

()

The function f z has simple poles at z = ±1 and z = ± i .

Solution:

From Proposition 10.1-4 we have:

()

The function f z has simple poles at z = ±1 . From Proposition 10.1-4 we have: !

()

Res ⎡⎣ f z , z0 ⎤⎦ =

!

( ) q′ ( z0 ) p z0

()

Res ⎡⎣ f z , ±1⎤⎦ = lim

z = ±1

where!

z2 1 = 3= 4z q′ 4 z p

Therefore:

and so: !

( )! Res ⎡⎣ f ( z ) , z0 ⎤⎦ = q′ ( z0 ) p z0

ze

(

z

d 2 z −1 dz

)

z

1 1 Res ⎡⎣ f z ,1⎤⎦ = lim = z= 1 4 z 4

!

Res ⎡⎣ f z , − 1⎤⎦ = lim

z

ze e = lim z = ±1 2 z z = ±1 2

= lim

!

() ()

1 1 =− z = −1 4 z 4 495

!

1 i Res ⎡⎣ f z , i ⎤⎦ = lim =− z= i 4 z 4

!

1 i Res ⎡⎣ f z , − i ⎤⎦ = lim = z = −i 4 z 4

()

Example 10.1-15

()

()

Determine the residues of f z =

sin z . 2 1+ z

Solution:

()

Example 10.1-14

The function f z has simple poles at z = ± i . From

()

Determine the residues of f z =

1 . 2 2 z +a

Proposition 10.1-4 we have:

Solution:

()

The function f z has simple poles at z = ± ia . From

()

Res ⎡⎣ f z , ± i a ⎤⎦ = lim

z = ± ia

(

1

d 2 2 z +a dz

)

1 = lim z = ± ia 2 z

sin z ! Res ⎡⎣ f z , i ⎤⎦ = 2z

()

!

()

1 i = =− i 2a 2a z =ia

()

Res ⎡⎣ f z , − i a ⎤⎦ =

1 2z

=− z = −i a

1 i = i 2a 2a

)

sin i 1 e−1 − e 1 e − e−1 1 = = = = sinh1 2 i 2 i 2 i 2 2 2 z =− i

sin z Res ⎡⎣ f z , − i ⎤⎦ = 2z

()

!

= z=i

( )=

sin − i −2i

1 e − e−1 1 e − e−1 = −2i 2i 2 2

or

and !

(

d 1+ z 2 dz

sin z z=±i 2 z

= lim

and

and so:

1 Res ⎡⎣ f z , i a ⎤⎦ = 2z

z=±i

sin z

and so from equation (4.3-55):

Proposition 10.1-5 we have: !

()

Res ⎡⎣ f z , ± i ⎤⎦ = lim

!

!

()

Res ⎡⎣ f z , − i ⎤⎦ =

sin z 2z

= z=i

1 sinh1 2

The same result can be obtained using Proposition 10.1-2. 496

!

Example 10.1-16

()

Res ⎡⎣ f z , n π ⎤⎦ = lim

z = nπ

()

Determine the residue of f z = cot z . !

Solution:

d sin z dz

=

1 cos z

( )n

= −1 z = nπ

()

The logarithmic residue of a function f z at a point z0 is

defined to be:

The function:

() ()

⎡ f′ z ⎤ Res ⎢ , z0 ⎥ ! f z ⎢⎣ ⎥⎦

!

cos z f z = cot z = sin z

()

has simple poles at z = n π where n = 0, ± 1, ± 2, ± 3,! . From Proposition 10.1-4 we have: !

1

Proposition 10.1-6:

()

If a complex function f z is analytic at a point z = z0 , and if

cos z cos z Res ⎡⎣ f z , n π ⎤⎦ = lim = lim =1 z = nπ d z = n π cos z sin z dz

()

See Example 10.1-5.

(10.1-36)

() f ( z ) at z0 is k .

f z has a zero of order k at z0 , then the logarithmic residue of

Proof: !

Example 10.1-17

From Proposition 9.2-2 we can write:

() (

()

Determine the residue of f z =

1 . sin z

Solution:

()

The function f z has simple poles at z = n π where

n = 0, ± 1, ± 2, ± 3,! . From Proposition 10.1-4 we have:

)k ( )

f z = z − z0 g z !

!

(10.1-37)

()

where k is a unique integer, g z is analytic at z0 , and where

( )

g z0 ≠ 0 . Therefore we can write: !

()

(

f ′ z = k z − z0

)k−1 g ( z ) + ( z − z0 )k g ′ ( z ) !

(10.1-38)

and so: 497

(

( ) = k + g′ ( z ) ! f ( z ) z − z0 g ( z )

In the product of these two series, the coefficient of z − z0

f′ z

!

(10.1-39)

which has a simple pole at point z0 . Using Proposition 10.1-2 we have:

() ()

⎡ f′ z ⎤ ! Res ⎢ , z0 ⎥ = lim ⎢⎣ f z ⎥⎦ z → z0

() ()

)

a−1 b0 . Therefore we have:

() ()

Res ⎡⎣ g z f z , z0 ⎤⎦ = a−1 b0 !

!

()

(10.1-44)

( )

But Res ⎡⎣ f z , z0 ⎤⎦ = a−1 and g z0 = b0 . Therefore we have:

⎡ ⎛ k g′ z ⎞ ⎤ + ⎢ z − z0 ⎜ ⎟ ⎥ = k ! (10.1-40) z − z g z ⎝ ⎠ ⎥⎦ ⎢⎣ 0

(

)−1 is

() ()

( )

()

Res ⎡⎣ g z f z , z0 ⎤⎦ = g z0 Res ⎡⎣ f z , z0 ⎤⎦ ! (10.1-45)

! ■

■

Proposition 10.1-8: Proposition 10.1-7:

() and a complex function g ( z ) is analytic at z0 , then: Res ⎡⎣ g ( z ) f ( z ) , z0 ⎤⎦ = g ( z0 ) Res ⎡⎣ f ( z ) , z0 ⎤⎦ ! (10.1-41)

() and a complex function h ( z )

If a complex function f z has a simple pole at a point z = z0 ,

If a complex function f z has a simple pole at a point z = z0 , !

Proof: ! !

()

The Laurent series for f z is:

()

∞

f z =

∑ a (z − z ) n

n

0

n = −1

=

a−1 z − z0

(

)

+ a0 + a1 z − z0 + ! ! (10.1-42)

()

()

! g z =

∑ k =0

(

bk z − z0

()

)k =

(

then:

() ()

⎡f z ⎤ 1 Res ⎢ , z0 ⎥ = Res ⎡⎣ f z , z0 ⎤⎦ ! ⎢⎣ h z ⎥⎦ h z0

!

( )

()

Proof:

and, since g z is analytic at z0 , the Taylor series for g z is: ∞

( )

is analytic at z0 with h z0 ≠ 0 ,

)

(

)2

b0 + b1 z − z0 + b2 z − z0 + ! !(10.1-43)

!

()

(10.1-46)

()

Follows from Proposition 10.1-7 with h z = 1 g z .

■

Example 10.1-18

()

Determine the residue of f z =

cos z . z

Solution: 498

()

The function f z has a pole of order 1 at z = 0 . From

(

!

Proposition 10.1-7 we have:

⎡ 1⎤ Res ⎡⎣ f z , 0 ⎤⎦ = cos z z = 0 Re s ⎢ z ⎥ = 1 ⎣ z⎦

()

!

d k−1 ⎡ z − z0 dz k−1 ⎢⎣

)k f ( z )⎤⎥⎦ = ( k − 1) ! a−1 + k! a0 ( z − z0 ) k + 1) ! ( + a

!

(1 z − z0 )2 +! !

2!

(10.1-50)

Taking the limit as z → z0 , we obtain: Proposition 10.1-9:

()

If a complex function f z has a pole of order k ≥ 2 at a point

z = z0 , then:

d k−1 ⎡ lim z − z0 ⎢ z → z0 dz k−1 ⎣

(

!

)k f ( z )⎤⎦⎥ = ( k − 1) ! a−1 !

(10.1-51)

and so we have a−1 :

d k−1 ⎡ ! Res ⎡⎣ f z , z0 ⎤⎦ = lim z − z0 z → z0 k − 1 ! dz k−1 ⎢ ⎣

()

(

1

(

)

) () k

f z ⎤ ! (10.1-47) ⎥⎦

d k−1 ⎡ ! Res ⎡⎣ f z , z0 ⎤⎦ = lim z − z0 z → z0 k − 1 ! dz k−1 ⎢ ⎣

()

1

(

(

)

)k f ( z )⎤⎥⎦ !

(10.1-52)

■

Proof:

()

If f z

!

Example 10.1-19

has a pole of order k ≥ 2 at a point z = z0 , the

()

Laurent series about z0 for f z is given by equation (9.1-8):

()

f z =

! ! !

a− k

( z − z0 )

k

+

a− ( k −1)

( z − z0 )

k−1

+!+

(

where a− k ≠ 0 . Multiplying by z − z0 ! !

a−1 z − z0

)

(

)

+ a0 + a1 z − z0 + !

(

+ a0 z − z0

)

(

+ a1 z − z0

)

k+1

Differentiating k − 1 times, we have:

()

The function f z has a pole of order 2-1=1 at z = 0 as given

we obtain:

(

+ a2 z − z0

)

k+2

sin z . 2 z

Solution:

(10.1-48) k

using Proposition 9.3-3. From Proposition 10.1-9 we have:

( z − z0 )k f ( z ) = a− k + a− ( k−1) ( z − z0 ) +!+ a−1 ( z − z0 )k−1 k

()

Determine the residue of f z =

! + ! !(10.1-49)

1 d ⎡ 2 sin z ⎤ Res ⎡⎣ f z , 0 ⎤⎦ = lim = lim cos z = 1 ⎢z 2 ⎥ z → 0 2 − 1 ! dz z ⎣ ⎦ z→0

()

(

)

(see Example 10.1-4). 499

()

Res ⎡⎣ f z , − 1⎤⎦ = lim

!

Example 10.1-20

()

Determine the residue of f z =

5 z2

( z − i)

2

.

⎡ d ⎢ Res ⎡⎣ f z , 3⎤⎦ = lim z−3 z → 3 dz ⎢ ⎣

()

!

()

has a pole of order 2 at z = i . From

()

(

(

)

)2

⎤ 5z ⎥ = lim 10z = 10i 2 ⎥ z − i ⎦ z→i 2

(

−8 1 =− 16 2

(

)

2

(

⎤ ⎥ 2 z + 1 z − 3 ⎥⎦ 5z − 3

)(

)

or

Proposition 10.1-9 we then have:

⎡ 1 d ⎢ ! Res ⎡⎣ f z , i ⎤⎦ = lim z−i z → i 2 − 1 ! dz ⎢ ⎣

( z − 3)2

=

and from Proposition 10.1-9 we have:

Solution: The function f z

z → −1

5z − 3

d ⎡ 5z − 3 ⎤ 8 8 1 Res ⎡⎣ f z , 3⎤⎦ = lim = lim = = ⎢ z + 1 ⎥ z →3 2 z → 3 dz 16 2 ⎣ ⎦ z +1

()

!

(

)

)

Example 10.1-22 Example 10.1-21

()

Determine the residues of f z =

()

Determine the residue of f z =

5z −3

( z + 1) ( z − 3)

2

.

()

The function f z

()

The function f z has a simple pole at z = −1 and a pole of

⎡ 5z − 3 Res ⎡⎣ f z , − 1⎤⎦ = lim ⎢ z + 1 z → −1 ⎢ z +1 z − 3 ⎣

()

! or

(

)

(

)(

)

⎤ ⎥ 2 ⎥ ⎦

( z + 5)

3

.

Solution:

Solution:

order 2 at z = 3 . From Proposition 10.1-2 we have:

2 z ez

has a pole of order 3 at z = −5 . From

Proposition 10.1-9 we have:

⎡ d2 ⎢ Res ⎡⎣ f z , − 5⎤⎦ = lim z +5 z → −5 3− 1 ! dz 2 ⎢ ⎣

()

!

!

(

1

)

(

)

3

(

⎤ 2 z ez ⎥ 3 z + 5 ⎥⎦

)

1 d2 ⎡ d ⎡ z⎤ = lim 2 z e = lim 1+ z e z ⎤⎦ = − 4 e−5 2 ⎣ ⎦ ⎣ z → −5 2 dz z → −5 dz

(

)

500

Example 10.1-23

()

Determine the residue of f z = branch of z

1 2

.

(

z1

2

)

z z−3

for the principal

2

()

has a branch point at z = 0 . The function f z does

have:

⎡ 2 1 d ⎢ z1 2 Res ⎡⎣ f z , 3⎤⎦ = lim z−3 z → 3 2 − 1 ! dz ⎢ z z−3 ⎣

()

To determine the coefficient a−1 for a complex function

()

(

)

(

)

(

)

complex plane, we must use a contour C enclosing the singularity at infinity. This can be accomplished by choosing a circular contour having a very large radius z = R . To be consistent with the definition of a residue for a finite z0 , the singularity at infinity must be enclosed by the contour C . This

⎤ ⎥ 2 ⎥ ⎦

means that the singularity must be to the left as the contour C is traversed during the integration. It is therefore necessary for the integration in equation (10.1-53) to be performed in a

or

⎡ 1 1 ⎤ d ⎡ 1 ⎤ 1 Res ⎡⎣ f z , 3⎤⎦ = lim = lim − = − ⎢ ⎥ ⎢ ⎥ 3 2 z → 3 dz z 1 2 ⎣ ⎦ z→3 ⎣ 2 z ⎦ 6 3

()

where the positive square root is chosen since we are on the principal branch.

10.1.2! THE RESIDUE AT INFINITY !

(10.1-53)

f z having a singularity at infinity rather than in the finite

have a pole of order 2 at z = 3 . From Proposition 10.1-9 we

!

C

()

f z dz !

()

!

()

!

!∫

coefficient of the Laurent series expansion of f z .

The function f z does not have a simple pole at z = 0 since 2

1 2π i

where the contour C encloses the singularity, and a−1 is a

Solution:

z1

a−1 =

!

negative direction. Using Proposition 5.3-1, the residue at infinity is then given by:

1 ! Res ⎡⎣ f z , ∞ ⎤⎦ = 2π i

()

!∫

C

()

f z dz = −

1 2π i

!∫

C

()

f z dz ! (10.1-54)

and so we have: !

()

An isolated singularity of a complex function f z in the

complex plane has a residue given by equation (10.1-6):

!

()

Res ⎡⎣ f z , ∞ ⎤⎦ = − a−1 !

(10.1-55)

To determine the residue at infinity, we can use the fact

()

that a complex function f z will have a pole or zero at infinity 501

only if it has a corresponding zero or pole, respectively, at z = 0

Example 10.1-24

(see Proposition 2.7-3). We then want to make the substitution

()

z → 1 w so that: !

!∫

Determine the residue of f z =

()

f z dz = −

C

!∫

C

⎛ 1⎞ f⎜ ⎟ d 1 w = ⎝ w⎠

(

)

!∫

C

1 ⎛ 1⎞ f ⎜ ⎟ dw 2 w ⎝ w⎠

! !

Solution: This is a Laurent series having only one term, and so a−1 = 1 .

(10.1-56)

Therefore:

and the singularity is now at w = 0 . Changing the variable

w → z , the residue at infinity can be defined as: ⎡ 1 ⎛ 1⎞ ⎤ Res ⎡⎣ f z , ∞ ⎤⎦ = − Res ⎢ 2 f ⎜ ⎟ , 0 ⎥ ! ⎝ z⎠ ⎦ ⎣z

()

! !

!

()

Res ⎡⎣ f z , ∞ ⎤⎦ = − a−1 = −1

(10.1-57) Example 10.1-25

()

From equation (10.1-57) and Proposition 10.1-2 we can

Determine the residue of f z = e2

also write: !

⎡ ⎛ 1 ⎛ 1⎞ ⎞ ⎤ Res ⎡⎣ f z , ∞ ⎤⎦ = lim ⎢ z ⎜ − 2 f ⎜ ⎟ ⎟ ⎥ ! z→0 ⎢ ⎝ z ⎝ z ⎠ ⎠ ⎥⎦ ⎣

()

!

()

(10.1-58)

at the point z = ∞ .

From Example 90.1-4 we have the Laurent series for

()

f z = e2 z :

(10.1-59)

!

2 22 23 24 25 f z = 1+ + + + + +! ! z 2! z 2 3!z 3 4!z 4 5!z 5

()

z >0

so a−1 = 2 . Therefore:

or !

z

Solution:

and so:

⎡ 1 ⎛ 1⎞ ⎤ Res ⎡⎣ f z , ∞ ⎤⎦ = lim ⎢ − f ⎜ ⎟ ⎥ ! z→0 ⎣ z ⎝ z⎠ ⎦

1 at the point z = ∞ . z

()

()

Res ⎡⎣ f z , ∞ ⎤⎦ = lim ⎡⎣ − z f z ⎤⎦ ! z→∞

(10.1-60)

!

()

Res ⎡⎣ f z , ∞ ⎤⎦ = − 2

502

!

10.2! RESIDUE INTEGRATION !

()

Since the function f z

is analytic in D except at an

()

isolated singularity at a point z0 , we can represent f z

at

Rather than the singularities of a complex function

point z0 using the Laurent series in equation (10.1-1). A simple

increasing the difficulty in evaluating integrals of the function,

closed circular contour C enclosing z0 is chosen having a small

these singularities can actually aid in this evaluation. Residue

enough radius so as to lie within the circle of convergence of

integration has been found to be very useful in calculating not

the Laurent series representing f z

only integrals of complex functions, but also integrals of real-

(10.1-2) we can then write:

()

valued functions. Residues provide a method for evaluating integrals of functions that are analytic within a closed contour

!

1 2π i

a−1 =

except for some isolated singularities. It is the residues of the integrand at singular points within the contour that determine the value of the contour integral.

!∫

C

Singularity:

()

( ) dz = 1 f z dz ! !∫C ( z − z0 )−1+1 2π i !∫C ( ) f z

(10.2-2)

or !

Proposition 10.2-1, Cauchy’s Residue Theorem for an Isolated

at z0 . From equation

()

()

f z dz = 2 π i Res ⎡⎣ f z , z0 ⎤⎦ !

as given in equation (10.1-9).

(10.2-3)

■

If a complex function f z is analytic in a domain D except at

!

an isolated singularity at a point z = z0 , then:

far-reaching. If we are able by any method to determine the

!∫

!

C

()

()

f z dz = 2 π i Res ⎡⎣ f z , z0 ⎤⎦ !

(10.2-1)

The implications of this equation are rather profound and

()

coefficient a−1 of the Laurent series of a complex function f z

having an isolated singularity at point z0 , then we can evaluate

()

where C is any positively oriented simple closed contour lying

the integral of f z using a closed contour around z0 . This type

within D and enclosing z0 .

of integration is known as residue integration, and it can be

Proof:

considered an extension of the method of integration using Cauchy’s integral formula. Residue integration can often make 503

possible the evaluation of integrals even when an explicit

Solution:

antiderivative does not exist. Note that if a closed contour C

()

The function f z has a pole of order 2-1=1 at z = 0 as given

()

does not enclose any singularities of an analytic function f z ,

()

then we have Res ⎡⎣ f z , z0 ⎤⎦ = 0 for any point z0 within C , and equation (10.2-3) becomes the Cauchy-Goursat theorem given

using Proposition 9.3-3. From Example 10.1-19 we have: !

in Proposition 5.4-2.

From Proposition 10.2-1 we then have:

Example 10.2-1

cos z Integrate the function f z = 2 around the circle z = 1. z

()

!

()

1 1 z2 z4 f z = 2 − + − +! ! 2! 4! 6! z

()

()

z = 1 using:

z >0

1.! contour integration.

()

C

()

Integrate the function f z = 1 z around the unit circle

We see that a−1 = 0 = Res ⎡⎣ f z , 0 ⎤⎦ . From Proposition 10.2-1 we then have:

!∫

sin z dz = 2 π i a−1 = 2 π i 1 = 2 π i z2

Example 10.2-3

From Example 8.11-5 we have the Laurent series for f z :

!

!∫

C

Solution:

!

()

Res ⎡⎣ f z , 0 ⎤⎦ = 1

cos z dz = 2 π i a−1 = 2 π i 0 = 0 z2

()

2.! residue calculus. 3.! the antiderivative. Solution: ! 1.! We will use z = eiθ and dz = i eiθ dθ . We then have: !

Example 10.2-2

!∫

C

()

Integrate the function f z =

sin z around the circle z = 1. z2

1 dz = z

∫

2π

e−iθ i eiθ dθ = 2 π i

0

()

! 2.! The function f z = 1 z has a simple pole at z = 0 . 504

! ! !

Using Proposition 10.1-2 we have:

!∫

C

⎡ 1 dz = 2 π i lim ⎢ z z→ 0 z ⎣

From Proposition 10.1-9 for a pole of order n we have:

1⎤ = 2π i z ⎥⎦

( ) , z ⎤⎥ = lim 1 d n−1 ⎡⎢ z − z n f ( z ) ⎤⎥ ( 0) 0 n n z → z ( n − 1) ! dz n−1 ⎢ ⎥ ⎥ z − z ( ) ⎦ ( ) 0 ⎣ ⎦

⎡ f z Res ⎢ ⎢ z−z 0 ⎣

3.! We have: !

!∫

C

1 dz = z

or

!∫

C

( )

d ln z = ln z

+ i arg z

z =1

2π 0

= 2π i

( ) , z ⎤⎥ = lim 1 d n−1 ⎡ f z ⎤ ( )⎦ 0 n z → z ( n− 1)! dz n−1 ⎣ ⎥ ( ) ⎦

⎡ f z Res ⎢ ⎢ z−z 0 ⎣

Example 10.2-4

()

( ) , z ⎤⎥ = f ( n−1) ( z0 ) 0 n ( ) ⎥⎦ (n − 1)!

⎡ f z Res ⎢ ⎢ z−z 0 ⎣

at an isolated singularity at a point z = z0 . Evaluate the integral:

( ) dz !∫C ( z − z0 )n

and so:

f z

!

where n ∈! , and where C is any positively oriented simple

Using Proposition 10.2-1 we can write:

( ) dz = 2 π i Res ⎡⎢ f ( z ) , z ⎤⎥ !∫C ( z − z0 )n ⎢ ( z − z )n 0 ⎥ 0 ⎣ ⎦

( ) dz = 2 π i f ( n−1) z !∫C ( z − z0 )n ( n − 1)! ( 0 ) f z

We then also have:

closed contour enclosing z0 , and lying within D . Solution:

0

Therefore:

Let f z be analytic in a simply connected domain D except

f z

0

!

n − 1)! ( n−1) ( f ( z0 ) = 2 π i

( ) dz !∫C ( z − z0 )n f z

which is Cauchy’s integral formula for derivatives as given in Proposition 6.2-1

505

10.2.1! CAUCHY’S RESIDUE THEOREM Proposition 10.2-2, Cauchy’s Residue Theorem: Let C be a simple closed contour that lies within a simply

()

connected domain D . If a complex function f z is analytic in

D except at a finite number of isolated singular points z1 , z2 , ! , zk that lie inside C , then:

!∫

!

C

Proof: !

k

()

f z dz = 2 π i

∑ Res ⎡⎣ f ( z ), z ⎤⎦ ! n

(10.2-4)

n =1

()

Since f z is analytic in D except at a finite number of

()

isolated singular points z1 , z2 , ! , zk , we can represent f z

points in D with small circular subcontours that are centered at

Figure 10.2-1! Simple closed contour C within a simply connected domain D . Isolated singularities inside C are each enclosed by a subcontour.

the zk , and that lie inside a closed contour C as shown in

From Proposition 10.2-1 we have:

using a Laurent series. We can encircle each of the singular

Figure 10.2-1. Each of these subcontours is small enough to lie within the circle of convergence of the Laurent series expansion

()

of f z about the singularity zk . ! !

!∫

C

()

f z dz =

∑ !∫ n =1

Cn

()

f z dz !

Cn

()

()

f z dz = 2 π i Res ⎡⎣ f z , zn ⎤⎦ ! n = 1, 2, 3, ! , k !

(10.2-6)

Therefore:

We can now use the Proposition 5.5-2 to write: k

!∫

!

(10.2-5)

!∫

!

C

()

f z dz = 2 π i

k

∑ Res ⎡⎣ f ( z ), z ⎤⎦ ! n

(10.2-7)

n =1

■

506

!

!

Residue integration replaces integration over a closed

contour C with the sum of residues for all isolated singular

()

Integrate the function f z =

5− 2 z

(

()

!

)

!∫

C

around a simple closed 2.

path C situated such that the poles 0 and 1 are:

!

1.

()

!

()

(

!

!

(

)

!

⎤ ⎥ ⎥⎦

!

⎡ 5− 2z ⎤ + lim ⎢ z − 1 ⎥ z →1 z z − 1 ⎢⎣ ⎥⎦ !

!∫

Both poles 0 and 1 are inside C . Using Proposition

⎡ 5− 2z Res ⎡⎣ f z , 0 ⎤⎦ + Res ⎡⎣ f z , 1⎤⎦ = lim ⎢ z z→0 ⎢⎣ z z − 1

)

(

)

or

5− 2 z 5− 2 z Res ⎡⎣ f z , 0 ⎤⎦ + Res ⎡⎣ f z , 1⎤⎦ = lim + lim z→0 z − 1 z →1 z

()

()

( )

Pole 0 inside C and pole 1 outside C . Using

C

!

)

(

)

Therefore:

!

10.1-2, the residues are:

(

z z−1

dz = 2 π i − 2 = − 4 π i

()

!

Solution:

5− 2 z

⎡ 5− 2z ⎤ 5− 2z ⎡ ⎤ Res ⎣ f z , 0 ⎦ = lim ⎢ z = −5 ⎥ = lim z→0 z→0 z − 1 z z − 1 ⎢⎣ ⎥⎦

!

! 3.! both are outside C .

)

Propositions 10.1-2 and 5.4-2, the residue is:

! 1.! both inside C . ! 2.! 0 is inside and 1 is outside C .

( ) (

Therefore:

!

z z−1

()

Res ⎡⎣ f z , 0 ⎤⎦ + Res ⎡⎣ f z , 1⎤⎦ = −5 + 5 − 2 = − 2

!

points within C , each enclosed by a circular subcontour. Example 10.2-5

and so:

3.!

5− 2 z dz = 2 π i −5 = − 10 π i z2 − z

( )

Both poles 0 and 1 outside C . Using Proposition 5.4-2, we have:

!∫

C

5− 2 z

(

)

z z−1

dz = 0

Example 10.2-6

z ez Integrate the function f z = 2 around the circle z = 3 z −1 shown in Figure 10.2-2.

()

507

Example 10.2-7

z2 + 2 Integrate the function f z = 2 around the circle z = 2 . z +1

()

Solution: We have: !

z2 + 2 z2 + 2 f z = 2 = z +1 z+i z−i

()

The function

( )( ) f ( z ) has two simple poles at z = ± i . From

Proposition 10.1-2 the residues are: Figure 10.2-2! Contour C is the circle z = 3 encircling a function f z that has poles at z = ±1 .

()

!

()

(

Res ⎡⎣ f z , i ⎤⎦ = lim z − i z→i

)

Solution:

()

The function f z has simple poles at z = ±1 (see Figure

!

()

(

Res ⎡⎣ f z , − i ⎤⎦ = lim z + i z→−i

(

)

10.2-2). From Example 10.1-12 we know that this function

!

()

e−1 Res ⎡⎣ f z , − 1⎤⎦ = 2

()

!

!

!∫

C

⎛ e e−1 ⎞ ⎜ 2 + 2 ⎟ = 2 π i cosh1 ⎝ ⎠

!∫

C

Therefore:

z ez dz = 2 π i z2 − 1

z2 + 2 1 = lim = z→i z + i 2i z+i z−i

)(

(

z2 + 2

)

z2 + 2 1 = lim =− z→−i z − i 2i z+i z−i

)(

)

Therefore:

has residues:

e Res ⎡⎣ f z , 1⎤⎦ = ! 2

z2 + 2

!

⎛ 1 1⎞ z2 + 2 dz = 2 π i ⎜⎝ 2i − 2i ⎟⎠ = 0 2 z +1

()

If z = z0 is an essential singularity of a function f z , then

()

to obtain the residue of f z at z0 it is necessary to determine 508

()

the coefficient a−1 using the Laurent series expansion of f z in a neighborhood of z0 .

10.2.2! RESIDUE INTEGRATION AT INFINITY !

Example 10.2-8

()

Integrate the function f z = z 2 e1 z around the unit circle

z = 1.

In some cases where a complex function has many isolated

singularities in the z-plane, it can be unwieldy to use Cauchy’s residue theorem to integrate the function in the way previously discussed. Moreover in cases where a contour encloses a branch cut, it is not possible to directly apply the residue theorem.

Solution:

Instead a slightly different method known as residue

()

The function f z has an essential singularity at z = 0 . We

integration at infinity may be useful for these cases.

can write:

!

()

f z = z 2 e1

!

z

⎛ 1 1 1 1 1 1 1 ⎞ = z 2 ⎜ 1+ + + + +! ⎟⎠ z 2! z 2 3! z 3 4! z 4 ⎝

or !

()

We will now consider a complex function f z which is

analytic in the z-plane except for a finite number of singular points. The number of singular points can be quite large. We will encircle all these singularities with a positively oriented

()

contour C as shown in Figure 10.2-3. The function f z will

()

f z = z 2 e1 z = z 2 + z +

1 11 1 1 + + +! 2 3! z 4! z 2

then be analytic everywhere in the z-plane outside the contour

C.

Therefore: !

1 1 Res ⎡⎣ f z , 0 ⎤⎦ = = 3! 6

()

connected domain D . If a complex function f z is analytic in

⎛ 1⎞ π i z e dz = 2 π i ⎜ ⎟ = 3 ⎝ 6⎠ C

!∫

Let C be a simple closed contour that lies within a simply

()

and so: !

Proposition 10.2-3:

2 1z

D except at a finite number of isolated singular points z1 , z2 , ! , zk that lie inside C , then:

509

!∫

!

C

()

()

f z dz = − 2 π i Res ⎡⎣ f z , ∞ ⎤⎦ !

(10.2-8)

()

The function f z will then be analytic everywhere in the

() f ( z ) will then have an

z-plane outside the circle z = R . That is, f z will be analytic in the annulus R < z < ∞ . The function

and

!∫

!

!

C

⎡ 1 ⎛ 1⎞ ⎤ f z dz = 2 π i Res ⎢ 2 f ⎜ ⎟ , 0 ⎥ ! ⎝ z⎠ ⎦ ⎣z

()

isolated singular point at z = ∞ (see Section 10.1.2). (10.2-9)

Proof:

!

Finally, we will choose another positive number R0 where

R0 > R and we will let C0 be the circular contour z = R0 as shown in Figure 10.2-3. We will let R0 be very large. Visualizing

We can now choose a positive number R such that the

this contour mapped onto a Riemann sphere (see Section 2.6),

contour C lies inside the circle z = R (shown as a dashed line

we see that the mapped contour will be a very small circle

in Figure 10.2-3).

enclosing the north pole (the point at infinity). The orientation

!

of a contour in the z-plane is reversed when it is mapped onto the Riemann sphere. Therefore by letting C0 be negatively oriented in the z-plane, we ensure that the pole at z = ∞ always lies on the left as we traverse C0 . The pole at z = ∞ is then in the interior of C0 . ! !

We can define the residue of the singularity at infinity as:

!∫

C0

()

()

f z dz = 2 π i Res ⎡⎣ f z , ∞ ⎤⎦ !

(10.2-10)

We have from the deformation theorem (Proposition 5.5-1): !

!∫

C

Figure 10.2-3! Contours in the z-plane for residue integration at infinity.

()

f z dz =

!∫

− C0

()

f z dz = −

!∫

C0

()

f z dz !

(10.2-11)

Therefore: 510

!∫

!

C

!

()

()

f z dz = − 2 π i Res ⎡⎣ f z , ∞ ⎤⎦ !

Using equation (10.1-57) we then have:

!∫

!

C

which is the same result obtained from Example 10.2-2.

(10.2-12)

Proposition 10.2-4:

⎡ 1 ⎛ 1⎞ ⎤ f z dz = 2 π i Res ⎢ 2 f ⎜ ⎟ , 0 ⎥ ! ⎝ z⎠ ⎦ ⎣z

()

Let C be a simple closed contour that lies within a simply

(10.2-13)

()

connected domain D . If a complex function f z is analytic in

■

D except at a finite number of isolated singular points

Example 10.2-9

z1 , z2 , ! , zk that lie inside C , then the sum of all residues of

()

Integrate f z =

()

f z , including the residue at z = ∞ , is equal to zero:

sin z around the unit circle z = 1 . 2 z

k

∑ Res ⎡⎣ f ( z ), z ⎤⎦ + Res ⎡⎣ f ( z ), ∞ ⎤⎦ = 0 !

!

Solution:

n

(10.2-14)

n =1

()

The function f z has a pole of order 2-1=1 at z = 0 as given using Proposition 9.3-3. We will determine the residue at infinity. We have: !

⎞ 1 ⎛ 1⎞ 1 2 1 ⎛1 1 1 f = z sin = − + −! ⎜ ⎟ ⎟⎠ z ⎜⎝ z z 3 3! z 5 5! z2 ⎝ z ⎠ z2

Proof: !

we have: !

⎡ 1 ⎛ 1⎞ ⎤ Res ⎢ 2 f ⎜ ⎟ , 0 ⎥ = 1 ⎝ z⎠ ⎦ ⎣z

! !

From equation (10.2-13) we then have: !

!∫

C

!∫

C

Therefore: !

From the Cauchy’s residue theorem (Proposition 10.2-2)

()

∑ Res ⎡⎣ f ( z ), z ⎤⎦ ! n

(10.2-15)

f z dz = − 2 π i Res ⎡⎣ f z , ∞ ⎤⎦ !

(10.2-16)

n =1

From Proposition 10.2-3 we have:

!∫

C

⎡ 1 ⎛ 1⎞ ⎤ sin z dz = 2 π i Res ⎢ 2 f ⎜ ⎟ , 0⎥ = 2π i 1 = 2π i 2 ⎝ z⎠ ⎦ z ⎣z

()

f z dz = 2 π i

k

()

()

Therefore:

511

will now show examples of the use of residue integration in

k

!

∑ Res ⎡⎣ f ( z ), z ⎤⎦ + Res ⎡⎣ f ( z ), ∞ ⎤⎦

evaluating both definite and improper integrals of real-valued

n

n =1

=

!

1 ⎡ 2 π i ⎢⎣

!∫

C

()

f z dz −

!∫

C

⎤ f z dz ⎥ = 0 ! ⎦

()

functions. (10.2-17)

■

!

Proposition 10.2-5:

()

If a complex function f z

()

Definite integrals of real-valued rational trigonometric

functions have the form: is rational, then the sum of all

residues of f z , including the residue at z = ∞ , is equal to zero. Proof:

10.3.1! DEFINITE INTEGRALS OF TRIGONOMETRIC FUNCTIONS

()

( ) ∫0 ∫0 Q (cosθ , sinθ ) dθ ! (10.3-1) where P ( cosθ , sin θ ) and Q ( cosθ , sin θ ) are polynomials having real coefficients of sin θ and cosθ . If a function f ( cosθ , sin θ ) is 2π

!

(

2π

)

f cosθ , sin θ dθ =

P cosθ , sin θ

Since f z is rational, it will be analytic except at a finite

single-valued and has only poles as singularities, then the

number of isolated singularities. From Proposition 10.2-4 we

integral in equation (10.3-1) can be evaluated using residue

then see that the sum of all residues of f z , including the

integration. The first step in evaluating this integral is to

residue at z = ∞ , is equal to zero.

parameterize it as shown in the following proposition.

!

()

■

10.3! EVALUATING INTEGRALS OF REALVALUED FUNCTIONS USING RESIDUES !

Proposition 10.3-1: Certain definite integrals of real-valued rational trigonometric functions can be expressed in the form:

Residue integration of real-valued functions is done by

analytic continuation which involves replacing real variables

!

in the integrand with complex variables (see Section 9.8). We

!

∫

2π

0

!

(

)

f cosθ , sin θ dθ =

!∫

C

⎛1⎛ 1⎞ 1 ⎛ 1 ⎞ ⎞ dz f ⎜ ⎜ z + ⎟ , ⎜ z − ⎟⎟ z ⎠ 2i ⎝ z⎠⎠ iz ⎝2⎝ (10.3-2) 512

where the contour C is a closed unit circle in the complex plane.

Example 10.3-1

Proof: !

Evaluate the integral

Expressing the complex variable in exponential form

0

z = eiθ , we can rewrite the trigonometric functions cosθ and

cosθ =

1⎛ 1⎞ z + ! 2 ⎜⎝ z ⎟⎠

Using equations (10.3-3) and (10.3-5), and integrating over a

sin θ =

1 ⎛ 1⎞ z − ! 2i ⎜⎝ z ⎟⎠

closed contour C that is a unit circle, we have:

(10.3-3) !

We also have:

dz = i eiθ dθ = i z dθ !

!

(10.3-5)

0

As θ goes from 0 to 2 π , a point on the circle z = eiθ will

make one complete traversal of the unit circle contour in the positive direction. We then have: !

∫

2π

0

!!

!∫

C

dz i z ⎛ 1⎞ 5− 2 ⎜ z + z⎟ ⎝ ⎠

=

1 i

!∫

C

dz 5z − 2 z 2 − 2

or

dz dθ = ! iz

!

∫

2π

dθ = 5 − 4cosθ

(10.3-4)

so that: !

∫

2π

0

z = eiθ !

!

dθ . 5 − 4 cosθ

Solution:

sin θ using equation (1.13-51): !

∫

2π

(

)

f cosθ , sin θ dθ =

!∫

C

!∫

C

dz 2 z 2 − 5z + 2

=

i 2

!∫

dz

C

⎛ 1⎞ z − z−2 ⎜⎝ 2 ⎟⎠

(

)

()

The integrand f z has two simple poles: z = 1 2 and z = 2 . The pole z = 2 is outside the contour C (see Figure 10.3-1). The pole z = 2 then does not contribute to the integral in

⎛1⎛ 1⎞ 1 ⎛ 1 ⎞ ⎞ dz f ⎜ ⎜ z + ⎟ , ⎜ z − ⎟⎟ z ⎠ 2i ⎝ z⎠⎠ iz ⎝2⎝

residue integration. From Proposition 10.1-2 we have:

(10.3-6)

where C is the closed unit circle z = 1 in the complex plane.

dθ 1 =− 5 − 4cosθ i

■

⎡ ⎢⎛ ⎡ 1⎤ 1⎞ 1 Res ⎢ f z , ⎥ = lim ⎢⎜ z − ⎟ 2 ⎦ z →1 2 ⎢⎝ 2⎠ ⎛ 1⎞ ⎣ z − ⎢ ⎜⎝ ⎟⎠ z − 2 2 ⎣

()

(

)

⎤ ⎥ ⎥=−2 3 ⎥ ⎥ ⎦ 513

(10.3-5), and integrating over a closed contour C that is a unit circle, we have: !

∫

2π

∫

2π

0

dθ = a + b cosθ

!∫

C

dz i z 1⎞ 1 ⎛ a + b⎜ z + ⎟ z⎠ 2 ⎝

= − 2i

!∫

C

dz b z2 + 2 a z + b

or !

0

Figure 10.3-1! Contour C is a unit circle z = 1 . The poles are at

!

(

z2 = − a − a 2 − b2

Therefore:

∫

2π

0

⎛ 2 ⎞ 2π dθ i = 2π i ⎜ − ⎟ = 5 − 4 cosθ 2 3 ⎝ 3⎠

C

⎛ a − a 2 − b2 ⎞ ⎛ a + a 2 − b2 ⎞ ⎜z+ ⎟ ⎜z+ ⎟ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ b b

(

!

) b . We can write:

) b and

− a − a 2 − b2 − a + a 2 − b2 b2 z1 z2 = = 2 =1 b b b Therefore z1 z2 = 1 . Since z1 < z2 , we then must have z1 < 1

Example 10.3-2 Evaluate the integral

!∫

dz

The integrand has simple poles at z1 = − a + a 2 − b2

z = 1 2 and z = 2 on the x-axis. !

dθ = −2i a + b cosθ

∫

2π

0

dθ where a > b > 0 . a + b cosθ

(

and z2 > 1 . This means the pole z2 = − a − a 2 − b2

) b is

outside the contour C (a unit circle) and so does not

Solution:

contribute to the integral. From Proposition 10.1-5 the

Since a > b > 0 , the denominator of the integrand will not

residue for the simple pole z1 = − a + a 2 − b2

equal zero for any value of θ . Using equations (10.3-3) and

(

) b is:

514

⎡ − a+ a 2 − b2 ! Res ⎢ f z , b ⎢⎣

()

⎤ 1 ⎥= ⎥⎦ d b z2 + 2 a z + b dz

(

where the contour C is a closed unit circle in the complex plane.

)

⎛ ⎞ z = ⎜⎝ − a + a 2 −b2 ⎟⎠ b

Proof: !

z = eiθ , we can rewrite the trigonometric functions cosθ and

or ⎡ − a+ a 2 − b2 ! Res ⎢ f z , b ⎢⎣

()

⎤ ⎥= ⎥⎦ 2 b z + 2 a

sin θ using equation (1.13-54):

1

()

⎤ 1 ⎥= ⎥⎦ 2 a 2 − b2

!

∫

!

0

dθ 1 = − 2 i 2π i = 2 2 a + b cosθ 2 a −b

2π 2

a −b

2

!

a>b>0

functions can be expressed in the form:

0

!

!

(10.3-8)

(

)

f cos nθ , sin nθ dθ =

!∫

C

⎛ 1 ⎛ n 1 ⎞ 1 ⎛ n 1 ⎞ ⎞ dz f ⎜ ⎜ z + n ⎟ , ⎜ z − n ⎟⎟ ⎝2⎝ z ⎠ 2i ⎝ z ⎠⎠ iz (10.3-9) ■

10.3.2! IMPROPER INTEGRALS An integral that has one or both of its limits infinite, or

that has an integrand which becomes infinite at points within

Certain definite integrals of real-valued rational trigonometric

∫

1 ⎛ n 1⎞ z − n⎟! 2i ⎜⎝ z ⎠

where C is the closed unit circle z = 1 in the complex plane.

Proposition 10.3-2:

(

sin nθ =

!!

!

2π

∫

2π

0

Therefore: 2π

1⎛ n 1 ⎞ z + n⎟ ! 2 ⎜⎝ z ⎠

Following the proof of Proposition 10.3-1, we then have:

⎡ − a+ a 2 − b2 Res ⎢ f z , b ⎢⎣

!

cos nθ =

!

⎛ ⎞ z = ⎜ − a + a 2 −b2 ⎟ b ⎝ ⎠

and so:

!

Expressing the complex variable in exponential form

)

f cos nθ , sin nθ dθ =

!∫

C

⎛ 1 ⎛ n 1 ⎞ 1 ⎛ n 1 ⎞ ⎞ dz f ⎜ ⎜ z + n ⎟ , ⎜ z − n ⎟⎟ ⎝2⎝ z ⎠ 2i ⎝ z ⎠⎠ iz

the range of integration is called an improper integral. An infinite limit along the real line: !

∫

∞

−∞

()

f x dx !

(10.3-10)

(10.3-7) 515

()

where f x

is a continuous real-valued function is then an

improper integral. This integral is obtained by using limiting

10.3.3! CAUCHY PRINCIPAL VALUE

values of proper integrals:

!

!

∫

∞

−∞

()

f x dx = lim

R→ ∞

∫

0

−R

()

f x dx + lim

R→ ∞

∫

R

0

()

f x dx ! (10.3-11)

!

∫

()

f x dx = lim

0

R→ ∞

∫

R

()

f x dx !

0

∫

!

∞

−∞

()

f x dx !

(10.3-14)

is designated by PV , and is defined to be the number given by:

If the limit exists (is finite) for the improper integral: ∞

The Cauchy principal value of the improper integral:

PV

!

(10.3-12)

∫

∞

−∞

()

f x dx = lim

R→ ∞

∫

R

−R

()

f x dx !

(10.3-15)

then the integral is convergent; if the limit does not exist the

!

integral is divergent.

(10.3-15) exists even though the limits in equations (10.3-12)

!

and (10.3-13) do not (see Example 10.3-3). If the integrals in

!

Similarly, if the limit exists for the improper integral:

∫

0

−∞

()

f x dx = lim

R→ ∞

∫

0

−R

()

f x dx !

equation (10.3-11) exist, then the Cauchy principal value given (10.3-13)

in equation (10.3-15) will exist. The converse is not true,

then the integral is convergent; if the limit does not exist the integral is divergent. !

()

For an improper integral of a real-valued function f x

over the interval

(

− ∞, ∞

)

as given in equation (10.3-11) to

converge, the following two conditions need apply:

however; the existence of the Cauchy principal value given in equation (10.3-15) does not guarantee the existence of the limits in equation (10.3-11).

()

An integral of a real-valued function f x having a single

!

discontinuity at a point c in the integration interval a ≤ c ≤ b is given by:

! 1.!

The rate of convergence to each limit must be the same.

! 2.!

f x must be continuous over the real interval − ∞, ∞

()

In the case of certain functions, the limit in equation

(

)

!

∫

b a

()

f x dx = lim

ε →0

∫

c−ε

a

()

f x dx + lim

ε →0

∫

b

c+ε

()

f x dx !

(10.3-16)

516

The Cauchy principal value for the integral in equation (10.3-16) is defined to be:

∫

! PV

b a

⎡ f x dx = lim ⎢ ε →0 ⎢ ⎣

()

∫

c−ε

a

()

f x dx +

∫

b

c+ε

()

exist, while the Cauchy principal value in equation (10.3-17) will exist.

∫

0

−∞

x dx ,

∫

∞

x dx ,

∫

x dx , and

contour in the complex plane with integration over an interval

∞

x dx .

of the real line. These methods involve forming a closed contour C having the real line path segment as just one segment of the closed contour C .

0

∫

0

∫ x dx = lim ∫

R

!

− R2 x dx = lim x dx = lim = −∞ R → ∞ R → ∞ 2 −∞ −R ∞

R→ ∞

0

∫

!

!

0

x dx does not exist since lim

R→ ∞

The contour C chosen for the integral must be such that

contributions to the integral from the segment CR of the contour not along the real axis can be shown to approach zero

R2 x dx = lim =∞ R→ ∞ 2

∞

−∞

!

In this section we will consider improper integrals of real-

required that can relate residue integration over a closed

−∞

0

Solution:

!

!

variables are changed to complex variables. Methods are then

∞

−∞

∫

10.3.4! IMPROPER INTEGRALS OF REAL-VALUED RATIONAL FUNCTIONS

integration. Analytic continuation is used first, whereby real

Evaluate the integrals

!

∫

valued rational functions that can be evaluated using residue

Example 10.3-3

∫

∫

R

We see that the Cauchy principal value exists.

⎤ f x dx ⎥ ! (10.3-17) ⎥⎦

For certain functions the limits in equation (10.3-16) may not

PV

!

⎛ R2 R2 ⎞ PV x dx = lim x dx = lim ⎜ − ⎟ =0 R→ ∞ − R R→ ∞ ⎝ 2 2 ⎠ −∞ ∞

as the contour is expanded to infinity along the real axis (see

∫

0

R

x dx and lim

−R

R→ ∞

∫ x dx

do not exist, and so their sum does not exist.

0

Proposition 10.3-3). The contour integral can be written: !

!∫

C

()

f z dz = lim

R→ ∞

∫

R

−R

()

f x dx + lim

R→ ∞

∫

CR

()

f z dz ! (10.3-18)

517

The integral over the real axis then becomes equal to the

Factoring the highest power of z from each polynomial:

integral over the closed contour C calculated using residue integration (see Propositions 10.3-3 and 10.3.4). The curve often chosen for the segment CR is a semicircle in either the upper

!

()

f z =

half or the lower half of the complex plane.

()

Let f z be a complex rational function having the form:

( )! Q(z) where both P ( z ) and Q ( z ) ()

f z =

lim

∫

CR

()

(10.3-20)

!

(10.3-22)

+

zn

z

m

a1 z n−1

+

+ !+

b1 z

m−1

an−1

+!+

z

+ an !

bm−1 z

(10.3-23)

+ bm !

(10.3-24)

()

f z =

!

( )! z m Q1 ( z ) z n P1 z

(10.3-25)

Using the semicircular contour CR shown in Figure 10.3-2,

5.3-7 we can write:

We have:

( ) = a0 + a1 z + a2 z + !+ an−1 z + an z ! ( ) Q z b + b z + b z 2 +!+ b z m−1 + b z m () 0 1 2 m−1 m

! f z =

bm−1

and letting z = R e iθ so that dz = i R e iθ dθ , from Proposition

Proof: !

b1

We have: !

f z dz = 0 !

b0

Q1 z =

are polynomials in z of degree n

0 ≤ θ ≤ π , then:

()

!

(10.3-19)

()

()

a0

P1 z =

consists of a semicircular contour given by z = R e iθ where

R→ ∞

⎞ + +!+ + b m⎟ ⎜⎝ z m z m−1 z ⎠

!

P z

and m , respectively, and where m ≥ n + 2 and Q z ≠ 0 . If CR

!

() = ⎛b Q(z) zm 0 P z

Letting:

Proposition 10.3-3:

!

a1 an−1 ⎛ a0 ⎞ z n ⎜ n + n−1 + !+ + an ⎟ z ⎝z ⎠ z

P z

2

n−1

!

n

∫

CR

(10.3-21)

()

f z dz ≤

∫

CR

( ) R m Q1( R ) R n P1 R

i R dθ

!

(10.3-26)

or 518

∫

!

()

f z dz ≤

CR

∫

CR

( ) R m Q1( R ) R n P1 R

R dθ

!

(10.3-27)

Proposition 10.3-4:

()

Let f x be a real-valued rational function having the form:

()

f x =

!

( )! Q( x) P x

(10.3-30)

()

()

where both P x and Q x are polynomials in x of degree n

()

and m , respectively. If m ≥ n + 2 and Q x ≠ 0 , then the Cauchy principal value will exist, and we have:

PV

!

∫

∞

k

()

f x dx = 2 π i

−∞

∑ Res ⎡⎣ f ( z), z ⎤⎦ ! j

(10.3-31)

j =1

()

where zj are the k points at which f z has poles in the upper half of the complex plane.

Figure 10.3-2! Semicircle contour CR in the z-plane.

( )

Proof:

( )

As R → ∞ we have P1 R → an and Q1 R → bm . Since m ≥ n + 2 we have:

R→ ∞

( ) R= R m Q1( R ) R n P1 R

an

1 lim = 0! bm R → ∞ R m − ( n+1)

the complex plane (see Proposition 9.3-19). The locations of (10.3-28)

lim

R→ ∞

■

()

these poles occur where Q z = 0 . To enclose those poles, we will use the contour C shown in Figure 10.3-3 which consists of

Therefore: !

()

()

Since f x is a rational function, its analytic continuation

f z can have only a finite number of poles in the upper half of lim

!

!

a semicircular contour CR having radius R and the interval of

∫

CR

()

f z dz = 0 !

the real axis from − R to R . By making R large enough, we can (10.3-29)

()

enclose all the poles of f z that exist in the upper half of the complex plane. 519

From equations (10.3-32) and (10.3-33) we then have:

∫

!

R

−R

∫

()

f x dx +

k

()

f z dz = 2 π i

CR

∑ Res ⎡⎣ f ( z), z ⎤⎦ j

! (10.3-34)

j =1

As R → ∞ the first integral on the left corresponds to the

!

integral in equation (10.3-31), and so is the desired integral. From Proposition 10.3-3 we have for the second integral on the left:

lim

! Figure 10.3-3! Closed contour C consisting of semicircle CR and real axis interval ⎡⎣ − R, R ⎤⎦ in the z-plane. !

!∫

C

()

∑ Res ⎡⎣ f ( z), z ⎤⎦ ! j

()

f z dz = 0 !

CR

(10.3-35)

lim

!

R→ ∞

∫

R

−R

()

k

f x dx = 2 π i

∑ Res ⎡⎣ f ( z), z ⎤⎦ ! j

(10.3-36)

j =1

or

k

f x dx = 2 π i

∫

and so equation (10.3-34) becomes as R → ∞ :

From Cauchy’s residue theorem (Proposition 10.2-2) we

have: !

R→∞

(10.3-32)

j =1

()

where zj are the k isolated singular poles of f z inside the

PV

!

∫

∞

−∞

()

f x dx = 2 π i

k

∑ Res ⎡⎣ f ( z), z ⎤⎦ ! j

(10.3-37)

j =1

■

contour C . ! !

Using the contour in Figure 10.3-3 we can write:

!∫

C

()

f z dz =

∫

R

−R

()

f x dx +

∫

CR

()

f z dz !

(10.3-33)

Proposition 10.3-5:

()

Let f x be a real-valued rational function having the form:

520

!

( )! Q( x) where both P ( x ) and Q ( x ) ()

f x =

P x

(10.3-38) are polynomials in x of degree n

()

and m , respectively. If m ≥ n + 2 and Q x ≠ 0 , then the Cauchy principal value will exist and we have:

PV

!

∫

∞

−∞

k

()

f x dx = − 2 π i

∑ Res ⎡⎣ f ( z), z ⎤⎦ ! j

(10.3-39)

j =1

()

where zj are points at which f z has poles in the lower half of the complex plane. Proof: !

()

()

Since f x is a rational function, f z can have only a

finite number of poles in the lower half of the complex plane (see Proposition 9.3-19). To enclose those poles, we will use the contour C shown in Figure 10.3-4 which consists of a semicircular contour CR having radius R and the interval of the real axis from − R to R . By making R large enough, we can

()

enclose all the poles of f z that exist in the lower half of the

Figure 10.3-4! Closed contour C consisting of semicircle CR and real axis interval − R, R in the z-plane.

(

!

the integration around the semicircular contour CR in the lower half-plane must be in the negative direction. Therefore the Cauchy integral equation has a negative sign, and so we have: !

∫

R

−R

Using the contour in Figure 10.3-4 we can write:

!∫ f ( z ) dz = ∫ C

To have the contour on the real axis travel from − R to R ,

!

complex plane. !

R

−R

()

f x dx +

∫

CR

()

f z dz !

(10.3-40)

)

!

()

f x dx +

∫

CR

()

f z dz = − 2 π i

k

∑ Res ⎡⎣ f ( z), z ⎤⎦ j

! (10.3-41)

j =1

As R → ∞ the first integral on the left corresponds to the

integral in equation (10.3-39), and so is the desired integral.

521

From Proposition 10.3-3 we have for the second integral on the left:

lim

!

R→∞

∫

()

f z dz = 0 !

CR

(10.3-42)

and so equation (10.3-41) becomes as R → ∞ :

PV

!

∫

∞

()

()

!

PV

∫

∞

−∞

∑ Res ⎡⎣ f ( z), z ⎤⎦ j

!

(

Example 10.3-4

Solution: We will evaluate:

Solution: !

We will evaluate: !

∫

−∞

()

f z dz !

∫

∞

−∞

()

where f z =

1 z2 + 1

()

f z dz !

()

where f z =

1 z2 + 1

Inside the closed contour shown in Figure 10.3-4 this function is rational, having one singular point in the lower half-plane at z = − i . Using Proposition 10.1-2 its residue is:

which has singularities at z = ± i in the form of simple poles.

()

Inside the closed contour shown in Figure 10.3-3 f z is a rational function having only one singular point in the upper half-plane. This point is z = i . Using Proposition 10.1-2, its residue is:

dx using the closed contour x2 + 1

shown in Figure 10.3-4.

dx using the closed contour x2 + 1

shown in Figure 10.3-3.

∞

∫

∞

−∞

−∞

)

Example 10.3-5 Evaluate the integral

∫

)(

dx 1 = 2 π i =π 2i x2 + 1

■

Evaluate the integral

(

(10.3-43)

j =1

∞

)

⎤ 1 ⎥= z − i z + i ⎥⎦ 2i 1

We then have from equation (10.3-31):

k

f x dx = − 2 π i

−∞

!

⎡ ⎡ ⎤ Res ⎣ f z , i ⎦ = lim ⎢ z − i z→i ⎢⎣

!

⎡ ⎡ ⎤ Res ⎣ f z , − i ⎦ = lim ⎢ z + i z→−i ⎢⎣

()

(

)(

⎤ 1 ⎥ =− 2i z − i z + i ⎥⎦ 1

)(

)

We then have from equation (10.3-39): 522

PV

!

∫

∞

−∞

⎛ 1⎞ dx = −2 π i ⎜⎝ − 2i ⎟⎠ = π x2 + 1

!

which is the same result as obtained in Example 10.3-4

⎡ 1⎢ 1 Res ⎡⎣ f z , i ⎤⎦ = lim 12 z→i 2 ⎢ z+i ⎣

()

(

⎤ ⎥= 6 5 ⎥ 2i ⎦

( )

)

5

=−

3i 16

We then have from equation (10.3-31): Example 10.3-6

∞

∞

Evaluate the integral

!

∫ ( x + 1) −∞

dx

−∞

using the closed contour

3

2

∫ (

PV

shown in Figure 10.3-3.

⎛ 3i ⎞ 3π = 2 π i ⎜⎝ − 16 ⎟⎠ = 8 3 2 x +1 dx

)

Example 10.3-7 ∞

Solution:

Evaluate the integral

We will evaluate:

∫

!

∞

−∞

()

f z dz !

()

where f z =

1

( z + 1)

∞

!

upper half-plane. This point is the third order pole at z = i . Using Proposition 10.1-9, its residue is:

or

(

)

(

z−i

)( 3

⎤ ⎥ 3 z + i ⎥⎦

)

2

2

2

1 = 2

∞

∫ (x + a ) −∞

dx

2

2

2

We will evaluate: !

1

∫ (x + a ) dx

0

is a rational function having only one singular point in the

3

.

Since the integrand is an even function of x , we have:

()

⎡ 1 d2 ⎢ Res ⎡⎣ f z , i ⎤⎦ = lim z−i z → i 2 dz 2 ⎢ ⎣

2

2

Solution:

3

2

poles. Inside the closed contour shown in Figure 10.3-3 f z

()

2

0

which has singularities at z = ± i in the form of third order

!

∫ (x + a ) dx

1 2

∫

∞

−∞

()

f z dz !

()

where f z =

(

1 z 2 + a2

)

2

which has singularities at z = ± a i in the form of second order poles. Inside the closed contour shown in Figure 523

()

10.3-3 f z is a rational function having only one singular

which is a rational function having four singular points

point in the upper half-plane. This point is the second order

(corresponding to the four roots of z 4 + 1 ). From Example

pole at z = a i . Using Proposition 10.1-9, its residue is:

1.14-10 we know these roots are on a unit circle:

⎡ d ⎢ Res ⎡⎣ f z , a i ⎤⎦ = lim z − ai z → ai dz ⎢ ⎣

()

!

(

)2

(

⎤ 1 ⎥ 2 2 z − a i z + a i ⎥⎦

)(

)

!

z1 = ei π 4 !

z2 = ei 3π

4

!

z3 = e i 5π 4 !

z4 = ei 7 π

4

or !

We will use the closed contour shown in Figure 10.3-5 to

⎡ −2 Res ⎡⎣ f z , a i ⎤⎦ = lim ⎢ z→ ai ⎢ ⎣ z + ai

()

(

⎤ ⎥ = −2 3 ⎥ 2ai ⎦

)

(

We then have from equation (10.3-31): ∞

!

∫ (x + a ) 0

dx

2

2

2

=

1 2

⎡ ⎛ i ⎞⎤ π 2 π i − ⎢ ⎜⎝ 4 a 3 ⎟⎠ ⎥ = 4 a 3 ⎢⎣ ⎥⎦

Example 10.3-8 Evaluate the integral

∫

∞

−∞

dx . x4 + 1

)

3

()

evaluate this integral. Inside this contour the function f z

−i = 4 a3

is a rational function having two singular points in the upper half-plane: !

∫

−∞

()

f z dz !

4

()

are: !

1 1 Res ⎡⎣ f z , z1 ⎤⎦ = lim = = z→ z1 q′ z 4 z13 4ei 3 π 1

!

1 1 Res ⎡⎣ f z , z2 ⎤⎦ = 3 = i 9 π 4 z2 4e

() ()

1

( )

4

e−i9 π = 4

4

e− i3 π = 4

4

4

We then have from equation (10.3-31):

We will evaluate: !

z2 = ei 3π

Using Proposition 10.1-5 with q z = z 4 + 1 , their residues

Solution:

∞

z1 = ei π 4 !

1 where f z = 4 z +1

()

!

PV

∫

∞

−∞

⎛ e− i 3π 4 e−i 9π 4 ⎞ 2 π i − i 3π 4 −i 9π dx = 2π i ⎜ + = e +e 4 ⎟⎠ 4 x4 + 1 ⎝ 4

(

4

) 524

Using equations (1.13-12) and (1.13-9): !

e± 2π i = 1 !

Therefore we have:

e π i = −1

!

we have: ! !

e− i3 π e

4

−i 9π 4

= e2 π i− i 3 π 4 = ei 5π 4 = eπ i ei π =e

− 2π i

e

− iπ 4

=e

4

= − ei π

4

− iπ 4

∫

PV

∞

−∞

(

dx π i iπ 4 − i π = − e −e 2 x4 + 1

4

)

=

(

π iπ 4 − i π e −e 2i

4

)

or using equation (4.3-1): !

∫

PV

∞

−∞

dx π π π 2 = π sin = = 4 2 x4 + 1 2

Example 10.3-9 Evaluate the integral

∫

∞

0

x 2 dx . 6 x +1

Solution: We will use the closed contour shown in Figure 10.3-3 to ∞

x 2 dx evaluate the integral . Since the integrand is an 6 0 x +1 even function of x , we have:

∫

!

∫

∞

x 2 dx 1 = x6 + 1 2

0

Figure 10.3-5! Closed contour C in the z-plane. Poles exist at

z1 = ei π 4 , z2 = ei 3π 4 , z3 = e i5π 4 , and z4 = e i 7 π 4 .

∫

∞

−∞

x 2 dx x6 + 1

We will evaluate: !

1 2

∫

∞

−∞

()

f z dz !

z2 where f z = 6 z +1

()

525

which is a rational function having six singular points (1.14-12) we find that this function has three singular points

10.3.5! IMPROPER INTEGRALS OF TRIGONOMETRIC FUNCTIONS

in the upper half-plane:

!

(corresponding to the six roots of z 6 + 1 ). Using equation

!

z1 = e i π 6 !

z2 = e i π 2 !

z3 = e i 5 π

form:

6

()

Using Proposition 10.1-4 with q z = z 6 + 1, their residues are: !

()

Res ⎡⎣ f z , z1 ⎤⎦ = lim z→ z1

( ) z2 1 1 e− i π 2 = lim 5 = 3 = i 3 π 6 = 6 q′ ( z1 ) z→ z 6 z 6 z1 6e p z1

1

!

1 1 Res ⎡⎣ f z , z2 ⎤⎦ = 3 = i 3 π 6 z2 6e

!

1 1 Res ⎡⎣ f z , z3 ⎤⎦ = 3 = i 15 π 6 z3 6e

() ()

2

6

e− i 3π = 6 e− i π = 6

! PV

∫

−∞

!

∫

0

−∞

()

f x cos a x dx !

∫

−∞

()

f x sin a x dx ! (10.3-44)

()

integrals are just the real and imaginary parts of the integral: !

∫

∞

()

f x e i a x dx !

(10.3-45)

and so they are classified as Fourier transforms. !

2

To evaluate these integrals a closed contour in the complex

plane must be constructed that includes an interval along the real axis. When analytic continuation is used to replace the realvalued functions in the integrand of equation (10.3-44) with

⎛ e− i π 2 e−i 3π 2 e−i π 2 ⎞ 2 π i x 2 dx = 2π i ⎜ + + = −i + i − i 6 6 ⎟⎠ 6 x6 + 1 ⎝ 6

x 2 dx 1 = PV 6 x +1 2

∫

∞

where f x is a rational real-valued function and a > 0 . These

2

(

)

complex functions, the trigonometric functions in the integrand will not be bounded. Therefore it is necessary that certain

()

restrictions apply to the function f z as given in the following

and so: ∞

!

∞

−∞

We then have from equation (10.3-31): ∞

Improper integrals of trigonometric functions have the

∫

∞

−∞

x 2 dx 1 2 π π = = 6 2 6 6 x +1

two propositions. This is the rational that leads to the Laplace transform.

526

Since i eiθ = 1 and e iaRcosθ = 1, we have:

Proposition 10.3-6:

()

Let f z be analytic in the upper half-plane except for a finite number of poles, and let CR consist of a semicircular contour of

()

∫

!

CR

radius R . If f z ≤ M and if M → 0 as R → ∞ , then:

lim

!

R →∞

∫

()

f z e

CR

iaz

dz = 0 !

∫

CR

(10.3-51)

()

f z e iaz dz ≤

∫

(

()

(10.3-47)

∫

!

CR

()

z = Re iθ !

)

about θ = π 2 since sin π − θ = sin θ . We can therefore let

f z e ia z dz !

CR

In the range 0 ≤ θ ≤ π , the integrand e− a R sinθ is symmetric

!

We are given that f z ≤ M . Letting: !

0

the complex plane, as shown in Figure 10.3-2.

Using Proposition 5.3-7 we can write:

!

∫

e− a R sinθ dθ !

direction along the semicircular contour CR in the upper half of

Proof: !

dz ≤ R M

π

If a > 0 , integration must proceed in a counterclockwise

!

(10.3-46)

()

f z e

ia z

dz = R ie iθ dθ !

!

(10.3-48)

(

)

(

()

f z e

ia z

dz ≤ 2 RM

The chord of

)

∫

π 2

0

y = sin θ

e− a R sinθ dθ !

(10.3-52)

in the range 0 ≤ θ ≤ π 2

is

y = 2 π θ = 2θ π . In the range 0 ≤ θ ≤ π 2 we will have sin θ

for the semicircular contour, and using z = R cosθ + i sin θ ,

greater than or equal to the chord connecting the endpoints of

equation (10.3-47) becomes:

this range (see Figure 10.3-6).

!

∫

CR

()

f z e

ia z

dz ≤

∫

π

Me

ia ( Rcosθ +i Rsin θ )

R ie

0

iθ

dθ !

! (10.3-49) !

or !

∫

CR

()

f z e ia z dz ≤ R M

∫

π

0

e

ia ( Rcosθ +i Rsin θ )

Therefore we have:

ie iθ dθ ! (10.3-50)

sin θ ≥

2θ ≥ 0! π

(10.3-53)

This is known as Jordan’s inequality. We then obtain: !

e− a R sinθ ≤ e− 2 a R θ π !

(10.3-54)

527

! !

lim

R →∞

∫

()

f z e ia z dz = 0 !

CR

(10.3-57)

If a < 0 , integration must proceed in a counterclockwise

direction along the contour CR in the lower half of the complex plane, as shown in Figure 10.3-7. We then will have sin θ < 0 , and so the exponential decreases as R → ∞ .

■

Figure 10.3-6! Chord connecting the endpoints of sin θ in the range 0 ≤ θ ≤ π 2 . We can now write: !

∫

CR

()

f z e

ia z

dz ≤ 2 R M

∫

π 2

e− 2 a R θ π dθ !

0

(10.3-55)

and so: !

∫

CR

()

f z e ia z dz ≤

(

)

πM 1− e− a R ! a

We have M → 0 as R → ∞ . Therefore:

Figure 10.3-7! Semicircle contour CR in the z-plane. Proposition 10.3-7:

()

(10.3-56)

Let f x be a real-valued rational function having the form: !

()

f x =

( )! Q( x) P x

(10.3-58) 528

()

()

where both P x and Q x are polynomials in x of degree n

()

and m , respectively. If m ≥ n + 2 and Q x ≠ 0 , then the Cauchy

!

C

principal value will exist and we have: !

PV

∫

∞

−∞

()

k

f x ei ω x dx = 2 π i

∑ j =1

()

∫

−∞

k

()

f x cos ω x dx = −2 π

∑ j =1

!!

!

PV

∫

−∞

()

Res ⎡⎣ f z ei ω z , zj ⎤⎦ ! (10.3-62)

From Proposition 10.3-6 we then have:

()

lim

!

R →∞

∫

()

f z ei ω z dz = 0 !

CR

(10.3-63)

Therefore:

Im Res ⎡⎣ f z ei ω z , zj ⎤⎦

! PV

∫

∞

()

f x e

−∞

()

∑ Re Res ⎡⎣ f ( z ) e j =1

!

iω z

, zj ⎤⎦

!

k

iω x

dx = 2 π i

∑ j =1

()

Res ⎡⎣ f z ei ω z , zj ⎤⎦ !

(10.3-64)

Equating the real and imaginary parts of both sides of

equation (10.3-64), we have:

(10.3-61)

PV

! We will use the contour in Figure 10.3-3, in which the

segment CR consists of a semicircular contour given by z = R e

∫

∞

−∞

Proof: !

∑

= 2π i

()

k

f x sin ω x dx = 2 π

! !

∫

f z ei ω z dz

CR

j =1

(10.3-60) ∞

−R

()

f x ei ω x dx +

Res ⎡⎣ f z ei ω z , zj ⎤⎦ ! (10.3-59)

upper half of the complex plane, and where ω > 0 . We also have:

PV

∫

!

()

!

dz =

R

k

where zj are the k points at which f z ei ω z has poles in the

∞

!∫

()

f z e

iω z

k

()

f x cos ω x dx = −2 π

∑ Im Res ⎡⎣ f ( z ) e

iω z

j =1

!!

(10.3-65)

iθ

where 0 ≤ θ ≤ π . From Cauchy’s residue theorem (Proposition

, zj ⎤⎦

PV

!

10.2-2) we then have:

∫

∞

−∞

! ! ■

k

()

f x sin ω x dx = 2 π !

∑ Re Res ⎡⎣ f ( z ) e j =1

iω z

, zj ⎤⎦ (10.3-66)

529

Example 10.3-11

Example 10.3-10

Solution:

Solution:

We will use the closed contour shown in Figure 10.3-3. We

We will use the closed contour shown in Figure 10.3-3. We

will evaluate:

will evaluate:

∫

∞

()

f z e

iω z

1 where f z = 2 z +1

()

dz !

!

−∞

()

()

where f z =

f z sin z dz !

z z2 + 4

Inside the closed contour this function has only one pole in

Inside the closed contour this function has only one pole in

the upper half-plane. This is the simple pole at z = i . Using

the upper half-plane. This is the simple pole at z = 2i . Using

equation (10.3-65) we have:

equation (10.3-66) we have:

∞

cos ω x PV dx = −2 π 2 x + 1 −∞

∫

k

∑ n =1

()

Im Res ⎡⎣ f z ei ω z , i ⎤⎦

!

(

)

(

)(

)

k

∑ n =1

()

Re Res ⎡⎣ f z ei z , 2i ⎤⎦

!

⎛ ⎞ 2i e− 2 z ei z 1 iz ⎡ ⎤ Res ⎣ f z e , 2i ⎦ = lim ⎜ z − 2i = = ⎟ z→2i 4i z − 2i z + 2i ⎠ 2e 2 ⎝

()

(

)

(

)(

)

and so: ∞

−ω

(

)

cos ω x e PV dx = − 2 π Im = π Im e− ω i = π e− ω 2 2i −∞ x +1

∫

∫

x sin x dx = 2 π 2 x +4

where from Proposition 10.1-2:

iω z iω z −ω ⎛ ⎞ e e e i ω z Res ⎡⎣ f z e , i ⎤⎦ = lim ⎜ z − i = ⎟ = lim z→i z − i z + i ⎠ z → i z + i 2i ⎝

()

PV

∞

−∞

and so: !

∫

∞

−∞

where from Proposition 10.1-2: !

x sin x dx . x2 + 4

Evaluate the integral

−∞

!

∫

∞

cos ω x Evaluate the integral dx with ω > 0 . 2 −∞ x +1

∫

!

∞

!

PV

∫

∞

−∞

x sin x 1 π dx = 2 π Re = x2 + 4 2e2 e2

530

10.3.6! IMPROPER INTEGRALS OF FUNCTIONS WITH REAL EXPONENTIALS !

!

2 =1 1+ 1

()

Therefore as R → ∞ we cannot have f z → 0 , and so we

A contour that includes a semicircular curve CR as shown

cannot eliminate the contribution of a semicircular contour

in Figure 10.3-3 is not useful for evaluating improper integrals

CR to the integral:

having an integrand that does not approach zero at infinity (see Example 10.3-12). A rectangular contour that takes advantage of

()

f z ≥

!

∫

∞

−∞

the periodicity of the complex function can sometimes be

1 dz cosh a z

helpful in these cases (see Example 10.3-13). Example 10.3-13

Example 10.3-12 Show that the improper integral

∫

∞

−∞

eax Evaluate the integral dx where 0 < a < 1 . x 1+ e −∞

∫

1 dx does not go cosh a x

Solution:

to zero on the contour CR shown in Figure 10.3-3.

We will evaluate:

Solution: !

We will evaluate: !

∫

∞

−∞

()

f z dz !

()

()

f z =

2 ea Rcosθ ei a Rsinθ + e− a Rcosθ e−i a Rsinθ

≥

On the imaginary axis we have for θ = π 2 :

ea Rcosθ

()

(

)

imaginary axis where z = i π + 2 k π so that: !

2 + e− a Rcosθ

()

f z dz !

e az where f z = 1+ e z

This function has an infinity of simple poles along the

1 2 = az − az cosh a z e + e

On the contour CR with radius R we have: !

∫

∞

−∞

where f z =

∞

i π +2 k π ) ez = e ( = −1!

k = 0, ± 1, ± 2, !

(see Example 1.13-1). We will choose a contour that encloses the least number of poles. This is the rectangular contour

C = C1 ∪ C2 ∪ C3 ∪ C4 shown in Figure 10.3-8 that encloses 531

only the pole z = i π . The vertical dimensions of this contour are such that 0 ≤ y ≤ 2 π .

From Proposition 10.1-4 we have: !

⎡ eaz ⎤ ⎡ eaz ⎤ ei aπ ei aπ Res ⎡⎣ f z , i π ⎤⎦ = lim ⎢ = lim ⎢ z ⎥ = i π = = − e i aπ ⎥ z z → i π 1+ e −1 ⎣ ⎦ z → iπ ⎣ e ⎦ e

()

Therefore:

()

2 π i Res ⎡⎣ f z , i π ⎤⎦ = − 2 π i ei aπ

!

Evaluating the integrals along the contours C1 , C2 , C3 , and

C4 , we have: Along C1 we have z = x :

eaz dz = z C1 1+ e

∫

! Figure 10.3-8! Contour C = C1 ∪ C2 ∪ C3 ∪ C4 enclosing a pole at z = i π in the z-plane. We then can write:

!∫

!

C

()

f x dx = 2 π i Res ⎡⎣ f z , i π ⎤⎦

∫

C1

!

()

f z dz +

∫

C2

()

f z dz +

∫

C3

()

f z dz +

∫

eaz dz = z C2 1+ e

∫

∫

2π

0

a L+i y ) e (

1+ e L+i y

i dy =

∫

2π

0

ea L 1+ e L

i dy =

ea L 1+ e L

∫

2π

i dy

0

and

or !

eax dx x − L 1+ e

Along C2 we have z = L + i y : !

()

L

∫

C4

!

()

f z dz =

()

2 π i Res ⎡⎣ f z , i π ⎤⎦

a−1 L ea L ea L ea L e− L e( ) ≤ L = = 1+ e L e − 1 1− e− L 1− e− L

Letting L → ∞ , we have since 0 < a < 1: 532

eaz ea L dz = z 1+ e 1+ e L C2

∫

!

∫

2π

0

a−1 L e( ) i dy ≤ 2 π lim =0 L→ ∞ 1− e− L

!∫ f ( x ) dx = 2π i Res ⎡⎣ f ( z ), i π ⎤⎦

!

C

becomes: Along C3 we have z = x + 2π i : !

az

∫

e dz = z 1+ e C3

∫

−L L

a ( x+2π i )

e 2a π i dx = e 1+ e x+2π i

∫

−L L

−L

ax

e dx 1+ e x

or reversing limits: !

∫

L

eax dx x 1+ e −L

∫

Along C4 we have z = − L + i y : !

∫

∫

a − L+i y ) e ( i dy = − L+i y ) ( 1+ e

2π

∫

2π

0

e− a L 1+ e

−L

i dy

and !

e− a L 1+ e− L

!

∫

e dz = z C4 1+ e

∫

−L

eax dx = − 2 π i ei aπ x 1+ e

(1− e ) ∫ 2a π i

∞

eax i aπ dx = − 2 π i e x − ∞ 1+ e

∞

eax e i aπ 2π i π ! dx = − 2 π i = = x 2a π i aπ i − aπ i sin a π 1− e e −e − ∞ 1+ e

∫

0 < a 1 and M are constants 549

!∫ π cot (π z ) f ( z ) dz

!

→ 0!

(10.4-22)

CN

and so from equation (10.4-20):

From Proposition 10.1-2 we have: π cot ( π z) π cot ( π a i) ! Res ⎡⎣ g ( z ) f ( z ) , a i⎤⎦ = lim ( z − a i ) = z→ ai 2ai ( z − a i) ( z + a i)

∞

∑ f ( n) = −∑ Res ⎡⎣π cot π z f ( z ) ⎤⎦

!

!

poles of f ( z ) within CN

n = −∞

or

(10.4-23)

■

!

Show that

∑

1 π = coth π a where a > 0 . 2 2 a n +a

( )

n=−∞

!

!

Let:

()

f z =

1 ! z 2 + a2

()

g z = π cot π z

() ()

g z f z =

π cot π z z 2 + a2

∑

n = −∞

∑

1 =− Res ⎡⎣ g z f z , ± a i⎤⎦ 2 2 n +a z = ± ai

() ()

( )

() ()

and so from Proposition 10.4-2: !

we have: ∞

π coth( π a ) 2a

π Res ⎡⎣ g z f z , ± a i⎤⎦ = − coth π a a

∞

which has simple poles at z = ± a i . From Proposition 10.6-2

!

∑

z = ± ai

We then have: !

() ()

Res ⎡⎣ g z f z , − a i⎤⎦ = −

Therefore:

Solution:

!

π coth( π a ) 2a

Similarly:

Example 10.4-2 ∞

() ()

Res ⎡⎣ g z f z , a i⎤⎦ = −

∑

⎛ π ⎞ π 1 coth π a = − − ⎜⎝ a ⎟⎠ = a coth π a 2 2 n + a n=−∞

( )

( )

Example 10.4-3 ∞

∑

2 1 π Show that . = 2 6 n n = −∞

Solution: 550

∞

Let: !

!

()

f z =

1 ! 2 z

()

() ()

g z f z =

Proposition 10.4-3:

π cot π z z2

The Laurent series expansion of cot z about z = 0 is: !

π cot π z 1 π 2 π 4 z = 3− − −! 3z 45 z2 z which we see has a pole of order 3 at z = 0 with a residue of

− π 3 . From Proposition 10.4-2 we then have: 2

∞

!

⎛ π2⎞ π2 1 = − ⎜− ⎟ = 2 3 n ⎝ 3⎠ n=−∞

∑

()

Since f z is even, we can write: ∞

!

On the square contour CN having vertices at the points

( N + 1 2)( ±1± i ) where N

2

∑ n =1

and so:

1 π2 = 3 n2

is an integer, we have:

csc π z < A !

!

1 z z3 cot z = − − −! z 3 45 Therefore we have:

!

n =1

g z = π cot π z

We then have: !

∑

1 π2 = 2 6 n

(10.4-24)

where A is a constant. Proof: !

Let ω = N + 1 2 . We will then have:

!

cos ω π = 0 !

!

We will first consider the line segments L1 and L3 of the

( )N !

sin ω π = −1

for all ω ! (10.4-25)

contour CN where z = ± ω + i y . Using equation (4.3-32) we can write: !

csc π z =

1 1 = sin π z ± sin ω π cosh π y + i cos ω π sinh π y

!!

(10.4-26)

or using equation (10.4-25): !

csc π z =

1 < A1 ! cosh π y

(10.4-27) 551

csc π z < A1 !

!

(

)(

)

having vertices at the points N + 1 2 ±1± i , then the sum of

Therefore: for all ω !

the series f ( n ) is given by:

(10.4-28)

∞

!

We will next consider the line segments L2 and L4 of the

n −1 ( ) ∑ f ( n) = −∑ Res ⎡⎣π csc π z f ( z ) ⎤⎦

!

contour CN where z = x ± iω . We then have from equation

!(10.4-32)

poles of f ( z ) within CN

n = −∞

(4.3-2): !

2i 2i ! csc π z = π i z − π i z = π i x ∓ π ω e −e e − e− π i x ± π ω

Proof: (10.4-29)

!

and so: !

We have from equations (10.4-3), (10.4-8), and (10.4-10):

!∫ π csc (π z ) f ( z ) dz = CN

csc π z ≤

2i π ix ∓π ω

− e

e

−π i x ± π ω

=

2 eπ ω − e− π ω

∞

!

(10.4-30)

!

n −1 ( ) ∑ f ( n) + 2π i ∑ Res ⎡⎣π csc π z f ( z ) ⎤⎦

2π i

poles of f ( z ) within CN

n = −∞

!

or !

!

1 csc π z ≤ = A2 ! sinh π ω

(

()

We will now choose N large enough so that all poles of f z (10.4-31)

)

Letting A = max A1, A2 we have csc π z < A on CN where A is a constant.

■

are inside the contour CN . Each line segment of CN will have a

(

!

Proposition 10.4-4, Summation of Series: Let f z be a meromorphic function none of whose poles are

()

integers. If f z < M z

k

!∫

π csc (π z ) f ( z ) dz ≤

π AM Nk

(8 N + 4 ) !

(10.4-34)

where A is a constant, and where the length of the contour CN is 8 N + 4 . As N → ∞ we obtain:

where k > 1 and M are constants

independent of the integer N along the square contour CN

)

length of z ≥ 2 N + 1 2 . We can write using equation (10.4-24):

CN

()

(10.4-33)

!

!∫

CN

π csc (π z ) f ( z ) dz → 0 !

(10.4-35) 552

()

and so from equation (10.4-33): ∞

∑ (−1) f ( n) = −∑ Res ⎡⎣π csc π z f ( z ) ⎤⎦

!

n

Since f z is even, we can write: ∞

!(10.4-36)

!

poles of f ( z ) within CN

n = −∞

or ∞

Example 10.4-4 ∞

Show that

∑

!

( −1 )n

n =1

n2

π2 . =− 12

Solution:

()

f z =

Let!

1 z2

∑ n =1

1 π2 =− 12 n2

10.5! MITTAG-LEFFLER’S EXPANSION THEOREM !

Mittag-Leffler’s expansion theorem provides a means of

()

representing a meromorphic function f z as an expansion in

The Laurent series expansion of csc z about z = 0 is:

()

π csc π z 1 π 7π z = + + +! 2 3 6 z 360 z z 4

which we see has a pole of order 3 at z = 0 with a residue of

will consider the special case in which a meromorphic function can be represented as the sum of partial fractions. Proposition 10.5-1, Mittag-Leffler’s Partial-Fraction Expansion

π 2 6 . From Proposition 10.4-4 we then have: !

⎛π2⎞ 1 π2 =−⎜ ⎟ =− 2 6 n ⎝ 6⎠ n=−∞

∑

()

about only one singular point of f z . We will assume that the function f z has a countably infinite number of poles, and we

2

∞

()

the form of a sum of polynomials involving the poles of f z . This differs from a Laurent expansion which, by definition, is

1 z 7 z3 csc z = + + +! z 6 360 Therefore we have:

!

∑ n =1

■

!

2

1 π2 =− 6 n2

Theorem: Let the only singularities of a bounded meromorphic function

()

f z

be simple poles a1 , a2 , a3, ! arranged in order of 553

()

increasing absolute value. Let the residues of f z at the points

a1 , a2 , a3, ! be b1, b2 , b3, ! . We then have:

()

∞

()

f z = f 0 +

!

∑ n =1

⎡ 1 1⎤ bn ⎢ + ⎥! ⎣ z − an an ⎦

will have poles at points a1 , a2 , a3, ! , z0 . The

( ) ( z − z0 )

residues of f z

at the points a1 , a2 , a3, ! can be

obtained using Proposition 10.1-2:

( z→a

lim z − an

!

n

)

()

f z

(

z − z0

=

) (

bn an − z0

)

!

(10.5-2)

( ) ( z − z0 ) at the point z0 is given by: f (z) lim ( z − z0 ) = f ( z0 ) ! (10.5-3) z→ z z − z ( 0) 0

!

()

f z

Let the circle CN of radius RN contain the first N poles of

()

without going through any singularity of f z . From

Cauchy’s residue theorem we then have for all poles of

( ) ( z − z0 )

f z

inside the circle CN :

is analytic at z = 0 , we can let z0 = 0 . Equation

CN

z − z0

( )

dz = f z0 +

n

n =1

!

(10.5-4)

0

( ) dz = f

f z

!∫

CN

z

()

N

0 +

∑a ! bn

n =1

(10.5-5)

n

The difference of equations (10.5-4) and (10.5-5) then yields:

1 ! 2π i !

()

()

⎡f z f z ⎤ − ⎢ ⎥ dz = f z0 − f 0 + z − z z ⎢⎣ ⎥⎦ 0

!∫

CN

!

( )

()

!

N

∑ n =1

bn ⎤ ⎡ bn − ⎥ ⎢ a − z ⎣ n 0 an ⎦ (10.5-6)

or

Similarly, the residue of f z !

()

1 2π i

()

( ) ( z − z0 )

∑ a −z bn

(10.5-4) then becomes:

Choosing a point z0 that is not a pole of f z , the function

f z

N

!∫

If f z

!

!

Proof: !

(10.5-1)

!

()

f z

1 2π i

1 2π i

! ! !

!∫

CN

(

()

f z

z z − z0

!

)

( )

()

dz = f z0 − f 0 +

N

∑ n =1

!

()

⎡ 1 1⎤ bn ⎢ − ⎥ ⎣ an − z0 an ⎦ (10.5-7)

()

Since f z is bounded, we have f z < M . For z on the

circle CN we have: !

z − z0 ≥ z − z0 = RN − z0 > 0 !

(10.5-8)

Therefore: 554

!∫

!

()

f z

CN

(

z z − z0

)

dz ≤

M 2 π RN

(

RN RN − z0

)

=

M 2π RN − z0

!

(10.5-9)

Letting N → ∞ , then RN → ∞ , and we have:

lim

!

N →∞

!∫

CN

()

f z

(

z z − z0

dz = 0 !

)

(10.5-10)

and so equation (10.5-7) becomes:

( )

()

∞

f z0 = f 0 +

!

∑ n =1

⎡ 1 1⎤ bn ⎢ + ⎥! ⎣ z0 − an an ⎦

(10.5-11)

Letting z0 = z be a general point that is not a pole, we have:

()

()

f z = f 0 +

!

∞

∑ n =1

⎡ 1 1⎤ bn ⎢ + ⎥! ⎣ z − an an ⎦

(10.5-12)

■

555

Chapter 11 Conformal Mapping

az+b w= cz + d

556

()

In this chapter we will discuss the mapping by analytic

rule is referred to as a mapping or transformation by f z of a

functions of point sets. The objective of this mapping generally

point in the z-plane to a point in the w-plane. Points in both the

is to transform a region having a complicated boundary into a

z-plane and the w-plane are specified by complex numbers.

region having a simpler boundary (such as that of a circular

!

disk). Physical problems involving boundary conditions can

the domain of definition of f z

then be solved using the simpler boundary, and then, using

image set of points Ω in the w-plane. The set of points Ω in the

inverse mapping, the solution of the original problem can be

w-plane is also known as the range of f z . The set of points in

obtained.

the z-plane is the inverse image of the image in the w-plane.

!

!

!

Such mappings provide powerful tools for solving many

As noted in Section 2.1.2, mapping of a set of points S in

()

in the z-plane defines an

()

When using the mapping of analytic functions to solve

two-dimensional potential theory problems involving Laplace’s

physical problems, it is important that a single-valued inverse

equation that arise in engineering and physics. We will present

z = f −1 w exist for the transformation w = f z

a number of example mappings that have proven particularly

mapping from the z-plane to the w-plane by w = f z , and then

useful in this regard. We will also discuss a three-dimensional

back to the z-plane by z = f −1 w will all be one-to-one (see

construction known as a Riemann surface that is designed to

Figure 11.1-1). Each point in the image will then correspond to a

geometrically represent multivalued functions.

single unique point in the inverse image. We will then have

( )

Complex functions of a single complex variable z = x + i y

() ( )

( )

(11.1-1)

()

where the complex function w = f z

()

when z1 ≠ z2 . The necessary condition for the

Proposition 11.1-1. !

w = f z = u x, y + i v x, y !

( )

so that the

existence of such a single-valued inverse is presented in

have the form: !

()

( )

f z1 ≠ f z2

11.1! DEFINITION OF MAPPING !

( )

defines the rule for

obtaining a complex number w from a complex number z . This

()

A mapping w = f z of a set S into a set Ω is continuous

if, for any point z0 contained in S and for a neighborhood

( )

Dδ w0 of its image w0 in Ω , there exists a neighborhood of z0

()

( )

mapped by w = f z into Dδ w0 .

557

!

The necessary condition for any transformation to have a

single-valued inverse at a point z0 is that the determinant of the Jacobian of the transformation be nonzero in the neighborhood

()

of the point. For the transformation w = f z

we then must

have in the neighborhood of z0 :

!

⎛ u, v ⎞ J⎜ = ⎝ x, y ⎟⎠

∂u ∂x

∂u ∂y

∂v ∂x

∂v ∂y

≠ 0!

(11.1-2)

()

Since f z is analytic, u and v will satisfy the Cauchy-Riemann Figure 11.1-1! Complex point z0 in the z-plane mapped into

()

equations. Therefore equation (11.1-2) can be written as:

point w0 in the w-plane by a function w = f z . Complex point w0 in the w-plane mapped into point z0 in the z-plane by the inverse function

!

( )

z = f −1 w .

∂v ∂x ≠ 0! ∂u ∂x

−

(11.1-3)

and

Proposition 11.1-1:

()

()

If f z is an analytic function in a domain D , w = f z will have a single-valued inverse z = f

( )

any point z0 at which f ′ z0 ≠ 0 . Proof:

∂u ⎛ u, v ⎞ ∂x J⎜ = ⎟ ⎝ x, y ⎠ ∂v ∂x

−1

( w) in the neighborhood of

!

⎛ u, v ⎞ J⎜ = ⎝ x, y ⎟⎠

∂v ∂y

∂u ∂y

∂u − ∂y

∂v ∂y

≠ 0!

(11.1-4)

We then have: 558

2

2

2

⎛ u, v ⎞ ⎛ ∂u ⎞ ⎛ ∂v ⎞ ∂u ∂v J⎜ = + = + i ≠ 0! ∂x ∂x ⎝ x, y ⎟⎠ ⎜⎝ ∂x ⎟⎠ ⎜⎝ ∂x ⎟⎠

!

C z . If C z is mapped onto a simple closed contour C w in the w-

(11.1-5)

()

plane by w = f z traversing C z once, then all points in the

and 2

2

univalent.

2

⎛ u, v ⎞ ⎛ ∂u ⎞ ⎛ ∂v ⎞ ∂u ∂v J⎜ = + = + i ≠ 0! ∂y ∂y ⎝ x, y ⎟⎠ ⎜⎝ ∂ y ⎟⎠ ⎜⎝ ∂ y ⎟⎠

!

(11.1-6)

⎛ u, v ⎞ J⎜ = f ′ z0 ⎝ x, y ⎟⎠ z

( )

!

2

Proof: !

Therefore we have from equation (3.4-25):

Let z0 be a point within the contour C z , and let the

()

≠ 0!

(11.1-7)

has no poles in D . From equation (9.4-4) we then can write for

w0 : !

We also have:

!∫ ( f ( z ) − w ) dz = 2π i N ! 1

C

JJ

!

−1 z0

= 1!

(11.1-8)

()

We can conclude that if f z is analytic at a point z0 , and

( )

if f ′ z0 ≠ 0 , then a single-valued inverse z = f

()

−1

( w)

of the

transformation w = f z will exist in some neighborhood of

()

point z0 . The transformation w = f z will provide a one-toone mapping of a neighborhood of point z0 . The function

()

w = f z is therefore univalent in D .

■

(11.1-9)

0

()

where f z ≠ w0 on C w . This equation gives us the number N

()

of zeros of f z − w0 inside C w , which is the number of times

()

we have f z = w0 . We can rewrite equation (11.1-9) as: !

!∫

C

!

1

w

( w − w0 )

dz = 2 π i N !

(11.1-10)

Using residue integration over the simple closed contour

outside C w . Therefore every point within C z is mapped only

Let f z be an analytic function in a domain D in the z-plane, z

w

C w , we find that N = 1 if w0 is inside C w , and N = 0 if w0 is

Proposition 11.1-2:

()

()

function w = f z map z0 to w0 . Since f z is analytic in D , it

0

!

()

interior of C z are mapped onto the interior of C w , and f z is

()

and C be a simple closed contour in D . Let f ′ z ≠ 0 inside

()

once to a point inside C w by f z , and no points within C z are

()

()

mapped to any point outside C w by f z . This means that f z is univalent.

■

559

Let C1z and C2z be two smooth curves in D that intersect at

!

11.2! CONFORMAL MAPPING

the point z0 in the z-plane. Let C1w and C2w be the images of

!

A mapping by a complex function is called conformal if it

these curves when mapped by w = f z into the w-plane. The

preserves local angles in both magnitude and sense, and the

curves C1w and C2w will intersect at the point w0 where

function is known as a conformal function. A function that is

w0 = f z0 . If the curves C1z and C2z are parametrized as z1 t

analytic in a domain will be conformal at any point in the domain where its derivative is nonzero (see Proposition 11.2-1). !

()

If the mapping by a function w = f z of a set of points S

in the z-plane into a set of points Ω in the w-plane is conformal

()

for all points in S , then w = f z is considered a conformal

()

( ) and z2 ( t ) , respectively, we then have at point z0 : ! z1 ( t0 ) = z2 ( t0 ) = z0 !

(

( ) ) and tan (arg z2′ (t0 ) ) , respectively:

tan arg z1′ t0

()

dy1 t

of S . ! Proposition 11.2-1:

(

( )) =

tan arg z1′ t0

dt dx1 dt

If w = f z is an analytic function in a domain D , and z0 is a

( )

( )

Proof: !

( )

()

=

dy1 dx1

!

(11.2-2)

( ) ( )

x1 t0 , y1 t0

t = t0

dy2 t !

Since f ′ z0 ≠ 0 we know from Proposition 11.1-1 that

t = t0

()

point in D at which f ′ z0 ≠ 0 , then the mapping w0 = f z0 will be conformal.

(11.2-1)

and the tangents to the curves C1z and C2z at point z0 are

mapping of S onto Ω , and Ω is said to be a conformal image

()

()

(

( )) =

tan arg z2′ t0

dt dx2 dt

t = t0

=

dy2 dx2

( ) ( )

!

(11.2-3)

x2 t0 , y2 t0

t = t0

w = f z will have a single-valued inverse in the neighborhood

The derivatives will exist since the curves are smooth. Let the

of the point z0 . Therefore w = f z will be one-to-one in the

polar angles of the two curves C1z and C2z at z0 be θ1 and θ 2 ,

neighborhood of point z0 .

respectively. We then have:

()

560

θ1 = arg ⎡⎣ z1′ ( t0 ) ⎤⎦ !

!

(11.2-4)

and

!

(11.2-9)

From equations (11.2-8), and (11.2-9), we can write:

θ 2 = arg ⎡⎣ z2′ ( t0 ) ⎤⎦ !

!

(11.2-5)

!

and the angles θ1 and θ 2 , then represent the respective angles

and

the tangents to the curves C1z and C2z make with the x-axis.

!

()

Mapping the two curves C1z and C2z using f z , their

!

φ2 = arg ⎡⎣ w2′ ( t0 ) ⎤⎦ = arg ⎡⎣ f ′ ( z0 ) z2′ ( t0 ) ⎤⎦ !

images C1w and C2w in the w-plane will also be smooth curves.

()

Since f z

( ) ( ( ))

is analytic in D , then both w1 t = f z1 t

( ( ) ) are differentiable, and are given by:

()

and

w2 t = f z2 t

()

! w1′ t

( ( ))

d = f z1 t t0 dt

( ( )) z ′ ( t ) = f ′ ( z ) ( )

= f ′ z1 t0 t0

1

0

0

z1′ t0 ! (11.2-6)

φ1 = arg ⎡⎣ f ′ ( z0 ) ⎤⎦ + arg ⎡⎣ z1′ ( t0 ) ⎤⎦ !

(11.2-10)

φ2 = arg ⎡⎣ f ′ ( z0 ) ⎤⎦ + arg ⎡⎣ z2′ ( t0 ) ⎤⎦ !

(11.2-11)

From equations (11.2-4), (11.2-5), (11.2-10), and (11.2-11), we have: !

φ1 = arg ⎡⎣ f ′ ( z0 ) ⎤⎦ + θ1 + 2 k1 π !

(11.2-12)

φ2 = arg ⎡⎣ f ′ ( z0 ) ⎤⎦ + θ 2 + 2 k 2 π !

(11.2-13)

and !

and

where k1 and k2 are integers. The tangents to the curves C1z and

d ! w2′ t = f z2 t t0 dt

C2z at point z0 are then each rotated by the mapping w0 = f z0

( ( ))

()

( )

( ( )) z′ ( t ) = f ′ ( z ) z′ ( t ) ! (11.2-7)

= f ′ z2 t0 t0

( )

2

0

0

2

0

( )

where f ′ z0 ≠ 0 , z1′ t0 ≠ 0 , and z2′ t0 ≠ 0 . !

Let the polar angles of the two curves C1w and C2w at w0 be

φ1 and φ2 , respectively. We then have: ! and

φ1 = arg ⎡⎣ w1′ ( t0 ) ⎤⎦ = arg ⎡⎣ f ′ ( z0 ) z1′( t0 ) ⎤⎦ !

(11.2-8)

( )

( )

through an angle of arg ⎡⎣ f ′ z0 ⎤⎦ . ! The angle between the tangents to the two curves C1z and

C2z in the z-plane at point z0 is: !

Δθ = θ 2 − θ1 !

(11.2-14)

and the angle between the tangents to the two curves C1w and

C2w in the w-plane at point w0 is: 561

!

Δφ = φ2 − φ1 !

(11.2-15)

Δφ = Δθ !

!

(11.2-18)

()

We see then that mapping by the function f z will be

(see Figure 11.2-1). Therefore equations (11.2-12), (11.2-13),

!

(11.2-14), and (11.2-15) give us:

conformal, and that the angle between two intersecting curves

!

(

)

Δφ = Δθ + 2 k2 − k1 π !

(11.2-16)

is preserved in both magnitude and sense (the sign is the same)

()

if f z is an analytic function in a domain D , and z0 is a point

( )

in D at which f ′ z0 ≠ 0 .

■

Proposition 11.2-2:

()

If a function f z is analytic and one-to-one in a domain D ,

()

then the mapping w = f z is conformal at every point in D . Proof: !

Follows from Proposition 11.2-1.

■

Proposition 11.2-3:

()

If a function f z is analytic and one-to-one inside and upon a

! !

will result in a simple closed curve in the w-plane.

Letting n = k2 − k1 where n is an integer, we have:

Δφ = Δθ + 2 n π !

()

simple closed contour C , then the mapping of C by w = f z

Figure 11.2-1! Conformal mapping of two curves.

(11.2-17)

Proof: !

()

Since f z is analytic on C , it is continuous on C , and so

()

where 0 ≤ Δθ < 2 π and 0 ≤ Δφ < 2 π . If n ≥ 1 then Δφ ≥ 2 π , and if

the mapping of C by w = f z results in a continuous curve in

n ≤ −1 then Δφ < 0 . Therefore we must have n = 0 , and so:

the w-plane. Since f z is one-to-one, as f z travels once over

()

()

562

()

the curve C , the function w = f z

never obtains the same

linear scale factor. This scale factor for mapping infinitesimal

value more than once. Therefore the mapped curve in the w-

line elements is simply f ′ z0 . The scale factor can vary with

plane can have no intersecting points. After f z

the point, and so is generally not a constant over any distance.

()

travels the

entire curve C once, it will return to is initial value. This means the mapped curve in the w-plane will also return to its initial

()

value. Therefore the analytic function w = f z maps a simple closed curve in the z-plane into a simple closed curve in the wplane. ! !

■

( )

Writing f ′ z0

( )

f ′ z0 = lim

( )

z − z0

( )

()

Is the mapping w = f z = e z conformal? Solution:

()

in the form:

()

we see that f ′ z0

Example 11.2-1

= lim

z → z0

w = f z = e z is conformal for the entire complex plane. Δw Δw = lim ! (11.2-19) z → z0 Δz Δz

provides a change in linear dimensions at

()

Example 11.2-2 Is the mapping w =

the point z0 as a result of the mapping w = f z . The modulus

( )

f ′ z0

()

Since e z is entire and f ′ z = e z ≠ 0 for all z , the mapping

f z − f z0

z → z0

( )

is therefore known as the dilation constant or linear

( )

1 conformal? z

Solution:

magnification factor at point z0 . If f ′ z0 = 0 then the point z0

If a function is differentiable, it must be independent of z

is called a critical point of f z , and the mapping w = f z will

(see Proposition 3.6-2). Therefore the mapping w = 1 z is not

not be conformal at z0 . Instead the angle between two curves

conformal. Note also that the sense of angles is not

will be magnified. Since critical points of f z

preserved by the mapping w = 1 z .

()

()

()

()

are zeros of

f ′ z , they are isolated. !

From equations (11.2-6) and (11.2-7) we see that the

mapping of a curve from the z-plane to the w-plane involves a 563

Proposition 11.2-4:

()

()

If f z is an analytic function in a domain D , and if f z has

()

w = f z will magnify all angles at z0 by the factor n .

!

! !

()

Since f z is an analytic function, it can be represented by

()

a Taylor series centered at the point z0 . Since f z has a zero of n−1 order n at z0 , we have f ′ z0 = ! = f ( ) z0 = 0 , and so z0 is a

() conformal at w0 = f ( z0 ) .

( )

( )

()

critical point of f z , and the mapping w = f z ! !

()

( )

(

f z = f z0 + an z − z0

)

n

(

+ an+1 z − z0

)

n+1

+! !

(11.2-20)

This series can be written in the form: !

()

( ) (

) ()

f z = f z0 + z − z0 g z !

! !

()

(

)

(

()

(11.2-23)

( n) z ( n+1) z ⎡ f f n 0 0 f z − f z0 = z − z0 ⎢ + ⎢ n! n +1 ! ⎣ !

()

( ) (

( )

)

⎤ ( ) ( z − z0 ) +!⎥⎥ ( ) ⎦ (11.2-24)

( ( ) ( ))

(

arg f z − f z0 = n arg z − z0

)

n+1 ⎛ f ( n) z f ( ) z0 0 + arg ⎜ + n! ⎜⎝ n +1 !

( )

⎞ ( ) ( z − z0 ) +!⎟⎟ ! ( ) ⎠

(11.2-25)

As z → z0 we have: !

(11.2-21) !

g z = an + an+1 z − z0 + an+2 z − z0

≠ 0!

z − z0 = Δz !

()

( )

f z − f z0 = w − w0 = Δw !

(11.2-26)

equation (11.2-25) becomes:

n

where !

n!

and so:

is not

The Taylor series centered at the point z0 is:

( )

From equations (11.2-21), (11.2-22), and (11.2-23) we obtain:

a zero of order n at a point z0 in D , then the mapping

Proof:

an =

!

n f ( ) z0

)2 +! ! ( )

(11.2-22)

⎛ f ( n) z 0 arg Δw = n arg Δz + arg ⎜ n! ⎜⎝

( )

( )⎞ !

( )

⎟ ⎟⎠

(11.2-27)

From equation (11.2-23) we see that equation (11.2-27) can be

and where g z in analytic at point z0 and g z0 = an ≠ 0 . We

rewritten as:

then have:

!

( )

( )

( )

arg Δw = n arg Δz + arg an !

(11.2-28) 564

We have:

If Δz1 and Δz2 are two infinitesimal curves that have an

!

angle Δθ , between them (see Figure 11.2-1), and if Δw1 and Δw2

!

are their images having an angle Δφ , between them, we have from equation (11.2-28):

(

)

( )

( )

(11.2-29)

(

)

(

( )

(11.2-30)

!

arg Δw1 = n arg Δz1 + arg an !

!

arg Δw2 = n arg Δz2 + arg an !

)

()

(

()

We then have f ′ 0 = 0 and so the mapping w = f z = z 2 is not conformal at z = 0 . Using z = ei θ and w = ei φ , we see that

()

with the mapping w = f z = z 2 we will have: !

Δφ = 2 Δθ !

()

()

Since f ′′ z = 2 ≠ 0 , the mapping w = f z = z 2 has a zero of

Therefore:

)

(

)

( ( )

( ))

arg Δw2 − arg Δw1 = n arg Δz2 − arg Δz1 !

!

()

f′ z =2z

order two at z = 0 . Therefore the result Δφ = 2 Δθ is in

(11.2-31)

agreement with Proposition 11.2-4.

or

Δφ = n Δθ !

!

(11.2-32)

()

We see then that if f z

()

Proposition 11.2-5, Conformal Mapping of Laplace’s Equation:

has a zero of order n at z0 , the

( )

If φ x, y is a harmonic function over a set of points S in the z-

mapping w = f z will magnify all angles at z0 by the factor n .

plane, then the conformal mapping of S with the analytic

■

() over which φ ( u, v ) is harmonic:

function w = f z will result in a set of points Ω in the w-plane Example 11.2-3 Consider two straight lines intersecting at the origin with the angle between the lines being Δθ . Let one of the lines lie along the x-axis. What is the angle Δφ between the lines as a

()

2

result of the mapping w = f z = z ? Solution:

∂2φ ∂2φ + 2 = 0! 2 ∂u ∂v

! Proof: !

(11.2-33)

()

The conformal mapping w = f z

( )

will transform the

( ( ) ( ))

harmonic function φ x, y into the function φ x u, v , y u, v . 565

!

To determine the form of the harmonic equation:

∂2φ ∂2φ + = 0! ∂x 2 ∂y 2

!

(11.2-34)

∂φ ∂φ ∂u ∂φ ∂v ! = + ∂x ∂u ∂x ∂v ∂x

(11.2-35)

and (11.2-36)

We then have:

∂ φ ∂ ⎛ ∂φ ⎞ ∂u ∂φ ∂ u ∂ ⎛ ∂φ ⎞ ∂v ∂φ ∂ v ! (11.2-37) = + + ⎜ ⎟ + ⎜ ⎟ ∂x 2 ∂x ⎝ ∂u ⎠ ∂x ∂u ∂x 2 ∂x ⎝ ∂v ⎠ ∂x ∂v ∂x 2 2

!

2

2

∂2φ ∂2φ ⎛ ∂u ⎞ ∂2φ ⎛ ∂v ⎞ ⎛ ∂2φ ∂2φ ⎞ ∂u ∂v ! = ⎜ ⎟ + 2 ⎜⎝ ∂x ⎟⎠ + ⎜ ∂u ∂v + ∂v ∂u ⎟ ∂x ∂x ∂x 2 ∂u 2 ⎝ ∂x ⎠ ∂v ⎝ ⎠ ∂φ ∂2 u ∂φ ∂2 v ! (11.2-41) + + 2 2 ∂u ∂x ∂v ∂x

! and

∂φ ∂φ ∂u ∂φ ∂v ! = + ∂ y ∂u ∂ y ∂v ∂ y

!

obtain: 2

when mapped to the w-plane, we calculate the derivatives: !

Using these operators on equations (11.2-37) and (11.2-38), we

2

2

2

∂2φ ∂2φ ⎛ ∂u ⎞ ∂2φ ⎛ ∂v ⎞ ⎛ ∂2φ ∂2φ ⎞ ∂u ∂v ! = ⎜ ⎟ + 2 ⎜⎝ ∂ y ⎟⎠ + ⎜ ∂u ∂v + ∂v ∂u ⎟ ∂ y ∂ y ∂ y 2 ∂u 2 ⎝ ∂ y ⎠ ∂v ⎝ ⎠ ∂φ ∂2 u ∂φ ∂2 v ! (11.2-42) + + ∂u ∂ y 2 ∂v ∂ y 2

!

Adding equations (11.2-41) and (11.2-42), we then have: and

∂2φ ∂ ⎛ ∂φ ⎞ ∂u ∂φ ∂2 u ∂ ⎛ ∂φ ⎞ ∂v ∂φ ∂2 v ! ! (11.2-38) = + + ⎜ ⎟ + ⎜ ⎟ 2 2 2 ∂ y ⎝ ∂u ⎠ ∂ y ∂u ∂ y ∂ y ⎝ ∂v ⎠ ∂ y ∂v ∂ y ∂y Equations (11.2-35) and (11.2-36) provide the operational forms: !

∂ ∂u ∂ ∂v ∂ ! = + ∂x ∂x ∂u ∂x ∂v

(11.2-39)

and !

∂ ∂u ∂ ∂v ∂ ! = + ∂ y ∂ y ∂u ∂ y ∂v

(11.2-40)

2 2 ∂2φ ∂2φ ∂2φ ⎡ ⎛ ∂u ⎞ ⎛ ∂u ⎞ ⎤ ∂2φ ! + 2 = 2 ⎢⎜ ⎟ +⎜ ⎟ ⎥+ 2 2 ∂x ∂y ∂u ⎢⎣ ⎝ ∂x ⎠ ⎝ ∂ y ⎠ ⎥⎦ ∂v

⎡ ⎛ ∂v ⎞ 2 ⎛ ∂v ⎞ 2 ⎤ ⎢⎜ ⎟ +⎜ ⎟ ⎥ ⎢⎣ ⎝ ∂x ⎠ ⎝ ∂ y ⎠ ⎥⎦

!

⎡ ∂2φ ∂2φ ⎤ ⎛ ∂u ∂v ∂u ∂v ⎞ +⎢ + + ⎥⎜ ∂u ∂v ∂v ∂u ∂x ∂x ∂ y ∂ y ⎟⎠ ⎝ ⎣ ⎦

!

∂φ + ∂u

⎛ ∂2 u ∂2 u ⎞ ∂φ ⎜ 2 + 2 ⎟ + ∂v ∂y ⎠ ⎝ ∂x

()

is analytic, u and v follow the Cauchy-

( )

Since w u, v = f z

⎛ ∂2 v ∂2 v ⎞ ⎜ 2 + 2 ⎟ ! (11.2-43) ∂y ⎠ ⎝ ∂x

Riemann equations. Therefore: 566

!

∂u ∂v ∂u ∂v ∂u ∂u ∂u ∂u + =− + = 0! ∂x ∂x ∂ y ∂y ∂x ∂y ∂y ∂x

(11.2-44)

()

and since the mapping is conformal, f ′ z ≠ 0 . Therefore: !

We also know from Proposition 6.9-1 that u and v are harmonic conjugates. Therefore we have: !

∂2 u ∂2 u + = 0! ∂x 2 ∂ y 2

!

∂2 v ∂2 v + =0! ∂x 2 ∂ y 2

(11.2-45)

!

⎡ ⎛ ∂v ⎞ 2 ⎛ ∂v ⎞ 2 ⎤ ⎢⎜ ⎟ +⎜ ⎟ ⎥ ⎢⎣ ⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎥⎦ (11.2-46)

(11.2-47)

From equation (3.4-25) we have: !

∂2φ ∂2φ + = f′ z ∂x 2 ∂ y 2

()

2⎛

∂2φ ∂2φ ⎞ ⎜ 2 + 2⎟! ∂v ⎠ ⎝ ∂u

(11.2-48)

!

■

11.3! MÖBIUS TRANSFORMATIONS !

()

The Möbius transformation w = T z is a special form of

complex variable mapping that is the ratio of two linear polynomials:

()

w=T z =

az +b ! cz + d

(11.3-1)

where a , b , c , and d are complex constants, and where the determinant of the transformation is not equal to zero: !

a b ≠ 0! c d

or!

a d − bc ≠ 0 !

(11.3-2)

The Möbius transformation is also known as the fractional

From equation (11.2-34) we have:

∂2φ ∂2φ + = 0! ∂x 2 ∂ y 2

() will result in a set of points Ω in the w-plane over which φ ( u, v )

!

2 2 ∂ φ ∂ φ ⎡ ⎛ ∂u ⎞ ⎛ ∂u ⎞ ⎤ ⎡ ∂2φ ∂2φ ⎤ + 2 = ⎢⎜ ⎟ +⎜ ⎟ ⎥ ⎢ 2 + 2 ⎥ 2 ∂x ∂y ∂v ⎦ ⎢⎣ ⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎥⎦ ⎣ ∂u 2

! !

( )

If φ x, y is a harmonic function over a set of points S in

is harmonic.

Using the Cauchy-Riemann equations again: 2

(11.2-50)

the z-plane, then the conformal mapping of S with w = f z

and so equation (11.2-43) becomes: 2 2 ∂2φ ∂2φ ∂2φ ⎡ ⎛ ∂u ⎞ ⎛ ∂u ⎞ ⎤ ∂2φ ! + 2 = 2 ⎢⎜ ⎟ +⎜ ⎟ ⎥+ 2 2 ∂x ∂y ∂u ⎢⎣ ⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎥⎦ ∂v !

∂2φ ∂2φ + = 0! ∂u 2 ∂v 2

linear transformation and the bilinear transformation. It is (11.2-49)

linear in both the variables w and z as can be seen by rewriting equation (11.3-1) in the form: 567

c w z + d w − a z = b!

!

()

(11.3-3)

The derivative T ′ z of the Möbius transformation is:

!

()

w′ = T ′ z =

!

(

) ( ( c z + d )2

a cz + d −c az +b

)

=

a d − bc

(

cz + d

)

2

!

The Möbius transformation then becomes:

()

w=T z =

!

(11.3-4)

az +b az +b a az +b a = = = ! bc c a z + b c cz + d cz + a

and so the Möbius transformation is a simple constant for all z if a d − bc = 0 .

The Möbius transformation is holomorphic at all points in the complex plane at which it exists (all points except z = − d c ). The Möbius transformation is conformal if a d − bc ≠ 0 ,

!

()

since then T ′ z ≠ 0 . If a d − bc = 0 , the Möbius transformation

Proposition 11.3-1, One-to-One Möbius Mapping in the Complex Plane:

will not be conformal, but will become a simple constant for all

z . Because w would then be a constant independent of z , the

Möbius transformations provide one-to-one mappings in the

entire z-plane would be mapped onto the same point in the w-

complex plane if and only if a d − bc ≠ 0 .

plane.

( )

Let w1 = T z1

( )

and w2 = T z2 . We need to show that if

Example 11.3-1

!

Show that the Möbius transformation is a simple constant

w1 = w2 then z1 = z2 only if a d − bc ≠ 0 .

for all z if a d − bc = 0 .

!

Solution: We have:

a d − bc = 0 !

! or !

Proof:

bc ! d= a

!

Let:

( )

w1 = T z1 =

a z1 + b c z1 + d

!

(11.3-5)

and !

( )

w2 = T z2 =

a z2 + b c z2 + d

!

(11.3-6)

568

We will suppose w1 = w2 so that:

a z1 + b

!

c z1 + d

=

a z2 + b c z2 + d

( )

z = T −1 w =

!

!

(11.3-7)

−d w+ b ! cw− a

(11.3-12)

which has the same form as equation (11.3-1). We must then have as in equation (11.3-2):

or

( a z1 + b) ( c z2 + d ) = ( a z2 + b) ( c z1 + d ) !

!

(11.3-8)

!

−d

b

c

−a

≠ 0!

or!

a d − bc ≠ 0 !

(11.3-13)

and so: !

a d z1 + bc z2 + a c z1 z2 + bd = a d z2 + bc z1 + a c z1 z2 + bd

!

!

Since the Möbius transformation function has an inverse, it is a

(11.3-9)

one-to-one mapping of the z-plane onto the w-plane as given in Proposition 11.3-1.

■

Therefore:

( a d − bc ) z1 = ( a d − bc ) z2 !

!

(11.3-10)

Proposition 11.3-2, One-to-One Möbius Mapping in the Extended Complex Plane:

We see that z1 = z2 if and only if we have a d − bc ≠ 0 as given in equation (11.3-2) so that this factor can be canceled out of

Möbius transformations provide one-to-one mappings in the

equation (11.3-10). Möbius transformations then provide one-

extended complex plane.

to-one mappings in the complex plane. !

■

The inverse Möbius transformation z = T

Proof: −1

( w)

can be

obtained by solving equation (11.3-1) for z : ! and so:

c z w+ d w = a z + b!

!

From

Proposition

11.3-1

we

know

that

Möbius

transformations provide one-to-one mappings in the complex plane.

(11.3-11)

!

From equation (11.3-1) we see that the point z = ∞ is

mapped into w = a c :

569

!

( ) ∞ = zlim →∞

w=T z

b a+ z d c+ z

=

The mapping function w = a z + b is clearly analytic, and w is

a ! c

(11.3-14)

into z = − d c :

!

()

f ′ z = a!

!

From equation (11.3-6) we see that the point w = ∞ is mapped

!

entire. The mapping w = a z + b is also conformal if a ≠ 0 since: (11.3-17)

A linear transformation can take the form of either a translation, a scaling, a rotation, or any combination of these

z = T −1

b −d + w −d ! w = lim = c ∞ w→ ∞ a c− w

( )

three changes. (11.3-15) 11.3.1.1!

From equations (11.3-14) and (11.3-15) we see that the

Möbius transformation is a one-to-one mapping of the

!

TRANSLATION

If a = 1 and b ≠ 0 , then equation (11.3-16) represents a pure

translation:

()

T z = z + b!

extended z-plane onto the extended w-plane. From equation

!

(11.3-1) we also see that z = − d c is an isolated singularity of

This transformation produces only a change in position without

the function that is the Möbius transformation. If c = 0 then both z = ∞ and w = ∞ , and the Möbius transformation is a linear mapping (see Section 11.3.1).

any change in orientation or size (see upper diagram in Figure 11.3-1).

■

11.3.1.2!

SCALING

11.3.1! LINEAR TRANSFORMATIONS

!

!

pure scaling:

The general linear transformation is a one-to-one

mapping obtained from the Möbius transformation when c = 0 and d = 1: !

()

w = T z = a z + b!

(11.3-18)

(11.3-16)

!

If a = a and b = 0 , then equation (11.3-16) represents a

()

T z = a z!

(11.3-19)

This transformation produces only a change in size without any change in orientation or position (see middle diagram in Figure 570

11.3-1). The scaling can take the form of a magnification or a contraction (dilation) according as to whether a > 1 or a < 1 .

11.3.1.3! !

ROTATION

If a = e i α and b = 0 , then equation (11.3-16) represents a

pure rotation about the origin through an angle α :

()

T z = e iα z = e

!

iarg ( a )

z!

(11.3-20)

This transformation produces only a change in orientation without any change in position or size (see lower diagram in Figure 11.3-1). This can be seen using the exponential form

z = r eiθ : i θ +α T z = e i α r eiθ = r e ( ) !

()

! 11.3.1.4! !

(11.3-21)

GENERAL EQUATION FOR A LINEAR TRANSFORMATION

Many transformations can be considered as composites of

a number of transformations. For example, a general linear

()

transformation T z

can be treated as three separate

()

()

transformations taken in sequence: T z = T1 ! T2 ! T3 z :

Figure 11.3-1! Linear mappings from the z-plane to the wplane: translation, scaling, and rotation. Upper diagram is a translation, middle diagram is a scaling, and lower diagram is a rotation.

!

1.!

!

2.! 3.

() arg a A rotation: T2 ( z ) = T1 e ( ) . A scaling: T3 ( z ) = T2 a !

A translation: T1 z = z + b a

A general equation for a linear transformation consisting of a translation, scaling, and rotation then is: 571

()

T z = a e

!

iarg ( a ) ⎛

b⎞ z + ! ⎜⎝ a ⎟⎠

(11.3-22)

Linear mappings preserve shapes since they preserve angles. A transformation that is linear can change the position, size, and orientation of a set of points in the complex plane, but cannot

Proof:

()

T2 z = a2 z + b2 where a1 ≠ 0 and a2 ≠ 0 . The composition of their consecutive mapping is given by:

change the basic shape of the set of points. A sequence of two or

!

more transformations will always involve some intermediate

or

complex planes.

!

Proposition 11.3-3, Compositions of Linear Mappings: A linear mappings is the composition of a translation, a rotation, and a scaling.

(T1 !T2 )( z ) = T1 ⎡⎣T2 ( z )⎤⎦ = a1 ( a2 z + b2 ) + b1 !

(11.3-23)

(T1 !T2 )( z ) = a1 a2 z + a1 b2 + b1 !

(11.3-24)

Letting c1 = a1 a2 and c2 = a1 b2 + b1 , we have: !

(T1 !T2 )( z ) = c1 z + c2 !

(11.3-25)

which is a linear mapping. Following the same logic, we see

Proof: !

()

We will consider two linear mappings T1 z = a1 z + b1 and

!

that any number of linear mappings performed successively

Follows from the above discussions.

■

Proposition 11.3-4, Composition of Successive Linear Mappings: The composition of successive consecutive linear mappings is

will be a linear mapping.

■

Example 11.3-2 Determine the image of the unit disk z = 1 under the

()

mapping T z = 2 ei π

also a linear mapping.

2

( z + 3) .

Solution: This transformation can be accomplished in three successive transformations: 572

We next scale the unit disk as shown in the middle diagram of Figure 11.3-2:

()

()

w2 = T2 z = 2 T1 z

!

Finally we rotate the unit disk about the origin as shown in the lower diagram of Figure 11.3-2:

()

()

()

w3 = T3 z = ei π 2 T2 z = iT2 z

!

Example 11.3-3

(

)( )

(

)( )

Compare the mappings T1 !T2 z and T2 !T1 z if

()

()

T1 z = a1 z + b1 and T2 z = a2 z + b2 . Solution: The composition of their consecutive mapping is given by: ! Figure 11.3-2! Linear mapping from the z-plane to the w-plane with the transformation: T z = 2 ei π 2 z + 3 . Upper diagram is a translation, middle diagram is a scaling, and lower diagram is a rotation.

()

(

)

and !

!

()

(T2 !T1 )( z ) = T2 ⎡⎣T1 ( z )⎤⎦ = a2 ( a1 z + b1) + b2 = a1 a2 z + a2 b1 + b2 We see that the mapping can change if the order of the transformation composition changes. If we have:

We first translate the unit disk as shown in the upper diagram of Figure 11.3-2:

(T1 !T2 )( z ) = T1 ⎡⎣T2 ( z )⎤⎦ = a1 ( a2 z + b2 ) + b1 = a1 a2 z + a1 b2 + b1

!

(T1 !T2 )( z ) = (T2 !T1 )( z )

w1 = T1 z = z + 3 573

()

()

then the mappings T1 z and T2 z are said to commute.

2!

(see Example 11.3-4).

11.3.2! INVERSION TRANSFORMATIONS !

Straight lines are mapped into circles or straight lines

The inversion transformation is a one-to-one mapping

3!

Points on the unit circle z = 1 in the z-plane will be

obtained from the Möbius transformation when a = d = 0 and

transformed into points on the unit circle w = 1 in the

b = c so that:

w-plane, and so the unit circle itself is not changed (see Figure 11.3-3).

1 w=T z = ! z

()

!

(11.3-26) 4!

All points that lie inside the unit circle in the z-plane

The mapping resulting from an inversion transformation is

will be transformed into points that lie outside the unit

called inversion mapping or reciprocal mapping.

circle in the w-plane.

!

Letting z = r eiθ and w = ρ eiφ , we obtain:

w = ρ eiφ =

!

1 1 = iθ ! z re

5!

All points that lie outside the unit circle in the z-plane will be transformed into points that lie inside the unit

(11.3-27)

circle in the w-plane, but with the origin deleted.

and so: 6!

1 ρ eiφ = e−iθ ! r

!

(11.3-28)

1 ρ= ! r

!

φ = −θ !

circle inside out. 7!

We then have for an inversion transformation:

The inversion transformation effectively turns a unit

A point on the top half of a unit circle centered at the origin in the z-plane will be reflected onto a point on

(11.3-29)

!

We have the following inversion mapping characteristics:

!

1!

the bottom half of a unit circle in the w-plane across the real axis since φ = − θ .

Circles are mapped into straight lines or circles. 574

Proposition 11.3-5, Composition of Möbius Mappings: Every Möbius transformation is composed of at most four mappings that are linear or inversion in form. Proof: !

We will begin with the general Möbius transformation:

()

w=T z =

!

az +b ! cz + d

(11.3-30)

We will consider two cases: c = 0 and c ≠ 0 . !

For c = 0 we must have d ≠ 0 . We then have:

() (

Figure 11.3-3! Mapping of a unit disk centered at z = 0 using

()

()

f z = T z =1 z . !

()

( )

( )

a function f z = g 1 z is analytic at z = 0 , then g w will be analytic at w = ∞ under the inversion transformation. If the point z = 0 is outside the region to be transformed, the transformed region in the w-plane will be bounded.

(11.3-31)

where:

()

a z! d

!

T1 z =

!

T2 z = z +

Under the inversion transformation, the image of the point

z = 0 will be the point w = ∞ in the extended complex plane. If

)( )

T z = T2 !T1 z !

!

()

b ! d

scaling and rotation!

(11.3-32)

translation!

(11.3-33)

and the mapping is: ! !

()

T z =

a b z+ ! d d

(11.3-34)

For c ≠ 0 we have: 575

() (

)( )

T z = T4 !T3 !T2 !T1 z !

!

(11.3-35)

conformal, and so Möbius transformations are conformal.

where:

d ! c

()

!

T1 z = z +

!

1 T2 z = ! z

translation!

()

T3 z

inversion!

a d − bc ) ( =− z!

()

c

T4 z = z +

!

(11.3-36)

()

T z =

2

a ! c

Möbius transformations map any circle into a circle or straight line, and any straight line into a circle or straight line.

scaling and rotation!

(11.3-38)

Proof: !

translation!

(11.3-39)

If the Möbius transformation takes the form of either a

translation or a scaling and rotation, a circle will clearly be mapped into a circle and a straight line will be mapped into a straight line. If the Möbius transformation takes the form of an

a bc − a d az +b ! + 2 = cz + d c c z+d c

(

)

(11.3-40)

■

inversion, then we can examine its mapping characteristics by considering the equation: !

Proposition 11.3-6, Möbius Mappings are Conformal: Every Möbius transformation is conformal. Proof: !

11.3.3! MÖBIUS MAPPING CHARACTERISTICS

(11.3-37)

and the mapping is: !

■

Proposition 11.3-7:

()

!

linear or inversion in form. Each of these forms of mapping is

From Proposition 11.3-5 we know that every Möbius

transformation is composed of at most four mappings that are

(

)

A x2 + y2 + B x + C y + E = 0 !

(11.3-41)

This equation represents a circle in the z-plane if A ≠ 0 , and a straight line in the z-plane if A = 0 and B and C are not both zero. !

(

Using x = z + z

)

(

2 and y = z − z

)

2i as given in equation

(1.4-12), we can rewrite equation (11.3-41) as: 576

!

Since y = a we have:

B C A z + z+z + z − z + E = 0! 2 2i

(

2

)

(

)

(11.3-42)

or

For the mapping inversion w = 1 z we then have: !

1 A w

2

B ⎛ 1 1⎞ C ⎛ 1 1⎞ + ⎜ + ⎟+ − ⎟ + E = 0! ⎜ 2 ⎝ w w⎠ 2 i ⎝ w w⎠

(11.3-43)

!

!

B C E w + w+ w − w − w + A = 0! 2 2i

(

)

(

)

! (11.3-44)

or using w = u + i v : !

(

)

(11.3-45)

which represents a circle in the w-plane if E ≠ 0 , and a straight line if E = 0 and B and C are not both zero.

(

w − w = 2i a w w = 2i a u 2 + v 2

)

and so: !

E u 2 + v 2 + Bu − C v + A = 0 !

1 1 − = 2i a w w Multiplying by w w :

Multiplying by w w : 2

z − z = 2i a

!

(

− 2i v = 2i a u 2 + v 2

)

Therefore: !

■

⎛ v⎞ u2 + ⎜ v 2 + ⎟ = 0 a⎠ ⎝ Completing the square, we have:

!

Note that a straight line can be considered to be a circle

having an infinite radius.

2

!

⎛ ⎛ 1 ⎞ 1 ⎞ u2 + ⎜ v + ⎟ = ⎜ ⎟ 2a⎠ ⎝ ⎝ 2a⎠

2

Example 11.3-4

and so the image of the line y = a is a circle with radius

Using the Möbius transformation w = 1 z , determine the

1 2 a centered on the v axis in the w-plane.

image of the line y = a .

( )

Solution: 577

or

11.3.4! MÖBIUS MAPPING FIXED POINTS !

The w-plane can be considered to be a modified z-plane. It

then makes sense to ask which points of the z-plane remain

(

!

z=

a−d ±

The transformation known as the identity or unit Möbius

() ()

!

(

)

2

d − a + 4bc 2c

designated as fixed points or invariant points. !

(11.3-47)

and so:

unmoved as the result of a mapping from the z-plane to the wplane so that w = z . Such points that map into themselves are

)

c z2 + d − a z − b = 0 !

!

!

(11.3-48)

If c ≠ 0 this equation has either two distinct solutions, and

transformation w = T z = I z = z will map all points into

so two fixed points exist, or this equation has one distinct

themselves, and so all z ∈! will be its fixed points. A Möbius

solution, and so one fixed point exists. If d − a + 4bc ≠ 0 , then

transformation that is not the identity transformation can have

two fixed points exist. If d − a + 4bc = 0 , then only one fixed

no more than two distinct points that are mapped into

point exists: z = a − d 2 c .

themselves.

!

(

)

(

(

)2

)2

If c = 0 from equation (11.3-46) we see that there is only

(

)

one distinct solution: z = b d − a . Proposition 11.3-8, Fixed Point Theorem:

!

A Möbius transformation has at most two fixed points unless it

() ()

is the identity mapping: w = T z = I z = z .

To find the fixed points we can use equations (11.3-1) and

(11.3-2) to write: !

()

w=T z = z =

az +b ! cz + d

it must then be the identity mapping.

■

Example 11.3-5

Proof: !

If a Möbius transformation has more than two fixed points

1 Determine the fixed points of the transformation T z = . z

()

Solution:

a d − bc ≠ 0 !

(11.3-46)

We have: 578

!

()

T z =

1 az +b = z cz + d

!

plane are mapped into the w-plane, they will be points on the

Therefore: !

a = 0!

boundary of the image region in the w-plane. Möbius

b = 1!

c = 1!

d =0

transformations will always map boundaries into boundaries.

Using equation (11.3-48): !

z=

If three distinct points of the boundary of a region in the z-

Proposition 11.3-9, Three Fixed Point Theorem:

± 4 = ±1 2

Three distinct points z1 , z2 , and z3 in the z-plane can always be mapped into three specified points w1 , w2 , and w3 in the w-

Therefore z = ±1 are the fixed points of the transformation

1 T z = . z

()

plane by a Möbius transformation w = T z that is unique:

()

w − w1 w2 − w3

!

w − w3 w2 − w1

=

z − z1 z2 − z3 z − z3 z2 − z1

!

(11.3-50)

11.3.5! MAPPING INTO SPECIFIED POINTS !

Using the Möbius transformation, it is always possible to

Proof:

map three distinct points in the z-plane into three specific

!

points in the w-plane. This is to be expected since the four

equations:

complex constants a, b, c, d in the Möbius transformation in

Using equation (11.3-1) we can write the following Möbius

az +b ! cz + d

!

w=

!

w1 =

!

w2 =

equation (11.3-1) can be reduced to three by forming ratios of three of the constants to the fourth: !

( a c) z + (b c) ! w = T (z) = z = z + ( d c)

(11.3-49)

Only three conditions are then needed to determine the Möbius transformation, and this transformation is unique.

a z1 + b c z1 + d

!

a z2 + b c z2 + d

!

a d − bc ≠ 0 !

(11.3-51)

a d − bc ≠ 0 !

(11.3-52)

a d − bc ≠ 0 !

(11.3-53)

579

! ! !

!

!

!

w3 =

a z3 + b c z3 + d

a d − bc ≠ 0 !

!

(11.3-54)

We can then write:

( a d − bc ) ( z − z1 ) ! w − w1 = − = c z + d c z1 + d ( c z + d ) ( c z1 + d ) a z1 + b

az +b

w − w3 =

az +b cz + d

w2 − w1 =

−

a z2 + b c z2 + d

a z3 + b c z3 + d

−

=

a z1 + b c z1 + d

( a d − bc ) ( z − z3 ) ! ( c z + d ) ( c z3 + d ) =

( a d − bc ) ( z2 − z1 ) ! ( c z2 + d ) ( c z1 + d )

( a d − bc ) ( z2 − z3 ) ! w2 − w3 = − = c z2 + d c z3 + d ( c z2 + d ) ( c z3 + d ) a z2 + b

a z3 + b

(11.3-55)

!

w − w3

=

w2 − w1 w2 − w3

=

( c z3 + d ) ( z − z1 ) ! ( c z1 + d ) ( z − z3 ) ( c z 3 + d ) ( z 2 − z1 ) ! ( c z1 + d ) ( z2 − z3 )

w − w1 w2 − w3

!

w − w3 w2 − w1

=

z − z1 z2 − z3 z − z3 z2 − z1

!

(11.3-61)

We know that the transformation in equation (11.3-61) is

an identity Möbius transformation from Proposition 11.3-8 (11.3-56)

since the transformation fixes more than two points. We then have: !

(11.3-57)

(11.3-58)

(11.3-59)

() ()

w=T z = I z = z!

(11.3-62)

If we assume that there are two such Möbius transformations:

() ()

(11.3-63)

() ()

(11.3-64)

!

w = T1 z = I z = z !

!

w = T2 z = I z = z !

then using Proposition 11.3-4 we can write: !

(11.3-57) by equation (11.3-58) we get:

w − w1

finally the Möbius transformation: !

Dividing equation (11.3-55) by equation (11.3-56) and equation

!

Dividing equation (11.3-59) by equation (11.3-60) we obtain

() ()

( ( ))

()

T1 z !T2 z = T1 T2 z = T1 z = z !

(11.3-65)

Therefore: !

()

()

(11.3-66)

()

(11.3-67)

T1 z = T2−1 z !

We also have: (11.3-60)

!

()

T2 z = T1 −1 z !

580

and so for all z :

()

!

()

T1 z = T2 z !

!

(11.3-68)

()

The Möbius transformation w = T z is then unique. !

■

z − z1 z2 − z3

!

z − z3 z2 − z1

(11.3-69)

k = 1, 2, 3!

( )

( )

T1 −1 !T2 z k = T1 −1 ⎡⎣T2 z k ⎤⎦ = T1 −1 ⎡⎣ wk ⎤⎦ = z k ! !

!

(11.3-71)

than two fixed points, by Proposition 11.3-8 we must have:

() ()

T1 −1 !T2 z = I z !

!

Proposition 11.3-10, Invariant Möbius Transformations:

Therefore T1 z = T2 z are identical.

()

()

!

(11.3-72)

■

The cross ratio of four points is invariant under the Möbius

!

transformation.

numerator and denominator containing infinity can be set

If a point in equation (11.3-50) is the point at infinity, the

equal to 1. To see why this is so, we will let z1 = ∞ . Equation

Proof: Follows from Proposition 11.3-9 and equation (11.3-61).

■

()

A sufficient condition for two Möbius transformations T1 z

()

()

()

and T2 z to be identical is that T1 z = T2 z for three distinct points: z1 , z2 , and z3 . Proof:

(11.3-50) then becomes: !

Proposition 11.3-11, Identical Möbius Transformations:

!

k = 1, 2, 3

and so T1 −1 !T2 has three fixed points. Since T1 −1 !T2 has more

is known as the cross ratio of the points z , z1 , z2 , and z3 .

!

(11.3-70)

()

! !

( )

We then can write use the inverse of T1 z to write: !

The expression:

( )

wk = T1 z k = T2 z k !

w − w1 w2 − w3

= lim

z − z1 z2 − z3

!

(11.3-73)

z −1 z1 z2 − z3 w − w1 w2 − w3 ! = lim z z → ∞ w − w3 w2 − w1 1 z − z3 2 −1 z1

(11.3-74)

w − w3 w2 − w1

z1→ ∞

z − z3 z2 − z1

or !

We are given: 581

and so:

!

−1 z 2 − z3 = ! w − w3 w2 − w1 z − z3 −1 w − w1 w2 − w3

!

(11.3-75)

with infinite radius). Therefore a circle can always be mapped into a circle. Taking the reciprocals we can write equation (11.3-50) in the form:

or

1 z 2 − z3 z 2 − z3 ! = = w − w3 w2 − w1 z − z3 1 z − z3 w − w1 w2 − w3

!

(11.3-76)

Since boundaries are always mapped into boundaries, this mapping can be accomplished by mapping the upper half z-

w + i 0 −1 z − i i + 1 − ∞ = = z−i w − 1 0 + i z − ∞ i + 1− i

plane boundary points z1 = −1 , z2 = 0 , and z3 = 1 into the boundary points w1 = − i , w2 = 1 , and w3 = i of the unit circle.

or

Using equation (11.3-77) we have:

w − i 1+ i z − 1 0 + 1 = w + i 1− i z + 1 0 − 1

!

and so we have the Möbius transformation:

1− z + i 2i − z

(11.3-77)

Solution:

Using equation (11.3-50) we have:

w=

!

center at the origin.

Solution:

!

z − z1 z2 − z3

half z-plane onto a unit circle in the w-plane having its

and w3 = 1 , respectively.

w−1

z − z3 z2 − z1

Determine the Möbius transformation that maps the upper

z1 = i , z2 = i + 1 , and z3 = ∞ into the points w1 = − i , w2 = 0 ,

( w + i) i = z − i

w − w1 w2 − w3

=

Example 11.3-7

Determine the Möbius transformation that maps the points

!

w − w3 w2 − w1

!

Example 11.3-6

!

Three points define either a circle or a straight line (a circle

or !

w − i 1+ i − z + 1 = w + i 1− i z +1 582

We then have:

a1

2 w z + 2 z = − 2 wi + 2i

!

!

and so:

w=

!

i− z i+ z

(11.3-81)

which is another Möbius transformation. Following the same logic, we see that any number of Möbius transformations performed successively will be a Möbius transformation.

Transformations:

■

The composition of successive Möbius transformations is also a

11.3.7! IDENTICAL MÖBIUS TRANSFORMATIONS

Möbius transformation.

!

Proof: We will consider two Möbius transformations:

()

T1 z =

a1 z + b1 c1 z + d1

!

()

T2 z =

a2 z + b2 c2 z + d2

!

()

()

()

T3 z = T1 !T2 z = T1 ⎡⎣T2 z ⎤⎦ !

Two Möbius transformations:

()

T1 z =

!

(11.3-78)

a1 z + b1 c1 z + d1

()

!

()

T2 z =

a2 z + b2 c2 z + d2

!

(11.3-82)

()

are identical, T1 z = T2 z , if we have for λ ≠ 0 : !

If they are applied in succession, we obtain:

or

a1 a2 + b1 c2 ) z + ( a1 b2 + b1 d2 ) ( ! T3 ( z ) = + d c b d z c a + + c d ( 1 2 1 2) ( 1 2 1 2 )

!

Proposition 11.3-12, Composition of Successive Möbius

!

(11.3-80)

and so:

11.3.6! SUCCESSIVE MÖBIUS TRANSFORMATIONS

!

( ) ( ) T3 ( z ) = = ! a2 z + b2 c1 ( a2 z + b2 ) + d1 ( c2 z + d2 ) c1 + d1 a1 a2 z + b2 + b1 c2 z + d2

c2 z + d2

See the upper diagram of Figure 11.5-2.

!

a2 z + b2 +b c2 z + d2 1

a1 = λ a2 !

b1 = λ b2 !

c1 = λ c2 !

d1 = λ d2 ! (11.3-83)

as can be seen from: (11.3-79)

()

! T1 z =

λ a2 z + λ b2 λ c2 z + λ d2

=

λ ( a2 z + b2 )

λ ( c2 z + d2 )

=

a2 z + b2 c2 z + d2

()

= T2 z !

(11.3-84) 583

Also see Proposition 11.3-11.

11.3.8! SYMMETRIC MÖBIUS TRANSFORMATIONS !

We will now consider two points z1 and z2 located so as to

!

Conversely, if a straight line or any circle through two

points z1 and z2 is orthogonal to a given straight line L , then z1 and z2 are symmetric with respect to L . !

Since a straight line is simply a circle having a radius that

be symmetric with respect to a given straight line L in the

is infinitely long, the line L need not be a straight line, but can

complex z-plane as shown in Figure 11.3-4. The line through

be a circle, and the above discussions will still hold. We are thus

points z1 and z2 will then be perpendicular to the line L . Any

led to the following proposition:

circle passing through both z1 and z2 will have its center on the line L .

Proposition 11.3-13, Symmetry Principle: Let z1 and z2 be any two points that are symmetric with respect to a given straight line Lz or circle Cz in the z-plane. Also let the z-plane be mapped onto the w-plane by the Möbius

()

transformation w = T z , where Lw and Cw are the images of

( )

Lz and Cz , respectively, in the w-plane. Then w1 = T z1 and

( )

w2 = T z2

will be two points symmetric with respect to the

images Lw and Cw in the w-plane. Proof: ! ! Figure 11.3-4! Two points z1 and z2 symmetric to a line L and on a circle C that passes through z1 and z2 .

We can write using the inverse Möbius transformation:

( )

z1 = T −1 w1 !

( )

z2 = T −1 w2 !

(11.3-85)

and !

( )

Cz = T −1 Cw !

(11.3-86) 584

( )

Now let Cw be a circle in the w-plane passing through the

for any given point z1 , there is only one point z2 = T −1 w2

two points w1 and w2 . The inverse image Cz of Cw in the z-

symmetric to z1 with respect to a given straight line or circle L .

!

()

plane under the Möbius transformation w = T z

must be a

straight line or circle that passes through both z1 and z2 . Since

■

these points are symmetric with respect to Cz , however, we

11.4! AUTOMORPHISMS OF A UNIT DISK

must have Cz orthogonal to Lz where it intersects Lz . Since

!

Möbius transformations are conformal, Cw must then also be

as an automorphism. An example of an automorphism of a

orthogonal to Lw . Therefore w1 and w2 will be symmetric with

unit disk is provided by Schwarz’s lemma, which pertains to

respect to Lw and Cw in the w-plane.

the mapping of an open unit disk onto a closed unit disk by a

■

Any conformal mapping of a region onto itself is known

()

holomorphic complex function f z . !

We see that linear transformations will always preserve

symmetry.

Proposition 11.4-1, Schwarz’s Lemma:

For any given point z1 , there is only one point z2 symmetric to

z1 with respect to a given straight line or circle L .

!

if

and

Proof: !

( ) is holomorphic in the open unit disk D1 (0) , and f ( 0 ) = 0 and f ( z ) ≤ 1 for all z ∈D1 ( 0 ) , then we have: f (z) ≤ z ! for z < 1 ! (11.4-1)

If f z

Proposition 11.3-14:

This proposition is obviously true when L is a straight

!

line. If L is a circle, it can always be transformed into a straight

()

( )

( )

will be symmetric to w1 = T z1 . Since the

Möbius transformation is one-to-one (see Proposition 11.3-1),

(11.4-2)

If equations (11.4-1) and (11.4-2) are equalities, then there exists

line using a Möbius transformation w = T z . Then only one point w2 = T z2

()

f ′ 0 ≤ 1! a θ ∈! such that:

!

()

f z = z eiθ !

for z < 1 !

(11.4-3) 585

Proof:

() open unit disk D1 ( 0 ) into the closed unit disk D1 ( 0 ) . Since f ( z ) is holomorphic in the open unit disk D1 ( 0 ) , and since f ( 0 ) = 0 , we can represent f ( z ) by a Maclaurin series: f ′′ ( 0 ) ! f ( z ) = f ′ (0) z + z + !! z < 1! (11.4-4) 2! !

()

Since f z ≤ 1 for all z < 1 , we see that f z maps the

2

Let

()

()

0 < z 0 we will have f ′ z ≠ 0 . Therefore

(

the mapping function w − w1 = z − x1

)α

1

π

()

the polygonal region shown in Figure 11.6-4. The function T z

is closely related to the mapping function given in equation (11.6-5).

is conformal for all z

where y > 0 .

Figure 11.6-3! Mapping of the upper half-plane y ≥ 0 by the

(

complex power function w − w1 = z − x1 !

)α

1

π

Figure 11.6-4! Polygonal region in the w-plane showing ! external angles and the vertices corresponding ! to points on the real axis of the z-plane.

.

()

We will now consider another mapping function T z that

is analytic for y > 0 and that maps the upper half-plane onto

! !

()

We will begin by specifying the derivative of T z to be:

()

(

T ′ z = A z − x1

)α

1

π −1

( z − x2 )α

2

π −1

!

(11.6-9) 592

where x1 , x2 , α 1 , and α 2 are all real with x1 < x2 , and where A is

()

complex. We will use a parameterization z t

Equation (11.6-11) then becomes:

to describe

intervals on the real axis in the z-plane. With this

()

( ( ))

()

!

parameterization the mapping w = T z becomes w t = T z t on the real axis. !

() α w′ ( t ) = A ( t − x1 )

We will let z t = t so that equation (11.6-9) becomes: 1

!

π −1

(t − x2 )α

2

π −1

!

( ( )) must have a constant value for each line segment of

Arg w′ t

the polygonal region. From equation (11.6-10) we can determine

()

the argument of w′ t :

( ( ))

!

( )

()

(

)

(

( ( ))

)

remains constant. When

( ( )) changes, w will head in a new direction and begin to

Arg w′ t

( ( ))

Arg w′ t

x1 −∞

( ) (

) (

)

= Arg A + α1 − π + α 2 − π !

!

(11.6-12)

(11.6-13)

!

Proceeding similarly, for the interval x1 < t < x2 we have:

( ( ))

Arg w′ t

x2 x1

( ) (

)

= Arg A + α 2 − π !

( ( ))

which is a change in the argument Arg w′ t

As z t = t moves along the x-axis, w will trace out a line

segment as long as Arg w′ t

( )

which is a constant value.

The slope of any given line segment must be constant and so

⎛α ⎞ ⎛α ⎞ ! Arg w′ t = Arg A + ⎜ 1 − 1⎟ Arg t − x1 + ⎜ 2 − 1⎟ Arg t − x2 ⎝π ⎠ ⎝ π ⎠ ! ! (11.6-11)

( ( ))

or !

(11.6-10)

⎛ α1 ⎞ ⎛ α2 ⎞ Arg w′ t = Arg A + ⎜ − 1⎟ π + ⎜ − 1⎟ π ! ⎝π ⎠ ⎝ π ⎠

! !

(11.6-14) x1 −∞

of π − α1 .

For the interval L2 : x2 < t < ∞ we have:

( ( )) x = Arg ( A) !

Arg w′ t

∞

(11.6-15)

2

( ( ))

which is a change in the argument Arg w′ t !

x2 x1

of π − α 2 .

Changes in the argument represent changes in direction

trace out a new line segment. In this way a polygonal region is

of the line segments, and so are the exterior angles for the

formed in the w-plane.

polygonal region. The interior angles are therefore α1 and α 2

!

We will first examine the interval L1 : − ∞ < t < x1 . We have:

(

)

(see Figure 11.6-4). !

!

t − x1 is negative so Arg t − x1 = π

!

t − x2 is negative since x1 < x2 so Arg t − x2 = π

(

)

By following the same procedure as described above, we

can map a set of points x1 , x2 , x3 , !, xn (where x1 < x2 < ! < xn ) into a polygonal region of the w-plane. The transformation 593

( )

()

wj = T xj will map the points xj to the vertices w1 , w2 , w3 , !, wn

elementary functions. In these cases T ′ z must be integrated

of the polygonal region. The indices of the vertices are specified

numerically.

such that they increase counterclockwise around the polygonal

!

region. An interval xj − xj+1 on the real axis is mapped into the

region is known to be 2 π :

line segment connecting the wj and wj+1 vertices. We then have

!

the following proposition.

The sum of the exterior angles of any bounded polygonal

(π − α1 ) + (π − α 2 ) +!+ (π − αn ) = 2 π !

(11.6-17)

Therefore we have for the exponents of equation (11.6-16): Proposition 11.6-1:

()

If T z is analytic in the upper half of the z-plane and has the derivative:

()

(

! T ′ z = A z − x1

)(α

1

π )−1

( z − x2 )(α

2

π )−1

(

! z − xn

)(α

n

π )−1

⎛ α1 ⎞ ⎛ α 2 ⎞ ⎛ αn ⎞ − 1 + − 1 +!+ ⎜⎝ π ⎟⎠ ⎜⎝ π ⎟⎠ ⎜⎝ π − 1⎟⎠ = − 2 !

!

(11.6-18)

If the polygonal region is bounded, the Schwarz-Christoffel ! (11.6-16)

where x1 < x2 < ! < xn , 0 < α j < 2 π , j = 1, 2, !, n , and A is a complex constant, then the upper half of the z-plane is mapped

()

by w = T z onto a polygonal region P of the w-plane having

transformation need only include n − 1 of the n interior angles of the region since only n − 1 of the n interior angles are independent. Proposition 11.6-2, The Schwarz-Christoffel Transformation:

interior angles α j where j = 1, 2, !, n .

()

If T z is analytic in the upper half of the z-plane and has the derivative:

Proof:

()

!

Follows from the procedures described above.

!

The function T z can be determined by integrating T ′ z

()

■

()

()

)(α

1

π )−1

( z − x2 )(α

2

π )−1

(

! z − xn

)(α

n

π )−1

! (11.6-19)

then we have where A, B ∈! :

as given in equation (11.6-16). The desired transformation is then the inverse of T z . For many polygonal regions no

!

antiderivative of T ′ z exists that can be expressed in terms of

! !

()

(

! T ′ z = A z − x1

()

T z =A

(α z − x1 ) ( ∫

1

π )−1

( z − x2 )(α

2

π )−1

(

! z − xn

)(α

n

π )−1

dz + B

(11.6-20) 594

Proof: !

Follows directly from equation (11.6-16).

■

Example 11.6-1 Determine the function that maps the upper half-plane

Im z > 0 onto the polygonal region having the form of a semi-infinite strip where − π 2 < w < π 2 . Solution: Let x1 = −1 , x2 = 1 , w1 = − π 2 , and w2 = π 2 so that α1 = π 2 and α 2 = π 2 as shown in Figure 11.6-5. We then have:

()

(

T ′ z = A z − x1

!

)α

1

π −1

( z − x2 )α

2

π −1

(

)−1 2 ( z − 1)−1 2

= A z +1

or !

()

T′ z =

(

A

Figure 11.6-5! Polygonal region in the form of a semi-infinite strip.

)

z2 − 1

12

()

The function T z is then given by: !

!

()

T z = Ai arcsin z + B We can use boundary conditions to determine A and B :

( )

()

T −1 = − π 2 and T 1 = π 2 . We then have:

−π − iπ =A +B 2 2 and

!

π iπ = A +B 2 2 Therefore B = 0 and A = − i , and so the function is: 595

!

()

T z = arcsin z

Example 11.6-2 Determine the function that maps the upper half-plane

Im z > 0 onto the polygonal region having the form of an equilateral triangle with two of its vertices at w = 0 and

w = 1. Solution: Since the polygonal region is equilateral, all of its interior angles are equal: !

α1 = α 2 = α 3 =

π 3

as shown in Figure 11.6-6. Let x1 = 0 , x2 = 1, w1 = 0 , and

w2 = 1 . We then have: !

()

(

T ′ z = A z − x1

)α

1

π −1

( z − x2 )α

2

π −1

()

The function T z is then given by: !

()

T z =A

∫

z

0

1 s

23

( s − 1)

23

(

)− 2 3

= A z−2 3 z − 1

ds + B

This integral can be solved numerically.

Figure 11.6-6! Polygonal region in the form of an equilateral triangle.

11.7! RIEMANN SURFACES !

A Riemann surface is a three-dimensional construction

designed to geometrically represent a multivalued function

()

()

w = f n z as a one-to-one function w = f z . A Riemann surface then provides a way to view a multivalued function as if it were a single-valued function. A Riemann surface makes it 596

possible to pass continuously from one branch of a function to another. !

Riemann surfaces consist of two or more levels or sheets

w2 = r

!

12 i

e

θ +2 π 2

= − r 1 2 ei θ 2 = − w1 !

(11.7-3)

That is, one point on the z-plane is mapped into two points on

called Riemann sheets, and so are not confined to a single

the w-plane.

complex plane. Each sheet corresponds to a branch of the

!

multivalued function, and each sheet covers an entire finite

traveling only from 0 to π . It then requires two branches for

complex plane. Adjacent sheets in a Riemann surface are

w = z to completely map z onto w : one from 0 ≤ φ ≤ π for

connected along the branch cut so that the upper lip of the cut

which w is positive and one from π ≤ φ ≤ 2 π for which w is

in the n − 1 sheet is attached to the lower lip of the n th sheet. In a

negative. These two branches are represented by two sheets in a

stack of interconnected Riemann sheets, the last sheet in the

Riemann surface (see Figure 11.7-1).

stack is reconnected to the first sheet. The origin is removed

!

from all the sheets. This construction of a Riemann surface is

upper sheet, after which a transfer down a ramp to the second

motivated by the fact that the value of a function f n z just

sheet occurs and the second circuit about w = 0 from π ≤ φ ≤ 2 π

above a branch cut is identical to the value of the function

is made. A third circuit will cause a transfer back up a ramp to

()

()

We will now consider the Riemann sheet for the complex

square root function. As noted in Chapter 4, for the function: !

()

w= f z = z !

The first circuit about w = 0 from 0 ≤ φ ≤ π occurs on the

the first sheet. After any complete circuit about w = 0 , a transfer

fn +1 z just below the branch cut. !

As z makes one complete circle from 0 to 2 π we have w

(11.7-1)

is made to another sheet. !

The dark line in Figure 11.7-1 is a branch cut which

extends from r = 0 to r = ∞ along the positive real axis. The branch cut exists in both sheets. The branch cut in the upper

where we choose a branch cut along the positive real axis, two

sheet is connected with the branch cut in the lower sheet at

values of w correspond to one value of z :

φ = 2 π . The branch cut in the lower sheet is also connected with

!

w1 = r eiθ = + r 1 2 eiθ 2 !

(11.7-2)

the branch cut in the upper sheet at φ = 0, 4 π . The function

()

w = f z = z thereby becomes single-valued. The dark circle in 597

the center of the sheets in Figure 11.7-1 is the branch point

has an infinity of sheets. Writing the complex logarithm

which is deleted.

function in the form: !

(

)

ln z = log e r + i θ + 2 k π !

k = 0, ± 1, ± 2,! !

(11.7-4)

we see that z is a point on the k th sheet.

Figure 11.7-1! Mapping of w = z into two sheets in the wplane. !

For complex functions having more than two branches,

more than two sheets are required for their Riemann surface: one for each branch. For some multivalued functions, their Riemann surface can consist of an infinite number of levels. For example, the Riemann surface for a complex logarithm function 598

Chapter 12 k

Infinite Products

Pk =

1+ b ( ∏ n) n =1

599

!

In this chapter we will define and review infinite products

of complex numbers and functions. We will show the close relation of infinite products to infinite series, and we will present some convergence conditions for infinite products.

()

states that every non-constant polynomial Pn z of finite degree

n ≥ 1: !

()

Pn z = a0 + a1 z + a2 z +!+ an−1 z

n−1

n

+ an z !

() (

Pn z = c z − z1

) ( z − z2 ) ( z − z3 ) ! ( z − zn ) !

n

!

Since it is not possible to multiply an infinity of terms, it is

sequence must be determined in the form of a limit. To obtain a limit that can represent the product of an infinite complex sequence, partial products of the sequence are used.

(12.0-1)

The product of the first k terms of an infinite sequence

{ an } is known as the k th partial product Pk

! (12.0-2)

Pk =

∏a ! n =1

!

P1 = a1

analytic functions as products, but rather than having a finite

2

number of factors, they will have infinitely many factors.

!

P2 =

12.1! INFINITE PRODUCT OF COMPLEX NUMBERS

!

!

is an infinite sequence of complex numbers or of

complex functions, then their infinite product is given by:

(12.1-2)

n

complex constant. It is also possible to similarly represent other

{ }

of the sequence:

k

We then have:

If an

(12.1-1)

3

where an is the nth term or factor of the product.

where the zeros zi can be complex numbers, and where c is a

!

1 2

n =1

!

can be completely factored so that: !

∏ a = a a a !!

complex sequence. Instead the product of an infinite complex

The fundamental theorem of algebra (Proposition 6.8-1)

2

!

not possible to directly obtain the product of an infinite

12.0! FINITE PRODUCT OF COMPLEX NUMBERS !

∞

!

∏a =a a ! n

1 2

n =1

(12.1-3)

! k

!

Pk =

∏ a = a a a !a n

1 2

3

k

n =1

600

!

For any given infinite sequence, the set of partial products

forms a sequence of partial products

{ Pk } .

This sequence

{ Pk } = P1 , P2 , P3 ,!, Pk !

product if some terms of the product are equal to zero. An infinite product having no terms equal to zero can still

consists of all the terms Pk as defined in equation (12.1-2): !

P = 0 provides essentially no information about the infinite

(12.1-4)

{ } has a finite limit, then

converge to zero, although a finite product cannot. !

If P = 0 the sequence of partial products is generally said

If the sequence of partial products Pk

to diverge to zero. An infinite product can then diverge by

the infinite product given in equation (12.1-1) is said to

having a limit of 0 , ∞ , or by continuously oscillating.

converge to a product (see Section 12.2). An infinite product is

!

product converges when the initial part of the product

therefore treated as a sequence of partial products.

containing these zero factors is removed, then the product is

12.2! CONVERGENCE AND DIVERGENCE OF INFINITE PRODUCTS !

{ }

A sequence of partial products Pk

is said to converge to

a limit P if we have:

defined to be convergent. In this case there must exist some minimal index n0 such that an ≠ 0 for all n ≥ n0 . In the following discussions we will assume that any zero factors existing in the initial part of a product have been removed.

k

!

lim P = lim

k→∞ k

k→∞

∏ a = P! n

Example 12.2-1 (12.2-1)

For a sequence of numbers:

n =1

{ }

where P ≠ 0 and P ≠ ∞ . The sequence of partial products Pk

is then called convergent, and P is taken as the limit of the infinite product. !

If a product has a finite number of zero factors, and if the

Only one term Pk of an infinite product need be zero for

!

{ an } = ⎧⎨ 1n ⎫⎬ ⎩ ⎭

{ } converges

determine if its sequence of partial products Pk to a limit.

the product to be zero. Therefore the removal of a few terms can change the result of an infinite product. Moreover, a limit of

Solution: 601

The partial product Pk is: k

!

Pk =

∏ n =1

!

1 111 1 = ! n 12 3 k

()

P = lim Pk = lim k = ∞ k→∞

k→∞

{ } diverges since

and so the sequence of partial products Pk it has a limit of ∞ .

We then have: !

⎛ 1⎞ P = lim Pk = lim ⎜ ⎟ = 0 k→∞ k→∞ ⎝ k ⎠

Example 12.2-3

{ } converges even

and so the sequence of partial products Pk

though it has a limit of 0 since no factor equals 0.

For a sequence of numbers: !

{ an } = {( −1)n }

{ } converges

determine if its sequence of partial products Pk

!

Example 12.2-2

to a limit.

For a sequence of numbers:

Solution:

{ an } = {n}

The partial product Pk is:

{ } converges

determine if its sequence of partial products Pk to a limit.

!

Pk =

n k −1 = −1 1 −1 ! −1 ( ) ( ) ( ) ( ) ( ) ∏ n =1

Solution:

The partial product oscillates between −1 and 1 , and does

The partial product Pk is:

not converge to a limit. Therefore the sequence of partial

k

!

k

Pk =

∏ n = (1)(2)(3) ! k

{ } diverges.

products Pk

n =1

We then have: 602

Example 12.2-4

12.2.1! CONVERGENCE REQUIREMENTS

For a sequence of numbers:

{ an } = {1+ ( −1)n−1 n }

!

Proposition 12.2-1, Product Convergence Requirement:

{ an } ,

For a sequence of complex numbers or functions

{ } converges

{ }

determine if its sequence of partial products Pk

the

sequence of its partial products Pk , where Pk > 0 for all k ,

to a limit.

will converge to a finite nonzero limit P if:

lim a = 1 !

!

Solution:

(12.2-2)

k→∞ k

The partial product Pk is: k

! Pk =

∏ n =1

( )

( )

n−1 k−1 ⎛ ⎛ −1 ⎞ ⎛ 1⎞ ⎛ −1 ⎞ 1 ⎞⎛ 1⎞ ⎜ 1+ ⎟ = 1+ ⎟ 1− ⎟ ⎜ 1+ ⎟ ! ⎜ 1+ n ⎟ ⎜⎝ 1⎟⎠ ⎜⎝ 2 ⎠⎝ 3⎠ k ⎟ ⎜⎝ ⎜ ⎠ ⎝ ⎠

We will let Pk and Pk−1 equal the partial products: k

Pk =

k−1

∏a !

Pk−1 =

n

n =1

k

Pk =

∏ n =1

( )

n−1 ⎛ −1 ⎞ 1 4 3 6 5 ⎛ 2 k ⎞ ⎛ 2 k − 1⎞ ⎜ 1+ ⎟=2 ! n ⎟ 2 3 4 5 6 ⎜⎝ 2 k − 1⎟⎠ ⎜⎝ 2 k ⎟⎠ ⎜⎝ ⎠

k

Pk =

∏ n =1

( )

n−1 ⎛ −1 ⎞ ⎜ 1+ ⎟ =1 n ⎜⎝ ⎟⎠

∏a ! n

(12.2-3)

n =1

If the sequence of partial products

{ Pk }

converges to a limit

P ≠ 0 as k → ∞ , we will have: !

Therefore: !

! !

or !

Proof:

lim P = lim Pk −1 = P !

k→∞ k

(12.2-4)

k→∞

We also have by definition of a partial product: k−1

{ } converges.

!

and the sequence of partial products Pk

Pk = ak

∏a = a P n

k k−1

!

(12.2-5)

n =1

and so: 603

Solution:

P ak = k ! Pk−1

!

(12.2-6)

Therefore using equation (12.2-4):

lim P Pk P k→∞ k lim ak = lim = = = 1! k→∞ k→∞ P lim Pk−1 P k−1

!

1.! We see by canceling adjacent terms that:

(12.2-7)

!

k→∞

■

!

Pk =

!

!

( ) f3 ( z ) f 4 ( z ) ! f1 ( z ) f 2 ( z ) f3 ( z )

f2 z

Therefore:

P = lim Pk = lim k→∞

k→∞

The converse to Proposition 12.2-1 is not true. It is possible

{ }

to have the k th term ak of the sequence of partial products Pk

{ }

converge to unity and yet not have Pk

!

Pk =

k→∞

convergence to a finite nonzero limit.

{ }

!

for the following:

k→∞

2.!

( ) ⎫⎪ ⎬ ( ) ⎪⎭

{an }

( ) ⎫⎪ ⎬ ( ) ⎪⎭

⎧⎪ f n z =⎨ ⎪⎩ f n+1 z

f1 z

Example 12.2-6

⎧⎪ f n+1 z an = ⎨ ⎪⎩ f n z

{ }

( ) = f1 ( z ) f k+1 ( z ) f k+1 ( z ) fk z

( ) = f z lim 1 1( ) k→∞ k→∞ f z f k+1 ( z ) k+1 ( )

P = lim Pk = lim

Determine the limit P of a sequence of partial products Pk

1.!

( ) f 2 ( z ) f3 ( z ) ! f 2 ( z ) f3 ( z ) f 4 ( z ) f1 z

Therefore:

Example 12.2-5

( ) = 1 lim f z k+1( ) f1 ( z ) f1 ( z ) k → ∞

f k+1 z

2.! We see by canceling adjacent terms that:

converge. Therefore

lim ak = 1 represents a necessary but not sufficient condition for

( ) = fk+1 ( z ) fk ( z ) f1 ( z )

f k+1 z

For a sequence of complex numbers: !

{ an } = ⎧⎨ n n+ 1 ⎫⎬ ⎩

⎭

{ } converges

determine if its sequence of partial products Pk to a limit.

604

Solution:

k

k

!

Pk =

∏ n =1

n +1 2 3 4 k +1 = ! n 1 2 3 k

k→∞

(

Proof:

)

!

{ }

diverges. We

If we have: k

!

also see that:

∏ a −1 < ε ! n

for all k > m > N !

(12.2-9)

n = m+1

k +1 =1 k→∞ k

then since:

lim ak = lim

k→∞

(12.2-8)

products.

and so the sequence of partial products Pk

!

for all k > m > N !

which is the Cauchy convergence criteria for infinite

P = lim Pk = lim k + 1 = ∞ k→∞

n

n = m+1

From Example 12.2-5 we have: !

∏ a −1 < ε !

!

The partial product Pk is:

showing that while lim ak = 1 is a necessary condition for a k→∞

sequence of partial products

{ Pk }

to converge to a finite

!

! Proposition 12.2-2, Cauchy Criteria for Infinite Products:

{ }

{ an } , the

sequence of its partial products Pk , where Pk > 0 for all k ,

∏a ! n

for all k > m !

(12.2-10)

for all k > m > N !

(12.2-11)

for all k ≥ N !

(12.2-12)

n = m+1

Pk −1 < ε ! Pm

For a given N let: !

Qk =

will converge to a finite nonzero limit if and only if for any real number ε > 0 there exists N ∈! such that:

k

we obtain:

nonzero limit, it is not a sufficient condition.

Given a sequence of complex numbers or functions

Pk = Pm

Pk ! PN

We then have: 605

!

Qk − 1 < ε !

for all k ≥ N !

(12.2-13)

or for some N > 0 : !

k

∏a −1 < ε !

!

n

(12.2-14)

{ } converges to a finite nonzero limit, let:

Conversely, if Pk

!

We can write equation (12.2-11) as: !

Pm PN

for all k > m > N !

(12.2-15)

for all k > m > N !

(12.2-16)

and so: !

for all k > m > N !

or

Pk −1 < ε ! Pm

(12.2-17)

Qk − Qm < 2 ε !

for all k > m > N !

(12.2-18)

Pk − Pm < 2 ε !

(12.2-23)

for all k > m > N !

(12.2-24)

k

∏ a −1 < ε !

!

n

n = m+1

We then have: !

for all k > m > N !

and so:

Using equation (12.2-14) we obtain: !

(12.2-21)

(12.2-22)

k→∞

!

Qk − Qm < Qm ε !

for all k > m !

lim Qk = 1!

!

Qk −1 < ε ! Qm

Pk ! Pm

Then from Proposition 12.2-1 we must have:

or !

Qk =

!

−1 < ε !

(12.2-20)

n = m+1

1− ε < Qk < 1+ ε < 2 !

Pk PN

for all k > m > N !

for all k > m > N !

and so the sequence of partial products

{ Pk }

■

(12.2-19) is a Cauchy

Proposition 12.2-3, Cauchy Criteria for Infinite Products:

{ an } , the

sequence, and so converges to a finite nonzero limit if for any

Given a sequence of complex numbers or functions

real number ε > 0 there exists N ∈! such that:

sequence of its partial products Pk , where Pk > 0 for all k ,

{ }

606

will converge to a finite nonzero limit if and only if for any real

!

number ε > 0 there exists N ∈! such that:

an = 1+ bn makes it possible to rewrite Proposition 12.2-1 in a

Pk −1 < ε ! Pm

!

for all k > m > N !

Writing the general term for complex products in the form

form similar to Proposition 8.2-1 for complex series. (12.2-25)

12.2.3! CONVERGENCE REQUIREMENTS

which is the Cauchy convergence criteria for infinite products. Proposition 12.2-4, Product Convergence Requirement:

Proof: !

(12.2-11).

{

}

Given a sequence of complex numbers or functions 1+ bk , the

Follows from Proposition 12.2-2 as shown in equation

sequence of its partial products

■

{ Pk }

where Pk > 0 for all k ,

will converge to a finite nonzero limit P if:

12.2.2! ANOTHER FORM OF INFINITE PRODUCTS !

Proof:

∞

!

∞

∏a =∏(

)

1+ bn !

n

n =1

(12.2-26)

n =1

! !

where:

We have:

(

Pk = 1+ bk

k−1

) ∏ (1+ bn ) = (1+ bk ) Pk−1 !

(12.2-30)

n =1

an = 1+ bn !

(12.2-27)

The k th partial product then becomes:

Pk =

∏ (1+ b ) ! n

n =1

Therefore: !

k

!

(12.2-29)

k→∞ k

Infinite products of complex numbers are often expressed

in the form:

!

lim b = 0 !

!

(12.2-28)

bk =

Pk − 1! Pk−1

(12.2-31)

We also have:

607

lim P = lim Pk −1 = P !

!

k→∞ k

(12.2-32)

k→∞

and so:

⎞ Pk ⎛ Pk ⎞ ⎛ klim ⎛P ⎞ →∞ lim bk = lim ⎜ − 1⎟ = ⎜ − 1⎟ = ⎜ − 1⎟ = 0 k→∞ k→∞ ⎝ P ⎠ ⎜⎝ lim Pk−1 ⎟⎠ ⎝ P ⎠ k−1 k→∞

! !!

Proof: !

Follows from Proposition 12.2-2 and equation (12.2-27).

!

Sometimes it can be advantageous to change an infinite

■

product into an infinite sum using logarithms. Proposition

(12.2-33)

12.2-6 shows that convergence properties remain unchanged.

■ ∞

!

If a product

∏ (1+ b ) has a finite number of b 's that are n

Proposition 12.2-6:

{

n

equal to −1 , and if the product converges when these zero-

k

sequence of its partial products Pk =

∏ (1+ b ) , where n

Pk > 0

n =1

factors are removed, the product is considered to be

for all k , converges to a nonzero limit if and only if the sequence

convergent.

k

of partial sums Sk =

{

}

Given a sequence of complex numbers or functions 1+ bk , the sequence of its partial products

{ Pk }

! !

Sk =

∑ Ln (1+ b ) ! n

(12.2-35)

n =1

converges to a finite limit S :

k

n = m+1

First we assume that the partial sum: k

number ε > 0 there exists N ∈! such that:

n

n

Proof:

where Pk > 0 for all k

will converge to a finite nonzero limit if and only if for any real

∏(1+ b ) − 1 < ε !

∑ Ln (1+ b ) converges. n =1

Proposition 12.2-5, Cauchy Criteria for Infinite Products:

!

}

Given a sequence of complex numbers or functions 1+ bk , the

n =1

for all k > m > N !

(12.2-34)

!

lim S = S !

k→∞ k

(12.2-36) 608

k

We can write: k

Pk =

!

∏(1+ b ) n

n =1

⎡ = exp ⎢ Ln ⎢ ⎣

k

⎤ 1+ bn ⎥ ! ⎥ ⎦

∏( n =1

k

)

(12.2-37)

∑ n =1

(

)

k

∑ n =1

⎤ Ln 1+ bn ⎥ ⎥ ⎦

(

)

S

Pk = e k !

(12.2-39)

Since the exponential function is continuous, we then have: S

lim Pk = lim e k = eS = P !

!

k→∞

k→∞

(12.2-40)

∏ (1+ bn ) n =1

the partial sum Sk =

converges to the nonzero limit P = e S if

k

∑ Ln (1+ b ) converges to S . This is to be n

k

Ln Pk + 2 π imk =

!

∑ Ln (1+ b ) = S ! n

k

(12.2-43)

n =1

where mk is an integer which is not necessarily zero since the principal values of the logarithms on each side of this equation may not be the same. This is because the logarithm of a product is not necessarily equal to the sum of the logarithms of the terms in the product. We now define:

!

αn = Arg (1+ bn ) !

βk = Arg Pk !

(12.2-44)

We then have from equation (12.2-43):

n =1

expected since the exponential of a finite sum will equal a finite

(12.2-42)

We can write from equation (12.2-41):

!

k

and so Pk =

lim P = P !

k→∞ k

(12.2-38)

where mk is an integer which is not necessarily zero. We have: !

(12.2-41)

n

converges to a finite limit P ≠ 0 : !

⎤ ⎡ Ln 1+ bn + 2 π i mk ⎥ = exp ⎢ ⎥ ⎢ ⎦ ⎣

∏ (1+ b ) ! n =1

or since e2 π i mk = 1 :

⎡ ! Pk = exp ⎢ ⎢ ⎣ ! !

Pk =

!

k

!

βk + 2 π mk =

∑α

n

!

(12.2-45)

n =1

product. and so: !

Conversely, assume that the partial product: 609

k

k−1

∑α − ∑α

!

n

n =1

n

(

)

(

)

= βk − βk−1 + 2 π mk − mk−1 ! (12.2-46)

( k→∞

)

lim mk − mk−1 = 0 −

1 β − β = 0! 2π

(

)

(12.2-53)

n =1

Since mk is an integer, it must then converge to some integer m

or

as k → ∞ for equation (12.2-53) to be true:

α k = ( βk − βk−1 ) + 2 π ( mk − mk−1 ) !

! !

!

(12.2-47)

Since we are given:

Pk =

∏ (1+ bn ) !

(12.2-48)

n =1

lim b = 0 !

(12.2-49)

k→∞ k

k→∞

(

)

lim β k = lim β k−1 = lim Arg Pk = Arg P = β ! k→∞

k

( k→∞

or as k → ∞ :

)

Sk =

∑ Ln (1+ b ) ! n

(12.2-56)

(12.2-51)

12.2.4! ABSOLUTELY CONVERGENT INFINITE PRODUCTS

k→∞

!

From equation (12.2-47) we have: ! lim mk − mk−1

(12.2-55)

converges.

{ }

k→∞

)

(12.2-50)

Since Pk converges to P , we obtain: !

k→∞

n =1

lim α k = lim Arg 1+ bk = Arg1 = 0 !

k→∞

(

lim S = lim Ln Pk + 2π i mk = Ln P + 2 π m !

k→∞ k

and so we can conclude that, if Pk converges, then !

Therefore: !

(12.2-54)

k→∞

equations (12.2-43) and (12.2-42): !

then by Proposition 12.2-4 we have: !

lim mk = lim mk−1 = m !

k→∞

Since Ln is continuous at the limit k → ∞ , we then have from

k

!

!

■

If the infinite product: ∞

1 1 = lim α k − lim β − βk−1 ! 2π k → ∞ 2π k → ∞ k

(

)

(12.2-52)

!

∏ (1+ b ) ! n

(12.2-57)

n =1

converges, then it is said to be absolutely convergent. If the infinite product: 610

k

∞

∏ (1+ bn ) !

!

(12.2-58)

!

Pk =

converges, but

∞

∏ (1+ b ) diverges, then ∏ (1+ b ) is said to n

n

n =1

n =1

be conditionally convergent.

n

and the partial sum

{

}

!

infinite product:

!

∞

(12.2-59)

Sk =

and the infinite series:

∑b

= b1 + b2 + ! + bk !

n

(

(12.2-60)

equation (12.2-59) is:

{ Pk }

(12.2-63)

(12.2-64)

We also have using the Maclaurin expansion of e

e

!

! The partial product

)

bk

= 1+ bk +

bk 2

2

+

bk

3

3!

+

bk

bk

:

4

4!

+! ≥ 1+ bk !

(12.2-65)

and so:

converge or diverge together.

!

) (

Sk ≤ Pk !

n =1

Proof:

)(

b1 + b2 + ! + bk ≤ 1+ b1 1+ b2 ! 1+ bk !

∞

∑

(12.2-62)

or

!

bn !

of the infinite series in equation

We have the inequality:

!

n =1

!

{Sk }

n =1

Given a sequence of complex numbers or functions 1+ bk , the

n

(12.2-61)

k

k

Proposition 12.2-7, Absolute Convergence:

∏ (1+ b ) !

2

(12.2-60) is !

!

1

n =1

n =1

∞

∏ (1+ b ) = (1+ b )(1+ b ) ! (1+ b ) !

of the infinite product in

(1+ b )(1+ b ) ! (1+ b ) ≤ e 1

2

k

b1

e

b2

!e

bk

!

(12.2-66)

!

(12.2-67)

or !

(1+ b )(1+ b ) ! (1+ b ) ≤ e 1

2

k

b1 + b2 + b3 + ! + bk

611

∞

Therefore:

Pk ≤ e !

!

∑b

! Sk

n

(12.2-68)

S Sk ≤ Pk ≤ e k !

converges. Proof:

(12.2-69)

{ } and the partial

!

are monotonic increasing real sequences. From

!

Since bn ≥ 0 for all n , the partial sum Sk

!

product

{ Pk }

equation (12.2-69) we see that

{ Pk }

is bounded above by e

Sk

and below by Sk . !

(12.2-71)

n=1

From equations (12.2-64) and (12.2-68) we have: !

!

Follows from Proposition 12.2-7. ∞

Note that if

∑b

n

converges only conditionally, then it

n=1

provides no information as to the possible convergence of ∞

Using Proposition 7.1-4 we see that the infinite product

∏ (1+ b ) . n

n =1

∞

∞

∏ (1+ b ) converges or diverges as the infinite series ∑ b n

n

n =1

Proposition 12.2-9:

n =1

converges or diverges.

{

the infinite product

Proposition 12.2-8:

{

∞

converges, then the product

n =1

n

Proof:

∞

(12.2-70)

!

Let Pk be the partial product:

n =1

converges if and only if the infinite series:

n

n =1

infinite product: n

∏(1+ b )

∏ (1+ b ) also converges.

}

Given a sequence of complex numbers or functions 1+ bk , the

∏ (1+ b ) !

}

Given a sequence of complex numbers or functions 1+ bk , if

■

∞

!

■

k

!

Pk =

∏ (1+ b ) ! n

(12.2-72)

n =1

612

{ } converges, we must have for any real number ε > 0

Since Pk

some integer N > 0 such that:

12.2.5! REARRANGEMENT OF A PRODUCT Proposition 12.2-10, Rearrangement of an Infinite Product:

k

∏ (1+ b ) − 1 < ε !

!

n

for all k > m > N !

{ } { } {cn } is any rearrangement of { bn } , and if the infinite product:

If bn and cn are two sequences of complex numbers where

(12.2-73)

n = m+1

which is the Cauchy criteria for infinite products (Proposition

∞

! !

converges, then the infinite products:

(12.2-74)

∞

(12.2-75)

and so: !

(1+bm+1 )(1+bm+2 )!(1+bm +k ) − 1 ≤ ε !

! !

for all k > m > N

!

(12.2-76)

From the Cauchy criteria for convergence (Proposition 12.2-5), we see that this product converges. Therefore an absolutely convergent product is also a convergent product. !

The converse is not true.

■

∏ (1+ c )!

n

(12.2-78)

n

n =1

(1+bm+1 )(1+bm+2 )!(1+bm +k ) − 1 ≤ (1+ bm+1 )(1+ bm+2 )!(1+ bm+k ) − 1 ! !

∞

∏ (1+ c ) !

!

or

(12.2-77)

n

n =1

(1+bm+1 )(1+bm+2 )!(1+bm+k ) ≤ (1+ bm+1 )(1+ bm+2 )!(1+ bm+k )

!

∏ (1+ b ) !

!

12.2-5). We also have the inequality:

n =1

are convergent. Proof: !

∞

∞

Since

∏ (1+ b ) n

converges, then

n =1

∑b

converges as

n

n=1

shown in Proposition 12.2-8. It follows from Proposition 8.2-10 ∞

that

∑c

n

∞

also converges to the same sum as ∞

n=1

Proposition 12.2-7 we see that

∏ (1+ c ) n

n =1

Finally, from Proposition 12.2-9 we know that converges.

∑b

n

. From

n=1

then converges. ∞

∏ (1+ c ) also n

n =1

■

613

12.3! UNIFORM CONVERGENCE OF AN INFINITE PRODUCT OF FUNCTIONS !

We will now consider a sequence of complex functions

{ f ( z ) } . An infinite product in the form:

where Pk > 0 for all k . Since

∑

!

∏ (1+ f ( z ) ) !

(12.3-1)

n

n =1

()

fn z !

n=1

{ ( ) } where:

number M and for all z ∈R :

sequence of partial products Pk z

()

Pk z =

!

k

∏ (1+ f ( z ) ) !

()

∏ (1+ f ( z ) ) n

n =1

!

converges uniformly in R .

k

Pk z ≤

(12.3-2)

n

n =1

∑

!

∑ f (z) n

is uniformly and absolutely

n=1

convergent in a bounded closed region R , then the infinite

n =1

()

()

fn z < ε !

k >m> N !

(12.3-6)

as given by the Cauchy criteria for infinite series (see Proposition 8.2-6). We can also write for all z ∈R :

∞

∏ (1+ f ( z ) ) is uniformly convergent in R . n

!

n =1

!

∑

⎞ fn z ⎟ ≤ M ! (12.3-5) ⎟⎠

n = m+1

∞

Proof:

k

all z ∈R :

Proposition 12.3-1, Uniform Convergence of an Infinite Product: If the function series

⎛ ≤ exp ⎜ ⎜⎝

For any real number ε > 0 there exists N ∈! such that for k

product

(12.3-4)

is uniformly convergent in R , we can write for some positive

is uniformly convergent in a bounded closed region R if the

!

(12.3-3)

n

n =1

∞

∞

∏ (1+ f ( z ) ) !

()

Pk z =

!

n

!

k

()

! !

()

We will define the partial product Pk z as:

()

()

Pk z − Pm z = Pm z

k

∏(1+ f ( z ) ) − 1 !

k >m> N

n

n = m+1

!

(12.3-7)

and so: 614

!

()

()

()

Pk z − Pm z ≤ Pm z

k

∏(1+ f ( z ) ) − 1 !

k >m> N

n

n = m+1

! !

!

(12.3-8)

!

()

fn z < Mn where Mn > 0 and where

Since

∑M

n

is

n=1

∞

()

()

()

Pk z − Pm z ≤ Pm z

Proposition 8.6-4 we know that the series

⎛ k ⎞ exp ⎜ fn z ⎟ − 1 ! ⎜⎝ n = m+1 ⎟⎠

∑

()

! !

k >m> N

∑ f ( z ) converges n

n=0

uniformly and absolutely. Therefore by Proposition 12.3-1 the ∞

!

(12.3-9)

infinite product

∏ (1+ f ( z ) ) is uniformly convergent in R . n

■

n =1

Therefore: !

∞

uniformly convergent (since it is not a function of z ), from

or !

Proof:

()

()

Pk z − Pm z ≤ M eε − 1 !

Example 12.3-1

k > m > N ! (12.3-10)

Determine if the infinite product having the partial

∞

and so the infinite product convergent in R .

∏ (1+ f ( z ) )

products:

is uniformly

n

n =1

■

!

()

Pk z =

∏ n =1

Proposition 12.3-2, M-Test for Uniform Convergence of an ∞

The infinite product of complex functions

∏(

Solution:

( ) ) is

1+ fn z

n =1

We can write:

uniformly convergent in a bounded closed region R if

()

∞

∑ n=1

series of positive constants.

⎛ z2 ⎞ ⎜ 1− 2 ⎟ ⎝ n ⎠

converges if z 2 < M where M > 0 .

Infinite Product:

fn z < Mn where Mn > 0 , and where

k

Mn is a convergent

!

z2 M ≤ = Mn n2 n2 The series: 615

∞

!

∑M

Proof:

n

From Proposition 8.6-6 we know that:

n =1

is uniformly convergent (does not depend upon z ).

()

f z =

!

Therefore the infinite product having the partial products: !

k

()

Pk z =

∏ n =1

⎛ z2 ⎞ ⎜ 1− 2 ⎟ ⎝ n ⎠

∞

∑ f (z) !

(12.3-13)

n

n =1

is holomorphic within the open region R . Differentiating equation (12.3-12), we have:

is uniformly convergent.

()

P′ z = lim

!

Proposition 12.3-3, Logarithmic Derivative of an Infinite

N

N→ ∞

N

∑ f ′( z ) ∏ (1+ f ( z ) ) ! n

n =1

()

N

Given a sequence of analytic functions

{ f (z) }

defined in an

n

∞

open simply connected region R so that

∑ f ( z ) converges

( ) = lim f ′ z () P( z ) N→ ∞ ∑ n n =1

P′ z

!

n

∞

N

∏ (1+ f ( z ) )

!

(12.3-15)

or

∞

(12.3-11)

where !

k

k≠n

k =1

uniformly on each compact subset of R , then we have:

f n′ ( z ) ( ) = d ( ln P ( z )) = ! ∑ dz P( z) f z 1+ n ( ) n =1

∏ (1+ f ( z ) ) k

n =1

P′ z

k≠n

Since P z ≠ 0 we can write:

Product:

!

(12.3-14)

k

( ) = ∞ f n′ ( z ) ! P ( z ) ∑ 1+ fn ( z ) n =1

P′ z

!

(12.3-16)

■

()

P z =

∞

∏ ( 1+ f ( z ) ) ≠ 0 ! n

(12.3-12)

n =1

616

12.4! ENTIRE FUNCTION EXPANSION AS AN INFINITE PRODUCT

()

We will consider an entire function f z

!

having zeros

{ }

only at points specified by a sequence of numbers an . Such a function can be represented in the form of an infinite product, each factor of which is zero for only a single value of an :

( ) ( ) ( z − a2 )!( z − an ) g ( z ) ! (12.4-1) where g ( z ) is then analytic and nonzero. We will let the sequence { an } be arranged in order of increasing absolute value. Since g ( z ) will not introduce any new zeros, g ( z ) can be f z = z − a1

!

!

() ()

h z =h 0 +

()

Taking the derivative of equation (12.4-1) and dividing by

g′ ( z ) ( ) 1 1 1 h( z ) = = + +!+ + f ( z ) ( z − a1 ) ( z − a2 ) ( z − an ) g ( z )

! !

(12.4-2)

()

( ) f (z)

The function h z = f ′ z

has simple poles at the points

a1 , a2 , a3, ! . From the Mittag-Leffler’s expansion theorem (Proposition 10.5-1) we can then write:

(12.4-3)

()

see that bn = 1 for all n . We then have: !

( ) = f ′ (0) + ∞ ⎡ 1 + 1 ⎤ ! ⎢ ⎥ z − a a f (z) f (0) ∑ n n ⎦ n =1 ⎣

f′ z

(12.4-4)

Integrating this equation from 0 to z along a path that does not pass through a pole, we have: !

()

()

Ln f z − Ln f 0 = z ∞

+

!

∑ n =1

f′ z

⎡ 1 1⎤ bn ⎢ + ⎥! ⎣ z − an an ⎦

where bn are the residues of f z at the points a1 , a2 , a3, ! . We

f z , we obtain: !

∑ n =1

expressed in the form of an exponential (see Proposition 9.3-8). !

∞

() f (0)

f′ 0

⎡ ⎤ z ! (12.4-5) Ln z − a + − Ln − a ⎢ n n ⎥ a ⎣ ⎦ n

(

)

(

)

Taking the exponential of this equation, we have: !

()

()

f z = f 0 e

(

z f ′( 0 ) f ( 0 )

)

∞

∏ n =1

⎛ z⎞ z 1− ⎜ a ⎟e ⎝ n ⎠

an

!

(12.4-6)

where we are taking an ≠ 0 for all n . Equation (12.4-6) is a special case of the more general theorem known as the Weierstrass factorization theorem.

617

Example 12.4-1 Determine an infinite product expansion for

sin z and sin z . z

Solution: The function sin z has zeros at points n π , where n takes on all integer values. We have:

sin z z ( f ′( 0 ) f ( 0 )) = cos0 = 1! f ′ 0 = 0! e =1 z→0 z From equation (12.4-6) for all points n π we then have:

()

()

f 0 = lim

!

sin z = z

∞

∏ n≠0

⎛ z ⎞ z ( nπ ) 1− ⎜⎝ n π ⎟⎠ e

or, writing this equation with n restricted to positive values: !

sin z = z

∞

∏ n =1

⎛ z ⎞ z ( nπ ) ⎛ z ⎞ − z ( nπ ) 1− e 1+ = ⎜⎝ n π ⎟⎠ ⎜⎝ n π ⎟⎠ e

∞

∏ n =1

⎛ z2 ⎞ ⎜ 1− 2 2 ⎟ ⎝ n π ⎠

and so: ∞

!

sin z = z

∏ n =1

⎛ z2 ⎞ ⎜ 1− 2 2 ⎟ ⎝ n π ⎠

618

Appendix A

THE GREEK ALPHABET

Iota!

ι!

Ι

Kappa!

κ!

Κ

Lambda!

λ!

Λ

Mu!

µ!

Μ

Nu!

ν!

Ν

Xi!

ξ!

Ξ

Omicron!

ο!

Ο

Pi!

π!

Π

Rho!

ρ!

Ρ

Sigma!

σ!

Σ

Tau!

τ!

Τ

Upsilon!

υ!

ϒ

Alpha!

α!

Α

Phi!

φ, ϕ !

Φ

Beta!

β!

Β

Chi!

χ!

Χ

Gamma!

γ!

Γ

Psi!

ψ!

Ψ

Delta!

δ!

Δ

Omega!

Ω

Epsilon!

ε!

ω!

Ε

Zeta!

ζ!

Ζ

Eta!

η!

Η

Theta!

θ!

Θ 619

()

f′ z =

!

Appendix B

!

d z e = ex = f x ! dx

()

()

f 0 = 1!

(B-2)

To show this, we will consider an entire complex function

() ( ) ( ) with f ( 0 ) = 1 , we have: ∂u ( x, y ) ∂v ( x, y ) ! +i = u + iv !

()

()

w = f z = u x, y + i v x, y . Since we are requiring f ' z = f z

∂x

∂x

( )

( )

u 0,0 + i v 0,0 = 1!

(B-3)

By equating real and imaginary parts of these equations, we obtain:

COMPLEX EXPONENTIAL FUNCTION

!

! !

In this appendix, we will show that the complex z

exponential function e is the only entire complex function that has the following two properties: !

1.!

d f ′ z = ez = ez = f z ! dz

()

! !

It is its own derivative:

2.!

()

()

()

f 0 = 1!

(B-1)

It agrees with the real-valued exponential function

( ) = u!

u 0,0 = 1 !

( ) = v!

v 0,0 = 0 !

∂u x, y ∂x

∂v x, y ∂x

( )

(B-4)

( )

(B-5)

Integrating these equations, we have: !

u = g y ex !

( )

g 0 = 1!

!

v = h y ex !

( )

h 0 = 0!

( )

where g y

( )

and h y

()

(B-6)

()

(B-7)

are a real-valued functions of y only

(since integration over x has already occurred). We then have:

f x = e x when z is real: 620

! !

∂u = g y ex ! ∂x

∂v = h y ex ! ∂x

( )

( )

()

Since the function f z

(B-8)

is entire, by definition it is

holomorphic over the whole complex plane. This means that the Cauchy-Riemann equations apply: !

∂u ∂v ! = ∂x ∂ y

!

( )

∂u ∂v =− ! ∂y ∂x

(B-9)

!

( )

∂v ∂u = h y e x = − ! (B-10) ∂x ∂y

( )

!

∂v = h′ y e x ! ∂y

( )

(B-11)

Using equations (B-10) and (B-11) we can write:

( )

( )

g y = h′ y !

( ) ( )

g ′′ y + g y = 0 !

( )

( )

h y = − g′ y !

(B-12)

( )

g y = c1 cos y + c2 sin y !

( )

(

)

(B-16)

( )

(

)

v = h y e x = − e x − c1 sin y + c2 cos y !

!

(B-17)

( ) ( )

h′′ y + h y = 0 !

(

) (

)

! e z = u + i v = e x ⎡⎣ c1 cos y + c2 sin y + i c1 sin y − c2 cos y ⎤⎦ ! (B-18) or letting k = c1 − i c2 :

(

!

)

e z = k e x cos y + i sin y !

(B-19)

()

Because we want the complex function f z to agree with

the real-valued exponential function e x for real z , we must have:

k ex = ex !

when y = 0 !

(B-20)

Therefore k = 1 = c1 − i c2 , and so c1 = 1 and c2 = 0 , and equation (B-19) becomes:

(B-13)

These differential equations have the solutions: !

(B-15)

u = g y e x = e x c1 cos y + c2 sin y !

!

!

Differentiating equation (B-12): !

)

Equations (B-6) and (B-14) give us:

!

where the prime mark indicates derivatives with respect to y . !

(

We then have:

From equations (B-6) and (B-7) we have:

∂u = g ′ y ex ! ∂y

( )

and equations (B-7) and (B-15) give us:

Therefore equations (B-8) becomes:

∂u ∂v ! = g y ex = ∂x ∂y

( )

h y = − g ′ y = − − c1 sin y + c2 cos y !

!

!

(

)

e z = e x cos y + i sin y = e x+i y !

(B-21)

This is just the complex exponential function as defined in (B-14)

equation (4.2-2). 621

Let C be a simple closed contour within the domain D , and z0 be any given point such that z0 + Δz lies inside the contour C .

Appendix C

Using Cauchy’s integral formula (Proposition 6.1-1) we can write: !

(

)

( )=

f z0 + Δz − f z0 Δz

1 2π i

!∫

C

DIFFERENTIATION OF THE CAUCHY INTEGRAL FORMULA

(

−

!

()

f z ⎛ ⎞ dz ⎜ ⎟ ⎝ z − z0 + Δz ⎠ Δz 1 2π i

)

!∫

C

()

⎛ f z ⎞ dz ⎜ z − z ⎟ Δz ! (C-2) ⎝ 0⎠

Using equation (C-1) we can write: !

1 f ′ z0 = lim Δz → 0 2π i

( )

!∫

C

()

()

f z f z ⎞ dz ⎛ ! (C-3) − ⎜ ⎟ z − z0 ⎠ Δz ⎝ z − z0 + Δz

(

)

or !

In this appendix, we will show that differentiation of the

1 f ′ z0 = lim Δz → 0 2π i

( )

!

Cauchy’s integral formula under the integral sign is justified.

()

Let f z be holomorphic on a simply connected domain D . By

( )= f′

df z0 dz0

( )

z0 = lim

Δz →0

(

)

( )

f z0 + Δz − f z0 Δz

!

(C-4)

Adding and subtracting a term in the integrand, we can write:

definition of a derivative we have: !

() !∫C ( z − z0 − Δz ) ( z − z0 ) dz ! f z

( )

!

f ′ z0 =

!

+ lim

(C-1) Δz →0

( ) dz !∫C ( z − z0 )2 ⎡ f (z) f (z) ⎤ 1 ⎢ ⎥ dz ! − 2 ∫C ( z − z0 − Δz ) ( z − z0 ) ( z − z0 ) ⎥⎦ 2 π i ⎢! ⎣

1 2π i

f z

(C-5)

622

or !

( )

f ′ z0 =

() !∫C ( z − z0 )2 f z

Δz + lim Δz → 0 2π i

! !

1 2π i

!

()

bounded in D . We then have for some integer M > 0 :

dz

() !∫C ( z − z0 − Δz ) ( z − z0 )2 f z

()

f z ≤ M!

!

dz !

(C-6)

and (C-10) we can write:

the right goes to zero when Δz → 0 . We will consider a circular

!

contour C1 of radius r and center z0 which lies inside the contour C . We then have from the deformation theorem

Δz 2π i

() !∫C ( z − z0 − Δz ) ( z − z0 )2 Δz 2π i

!

!

dz =

() !∫C ( z − z0 − Δz ) ( z − z0 )2 f z

!∫

() ( z − z0 − Δz ) ( z − z0 )2

Δz 2π i

() !∫C ( z − z0 − Δz ) ( z − z0 )2 dz

f z

C1

f z

!

dz ! (C-7) !

(C-8)

We will let Δz be such that Δz < r 2 so that z0 + Δz lies

inside C1 . We then have: !

z − z0 − Δz ≥ z − z0 − Δz = r − Δz > r −

( )

r 2 r2

)!

(C-11)

≤ Δz

2M ! 2 r

(C-12)

and so:

1

z − z0 = r !

2π

1

lim

Δz →0

Δz 2π i

where the contour C1 is given by: !

dz ≤

(

Δz M 2 π r

or

(Proposition 5.5-1):

f z

(C-10)

The total length of C1 is 2 π r . From equations (C-7), (C-8), (C-9),

We will now show that the value of the second integral on

Δz ! 2π i

()

Since f z is holomorphic on D , f z is continuous and

!∫

() ( z − z0 − Δz ) ( z − z0 )2 f z

C1

dz = 0 !

(C-13)

From equations (C-6), (C-7), and (C-13) we therefore have: !

1 f ′ z0 = 2π i

( )

( ) dz ! !∫C ( z − z0 )2 f z

(C-14)

r r = ! (C-9) 2 2 623

!

( )

The derivation of f ′′ z0 proceeds in a similar manner. By

definition we have:

( )

(

f ′′ z0 = lim

!

)

( )!

f ′ z0 + Δz − f ′ z0 Δz

Δ z →0

(C-15)

Using Cauchy’s integral formula (Proposition 6.1-1) and equation (C-14) we can write: !

1 f ′′ z0 = lim Δ z →0 2 π i

( )

()

( ) ⎤⎥ dz ! 2 ⎥ Δz ( ) ) ⎦

⎡ f z f z ⎢ − ⎢ z − z − Δz 2 z − z0 0 ⎣

!∫

(

C

!

1

1

− 2

=

2 Δz

− 3

2 Δz

1

+ 3

( z − z0 − Δz) ( z − z0 ) ( z − z0 ) ( z − z0 ) ( z − z0 − Δz) 2

−

!

1

( z − z0 )

2

!

2

(C-17)

2! 2π i

( ) dz !∫C ( z − z0 )3

1 + lim Δ z →0 2 π i

!

f z

( ) )(

⎡ 3 Δz z − z − 2 Δz 2 ⎤ 0 ⎥ dz ! (C-19) f z ⎢ 3 2 ⎢ z−z z − z0 − Δz ⎥⎦ 0 ⎣

!∫

()

C

(

(

)

)

()

where the product ⎡⎣ 3 z − z0 − 2 Δz ⎤⎦ f z is holomorphic and so is bounded. Therefore a bound M > 0 exists such that we have

(

)

3 z − z0 − 2 Δz ≤ M . We can then write:

(C-16)

Adding and subtracting a term in the integrand, we can write:

( )

f ′′ z0 =

!

!

2! f ′′ z0 = 2π i

( )

() !∫C ( z − z0 )3 f z

dz + lim

Δ z →0

( )! 2 r 3 ( r 2)

Δz M 2 π r 2π

(C-20)

As Δz → 0 we then obtain:

( ) dz ! !∫C ( z − z0 )3 f z

!

2! f ′′ z0 = 2π i

!

We have shown that:

( )

(C-21)

or !

( ) 2 3 ( z − z0 − Δz) ( z − z0 ) ( z − z0 ) ( z − z0 ) ( z − z0 − Δz)2 1

− 2

1

=

! Equation (C-16) then becomes:

2 Δz

+ 3

3 Δz 2 z − z0 − 2 Δz 3

(C-18)

!

n! n f ( ) z0 = 2π i

( )

( ) dz ! !∫C ( z − z0 )n +1 f z

(C-22)

where n = 1 and n = 2 . Continuing this process by mathematical induction for higher derivatives, we will now assume that 624

equation (C-14) holds for all n − 1 > 0 where n is an integer. By definition we have:

f ( n−1) ( z0 + Δz ) − f ( n−1) ( z0 ) n) ( f ( z0 ) = lim ! Δ z →0 Δz

!

(C-23)

Using Cauchy’s integral formula (Proposition 6.1-1) we can write: !

n − 1) ! ( n) ( f ( z0 ) = lim Δ z →0 2 π i

!∫

C

()

( ) ⎤⎥ dz ! n ) ( ) ⎥⎦ Δz

⎡ f z f z ⎢ − ⎢ z − z − Δz n z − z0 0 ⎣

(

!!

(C-24)

or

!

( n − 1) ! 2π i

Δ z →0

!∫

C

n n z − z0 ) − ( z − z0 − Δz ) ( f (z) n ( z − z0 − Δz)n ( z − z0 )

!!

dz ! Δz (C-25)

From the binomial expansion, we have:

( ( z − z ) − Δz ) = ( z − z ) − n ( z − z ) n

0

!

n

some positive power. ! !

Equation (C-25) then becomes: n f ( ) ( z0 ) = lim

Δ z →0

n! 2π i

!∫

C

() )(

⎡ ⎤ f z ⎢ +!⎥ dz n ⎢ z − z − Δz z − z ⎥ 0 0 ⎣ ⎦

(

)

(C-28)

Therefore for all integer n : !

n! n f ( ) z0 = 2π i

( )

( ) dz ! !∫C ( z − z0 )n +1 f z

(C-29)

This is Cauchy’s integral formula for derivatives.

n−1

0

0

+

(C-27)

where the dots represent terms having a factor of Δz raised to

!

n ! f ( ) ( z0 ) = lim

!

!

( z − z0 )n − ( z − z0 − Δz)n = n Δz n n z − z z − z − Δz ( ) ( 0 ) ( z − z0 − Δz)n ( z − z0 ) 0 2 n ( n − 1) ( Δz ) − + !! n 2 2 ( z − z0 − Δz ) ( z − z0 )

(

Δz

) ( z − z )n−2 (Δz )2 − ! ! (C-26)

n n −1 2

0

Therefore we have:

625

Appendix D

Proposition 1.8-3: Moduli of any two complex numbers z1 and z2 obey the inequality defined by:

z1 − z2 ≤ z1 + z2

!

Proposition 1.8-4: Moduli of any two nonzero complex numbers z1 and z2 obey the

SUMMARY OF PROPOSITIONS

law: !

z1 + z2 = z1 + z2 if and only if z1 = c z2 where c > 0 .

Proposition 1.8-1, Complex Numbers Cannot be Ordered: The set of complex numbers ! cannot be ordered.

Proposition 1.8-5, Reverse Triangle Inequality: Moduli of any two complex numbers z1 and z2 obey the reverse

Proposition 1.8-2, Triangle Inequality: Moduli of any two complex numbers z1 and z2 obey the

triangle inequality as defined by: !

z1 − z2 ≥ z1 − z2

triangle inequality as defined by: !

z1 + z2 ≤ z1 + z2

Proposition 1.8-6: Moduli of any two complex numbers z1 and z2 obey the reverse triangle inequality as defined by: 626

!

z1 + z2 ≥ z1 − z2

Proposition 1.12-1, de Moivre’s Theorem: de Moivre’s theorem states that: !

(cosθ + i sinθ )n = (cos nθ + i sin nθ ) where n is an integer and θ is a real variable.

Proposition 2.2-1: If z = z0 is a limit point of the set S , every deleted δ neighborhood of z0 contains infinitely many points of S . Proposition 2.2-2: If a point set S contains only a finite number of points, it cannot contain a limit point.

Proposition 2.2-5: The intersection of two open sets of points in the complex plane is an open set. Proposition 2.2-6: A point set S is closed if and only if its complement S C is open. Proposition 2.2-7: The empty point set ∅ is both open and closed. Proposition 2.2-8: The set of points z ∈! is both open and closed. Proposition 2.2-9: If the upper bound of a point set S is not an element of S , then

Proposition 2.2-3: A closed set contains all its limit points. Proposition 2.2-4: Every finite point set is closed.

it is a limit point of S . Proposition 2.3-1, Limit is Unique:

()

If a complex function f z has a limit at a given point z = z0 , then the limit is unique.

627

Proposition 2.3-2:

Proposition 2.3-5:

()

If a complex function f z has a limit at a given point z = z0 , the limit must be independent of the direction of approach to the

z → z0

Proposition 2.3-6:

Proposition 2.3-3:

()

z → z0

()

z → z0

() ()

()

z → z0

()

( )

lim f z = f z0 = w0

!

() ( )

lim u z = u z0 = u0

z → z0

z → z0

!

()

z → z0

()

() ()

()

()

lim ⎡⎣ f z ± g z ⎤⎦ = lim f z ± lim g z = A ± B z → z0 z → z0

z → z0

Proposition 2.3-8:

() ( )

()

()

If lim f z = A and lim g z = B , then:

lim v z = v z0 = v0

z → z0

()

If lim f z = A and lim g z = B , then

where w0 = u0 + i v0 if and only if: !

z → z0

Proposition 2.3-7:

possibly at a point z = z0 , then we will have: z → z0

()

lim c f z = c lim f z = c A

!

Proposition 2.3-4: If w = f z = u z + i v z is defined in a domain D except

()

If lim f z = A and if c is a complex constant, then:

If the limit lim f z = w0 exists, then lim f z = w0 .

!

()

z → z0

lim f z = w0 .

point z0 .

z → z0

()

If the limit lim f z = w0 exists where w0 is finite, then

z → z0

!

z → z0

() ()

()

()

lim ⎡⎣ f z g z ⎤⎦ = lim f z lim g z = A B z → z0 z → z0

z → z0

628

!

Proposition 2.3-9:

()

() lim f z ⎡ f (z) ⎤ z→ z ( ) A lim ⎢ = ! ⎥= z→ z g z lim g z () B ⎢⎣ ( ) ⎥⎦ z→ z

( ) is continuous at point z if g z ≠ 0 . ( 0) 0 g (z) f z

If lim f z = A and lim g z = B , then: z → z0

!

z → z0

0

Proposition 2.4-4:

() and if complex function f ( w) is continuous at a point w = w0 where w0 = g ( z0 ) , then f ( g ( z )) is also continuous at z0 .

If a complex function w = g z is continuous at a point z = z0 ,

B≠0

0

0

Proposition 2.4-1:

()

The complex function f z = z is continuous for all z ∈! .

Proposition 2.4-5:

()

If a complex function w = f z is continuous at a point z = z0 ,

()

Proposition 2.4-2:

then w = f z is continuous at z0 .

() () () point z0 = x0 + i y0 if and only if u ( x, y ) and v ( x, y ) continuous at the point ( x0 , y0 ) .

A complex function w = f z = u z + i v z is continuous at a are both

Proposition 2.4-6:

()

()

If w = f z is continuous and f z ≠ 0 at a point z = z0 , then

()

for some neighborhood of z0 we must have f z ≠ 0 .

Proposition 2.4-3:

()

()

If two complex functions f z and g z are continuous at a

Proposition 2.4-7:

()

If z = z0 is a limit point of S , and if w = f z is continuous at

point z0 , then: ! !

()

()

z0 such that:

f z ± g z is continuous at point z0 .

() ()

f z g z is continuous at point z0 .

!

()

( )

lim f z = f z0 = w0

z → z0

( ( )) is continuous at w

and if z = g f z

0

such that: 629

!

( ( )) ( )

lim g f z = g w0

w→ w0

( ( )) is continuous at z , and we have: ⎛ ⎞ lim g ( f ( z )) = g ⎜ lim f ( z )⎟ ⎝ ⎠

then g f z !

Proposition 2.7-1:

z → z0

0

If z0 is a point on the complex plane, then: !

z → z0

if and only if:

Proposition 2.4-8, Continuity of Polynomial Functions: !

All complex polynomials: !

()

2

P z = a0 + a1 z + a2 z +!+ an−1 z

n−1

+ an z

All complex rational functions:

()

)

()

f z

=0

If z0 is a point on the z-plane and w0 is a point on the w-plane, then: !

a0 + a1 z + a2 z 2 + !+ an−1 z n−1 + an z n

()

lim f z = w0

z →∞

if and only if:

b0 + b1 z + b2 z 2 +!+ bm−1 z m−1 + bm z m

()

where an ≠ 0 and bm ≠ 0 , and where the degree of f z is

(

z → z0

1

Proposition 2.7-2:

Proposition 2.4-9, Continuity of Complex Rational Functions:

f z =

lim

n

are continuous over the finite complex plane.

!

()

lim f z = ∞

z → z0

max m, n , are continuous over the finite complex plane.

!

⎛ 1⎞ lim f ⎜ ⎟ = w0 z→0 ⎝ z ⎠

630

Proposition 2.7-3:

Proposition 3.3-1, Differentiable Functions are Continuous:

()

!

()

continuous at z0 .

lim f z = ∞

z → z0

Proposition 3.3-2:

if and only if: !

()

If f z

⎛ 1⎞ lim f ⎜ ⎟ = 0 z → z0 ⎝ z ⎠

is differentiable at a point z = z0 , then the real and

()

imaginary parts of f z are continuous at z0 . Proposition 3.3-3:

Proposition 2.8-1:

()

()

If f z

()

If the limit lim f z = w0 exists where w0 is finite, then f z z → z0

()

If f z is differentiable at a point z = z0 , then f z is

If z0 is a point on the complex plane, then:

is differentiable at a point z = z0 , then the real and

()

imaginary parts of f z are differentiable at z0 .

is a bounded function in some neighborhood of the point z = z0 .

Proposition 3.3-4:

Proposition 2.8-2:

()

If w = f z is differentiable at a point z = z0 , and the inverse

()

If a complex function f z has a limit at the point z = z0 , then

()

( )

f z must be bounded in a neighborhood of z0 . Proposition 2.9-1, Bolzano-Weierstrass Theorem for Point Sets in the Complex Plane:

( )

( )

function z = g w exists and is differentiable at w0 = f z0 with

g ′ w0 ≠ 0 , then we have: !

( )

f ′ z0 =

1

( )

g ′ w0

If S is a bounded infinite set of points in the complex plane, then

S has a limit point. 631

Proposition 3.4-1, Cauchy-Riemann Equations:

Proposition 3.5-2:

() ( )

( )

The necessary and sufficient conditions for a complex function

If at a point z a complex function w = f z = u r, θ + i v r, θ

w = f z = u x, y + i v x, y

has a derivative f ′ z , then the polar exponential form of the

( ) ( ) ( ) to be differentiable z ( x, y ) are that the Cauchy-Riemann equations: ∂u ∂v = ! ∂x ∂ y

!

()

at a point

derivative is:

∂u ∂v =− ∂y ∂x

!

( )

()

f′ z =e

− iθ

⎡ ∂u ∂v ⎤ e− iθ ⎡ ∂v ∂u ⎤ + i = − i ⎢ ∂r ∂r ⎥⎦ r ⎢⎣ ∂θ ∂θ ⎥⎦ ⎣

hold at the point z x, y , and that the partial derivatives ∂u ∂x ,

∂u ∂ y , ∂v ∂x , and ∂v ∂ y all exist and are continuous in a

Proposition 3.6-1:

( )

neighborhood of the point z x, y .

The complex variables z and z are independent.

Proposition 3.5-1, Polar Form of Cauchy-Riemann Equations:

() ( )

( )

Proposition 3.6-2:

()

If the complex function f z = u r,θ + i v r,θ has a derivative

()

f′ z

For a complex function w = f z to be differentiable at a point

iθ

in some neighborhood of a point z = r e , then the

z , it is necessary that:

( )

Cauchy-Riemann equations in polar form at the point z r, θ

!

are: !

∂u 1 ∂v ! = ∂r r ∂θ

1 ∂u ∂v =− r ∂θ ∂r

( )=0

∂f z ∂z

Proposition 3.6-3:

()

Any complex function f z will satisfy the Cauchy-Riemann equations if and only if it is independent of z .

632

Proposition 3.6-4:

!

Proposition 3.7-4, Zero Derivative Theorem:

()

The Laplace operator can be written in terms of Wirtinger

Let f z be a holomorphic function in a domain D . Then if and

derivatives:

only if f ′ z = 0 for all z ∈D will f z be a constant in D .

∂2

∂2

∂2 + =4 ∂z ∂z ∂x 2 ∂ y 2

()

()

Proposition 3.7-5:

()

If f z is a holomorphic function in a domain D , and if the real Proposition 3.7-1, Continuous Holomorphic Functions

()

If a complex function f z is holomorphic on a domain D , then

()

f z is continuous in D . Proposition 3.7-2, Operations for Holomorphic Functions:

() () f ( z ) ± g ( z ) , f ( z ) g ( z ) , and f ( z ) g ( z )

If the complex functions f z and g z are both holomorphic on a domain D , then

()

with g z ≠ 0 are also all holomorphic on D . Proposition 3.7-3, Composition for Holomorphic Functions:

() () on a domain D , then g ( f ( z )) is also holomorphic on D .

If the complex functions f z and g z are both holomorphic

()

()

or imaginary part of f z is constant for all z ∈D , then f z is a constant in D . Proposition 3.7-6:

()

()

If f z and its conjugate f z are holomorphic functions in a

()

domain D , then f z is a constant in D . Proposition 3.7-7, Constant Modulus Theorem:

()

()

If f z is a holomorphic function in a domain D , and if f z

()

is constant for all z ∈D , then f z is a constant in D . Proposition 3.7-8:

()

()

If f z is a holomorphic function in a domain D , and if f z is

()

a real function for all z ∈D , then f z is a constant in D .

633

Proposition 3.7-9, Constant Argument Theorem:

()

Proposition 4.5-1:

If f z is a holomorphic function in a domain D , and if a

For z = r eiθ a complex logarithm Ln z is holomorphic when

branch of arg f z is constant for all z ∈D , then f z is a

− π < θ ≤ π and r > 0 .

()

()

constant in D . Proposition 5.3-1, Direction of Integration:

()

Proposition 3.9-1, L’Hôpital’s Rule:

()

!

()

If f z is an integrable complex function over a simple closed contour C in the positive direction, then f z is integrable over

a point z = z0 , and if f z0 = g z0 = 0 and g ′ z0 ≠ 0 , then:

the contour C in the negative direction, and we have:

lim

z → z0

( ) ( ) f (z) f ′(z) = lim g ( z ) z → z g′ ( z )

( )

!

!∫

()

f z dz

C

Complex Functions:

()

If a complex function f z is holomorphic in a simply

where k is an integer:

()

connected domain D , and if f z has an antiderivative

()

e z = e z+ i 2k π

e z = 1 if and only if z = 2 k π i where k is an integer.

()

f z dz = −

Proposition 5.3-2, Fundamental Theorem of Calculus for

The periodicity of the complex exponential function e z is 2 k π i

Proposition 4.2-2:

!∫

C

0

Proposition 4.2-1:

!

()

If f z and g z are complex functions that are holomorphic at

F z which is holomorphic in D so that: !

()

()

f z = F′ z =

()

dF z dz

then in the domain D :

634

∫

!

z2 z1

()

f z dz =

∫

z2 z1

2

dz

1

Proposition 5.3-6:

()

If a complex function f z

and !

( ) dz = F ( z ) − F ( z )

dF z

∫

()

f z dz =

∫

( ) dz = F ( z ) + c

dz

which is holomorphic in D , then along every simple closed contour C in D we must have:

Proposition 5.3-3:

!∫

()

()

()

domain D , and if f z has an antiderivative F z that is

()

()

f z dz = 0

C

If a complex function f z is continuous in a simply connected

Proposition 5.3-7, Contour Integral Inequality:

()

If C is a simple contour and f z is a complex function

holomorphic in D , then f z is holomorphic in D .

continuous over C , then:

Proposition 5.3-4: If F z and G z are distinct antiderivatives of a complex

()

()

connected domain D , and if f z has an antiderivative F z

!

()

()

dF z

where c is a complex constant.

()

is holomorphic in a simply

() simply connected domain D , then F ( z ) and G ( z ) differ by

∫

!

()

()

f z dz ≤

C

∫

C

()

f z

dz !

function f z , and if F z and G z are holomorphic in a

Proposition 5.3-8, ML-Inequality or Darboux Inequality:

some constant c .

()

()

The contour integral of a function f z that is holomorphic in a simply connected domain D will be independent of path in D if

()

f z has an antiderivative F z which is holomorphic in D .

be a

continuous complex function on C . If f z is bounded by a real

Proposition 5.3-5, Independent of Path:

()

()

Let C be a simple contour of length L , and let f z number M along C :

()

f z ≤M

! then:

635

∫

!

()

f z dz ≤ M L

C

Proposition 5.4-4, Independent of Path:

()

If a complex function f z is continuous in a simply connected Proposition 5.4-1, Cauchy’s Integral Theorem:

()

If f z is a holomorphic complex function with continuous first

domain D , and if !

order derivatives in a simply connected domain D , then along

!∫

()

f z dz = 0

C

C

for every simple closed contour C within D , then

every simple closed contour C that lies within D we must have: !

!∫ f ( z ) dz = 0

!

∫ f ( z ) dz is independent of path within D .

Proposition 5.4-2, Cauchy-Goursat Theorem: If

!

()

f z

is a holomorphic complex function in a simply

Proposition 5.4-5, Antiderivative Theorem:

()

connected domain D , then for every simple closed contour C

If a complex function f z is continuous in a simply connected

that lies within D we must have:

domain D , and if

!∫

C

()

f z dz = 0

!

!∫ f ( z ) dz = 0 C

()

for every simple closed contour C within D , then f z has an Proposition 5.4-3:

()

If f z is an entire function, then along every simple closed contour C in the complex plane we must have: !

!∫

C

()

antiderivative F z which is holomorphic in D so that: !

( )= f

dF z dz

(z)

()

f z dz = 0 636

Proposition 5.4-6: If

()

f z

Proposition 5.5-2, Cauchy-Goursat Theorem for Multiply

is a holomorphic complex function in a simply

()

connected domain D , then f z

()

!

Let C be a simple closed contour within a domain D , and let

will have an antiderivative

Ck where k = 1, 2, !, n be disjoint simple closed contours

F z which is holomorphic in D if and only if we have:

!∫

!

()

interior to the contour C . If f z is holomorphic at all points

()

f z dz = 0

C

Connected Domains:

within or on C , and outside or on each Ck , then:

for every simple closed contour C within D .

!

!∫

C

Proposition 5.4-7, Independent of Path: If

()

f z

is a holomorphic complex function in a simply

∫ f ( z ) dz !

is a point inside the contour C , then:

points within or on C1 , and outside or on C2 , then:

()

!∫

C2

()

1 dz = 2 π i z − z0

If C is a simple closed contour within a domain D , and if z = z0

()

C1

()

f z dz = 0

Proposition 5.5-4:

D , and let C2 be interior to C1 . If f z is holomorphic at all

f z dz

!∫

C

Let C1 and C2 be two simple closed contours within a domain

!∫

k =1

Ck

a point inside the contour C , then:

Proposition 5.5-1, Deformation Theorem:

f z dz =

∑ !∫

If C is a simple closed contour within a domain D , and if z0 is

is independent of path within D .

!

f z dz =

Proposition 5.5-3:

connected domain D , then: !

()

n

!

!∫

C

1

( z − z0 )n

⎧⎪ 2 π i n = 1 dz = ⎨ n ≠1 ⎪⎩ 0 637

where n is an integer.

!

∫ (

)

1 v z0 = 2π

2π

)

( )

Proposition 6.1-1, Cauchy’s Integral Formula:

()

Let f z

be a holomorphic complex function in a simply

!

( )

connected domain D . If C is a simple closed contour that lies

!

within D , and if z = z0 is a fixed point inside C , then:

Proposition 6.1-4:

( ) dz

Let f z

1 f z0 = 2π i

( )

!∫

C

()

f z

z − z0

0

be a holomorphic complex function in a simply

and if n loops are made around the contour C , then:

() if Dr ( z0 ) is a disk of radius r

centered at point z = z0 and

!

(

) ( )

n C, z0 f z0 =

contained within D , then:

( )

∫ (

v z0 + r ei θ dθ

within D , and if z = z0 is a fixed point inside the contour C ,

If f z is a holomorphic complex function on a domain D , and

!

0

u z0 + r ei θ dθ

connected domain D . If C is a simple closed contour that lies

Proposition 6.1-2:

1 f z0 = 2π

2π

1 u z0 = 2π

2π

(

!∫

C

()

f z

z − z0

dz

)

where n C, z0 is the winding number of C around z0 .

∫ f ( z + r e ) dθ 0

1 2π i

iθ

0

Proposition 6.1-5:

()

Proposition 6.1-3:

() a domain D , and if Dr ( z0 ) is a disk of radius r

Let f z

() ()

be a holomorphic complex function in a simply

connected domain D . If C is a simple closed contour that lies

If w = f z = u z + i v z is a holomorphic complex function on

within D , and if z = z0 is a fixed point that lies on the contour

centered at

C , then:

point z = z0 and contained within D , then: !

!∫

C

()

f z

z − z0

( )

dz = π i f z0

638

Proposition 6.1-6:

()

Let f z

!

Proposition 6.2-1, Cauchy’s Integral Formula for Derivatives:

()

Let f z

be a holomorphic complex function in a simply

connected domain D . If C1 and C2 are two concentric circles

connected domain D . If C is a simple closed contour within the

within D , and if z = z0 is a fixed point between the two circles,

domain D , and if z0 is any given point inside the contour C ,

then:

then f z has derivatives of all orders at the point z0 given by:

1 f z0 = 2π i

( )

!∫

C1

()

f z

1 dz − z − z0 2π i

!∫

C2

()

f z

z − z0

dz

Let

()

f z

()

! !

Proposition 6.1-7:

()

and g z

be two complex functions that are

()

if C is a simple closed contour C that lies within D , then:

!∫

C

simple closed contour C , then we have:

!∫

C

!If a complex function f z is holomorphic on a domain D , and

⎧ 0 ⎪ f z ⎪ dz = ⎨ 2 π i f z0 z − z0 ⎪ ⎪⎩ π i f z0

()

where n ∈! .

()

!

Proposition 6.1-8:

( ) ( )

z0

outside C

z0

inside C

z0

on C

( ) dz !∫C ( z − z0 )n+1 f z

If a complex function f z is holomorphic inside and on a

() () f ( z ) equals the value of g ( z ) everywhere within C .

value of f z on C equals the value of g z on C , then the value of

n! n f ( ) ( z0 ) = 2π i

Proposition 6.2-2:

holomorphic within and on a simple closed contour C . If the

!

be a holomorphic complex function in a simply

!

( ) dz = n!

n f() z

z − z0

( ) dz !∫C ( z − z0 )n+1 f z

where n ∈! .

639

Proposition 6.3-1, Higher Order Derivatives of a Holomorphic

Proposition 6.3-4, Morera’s Theorem:

()

Function Exist and are Holomorphic:

() ()

If a complex function f z is holomorphic at a point z = z0 , n then its derivatives f ( ) z of all orders n exist and are holomorphic at the point z0 .

If f z is a continuous complex function in a simply connected domain D , and if: !

!∫

()

f z dz = 0

C

()

for every simple closed contour C within D , then f z is

Proposition 6.3-2:

() ( )

holomorphic in D .

( )

If f z = u x, y + i v x, y is a holomorphic complex function at a point z = z0 in a domain D , then its real and imaginary parts

( )

( )

u x, y and v x, y

Proposition 6.4-1, Cauchy’s Inequality:

()

have continuous partial derivatives of all

Let f z

orders at the point z0 .

connected domain D , and let C be a circular contour within D of radius r = z − z0

Proposition 6.3-3:

() F ( z ) holomorphic in a simply connected domain D such that: dF ( z ) f (z) = F′(z) = dz ()

centered at z = z0 . If for some M > 0 ,

()

where M is a function of r , we have f z ≤ M for all points

If f z is a continuous complex function with an antiderivative

!

be a holomorphic complex function on a simply

on C , then: ! !

n! M n f ( ) z0 ≤ n r

( )

where n ∈! .

then f z is holomorphic in the domain D : Proposition 6.5-1, Liouville’s Theorem:

()

If an entire function f z is bounded, it must be a constant over the complex plane. 640

Proposition 6.5-2:

Proposition 6.7-3, Minimum Modulus Theorem:

()

A complex function bounded at infinity and not constant must

If f z ≠ 0 is a holomorphic and non-constant complex function

have at least one singularity in the finite complex plane.

within a closed bounded region R , and if f z is continuous on

R , then the minimum value of Proposition 6.6-1, Gauss’s Mean Value Theorem:

()

( ) is equal to f ( z ) around the boundary of any disk Dr ( z0 )

If f z is holomorphic on a domain D , then f z0 the mean value of

()

Proposition 6.8-1, Fundamental Theorem of Algebra:

()

Every non-constant polynomial Pn z : !

( )

() ( ) f ( z ) is constant on Dδ ( z0 ) .

that are complex constants with an ≠ 0 , will have n roots (which

()

are not necessarily distinct). Therefore Pn z can be completely factored so that:

Proposition 6.7-2, Maximum Modulus Theorem:

()

If f z

is a holomorphic and non-constant complex function

()

Pn z = a0 + a1 z + a2 z 2 +!+ an−1 z n−1 + an z n of finite degree n ≥ 1 , and having coefficients a0 , a1 , !, an−1 , an

( )

If f z is holomorphic on a disk Dδ z0 , and if f z ≤ f z0 for all z within Dδ z0 , then

must occur on the

boundary of R .

of radius r centered at point z = z0 and contained within D . Proposition 6.7-1, Modulus Theorem:

()

f z

()

!

() (

Pn z = c z − z1

) ( z − z2 ) ( z − z3 ) ! ( z − zn )

within a closed bounded region R , and if f z is continuous on

()

where the zeros zi can be complex numbers, and where c is a

R , then the maximum value of

must occur on the

complex constant.

()

f z

boundary of R . Proposition 6.8-2: Complex roots of polynomial equations having real coefficients only occur in conjugate pairs. 641

for any given k are orthogonal, where ck and dk are constants.

Proposition 6.9-1, Harmonic Function Theorem:

()

A complex function w = f z

is holomorphic on a simply

Proposition 6.9-5, Harmonic Conjugate:

()

( )

connected domain D if and only if f z is harmonic on D . The

If a real-valued function u x, y

imaginary part v x, y of f z is then the harmonic conjugate

connected domain D , then there exists a real-valued harmonic

( ) () of the real part u ( x, y ) of f ( z ) .

is harmonic in a simply

( ) ( ) f ( z ) = u + i v is holomorphic on D .

conjugate function v x, y of u x, y , defined up to an arbitrary constant, such that

Proposition 6.9-2, Harmonic Functions are Infinitely Differentiable:

Proposition 6.9-6:

()

( )

A harmonic function f z holomorphic on a simply connected

If a real-valued function u x, y

domain D is infinitely differentiable.

connected domain D , then any two harmonic conjugates of

is harmonic in a simply

( )

u x, y differ only by a constant. Proposition 6.9-3, Mean Value of a Harmonic Function:

()

If a function f z is harmonic on a simply connected domain

( )

Proposition 6.9-7:

()

D , and if z = z0 is a point within D , then f z0 is equal to the

If a complex function f z

mean value of f z around the boundary of any disk Dr z0 of

connected domain D , then the Laplacian operator on f z can

radius r centered at point z0 and contained within D .

be represented using Wirtinger derivatives:

()

( )

Proposition 6.9-4, Orthogonal Equipotential Curves:

( )

( )

!

is holomorphic on a simply

()

⎛ ∂2 ∂2 ⎞ ∂ ∂ ⎡ ⎤ ⎡⎣ f z ⎤⎦ + f z = 4 ⎜ 2 2⎟ ⎣ ⎦ ∂z ∂z ∂y ⎠ ⎝ ∂x

()

()

If u x, y and v x, y are conjugate harmonic functions, then the set of equipotential curves !

( )

u x, y = ck !

( )

v x, y = dk 642

Proposition 6.9-8:

Proposition 7.1-3, Convergent Sequences are Bounded:

()

If a complex function f z is holomorphic on a simply

{ }

If a complex sequence zn converges, then it is bounded.

connected domain D , then we have:

()

f′ z

!

2

=

∂ ∂ ⎡⎣ f z ⎤⎦ ∂z ∂z

()

Proposition 7.1-4, Monotonic Sequence Convergence: A monotonic real-valued sequence

{ an }

is convergent if and

only if it is bounded.

Proposition 6.9-9:

()

If a complex function f z is holomorphic and non-constant in

( )

a simply connected domain D , then the real part u x, y

()

of

Proposition 7.1-5, Subsequences Converge to the Same Limit as the Sequence:

f z cannot have a maximum or minimum value within D .

If a complex sequence Proposition 7.1-1, Convergent Sequence Requirement:

{ } converges to a limit z0

A complex sequence zn

if and only if

infinite subsequences

{ zn }

converges to a limit z0 , then all

{ z } of { z } also converge to the limit nk

n

z0 .

we have: !

Proposition 7.1-6, Bolzano-Weierstrass Theorem for Complex

lim zn − z0 = 0

n→ ∞

Sequences:

Proposition 7.1-2, Convergent Sequence has a Unique Limit:

{ }

If a complex sequence zn converges, then it has a unique limit.

Every bounded sequence of complex numbers has a convergent subsequence.

643

Proposition 7.1-7:

Proposition 7.1-13:

{ }

{

A point z0 is a limit point of an infinite bounded sequence zn

A complex sequence zn = xn + i yn

if there is a subsequence zn

z0 = x0 + i y0 so that:

{ } of { z } that converges to z . k

n

0

Proposition 7.1-8, Cauchy’s Convergence Criteria:

lim z = z0

!

n→∞ n

{ }

{ } converges if and only if it is a Cauchy

if and only if xn

A complex sequence zn

n→∞ n

Proposition 7.1-9:

!

lim

n,k → ∞

{ yn } converges

lim x = x0

!

{ }

converges to a limit x0 and

to a limit y0 so that:

sequence.

A complex sequence zn converges if and only if we have:

} converges to a limit

and !

zn − zk = 0

lim y = y0

n→∞ n

Proposition 7.1-14: Proposition 7.1-10, Cauchy Sequence Converges: Every complex Cauchy sequence converges.

{ }

If a sequence of complex numbers zn converges to a limit z0 : !

Proposition 7.1-11, Unique Limit of Cauchy Sequence: Every complex Cauchy sequence has a unique limit.

lim z = z0

n→∞ n

then: !

lim zn = z0

n→ ∞

Proposition 7.1-12, Cauchy Sequence Bounded: Every complex Cauchy sequence is bounded. 644

Proposition 7.1-15:

Proposition 7.1-17:

{ }

{ } and {ζ n } are two sequences of complex numbers, and if { zn } converges to a limit z0 :

A sequence of complex numbers zn converges to a limit

If zn

z0 = x0 + i y0 : !

lim z = z0

n→ ∞ n

!

if and only if: !

{

lim zn = z0

n→ ∞

!

{ }

For a sequence of complex numbers zn we have:

lim z = 0

(

)

lim ζ n − zn = 0

n→ ∞

{ }

then ζ n converges to a limit z0 : !

lim ζ n = z0

n→ ∞

n→ ∞ n

if and only if: !

}

and ζ n − zn converges to a limit 0:

Proposition 7.1-16:

!

lim z = z0

n→ ∞ n

Proposition 7.1-18:

lim zn = 0

If a sequence of complex numbers

n→ ∞

{ zn }

converges to a limit

z0 ≠ 0 : !

lim z = z0

n→ ∞ n

{

}

then the sequence 1 zn converges to a limit 1 z0 : !

lim 1 zn = 1 z0

n→ ∞

645

Proposition 7.2-1:

Proposition 7.2-4:

{ ( ) } and { b ( z )} are two convergent sequences of complex numbers, then { a ( z ) + b ( z ) } is also a convergent sequence of If an z

{ ( ) } and { b ( z )} are two convergent sequences of complex numbers, then { a ( z ) b ( z ) } is also a convergent sequence of If an z

n

n

n

n

()

()

()

()

{ ( )} is not zero. We will then have:

()

lim ⎡⎣ an z ± bn z ⎤⎦ = lim an z ± lim bn z n→∞ n→∞

n→ ∞

limit of bn z !

Proposition 7.2-2: The sum or difference of two null sequences is a null sequence.

n

n

n

complex numbers with:

() ()

()

()

lim ⎡⎣ an z bn z ⎤⎦ = lim an z lim bn z n→ ∞ n→ ∞

n→ ∞

()

lim b z

n→ ∞ n

Proposition 7.2-5:

sequence.

{ ( ) } and { b ( z )} are two convergent sequences of complex numbers, then { a ( z ) b ( z )} is also a convergent sequence of !

lim an ( z ) ( ) bn ( z ) ⎤⎦ = n→ ∞

lim ⎡ a z

n→ ∞ ⎣ n

The product of a bounded sequence and a null sequence is a null

Proposition 7.2-3: If an z

n

complex numbers providing that bn z ≠ 0 for all n > 0 and the

complex numbers with: !

n

Proposition 7.3-1:

()

A complex function f z will be continuous at a point z = z0 in a region R if and only if for every sequence

{ zn }

in R that

{ ( )} converges to f ( z ) .

converges to z0 , the sequence f zn

0

646

Proposition 7.3-2, Cauchy’s Convergence Criteria for Complex

Proposition 7.4-1, Uniformly Convergent Sequence:

Functions: A sequence of complex functions

A sequence of single-valued complex functions

{ f ( z )} converges pointwise

()

for any real number ε > 0 there exists N = N ε ∈! such that:

such that:

()

()

!

for all n > k > N

()

if and only if the sequence is uniformly Cauchy within R so that

Cauchy. For any real number ε > 0 , there then exists N ∈!

fn z − fk z < ε !

n

converges uniformly within its region of convergence R to f z

n

on a domain D if and only if in D the sequence is pointwise

!

{ f ( z )}

()

()

fn z − fk z < ε ! for all z ∈R when n, k > N

Proposition 7.4-2:

Proposition 7.3-3: If a sequence of single-valued complex functions continuous at a point z = z0 , then the sequence

{ f ( z )} is

If a sequence of continuous single-valued complex functions

n

its region of convergence R , then f z is continuous within R .

{ f ( z )} converges uniformly to f ( z ) at every point z

n

{ f ( z ) } is also

n

within

()

continuous at point z0 . Proposition 7.4-3: Proposition 7.3-4:

If a sequence of continuous single-valued complex functions

{ f ( z )} converges uniformly to f ( z ) at every point within its

If a sequence of continuous single-valued complex functions

{ f ( z )} converges pointwise at a point z = z within its region of convergence R , then { f ( z )} converges to f (z0): n

region of convergence R , and if z = z0 is a point within R , then

0

n

!

n

()

( )

lim f z = f z0

n→∞ n

we have: !

⎛ ⎞ lim ⎛ lim fn z ⎞ = lim ⎜ lim fn z ⎟ ⎠ n→ ∞ ⎝ z → z0 z → z0 ⎝ n→ ∞ ⎠

()

()

647

Proposition 7.4-4:

Proposition 7.4-7:

{ f ( z )} be a sequence of holomorphic functions in a simply connected domain D . If { f ( z )} converges uniformly to f ( z )

If a sequence of continuous single-valued complex functions

{ f ( z )}

()

converges uniformly to f z

n

Let

within its region of

{

convergence R , and if within R the terms of the sequence are all

()

()

converges uniformly to f z

uniform convergence:

lim

∫

C

()

()

fn z dz =

()

{ zk } have the limit zero:

!

∫

()

lim f z dz =

n C n→ ∞

∫

()

C

If a sequence of holomorphic functions

lim zk = 0

k→∞

Proposition 8.2-2, Series Remainder Theorem:

{ f ( z )} n

{ ( ) } will converge to a

A complex sequence of partial sums Sk z

()

sum S z

converges

()

uniformly to f z in a simply connected domain D , then f z is holomorphic in D .

lim z = 0 !

k→∞ k

f z dz

Proposition 7.4-6:

()

{ ( ) } to converge to a

sum S z , it is necessary that the k th term zk of the sequence

within its region of

convergence R , then for any contour C within R we have the

n→ ∞

k f ( ) z in D .

For a complex sequence of partial sums Sk z

If a sequence of continuous single-valued complex functions

!

( ) } converges uniformly to

Proposition 8.2-1, Series Convergence Requirement:

Proposition 7.4-5:

{ ( )}

n

k in D , then f n( ) z

bounded, then f z is bounded within R .

fn z

n

()

if and only if the remainder Rk z

of the series

converges so that: !

()

lim Rk z = 0

k→∞

648

Proposition 8.2-3:

Proposition 8.2-7, Omit beginning Terms of a Series:

{ ( ) } and the remainders { R ( z ) } will

An infinite complex series that is convergent will remain

For any k the complex sequence of partial sums Sk z corresponding complex sequence of

convergent if a finite number of terms are omitted from the

k

beginning of the series.

converge or diverge together.

Proposition 8.2-8: Proposition 8.2-4: If a complex sequence of partial sums

()

{ S ( z )} k

converges to a

()

k

!

()

()

k > N

ε > 0 there exists N ∈! such that: n

!

∑z

j

k > N

∑ Re ( S ( z )) !

∑ Im ( S ( z ))

k

k

n =1

converge. Proposition 8.2-9, Addition and Subtraction of Series:

{ }

{ }

If an and bn are two sequences of complex numbers, and if:

Proposition 8.2-6, Cauchy’s Series Convergence Criteria: A complex series converges if and only if, for any real number

k

n =1

ε > 0 there exists N ∈! such that: Sn z − Sk z

()

sums:

A complex series converges if and only if, for any real number

!

()

S z = Re S z + i Im S z if and only if the sequences of partial

sum S z , then the sequence of partial sums is bounded. Proposition 8.2-5, Cauchy’s Series Convergence Criteria:

{ ( ) } will converge to a sum

A sequence of partial sums Sk z

∞

!

∑a ! n

n =1

!

∞

∑b

n

n =1

are convergent infinite series, then:

j = k+1

649

∞

∑( a ± b )

!

n

Proposition 8.3-1, nth Term Test for Divergence:

n

n =1

∞

A complex series

is also a convergent series and ∞

∑(

!

n =1

⎡ ∞ ⎤ ⎡ ∞ ⎤ an ± bn = ⎢ an ⎥ ± ⎢ bn ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ n =1 ⎦ ⎣ n =1 ⎦

∑

)

∑

∑z

n

diverges if lim zn ≠ 0 .

n =1

n→ ∞

Proposition 8.3-2, Comparison Test: If zn ≤ Mn where Mn are nonnegative real numbers, and where

Proposition 8.2-10, Rearrangement of Series:

{ } { } { bn } is any rearrangement of { an } , and if:

If an and bn are two sequences of complex numbers where ∞

∞

!

n

! ∞

n

and

n =1

∑b

n

∑z

n

will converge absolutely in ! .

converge to the same sum:

n =1

∞

!

converges in ! , then the series:

n =1

is an absolutely convergent infinite series, then the two series

∑a

n

∞

n =1

∞

∑M n =1

!

∑a

!

the series:

∞

∑ a =∑ b n

n =1

Proposition 8.3-3:

{ converge absolutely to S ( z ) = X ( x ) + i Y ( y ) where:

n

The sum of a sequence of complex numbers zn = xn + i yn

n =1

!

()

S z =

} will

∞

∑z

n

n =1

650

!

if and only if: ∞

()

X x =

!

∑

xn !

∞

( )

Y y =

!

q > 1 or q = ∞ , the series diverges

!

!

q = 1, the test is inconclusive

n =1

n =1

!

∑

yn

!

Proposition 8.3-6, Cauchy’s Root Test:

both converge absolutely.

Given a series: Proposition 8.3-4, An Absolutely Convergent Series is Convergent:

∞

∑z

!

n

n=0

If a complex series converges absolutely, then it converges. That ∞

is, if

∑

∞

zj converges, then

j =1

∑z

j

also converges.

j =1

!

∑

zn

n =1

!

where:

lim

!

n→ ∞

zn+1 zn

n→ ∞

zn = q

!

q < 1 , the series converges absolutely

!

!

q > 1 or q = ∞ , the series diverges

!

!

q = 1, the test is inconclusive

Proposition 8.5-1:

()

=q

then if: !

n

then if:

Given a series: !

lim

!

Proposition 8.3-5, Ratio Test:

∞

where:

If a series of complex functions converges to a sum f z : !

q < 1 , the series converges absolutely

()

f z =

∞

∑ f (z) n

n=0

then: 651

()

k→∞

()

sum f z must be a continuous function within R .

Proposition 8.6-1, Cauchy Criteria for Uniform Convergence:

Proposition 8.6-3:

A complex function series:

If a sequence of continuous single-valued complex functions

{ f ( z )} converges uniformly to f ( z ) at every point within its

∞

∑ f (z)

!

()

converges uniformly to f z at all points z within R , then the

lim f k z = 0

!

n

region of convergence R , and if z = z0 is a point within R ,then

n

n=0

we have:

converges uniformly and absolutely to a limit for every point z within its region of convergence R if and only if the sequence is

∞

lim

!

z→z0

uniformly Cauchy within R such that:

∞

∑ f ( z ) =∑ lim f ( z) n

n=0

n=0

z→z0 n

n

∑ f (z) < ε !

!

j

for all z ∈R and n > k > N

Proposition 8.6-4, Weierstrass’s M-Test:

j = k+1

Let

{ f ( z )} be a sequence of single-valued complex functions n

!

n

()

is convergent, and if fn z ≤ Mn independent of z in a domain

D , then the complex function series:

∞

∞

n

n=0

∑M n=0

convergence R of the sequence. If the complex function series:

∑ f (z)

n

∞

each of which is continuous at all points z within the region of

!

n

sequence of positive constants. If the series:

Proposition 8.6-2: Let

{ f ( z )} be a sequence of complex functions and { M } be a

!

∑ f (z) n

n=0

652

converges uniformly and absolutely in D for all n > N where Proposition 8.6-7:

N ∈! .

()

If a series of continuous complex functions fn z

()

Proposition 8.6-5, Term-by-Term Integration:

()

If a series of continuous complex functions f k z

()

converges

uniformly to the sum f z in a domain D , and if the series is

uniformly to the sum f z in a domain D , then for any simple

()

()

multiplied by any function g z bounded in D , the resulting

converges

series will also converge uniformly in D .

contour C that lies within D the integral of f z along C can be determined using term-by-term integration of the series: n

!

lim

n→ ∞

n

∫ ∑ f ( z ) dz = lim ∑ ∫ C

k

k=0

n→ ∞

k=0

C

Proposition 8.6-8, Term by Term Differentiation:

()

If a series of continuous complex functions fn z

()

fk z dz

()

converges

uniformly to the sum f z in a domain D , then the derivative

()

of f z at any point z in D can be determined by term-by-term differentiation of the series:

Proposition 8.6-6:

{ ( ) } is a sequence of holomorphic functions in a simply

If fk z

!

k f ( ) ( z) =

∞

∑

k f n( ) z

()

n=0

connected domain D , and if the series: ∞

!

∑ f (z)

Proposition 8.7-1, Circle of Convergence of a Power Series:

k

k= 0

()

Every power series:

()

converges uniformly to f z in D , then f z is holomorphic on D .

∞

!

∑

(

an z − z0

)n

n=0

has a circle of convergence z − z0 = r with radius given by:

653

!

an r = lim n→ ∞ an+1

Proposition 8.7-4, Radius of Convergence: The radius of convergence r of a power series:

if this limit exists. Within the circle of convergence the power series will converge absolutely, and outside this circle it will

∞

!

diverge.

∑

(

an z − z0

)n

n=0

Proposition 8.7-2, Infinite Radius of Convergence: ∞

If for a power series

∑a (z− z ) n

0

n

we have:

is given by: !

1

r=

lim

n=0

!

lim

n→ ∞

an+1 an

n→ ∞

=0 !

converges for all z (for the entire z-plane). Proposition 8.7-3, Zero Radius of Convergence: ∞

!

lim

n→ ∞

an+1 an

an

and by:

then the radius of convergence is r = ∞ , and the power series

If for a power series

n

∑

(

an z − z0

converges only at the point of expansion z0 .

an+1

Proposition 8.7-5, Power Series Converges at its Center: Every power series:

n=0

then the radius of convergence is r = 0 , and the power series

n→ ∞

an

if the limits exist.

)n we have:

=∞

r = lim

∞

!

∑

(

an z − z0

)n

n=0

converges at its center z = z0 .

654

!

Proposition 8.7-6, Abel’s Theorem: If a power series: ∞

!

∑

(

an z − z0

(

lim an z − z0

n→ ∞

)n = 0

Proposition 8.7-9, Uniform Convergence of a Power Series:

)n

()

Every function f z that can be represented by a power series:

n=0

converges at some point z = z1 where z1 ≠ z0 , then it converges

!

()

f z =

∞

∑

(

an z − z0

)n

n=0

absolutely for every point z inside the circle r = z1 − z0 .

converges uniformly inside its circle of convergence z − z0 = r . Proposition 8.7-7, Divergence of a Power Series: Proposition 8.7-10, Continuity of a Function Represented by a

If a power series: ∞

!

∑

(

an z − z0

Power Series:

)n

()

If a power series represents a function f z :

n=0

diverges at a point z = z1 where z1 ≠ z0 , then it diverges for every point z outside the circle r = z1 − z0 .

!

()

f z =

∞

∑

(

an z − z0

)n

n=0

()

then f z will be continuous at all points within its circle of Proposition 8.7-8:

convergence z − z0 = r .

()

If a power series converges to a sum f z :

()

f z =

!

∞

∑

(

an z − z0

)n

n=0

then: 655

Proposition 8.7-13, Uniqueness of Power Series Representation:

Proposition 8.7-11, Term by Term Integration: A power series:

()

f z =

!

A power series provides a unique representation of a function

∞

∑ a (z − z ) n

()

f z within its circle of convergence z − z0 = r .

n

0

Proposition 8.7-14:

n=0

!

can be integrated term by term at every point along any simple

The integral of a power series along any simple closed contour

contour C lying within its circle of convergence:

C lying inside the circle of convergence of the series is given by:

∫

C

∞

()

f z dz =

∫∑ C

(

an z − z0

)

n

n=0

∞

dz =

∑∫ n=0

C

(

an z − z0

)

∞

n

dz

Proposition 8.7-12, Radius of Convergence of Integrated !

()

f z =

∑ a (z − z ) n

n=0

the radius of convergence z − z0 = r is the same as that of the

)n dz = 0

n=0

()

If a function f z can be represented by a power series:

()

f z =

∞

∑

(

an z − z0

)n

n=0

n

0

(

an z − z0

Proposition 8.7-15, Power Series is Holomorphic:

!

∞

!∫ ∑ C

Power Series: For a power series:

!

!

()

then f z is holomorphic at every point inside its circle of convergence z − z0 = r

series obtained by integrating the power series: !

∫

C

()

f z dz =

∞

n+1 z − z ( ) ∑ n +1 0

an

n=0

656

Proposition 8.7-16, Term by Term Differentiation:

Proposition 8.7-18, Derivatives of All Orders:

A power series: !

()

f z =

Derivatives of all orders exist for a power series:

∞

∑ a (z − z ) n

n

!

0

n=0

∞

()

∑na (z − z ) n

(

an z − z0

)n

at every point inside its circle of convergence so that:

of convergence z − z0 = r giving the derivative of the sum:

f′ z =

∑

()

f z =

n=0

can be differentiated term by term at every point inside its circle

!

∞

n−1

!

k f( ) z =

()

∞

∑ ( n − k )! n!

(

an z − z0

)n−k

n= k

0

n =1

Proposition 8.7-19: Proposition 8.7-17, Radius of Convergence of the Derivative of a

()

()

all integrals of f z are also entire functions.

The derivative of a power series: !

()

∞

f′ z =

∑ na ( z − z ) n

n−1

Proposition 8.7-20, Convergence of Sums and Products of Two Series:

0

n= 1

()

!

()

f z =

∑ n=0

()

If f z and g z are two functions that can be represented by

has the same radius of convergence as the original power series: ∞

()

If f z is an entire function, then all derivatives of f z and

Power Series:

the converging series:

(

an z − z0

)n

!

()

f z =

∞

∑

(

an z − z0

)n

n=0

657

!

∞

()

g z =

∑

(

bn z − z0

)n

Proposition 8.8-2, Existence of Taylor Series:

n=0

() ()

f z +g z

() ()

f z g z

and

()

exist at a point z0 if and only if f z is infinitely differentiable

both also have power series

at z = z0 .

representations with a radius of convergence not less than

(

)

r = min r1 , r2 .

Proposition 8.8-3, Existence of Taylor Series:

Proposition 8.8-1, Taylor Series:

()

()

series:

()

f z =

exist on a domain D if and only if f z is holomorphic on D . Proposition 8.8-4:

∞

∑

()

A Taylor series representation of a complex function f z will

If a complex function f z can be represented by the power

!

()

A Taylor series representation of a complex function f z will

whose circle of convergence radii are r1 and r2 respectively, then

(

an z − z0

)n

()

If a complex function f z can be represented as a Taylor series

n=0

()

about a point z = z0 , then f z will be divisible by z − z0 if and

( )

inside its circle of convergence z − z0 = r , then the coefficients

()

only if f z0 = 0 so that z0 is a zero of f z .

of the series are given by: !

an =

n f ( ) ( z0 )

n!

=

1 2π i

( ) ds !∫C ( s − z0 )n+1 ) f s

where n = 0, 1, 2, ! , and the series is a Taylor series.

Proposition 8.8-5, Uniform Convergence of Taylor Series: If a Taylor series: !

()

f z =

∞

∑ n=0

n f ( ) z0

( )

n!

( z − z0 )n

has a radius of convergence r , then the series converges uniformly on any closed disk of radius 0 < ρ < r . 658

Proposition 8.8-6, Taylor Series Representation of Holomorphic Functions:

()

If a complex function f z

Proposition 8.8-9, Uniqueness of Taylor Series:

( )

is holomorphic in a simply

points in a disk Dr z0 , the coefficients an of the series:

connected domain D , and if z = z0 is some given point in D ,

()

then f z can be expanded as a Taylor series centered at z0 : !

()

f z =

∞

∑

(

an z − z0

)n

∞

=

n=0

∑ n=0

n f ( ) ( z0 )

n!

!

()

f z =

∞

∑

(

an z − z0

)n

n=0

( z − z0 )n

()

( )

are unique for an expansion of f z in Dr z0 .

( )

that converges on a disk Dr z0 having a radius r such that the

Proposition 8.8-10, Term by Term Differentiation of a Taylor Series:

disk is the largest that can be contained within D .

A Taylor series:

Proposition 8.8-7, Singularity on Circle of Convergence: The circle of convergence of a Taylor series representing a

()

()

If a Taylor series about a point z = z0 converges to f z for all

!

()

()

f z =

∞

∑

(

an z − z0

)n

n=0

function f z contains at least one singularity of f z on the

can be differentiated term by term at every point inside its circle

circumference of the circle.

of convergence z − z0 = r , giving the derivative of the sum: Proposition 8.8-8:

()

Every entire complex function f z can be represented as a Taylor series.

!

()

f′ z =

∞

∑

(

n an z − z0

)n−1

n=1

659

Proposition 8.8-11, Derivatives of All Orders of a Taylor Series:

Proposition 8.8-14, Taylor Series Inequality:

()

Let f z be holomorphic on a simply connected domain D and

Derivatives of all orders exist for a Taylor series: !

()

f z =

∞

∑

n f ( ) z0

( )

n!

n=0

have a circle C of convergence z − z0 = r inside D . If for some

( z − z0 )n

()

M > 0 , where M is a function of r , we have f z ≤ M for all points inside and on C , then the coefficients of the Taylor series

()

representing f z are limited by:

at every point inside its circle of convergence. !

Proposition 8.8-12:

() () () ()

If f z and g z are holomorphic in a domain D , then n n f z ≡ g z if and only if f ( ) z ≡ g ( ) z for any integer n .

()

()

!

()

f′ z =

∑ na ( z − z ) n

0

n−1

!

()

f z =

∑ a (z − z ) n

n=0

0

()

expansions about the same point where: !

()

f z =

∞

∑

(

an z − z0

)n

n=0

!

has the same radius of convergence as the original Taylor series: n

n = 0, 1, 2, !

!

()

with a circle of convergence z − z0 = r1 , and where:

n=1

∞

rn

If two complex functions f z and g z have Taylor series

Taylor Series:

∞

M

Proposition 8.10-1, Taylor and Maclaurin Series Properties:

Proposition 8.8-13, Radius of Convergence of the Derivative of a

The derivative of a Taylor series:

an ≤

!

()

g z =

∞

∑

(

bn z − z0

)n

n=0

with a circle of convergence z − z0 = r2 , then we have

660

the following properties for a circle of convergence of

(

)

Proposition 8.11-1, Convergence of Laurent Series:

r = min r1, r2 : !

1.!

∞

() ()

f z ±g z =

∑( a ± b ) ( z − z ) n

n

()

If a complex function f z has an isolated singularity at a point

n

n=0

!

2.!

() ()

f z g z =

∞

∑

z0 will converge for all points in an annular domain centered at

(

cn z − z0

z0 , and will converge absolutely and uniformly in every sub-

)n

()

annulus of this domain. The function f z will be analytic in

n=0

the annular domain.

where

Proposition 8.11-2, Laurent Series:

n

cn =

!

∑a b k

()

n−k

If a complex function f z is analytic in an annular domain D

k =0

!

3.!

( ) = c z−z n ! n( 0) g (z) ∑ n=0 ∞

f z

where !

an =

∑

that is centered at an isolated singularity z = z0 and defined by

( )

r < z − z0 < R , where 0 ≤ r < R ≤ ∞ , then

g z0 ≠ 0

()

f z

can be

represented by the series: !

n

()

z = z0 , the Laurent series representing the function f z about

0

()

∞

f z =

∑ (z − z ) ∑ n

n=1

ck bn−k

k =0

∞

bn

0

+

(

cn z − z0

)n

n=0

where !

1 bn = 2π i

( ) ds ! !∫C (s − z0 )− n +1 f s

n =1, 2, 3, !

661

!

cn =

1 2π i

( ) ds ! !∫ C (s − z0 )n +1 f s

n = 0, 1, 2, !

Proposition 8.11-4, Uniqueness of Laurent Series: If the Laurent series about an isolated singularity z0 converges

and where C is any positively oriented simple closed contour

()

to f z for all points in an annular domain D centered at

lying within the annular domain D surrounding z = z0 .

z = z0 , the coefficients an of the series:

Proposition 8.11-3, Coefficients of the Laurent Series:

()

If a complex function f z is analytic in an annular domain D

!

()

f z =

∑

(

an z − z0

∑

)n ()

Proposition 8.11-5, Cauchy’s Estimate for the Coefficients of a

)n

Laurent Series:

n=−∞

()

If a Laurent series about a point z0 converges to f z for all

are given by: !

(

an z − z0

are unique, and so is the Laurent series representing f z .

Laurent series representing the function: !

()

f z =

n=−∞

centered at an isolated singularity z = z0 , the coefficients of the ∞

∞

( ) ds ! 1 an = ∫C (s − z0 )n+1 2π i ! f s

points in the annulus ρ1 < z − z0 < ρ2 , the absolute value of the

n = 0, ± 1, ± 2, !

where C is any positively oriented simple closed circular

coefficients an of the series: !

()

∞

f z =

∑

(

an z − z0

)n

n=−∞

contour centered at z0 and lying within the annular domain

can be estimated by:

surrounding z0 . !

an ≤

M ! n ρ

n = 0, ± 1, ± 2, !

()

where M = max f z

and ρ = z − z0 . 662

Proposition 9.1-1, Riemann’s Theorem on Isolated Removable

Proposition 9.1-5, Pole not Bounded:

Singularities:

( )

Dδ* z0 , and that has a pole at a point z = z0 cannot be bounded.

If a point z = z0 is an isolated singularity of an analytic function

() () f ( z ) is bounded in some neighborhood of z0 .

f z , then z0 is a removable singularity of f z if and only if

Proposition 9.1-6:

( )

Dδ* z0 has a pole of order k ∈! at a point z = z0 , then:

If a point z = z0 is an isolated singularity of an analytic function

()

f z , then z0 is a removable singularity of f z if and only if: !

( z→ z

!

lim z − z0

( z→ z

)k f ( z ) ≠ 0

( z→ z

)k+1 f ( z ) = 0

0

) ()

and

lim z − z0 f z = 0 0

()

If a complex function f z that is analytic in a punctured disk

Proposition 9.1-2, Isolated Removable Singularity:

()

()

A complex function f z that is analytic in a punctured disk

!

lim z − z0 0

Proposition 9.1-3, Isolated Removable Singularity: If a point z = z0 is an isolated singularity of an analytic function

()

Proposition 9.1-7:

()

()

f z , then z0 will be a removable singularity of f z if and

()

If a complex function f z that is analytic in a punctured disk

only if lim f z exists and is finite. z → z0

( ) has a pole of order k ∈! at point z = z0 , then ( z − z0 ) k f ( z ) will not have a pole at z0 , but ( z − z0 )k−1 f ( z ) will Dδ* z0

Proposition 9.1-4, Isolated Removable Singularity:

() () g ( z ) have a zero of order n

() f (z) g (z)

have pole at z0 . We will also then have:

If f z and g z are analytic functions, and if both f z and at a point z = z0 , then

has a removable singularity at z0 .

!

()

lim f z = ∞

z → z0

663

Proposition 9.1-8:

Proposition 9.1-11, Meromorphic Function:

()

()

( ) if f ( z ) can be represented by the Laurent series: g (z) f (z) = ( z − z0 )k where g ( z ) is analytic at z0 , and g ( z0 ) ≠ 0 .

Dδ* z0 will have a pole of order k at the point z = z0 if and only

!

() f (z).

If f z is a meromorphic function, then f ′ z will also be a

A complex function f z that is analytic in a punctured disk

meromorphic function with the same poles as Proposition 9.1-12:

() curve C , then the poles of f ( z ) inside C are finite in number.

If a complex function f z is analytic inside a simple closed

Proposition 9.1-13: Proposition 9.1-9:

()

()

()

An analytic function f z can have an infinite number of poles

If f z and g z are complex functions that are analytic in a

( )

only in an open or unbounded region.

punctured disk Dδ* z0 , and that have poles of orders j and k ,

() ()

respectively, at a point z = z0 , then f z g z will have a pole of

Proposition 9.1-14, Casorati-Weierstrass Theorem:

order j + k at z0 .

()

If an analytic function f z has an isolated essential singularity at a point z = z0 , then in any δ neighborhood of z0 , the function

Proposition 9.1-10:

()

If a complex function f z

( )

()

f z will come arbitrarily close to any given complex value w0 . is analytic in a punctured disk

Dδ* z0 and has a pole of order k at the point z = z0 , then the n nth derivative f ( ) z of f z will have a pole of order k + n at

()

()

That is, given any complex value w0 , for every real number

ε > 0 we will have: !

()

f z − w0 < ε !

when 0 < z − z0 < δ

z = z0 . 664

!

Proposition 9.1-15:

()

If an analytic function f z

()

∞

f z =

∑

(

an z − z0

)n = ak ( z − z0 )k +

(

ak+1 z − z0

n=0

has an isolated singularity at a

point z = z0 , then the singularity z0 will be a pole if and only if

()

such that:

lim f z = ∞ .

!

an = 0 !

for n = 0, 1, 2,!, k − 1

Proposition 9.1-16:

!

an ≠ 0 !

for n = k

z → z0

()

If an analytic function f z

has an isolated singularity at a

point z = z0 , then this singularity is an essential singularity if

Proposition 9.2-2:

()

()

( )

A complex function f z that is analytic in a disk Dδ z0 will

and only if lim f z does not exist, even as infinity. z → z0

)k+1 + !

have a zero of order k at the point z = z0 if and only if its Taylor

Proposition 9.1-17:

series can be expressed in the form:

()

If a complex function f z is analytic in a domain D except for

!

an infinite set of points z1 , z2 , z3 , ! which are singularities of

()

)k ( ) ()

where k is a unique integer, and g z is a power series analytic

f z , and if these points have a limit point z = z0 in D , then

( )

( )

in the disk Dδ z0 with g z0 ≠ 0 .

()

this limit point must be a singularity of f z . Moreover this singularity must be a nonisolated essential singularity.

() (

f z = z − z0 g z

Proposition 9.2-3:

()

( )

Let f z be a complex function analytic in a disk Dδ z0 and

Proposition 9.2-1:

()

let it have a Taylor series that can be expressed in the form:

( )

A complex function f z that is analytic in a disk Dδ z0 will have a zero of order k at the point z = z0 if and only if it can be

!

() (

)k ( )

f z = z − z0 g z

represented by a Taylor series: 665

()

( ) ( )

where g z is analytic in Dδ z0 , g z0 ≠ 0 , and k ≥ 2 is an

()

integer. The function f z then has a zero of order k at z = z0 ,

()

and f ′ z has a zero of order k − 1 at z0 .

Proposition 9.2-6, Isolated Zero:

()

Let a complex function f z be analytic in a domain D , and let

( )

()

z = z0 be a point in D . If f z0 = 0 , then either f z

Proposition 9.2-4:

()

Let a complex function f z

is

identically zero in a neighborhood of z0 , or z0 is an isolated zero be holomorphic in a simply

()

of f z .

connected domain D . If C is a simple closed contour within the

()

domain D , and if f z has exactly one zero within C which lies at a point z0 , then:

1 z0 = 2π i

!

!∫

C

( ) dz f (z)

z f′ z

Proposition 9.2-5:

()

()

disk Dδ z0 and have zeros of orders j and k , respectively, at a

() ()

point z = z0 , then f z g z will have a zero of order j + k at

z0 .

()

Let a complex function f z be analytic in a domain D , and let

()

z = z0 be a point in D . If z0 is an isolated zero of f z , then

( )

there will exist a deleted disk Dδ* z0 with δ > 0 containing no

()

zeros of f z . Proposition 9.2-8:

If f z and g z are complex functions that are analytic in a

( )

Proposition 9.2-7, Isolated Zero:

()

{ }

Let a complex function f z be analytic in a domain D . If zn

is a sequence of distinct points in D which converges to a point

( )

()

z = z0 , and if f zn = 0 for all n , then f z is identically zero for all points z in D .

666

Proposition 9.2-12: Proposition 9.2-9:

()

If a complex function f z

that is not identically zero is

analytic inside a simple closed curve C , then the isolated zeros

()

of f z inside C are finite in number.

()

()

that is not identically zero is

can be expressed as the factored

Taylor series: !

() (

f z = z − z1

) ( z − z2 ) ( z − z3 ) k1

k2

k3

(

! z − zn

()

is analytic at z0 and has a zero of order k .

()

()

If the complex functions g z and h z are analytic at a point

analytic in a domain D , and if it has a finite number of isolated

()

()

Proposition 9.3-2:

Proposition 9.2-11, Factoring a Taylor Series:

zeros inside D , then f z

()

f z , then z0 is a pole of order k of f z if and only if 1 f z

that is not identically zero is

zeros of f z inside R are finite in number.

()

unbounded region.

If a point z = z0 is an isolated singularity of a complex function

analytic inside a closed bounded region R , then the isolated

If a complex function f z

an infinite number of isolated zeros only in an open or

Proposition 9.3-1:

Proposition 9.2-10: If a complex function f z

()

An analytic function f z that is not identically zero can have

) g (z) kn

( )

()

() ()

z = z0 , and if g z0 ≠ 0 , then f z = g z h z has a pole of

()

order k at z0 if and only if h z has a zero of order k at z0 . Proposition 9.3-3:

()

()

If f z and g z are complex functions that are analytic in a

( )

disk Dδ z0

and have zeros of orders j and k

( ) g ( z ) will have:

( j ≠ k ),

where the zeros zi are complex numbers, ki are integers, and

respectively, at a point z0 , then f z

g z ≠ 0 is a power series.

!

– an isolated removable singularity at z0 if j − k = 1

!

– a zero of order j − k at z0 if j − k > 1.

!

– a pole of order k − j at z0 if j < k .

()

667

Proposition 9.3-4:

Proposition 9.3-8:

()

If and only if a transcendental entire complex function f z has

Singularities of rational functions are always isolated

()

no zeros can f z be expressed in the form: Proposition 9.3-5:

!

If a point z = z0 is an isolated singularity of a complex function

()

()

f z , then z0 is a pole of order k if and only if lim f z = ∞ . z → z0

() where g ( z ) is an entire function.

Proposition 9.3-9:

Proposition 9.3-6:

()

If a transcendental analytic function f z has no zeros in a

()

If a complex function f z that is analytic in a punctured disk

simply connected domain D , then it is possible to define a

( ) has a pole of order k ∈! at point j lim ( z − z0 ) f ( z ) = ∞ where j = 0, 1, !, k − 1 . z→ z

branch of ln f z

Dδ* z0

( ( )) in D .

z = z0 , then

Proposition 9.3-10:

0

()

If a transcendental entire complex function f z has distinct

Proposition 9.3-7:

()

If a transcendental complex function f z

()

can be expressed in the form: g z f z =e ( )

() where g ( z ) is analytic in D .

()

zeros zj of order mk , where k = 1, 2, 3,!, n , then f z can be

is analytic in a

expressed in the form:

()

simply connected domain D , and if f z ≠ 0 in D , then f z

!

g z f z =e ( )

!

()

f z =

n

m z − z ( ) ∏ k

k

g z e ()

k =1

()

where g z is an entire function.

668

Proposition 9.3-11:

Proposition 9.3-15:

()

()

If a transcendental complex function f z has no zeros and no

If f z is a non-constant entire complex function having a pole

poles in a simply connected domain D , then it is possible to

of order n at infinity, then f z must be a polynomial of degree

( ( ))

define an analytic branch of f z

1 n

n.

in D where n ∈! .

Proposition 9.3-16:

Proposition 9.3-12:

()

If a complex function f z

()

If a complex function f z is rational, then it is meromorphic in

is meromorphic in the complex

Proposition 9.3-13:

()

A complex function f z is analytic in the extended complex

pole at infinity. Proposition 9.3-17:

()

If f z is a complex function that is analytic at infinity, then: !

()

f z =

∑z

()

If a complex function f z

plane if and only if it is a constant function. Proposition 9.3-14:

()

the extended complex plane, and f z is either analytic or has a

plane, then it can be expressed as the ratio of entire functions.

∞

()

()

has an essential singularity at

infinity, then f z must be transcendental. Proposition 9.3-18:

( ) is meromorphic in the extended f ( z ) is a rational function, and can be

If a complex function f z

an

complex plane, then

n

represented in the form:

n =1

()

where this series converges to f z in some region z > R .

!

()

()

f z = c + f∞ z +

n

∑ f (z) k

k =1

669

()

f∞ z

!where

is the principal part of the Laurent series

()

() f ( z ) in the complex

representation of f z at infinity, and fk z is the principal part of the Laurent series representation of

Proposition 9.4-2, Winding Number Around the Origin:

()

Let w = f z

plane for the k th pole, and c is a constant given by: !

()

domain D . If C is a simple closed contour in D along which

()

()

there are no poles or zeros of f z , then:

c = lim ⎡⎣ f z − f∞ z ⎤⎦ z→∞

!∫

!

Proposition 9.3-19:

( ) dz = i d arg f z = n C, 0 2 π i !∫C ( ) ( ) f (z)

f′ z

C

()

w

(

)

If a complex function f z is rational, then it can have only a

where n C, 0

finite number of poles in the complex plane.

C winds completely around the origin in the w-plane.

Proposition 9.4-1, Argument Principle:

()

Let f z

()

be a meromorphic function in a simply connected

()

there are no poles or zeros of f z , then:

1 2π i

!∫

C

( ) dz = N − P f (z)

()

If f z and g z are analytic functions in a simply connected domain D , and if C is a simple closed contour in D where:

()

()

f z > g z ≥0

!

f′ z

()

() ()

for all points on C , then f z and f z + g z have the same

()

where N is the number of zeros of f z including multiplicities

()

inside C , and P is the number of poles of f z multiplicities inside C .

is the number of times the image contour Cw of

Proposition 9.5-1, Rouché’s Theorem:

domain D . If C is a simple closed contour in D along which

!

be an analytic function in a simply connected

including

number of zeros inside C (counting multiple zeros). Proposition 9.6-1, Hurwitz’s Theorem: Let

{ f ( z )} be a sequence of continuous single-valued complex n

()

functions that converges uniformly to f z ≠ 0 at every point z inside and on a simple closed contour C . If the contour C does 670

() N > 0 such that every function fn ( z ) with n > N number of zeros inside C as the function f ( z ) .

not pass through any zeros of f z , then there exists an integer has the same

Proposition 9.8-3:

()

at all points on a line segment within D , then

Proposition 9.7-1, Open Mapping Theorem:

() f (z) = 0

If f z is an analytic function in a domain D , and if f z = 0 everywhere in D .

()

If f z is a non-constant function analytic in a domain D in the z-plane, then the image of the domain D produced by the

( )

mapping w = f D is a domain in the w-plane.

Let D1 and D2 be two intersecting domains, and let their union

()

there is at most only one function

be analytic in a domain D1 . If the radius of

()

()

z1 is r1 , and about a point z2 is r2 , then: r1 − r2 ≤ z1 − z2 !

z1, z2 ∈D1

D2

and equal to f1 z in D .

convergence of the Taylor series expansion of f z about a point

!

() f 2 ( z ) that is analytic in

be the domain D . If a function f1 z is analytic in D1 , then

Proposition 9.8-1: Let f z

Proposition 9.8-4, Unique Analytic Continuation:

Proposition 9.8-5:

()

If a function f x defined on some interval of the real axis can be analytically continued into the complex plane, then this

Proposition 9.8-2, Identity Theorem:

() () () g ( z ) have the same value at each of the points zk of a line segment contained in D , then f ( z ) = g ( z ) everywhere in D . If f z and g z are analytic in a domain D , and if f z and

continuation can only occur in one way. Proposition 9.8-6: The behavior of a function that is analytic in a domain D is completely determined by its behavior in a small neighborhood of an arbitrary point in that domain.

671

Proposition 9.8-7:

()

Proposition 9.9-1, Schwarz Reflection Principle:

()

()

Let f z be analytic in a domain D1 , and g z be an analytic

If f z is an analytic function defined in a domain D in the

continuation of f z into a domain D2 so that f z = g z

upper half-plane that includes an interval of the real axis, then

()

()

()

()

everywhere in D1 ∩ D2 . Then g ′ z is an analytic continuation

()

of f ′ z .

!

()

()

domain D2 , and let the two domains be adjacent to each other

()

()

with D1 ∩ D2 = ∅ . Let f1 z = f 2 z

Proposition 9.9-2:

()

If f z is an analytic function defined on the upper half-plane

along a segment of the

()

()

the real axis, then f x

and f 2 z be continuous on this segment. Let C1 be a closed

()

common border segment. Then f1 z

()

and f 2 z

are analytic

can be continued analytically to the

entire complex plane.

contour within D1 that includes this common border segment, and let C2 be a closed contour within D2 that includes this

()

including the real axis, and if f x is a real-valued function on

common border between the two domains, and let both f1 z

()

() if and only if f ( x ) is real-valued function for each point of the real axis contained in D .

Proposition 9.8-8, Analytic Continuation Across a Boundary: Let f1 z be analytic in a domain D1 , and f 2 z be analytic in a

( )

f z = f z

Proposition 10.1-1:

()

If a complex function f z has a removable singularity at a

continuations of each other.

point z = z0 , then: !

()

Res ⎡⎣ f z , z0 ⎤⎦ = 0

672

Proposition 10.1-2:

Proposition 10.1-4:

()

()

() ()

Let f z = p z q z where the complex functions p z and

then:

q z are both analytic at a point z = z0 . If p z0 ≠ 0 and q z

()

(

()

) ()

then have: !

Proposition 10.1-3:

()

Res ⎡⎣ f z , z0 ⎤⎦ =

()

If a complex function f z has the form:

()

f z =

()

P z

A

=

+

B

+

C

( z − a ) ( z − b) ( z − c ) ( z − a ) ( z − b) ( z − c ) where P ( z ) is a polynomial of degree ≤ 2 , and a , b , and c are

!

( ) ⎡ ⎤ A = Res ⎣ f ( z ) , a ⎦ = ( a − b) ( a − c )

!

( ) ⎡ ⎤ B = Res ⎣ f ( z ) , b ⎦ = ( b − a )( b − c )

!

P b

() )( )

P c C = Res ⎡⎣ f z , c ⎤⎦ = c−a c−b

()

(

()

()

( ) q′ ( z0 ) p z0

Proposition 10.1-5:

()

()

()

Let f z = 1 q z where the complex function q z is analytic

()

at a point z = z0 . If q z has a simple zero at z0 so that z0 is a

()

simple pole of f z , we then have:

distinct complex numbers, then:

P a

( )

has a simple zero at z0 so that z0 is a simple pole of f z , we

Res ⎡⎣ f z , z0 ⎤⎦ = lim ⎡⎣ z − z0 f z ⎤⎦ z → z0

!

!

()

If a complex function f z has a simple pole at a point z = z0 ,

!

()

Res ⎡⎣ f z , z0 ⎤⎦ =

1

( )

q′ z0

Proposition 10.1-6:

()

If a complex function f z is analytic at a point z = z0 , and if

() f ( z ) at z0 is k .

f z has a zero of order k at z0 , then the logarithmic residue of

673

Proposition 10.1-7:

Proposition 10.2-1, Cauchy’s Residue Theorem for an Isolated

() and a complex function g ( z ) is analytic at z0 , then: Res ⎡⎣ g ( z ) f ( z ) , z0 ⎤⎦ = g ( z0 ) Res ⎡⎣ f ( z ) , z0 ⎤⎦

Singularity:

If a complex function f z has a simple pole at a point z = z0 , !

()

If a complex function f z is analytic in a domain D except at an isolated singularity at a point z = z0 , then: !

Proposition 10.1-8:

C

() and a complex function h ( z )

( )

() ()

⎡f z ⎤ 1 Res ⎢ , z0 ⎥ = Res ⎡⎣ f z , z0 ⎤⎦ ⎢⎣ h z ⎥⎦ h z0

Proposition 10.2-2, Cauchy’s Residue Theorem:

()

( )

Let C be a simple closed contour that lies within a simply

()

connected domain D . If a complex function f z is analytic in

D except at a finite number of isolated singular points

Proposition 10.1-9:

z1 , z2 , ! , zk that lie inside C , then:

()

If a complex function f z has a pole of order k ≥ 2 at a point

z = z0 , then:

!

!∫

C

d k−1 ⎡ ! Res ⎡⎣ f z , z0 ⎤⎦ = lim z − z0 z → z0 k − 1 ! dz k−1 ⎢ ⎣

()

()

within D and enclosing z0 .

is analytic at z0 with h z0 ≠ 0 ,

then:

()

f z dz = 2 π i Res ⎡⎣ f z , z0 ⎤⎦

where C is any positively oriented simple closed contour lying

If a complex function f z has a simple pole at a point z = z0 ,

!

!∫

(

1

)

(

) f ( z )⎤⎥⎦ k

()

f z dz = 2 π i

k

∑ Res ⎡⎣ f ( z ), z ⎤⎦ n

n =1

674

Proposition 10.2-3:

Proposition 10.2-5: If a complex function f z

connected domain D . If a complex function f z is analytic in

residues of f z , including the residue at z = ∞ , is equal to

D except at a finite number of isolated singular points

zero.

()

z1 , z2 , ! , zk that lie inside C , then:

!∫

!

C

()

()

is rational, then the sum of all

Proposition 10.3-1:

()

f z dz = − 2 π i Res ⎡⎣ f z , ∞ ⎤⎦

Certain definite integrals of real-valued rational trigonometric functions can be expressed in the form:

and !

()

Let C be a simple closed contour that lies within a simply

!∫

C

⎡ 1 ⎛ 1⎞ ⎤ f z dz = 2 π i Res ⎢ 2 f ⎜ ⎟ , 0 ⎥ ⎝ z⎠ ⎦ ⎣z

()

∫

!

2π

0

(

)

f cosθ , sin θ dθ =

!∫

C

⎛1⎛ 1⎞ 1 ⎛ 1 ⎞ ⎞ dz f ⎜ ⎜ z + ⎟ , ⎜ z − ⎟⎟ z ⎠ 2i ⎝ z⎠⎠ iz ⎝2⎝

where the contour C is a closed unit circle in the complex plane. Proposition 10.2-4: Let C be a simple closed contour that lies within a simply

Proposition 10.3-2:

()

connected domain D . If a complex function f z is analytic in

Certain definite integrals of real-valued rational trigonometric

D except at a finite number of isolated singular points

functions can be expressed in the form:

z1 , z2 , ! , zk that lie inside C , then the sum of all residues of

()

f z , including the residue at z = ∞ , is equal to zero: k

!

∑ n =1

()

()

Res ⎡⎣ f z , zn ⎤⎦ + Res ⎡⎣ f z , ∞ ⎤⎦ = 0

!

∫

2π

0

(

)

f cos nθ , sin nθ dθ =

!∫

C

⎛ 1 ⎛ n 1 ⎞ 1 ⎛ n 1 ⎞ ⎞ dz f ⎜ ⎜ z + n ⎟ , ⎜ z − n ⎟⎟ ⎝2⎝ z ⎠ 2i ⎝ z ⎠⎠ iz

where the contour C is a closed unit circle in the complex plane.

675

!

Proposition 10.3-3:

()

Let f z be a complex rational function having the form: !

() () Qz () where both P ( z )

∫

()

and Q z are polynomials in z of degree n

()

Let f x be a real-valued rational function having the form:

()

!

()

f z dz = 0

CR

j =1

Proposition 10.3-5:

0 ≤ θ ≤ π , then: lim

j

half of the complex plane.

consists of a semicircular contour given by z = R e iθ where

R→ ∞

−∞

∑ Res ⎡⎣ f ( z), z ⎤⎦ ()

and m , respectively, and where m ≥ n + 2 and Q z ≠ 0 . If CR

!

()

f x dx = 2 π i

k

where zj are the k points at which f z has poles in the upper

P z

f z =

∫

PV

∞

() Q( x) where both P ( x ) and Q ( x ) ()

f x =

P x

are polynomials in x of degree n

()

and m , respectively. If m ≥ n + 2 and Q x ≠ 0 , then the Cauchy principal value will exist and we have:

Proposition 10.3-4:

()

Let f x be a real-valued rational function having the form: !

()

f x =

() Q( x) P x

()

!

PV

∫

∞

−∞

()

f x dx = − 2 π i

k

∑ Res ⎡⎣ f ( z), z ⎤⎦ j

j =1

()

where zj are points at which f z has poles in the lower half of the complex plane.

()

where both P x and Q x are polynomials in x of degree n

()

and m , respectively. If m ≥ n + 2 and Q x ≠ 0 , then the Cauchy principal value will exist, and we have:

Proposition 10.3-6:

()

Let f z be analytic in the upper half-plane except for a finite number of poles, and let CR consist of a semicircular contour of

()

radius R . If f z ≤ M and if M → 0 as R → ∞ , then: 676

lim

!

R →∞

∫

()

f z ei a z dz = 0

CR

Proposition 10.3-8:

()

Let the complex function f z have a simple pole at point x0 on

Proposition 10.3-7:

the x-axis. If Cr is a circular arc contour defined by z = x0 + r e iθ

()

where θ1 ≤ θ ≤ θ 2 , we then have:

Let f x be a real-valued rational function having the form: !

() Q( x) where both P ( x ) and Q ( x ) ()

f x =

P x

!

Cr

are polynomials in x of degree n

()

and m , respectively. If m ≥ n + 2 and Q x ≠ 0 , then the Cauchy

!

PV

∫

−∞

()

f x ei ω x dx = 2 π i

k

∑ j =1

∫

()

Res ⎡⎣ f z ei ω z , zj ⎤⎦

where 0 ≤ θ ≤ π , we then have:

−∞

!

PV

∫

∞

−∞

()

k

f x cos ω x dx = −2 π

∑ Im Res ⎡⎣ f ( z ) e j =1

()

f x sin ω x dx = 2 π

k

∑ j =1

()

()

!

()

iω z

, zj ⎤⎦

Re Res ⎡⎣ f z ei ω z , zj ⎤⎦

∫

Cr

upper half of the complex plane, and where ω > 0 . We also have:

PV

)

the x-axis. If Cr is a circular arc contour defined by z = x0 + r e iθ

()

!

(

Let the complex function f z have a simple pole at point x0 on

where zj are the k points at which f z ei ω z has poles in the

∞

()

f z dz = i θ 2 − θ1 Res ⎡⎣ f z , x0 ⎤⎦

Proposition 10.3-9:

principal value will exist and we have: ∞

∫

()

()

f z dz = i π Res ⎡⎣ f z , x0 ⎤⎦

Proposition 10.3-10:

()

Let f x be a real-valued rational function having the form: !

()

f x =

() Q( x) P x

677

()

! PV

()

()

where both P x and Q x are polynomials in x of degree n

where zj are the k points at which f z ei ω z has poles in the

and m , respectively. If m ≥ n + 2 then the Cauchy principal

upper half of the complex plane, and xj are the s points where

value will exist and we will have:

Q x has zeros on the x-axis. We also have:

∫

∞

−∞

k

()

f x dx = 2 π i

()

s

∑ Res ⎡⎣ f ( z ), z ⎤⎦ + i π ∑ Res ⎡⎣ f ( z ), x ⎤⎦ j

j

j =1

! PV

j =1

()

where zj are k points at which f z has poles in the upper half

()

of the complex plane, and xj are s points where Q x has zeros

∫

∞

−∞

k

()

f x cos ω x dx = − 2 π

j =1

−π

!

∑ Im Res ⎡⎣ f ( z ) e

!

() Q( x) where both P ( x ) and Q ( x ) are polynomials in x of degree n ()

f x =

P x

!

PV

∫

−∞

()

f x e

k

iω x

dx = 2 π i

∑ j =1

s

!

+π i

∑ j =1

()

Res ⎡⎣ f z e

()

iω z

, zj ⎤⎦

Res ⎡⎣ f z ei ω z , xj ⎤⎦

k

()

f x sin ω x dx = 2 π

∑ j =1

s

+π

!

∑ j =1

, xj ⎤⎦

()

Re Res ⎡⎣ f z ei ω z , zj ⎤⎦

()

Re Res ⎡⎣ f z ei ω z , xj ⎤⎦

Proposition 10.3-12:

()

and m , respectively, where m ≥ n + 2 . We will have: ∞

∫

∞

−∞

()

iω z

j =1

! PV

Let f x be a real-valued rational function having the form:

()

s

on the x-axis. Proposition 10.3-11:

∑

Im Res ⎡⎣ f z ei ω z , zj ⎤⎦

Let f x be a real-valued rational function having the form: !

()

f x =

( )! Q( x)

xα P x

()

0 1 and M are constants

having vertices at the points N + 1 2 ±1± i , then the sum of

Proposition 10.4-2, Summation of Series:

integers. If f z < M z

k

independent of the integer N along the square contour CN

where A is a constant.

()

()

integers. If f z < M z

k

where k > 1 and M are constants

!

n −1 ( ) ∑ f ( n) = −∑ Res ⎡⎣π csc π z f ( z ) ⎤⎦

n = −∞

poles of f ( z ) within CN

independent of the integer N along the square contour CN

(

)(

)

having vertices at the points N + 1 2 ±1± i , then the sum of the series f ( n ) is given by: ∞

!

∑ f ( n) = −∑ Res ⎡⎣π cot π z f ( z ) ⎤⎦

n = −∞

poles of f ( z ) within CN

679

Proposition 10.5-1, Mittag-Leffler’s Partial-Fraction Expansion Theorem:

()

If w = f z is an analytic function in a domain D , and z0 is a

Let the only singularities of a bounded meromorphic function

()

f z

be simple poles a1 , a2 , a3, ! arranged in order of

()

increasing absolute value. Let the residues of f z at the points

a1 , a2 , a3, ! be b1, b2 , b3, ! . We then have: !

()

()

f z = f 0 +

∞

∑ n =1

Proposition 11.2-1:

( )

( )

point in D at which f ′ z0 ≠ 0 , then the mapping w0 = f z0 will be conformal. Proposition 11.2-2:

()

⎡ 1 1⎤ bn ⎢ + ⎥ z − a an ⎦ ⎣ n

If a function f z is analytic and one-to-one in a domain D ,

()

then the mapping w = f z is conformal at every point in D . Proposition 11.2-3:

Proposition 11.1-1:

()

()

If f z is an analytic function in a domain D , w = f z will

( )

have a single-valued inverse z = f −1 w in the neighborhood of

( )

any point z0 at which f ′ z0 ≠ 0 .

()

If a function f z is analytic and one-to-one inside and upon a

()

simple closed contour C , then the mapping of C by w = f z will result in a simple closed curve in the w-plane. Proposition 11.2-4:

Proposition 11.1-2:

()

Let f z be an analytic function in a domain D in the z-plane,

()

and C z be a simple closed contour in D . Let f ′ z ≠ 0 inside

C z . If C z is mapped onto a simple closed contour C w in the w-

()

()

()

If f z is an analytic function in a domain D , and if f z has a zero of order n at a point z0 in D , then the mapping

()

w = f z will magnify all angles at z0 by the factor n .

plane by w = f z traversing C z once, then all points in the

()

interior of C z are mapped onto the interior of C w , and f z is univalent. 680

Proposition 11.2-5, Conformal Mapping of Laplace’s Equation:

( )

If φ x, y is a harmonic function over a set of points S in the z-

Proposition 11.3-3, Compositions of Linear Mappings: A linear mappings is the composition of a translation, a rotation, and a scaling.

plane, then the conformal mapping of S with the analytic

() over which φ ( u, v ) is harmonic:

function w = f z will result in a set of points Ω in the w-plane

!

∂2φ ∂2φ + =0 ∂u 2 ∂v 2

Proposition 11.3-1, One-to-One Möbius Mapping in the Complex Plane: Möbius transformations provide one-to-one mappings in the complex plane if and only if a d − bc ≠ 0 . Proposition 11.3-2, One-to-One Möbius Mapping in the

Proposition 11.3-4, Composition of Successive Linear Mappings: The composition of successive consecutive linear mappings is also a linear mapping. Proposition 11.3-5, Composition of Möbius Mappings: Every Möbius transformation is composed of at most four mappings that are linear or inversion in form. Proposition 11.3-6, Möbius Mappings are Conformal: Every Möbius transformation is conformal.

Extended Complex Plane: Möbius transformations provide one-to-one mappings in the extended complex plane.

Proposition 11.3-7: Möbius transformations map any circle into a circle or straight line, and any straight line into a circle or straight line.

681

Proposition 11.3-8, Fixed Point Theorem:

Proposition 11.3-12, Composition of Successive Möbius Transformations:

A Möbius transformation has at most two fixed points unless it

() ()

is the identity mapping: w = T z = I z = z .

The composition of successive Möbius transformations is also a Möbius transformation.

Proposition 11.3-9, Three Fixed Point Theorem: Three distinct points z1 , z2 , and z3 in the z-plane can always be mapped into three specified points w1 , w2 , and w3 in the w-

Let z1 and z2 be any two points that are symmetric with respect

plane by a Möbius transformation w = T z that is unique:

to a given straight line Lz or circle Cz in the z-plane. Also let

()

!

Proposition 11.3-13, Symmetry Principle:

w − w1 w2 − w3 w − w3 w2 − w1

=

the z-plane be mapped onto the w-plane by the Möbius

z − z1 z2 − z3

()

transformation w = T z , where Lw and Cw are the images of

z − z3 z2 − z1

( )

w2 = T z2

Proposition 11.3-10, Invariant Möbius Transformations: The cross ratio of four points is invariant under the Möbius transformation.

images Lw and Cw in the w-plane.

For any given point z1 , there is only one point z2 symmetric to

()

A sufficient condition for two Möbius transformations T1 z

()

()

and T2 z to be identical is that T1 z = T2 z for three distinct points: z1 , z2 , and z3 .

will be two points symmetric with respect to the

Proposition 11.3-14:

Proposition 11.3-11, Identical Möbius Transformations:

()

( )

Lz and Cz , respectively, in the w-plane. Then w1 = T z1 and

z1 with respect to a given straight line or circle L . Proposition 11.4-1, Schwarz’s Lemma:

( ) is holomorphic in the open unit disk D1 (0) , f ( 0 ) = 0 and f ( z ) ≤ 1 for all z ∈D1 ( 0 ) , then we have:

If f z

and if

682

()

f z ≤ z !

!

for z < 1 Proposition 11.6-1:

and !

()

If T z is analytic in the upper half of the z-plane and has the

()

f ′ 0 ≤1

derivative:

If equations (11.4-1) and (11.4-2) are equalities, then there exists a θ ∈! such that: !

()

f z = z eiθ !

()

2

π )−1

(

! z − xn

)(α

n

π )−1

interior angles α j where j = 1, 2, !, n .

a−z ! 1− a z

a 0 for all k ,

will converge to a finite nonzero limit P if:

will converge to a finite nonzero limit if and only if for any real number ε > 0 there exists N ∈! such that:

lim ak = 1

k→∞

! Proposition 12.2-2, Cauchy Criteria for Infinite Products: Given a sequence of complex numbers or functions

{ }

{ an } , the

sequence of its partial products Pk , where Pk > 0 for all k ,

Pk −1 < ε ! Pm

for all k > m > N !

which is the Cauchy convergence criteria for infinite products. Proposition 12.2-4, Product Convergence Requirement:

will converge to a finite nonzero limit if and only if for any real sequence of its partial products

k

∏

an − 1 < ε !

}

{ Pk }

where Pk > 0 for all k ,

will converge to a finite nonzero limit P if:

for all k > m > N

n = m+1

{

Given a sequence of complex numbers or functions 1+ bk , the

number ε > 0 there exists N ∈! such that: !

{ an } , the

!

lim b = 0

k→∞ k

which is the Cauchy convergence criteria for infinite products.

684

∞

!

Proposition 12.2-5, Cauchy Criteria for Infinite Products:

{

{ Pk }

}

and the infinite series:

where Pk > 0 for all k

will converge to a finite nonzero limit if and only if for any real number ε > 0 there exists N ∈! such that:

∞

!

∏(1+ bn ) − 1

m > N

n = m+1

Proposition 12.2-8:

{

Proposition 12.2-6:

{

infinite product:

}

Given a sequence of complex numbers or functions 1+ bk , the

∞

k

sequence of its partial products Pk =

∏ (1+ b ) , where P n

k

>0

!

for all k , converges to a nonzero limit if and only if the sequence

n

converges if and only if the infinite series:

k

∑ Ln (1+ b ) converges.

∏ (1+ b ) n =1

n =1

of partial sums Sk =

}

Given a sequence of complex numbers or functions 1+ bk , the

∞

n

n =1

!

∑b

n

n=1

Proposition 12.2-7, Absolute Convergence:

converges.

{

}

Given a sequence of complex numbers or functions 1+ bk , the infinite product: 685

Proposition 12.3-1, Uniform Convergence of an Infinite Product:

Proposition 12.2-9:

{

}

Given a sequence of complex numbers or functions 1+ bk , if ∞

the infinite product ∞

∏(1+ b )

converges, then the product

n

n =1

∏ (1+ b ) also converges.

∞

If the function series

is uniformly and absolutely

n=1

convergent in a bounded closed region R , then the infinite product

n =1

∏ (1+ f ( z ) ) is uniformly convergent in R . n

n =1

Proposition 12.2-10, Rearrangement of an Infinite Product:

{ } { } {cn } is any rearrangement of { bn } , and if the infinite product:

Proposition 12.3-2, M-Test for Uniform Convergence of an Infinite Product:

If bn and cn are two sequences of complex numbers where ∞

∏ (1+ b ) n

n =1

are convergent.

∏ (1+ f ( z ) ) is n

n =1

∞

fn z < Mn where Mn > 0 , and where

converges, then the infinite products:

∏ (1+ c ) !

The infinite product of complex functions

()

n

∞

∞

uniformly convergent in a bounded closed region R if

n =1

!

n

∞

n

!

∑ f (z)

series of positive constants.

∑M

n

is a convergent

n=1

∞

∏ (1+ c ) n

n =1

Proposition 12.3-3, Logarithmic Derivative of an Infinite Product: Given a sequence of analytic functions

{ f (z) }

defined in an

n

∞

open simply connected region R so that

∑ f ( z ) converges n

n =1

uniformly on each compact subset of R , then we have: 686

!

( ) = d ( ln P ( z )) = ∞ f n′ ( z ) ∑ dz P( z) 1+ fn ( z ) n =1

P′ z

where !

()

P z =

∞

∏ ( 1+ f ( z ) ) ≠ 0 n

n =1

687

Appendix E

ELEMENTARY COMPLEX FUNCTION IDENTITIES !

!

d z e = ez ! dz

!

d g (z) g z e = e ( ) g′ z ! dz

!

e 1e 2 = e 1

!

(e ) = e

In this appendix, we will list some of the identities for

elementary complex function. where k = 0, ± 1, ± 2,! and

!

z

z

e1 e

z +z2

z

z2

n

nx

z − z2

=e1

!

(E.1-3)

(E.1-4)

n x+i y ei n y = e ( ) = e n z !

(E.1-5)

!

(E.1-6)

!

1 = e− z ! z e

(E.1-7)

!

e z = e x = e x = eRe z !

(E.1-8)

!

z ez ≤ e !

(E.1-9)

n = 0, ± 1, ± 2,! .

E.1!

()

z

!

(E.1-2)

COMPLEX EXPONENTIAL FUNCTION z

e =e

x+i y

x iy

x

(

)

= e e = e cos y + isin y !

!

e z = e z ei 2 k π = e z+i 2 k π !

(E.1-10)

!

e z = e x e−i y = e x−i y = e z !

(E.1-11)

(E.1-1)

688

E.2!

!

!

!

COMPLEX TRIGONOMETRIC FUNCTIONS ei z + e−i z cos z = ! 2

tan z =

sec z =

ei z − e−i z sin z = ! 2i iz

−i z

sin z e −e = − i i z −i z ! cos z e +e

1 2i = i z −i z ! sin z e − e

z ≠ nπ !

!

1 ei z + e−i z cot z = = i i z −i z ! tan z e −e

!

!

⎛ 1⎞ z ≠ ⎜ n + ⎟ π ! (E.2-2) 2⎠ ⎝ ⎛ 1⎞ z ≠ ⎜ n + ⎟ π ! (E.2-3) 2⎠ ⎝

csc z =

z ≠ nπ !

d tan z = sec 2 z ! dz

⎛ 1⎞ z ≠ ⎜n+ ⎟ π ! 2⎠ ⎝

(E.2-8)

!

d sec z = sec z tan z ! dz

⎛ 1⎞ z ≠ ⎜n+ ⎟ π ! 2⎠ ⎝

(E.2-9)

!

d csc z = − csc z cot z ! dz

z ≠ nπ !

(E.2-10)

!

d cot z = − csc 2 z ! dz

z ≠ nπ !

(E.2-11)

!

cos −z = cos z !

!

tan −z = − tan z !

(E.2-13)

!

cos 2 z + sin 2 z = 1 !

(E.2-14)

!

cos z1 ± z2 = cos z1 cos z2 ∓ sin z1 sin z2 !

!

sin z1 ± z 2 = sin z1 cos z2 ± cos z1 sin z2 !

!

tan z1 ± z 2 =

(E.2-1)

1 2 = i z −i z ! cos z e + e

!

!

(E.2-4)

(E.2-5)

d cos z = − sin z ! dz

(E.2-6)

d sin z = cos z ! dz

(E.2-7)

( )

( )

sin −z = − sin z !

( )

(E.2-12)

(

)

(E.2-15)

(

)

(E.2-16)

(

)

tan z1 ± tan z2 ! 1∓ tan z1 tanz2

(E.2-17)

689

!

(

) (

)

(

) (

)

2 sin z1 + z2 sin z1 − z2 = cos 2 z2 − cos 2 z1 !

!

2 cos z1 + z2 sin z1 − z2 = sin 2 z1 − sin 2 z2 !

!

⎛ z2 + z1 ⎞ ⎛ z2 − z1 ⎞ ! cos z2 − cos z1 = − 2 sin ⎜ sin ⎜ ⎟ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ 2

!

sin z = sin x cosh y + i cos x sinh y !

(E.2-28)

(E.2-19)

!

tan z =

sin x cos x + i sinh y cosh y ! cos 2 x + sinh 2 y

(E.2-29)

(E.2-20)

!

cos z = cos 2 x cosh 2 y + sin 2x sinh 2 y !

(E.2-30)

!

sin z = sin 2 x cosh 2 y + cos 2 x sinh 2 y !

(E.2-31)

(E.2-18)

!

2

cos 2 z = cos z − sin z !

(E.2-21)

!

sin 2 z = 2 sin z cos z !

(E.2-22)

!

cos z = cos 2 x + sinh 2 y !

(E.2-32)

!

cos z + π = − cos z !

(E.2-33)

!

(

)

(E.2-23)

!

sin z = sin 2 x + sinh 2 y !

sin z + π = − sin z !

(

)

(E.2-24)

!

cos z =

!

⎛ π⎞ cos ⎜ z + ⎟ = sin z ! 2⎠ ⎝

(E.2-25)

!

cos z ≥ cos x !

(E.2-35)

!

sin z ≥ sin x !

(E.2-36)

!

⎛ π⎞ sin ⎜ z + ⎟ = cos z ! 2⎠ ⎝

(E.2-26) !

cos z ≥ sinh y !

(E.2-37)

cos z = cos x cosh y − i sin x sinh y !

(E.2-27) !

sin z ≥ sinh y !

(E.2-38)

!

cosh 2 y − sin 2 x !

(E.2-34)

690

! ! !

cos z ≤ cosh y ! 2

(E.2-39)

2

sin z + cos z ≥ 1 !

(

)

cos z = cos z + 2 π !

(E.2-42)

(

)

(E.2-43)

cosh y = cos i y !

(E.2-44)

!

tan z = tan z + π !

! !

( )

( )

sinh y = − i sin i y ! sin z = sin z !

(E.2-46)

!

cos z = cos z !

(E.2-47)

!

2

sin z = sin z sin z = sin z sin z ! 2

cos z = cos z cos z = cos z cos z !

!

sinh z e z − e− z ! tanh z = = cosh z e z + e− z

!

sech z =

!

e z − e− z sinh z = ! 2

(E.3-1)

⎛ 1⎞ z ≠ ⎜n+ ⎟ π i! 2⎠ ⎝

(E.3-2)

1 ! cosh z

⎛ 1⎞ z ≠ ⎜n+ ⎟ π i! 2⎠ ⎝

(E.3-3)

csch z =

1 ! sinh z

z ≠ nπ i !

(E.3-4)

!

coth z =

1 ! tanh z

z ≠ nπ i !

(E.3-5)

!

d d ⎛ e z + e− z ⎞ e z − e− z cosh z = ⎜ = = sinh z ! dz dz ⎝ 2 ⎟⎠ 2

(E.3-6)

!

d d ⎛ e z − e− z ⎞ e z + e− z sinh z = ⎜ = = cosh z ! dz dz ⎝ 2 ⎟⎠ 2

(E.3-7)

(E.2-45)

!

!

!

e z + e− z cosh z = ! 2

(E.2-41)

)

sin z = sin z + 2 π !

COMPLEX HYPERBOLIC FUNCTIONS

(E.2-40)

(

!

E.3!

(E.2-48) (E.2-49)

691

!

⎛ 1⎞ z ≠ ⎜n+ ⎟ π i ! 2⎠ ⎝

d tanh z = sech2z ! dz d sechz = − sechz tanh z ! dz

⎛ 1⎞ z ≠ ⎜n+ ⎟ π i! 2⎠ ⎝

d csch z = − csch z coth z ! dz

z ≠ nπ i !

!

d coth z = − csch2z ! dz

z ≠ nπ i !

!

cos z = cosh i z !

!

sin z = − i sinh i z !

!

cosh z = cos i z !

!

!

! ! !

( )

( )

( )

( )

!

cosh z1 ± z2 = cosh z1 cosh z2 ± sinh z1 sinh z2 !

!

sinh z1 ± z2 = sinh z1 cosh z2 ± cosh z1 sinh z2 !

!

tanh z1 ± z2 =

(E.3-11)

!

sinh 2 z = 2 sinh z cosh z !

(E.3-22)

(E.3-12)

!

cosh 2 z = cosh2 z− sinh 2 z !

(E.3-23)

(E.3-13)

!

tanh 2 z =

2 tanh z ! 2 1+ tanh z

(E.3-24)

!

1+ coth 2 z ! coth 2 z = 2 coth z

(E.3-25)

!

sech 2 z = 1− tanh 2 z !

(E.3-26)

!

csch 2 z = coth 2 z − 1 !

(E.3-27)

(E.3-8)

(E.3-9)

(E.3-10)

(E.3-18)

(

)

(E.3-19)

(

)

(E.3-20)

(

)

tanh z1 ± tanh z2 1± tanh z1 tanh z2

!

(E.3-21)

(E.3-15)

( )

tanh z = − i tan i z !

( )

cosh 2 z − sinh 2 z = 1 !

(E.3-14)

sinh z = − i sin i z !

cosh −z = cosh z !

!

(E.3-16)

( )

sinh −z = − sinh z !

(E.3-17) !

sinh z1 + sinh z2 = 2 sinh

z1 + z2 2

cosh

z1 − z2 2

!

(E.3-28) 692

!

z1 − z2

sinh z1 − sinh z2 = 2 sinh

2

!

cosh z1 + cosh z2 = 2 cosh

!

cosh z1 − cosh z2 = 2 sinh

!

z1 + z2 2 z1 + z2 2

cosh

z1 + z2

cosh

sinh

cosh z = cos y cosh x + i sin y sinh x !

2

!

z1 − z2

z1 − z2 2

!

cosh z = cosh z + 2 k π i !

)

(E.3-39)

!

sinh z = sinh z + 2 k π i !

(

)

(E.3-40)

(E.3-30)

!

tanh z = tanh z + i k π !

(E.3-41)

(E.3-31)

E.4!

COMPLEX LOGARITHMIC FUNCTION

! (E.3-32)

!

sinh z = cos y sinh x + i sin y cosh x !

(E.3-33)

!

cosh z = cosh 2x − sin 2 y !

(E.3-34)

! !

sinh z = cosh z =

2

2

sinh x + sin y ! sinh 2x + cos 2 y !

!

!

cosh z ≤ cosh x ! sinh z ≥ sinh x !

(

)

(

)

ln z = log e r + i θ + 2 k π !

(E.4-1)

()

()

ln z = log e z + i arg z = log e r + i arg z !

()

()

Ln z = log e r + i Arg z !

!

ln z = Ln z + i 2 k π !

!

ln z = log e r + i Arg z + i 2 k π !

(E.3-35)

(E.4-2)

− π < Arg z ≤ π !

!

(E.4-5)

( )

( )

(E.4-6)

)

(E.4-7)

ln z1 + ln z2 = log e z1 + i arg z1 + log e z2 + i arg z2 !

(E.3-37)

(

!

ln z1 + ln z2 = log e z1 z2 + i θ1 + θ 2 + 2 k π !

!

ln z1 z2 = Ln r1 r2 + i θ1 + θ 2 + 2 k π !

(E.3-38)

(

)

(E.4-3) (E.4-4)

()

(E.3-36) !

!

(

(E.3-29)

!

2

!

( ) (

)

(E.4-8) 693

⎛ z1 ⎞ ln ⎜ ⎟ = ln r1 − ln r2 + i θ1 − θ 2 + 2 k π ! ⎝ z2 ⎠

(

!

)

( ) = 1!

d ln z

!

dz

z

() ()

f′ z d ln f z = ! dz f z

(

!

E.5!

( ))

(

)

1 2⎤

!

!

⎡ i + z2 − 1 ⎛ 1⎞ ⎢ csc −1 z = sin −1 ⎜ ⎟ = − i ln ⎢ z ⎝ z⎠ ⎢⎣

⎥⎦ !

(

(

⎡ cos z = − i ln ⎢ z + z 2 − 1 ⎣ −1

)

1 2⎤

⎥⎦ !

)

⎤ ⎥! ⎥ ⎥⎦

⎤ ⎥ ⎥! ⎥⎦

z ≠ 0 ! (E.5-4)

1 i − z i 1+ z i 1− i z ! ln = ln = ln 2i i + z 2 1− z 2 1+ i z

(E.4-10)

tan −1 z =

(E.4-11)

!

1 1 z+i i z−i ! cot −1 z = tan −1 = ln = ln z 2i z − i 2 z + i

(E.5-6)

!

d 1 ! sin −1 z = 2 dz 1− z

(E.5-7)

!

d 1 ! cos −1 z = − 2 dz 1− z

(E.5-8)

!

d 1 ! tan −1z = dz 1+ z 2

(E.5-9)

!

d 1 ! csc −1 z = − 2 dz z z −1

(E.5-10)

!

d 1 ! sec −1 z = 2 dz z z −1

(E.5-11)

(E.5-1) 1 2

!

)

1 2

!

INVERSE COMPLEX TRIGONOMETRIC FUNCTIONS ⎡ sin −1 z = − i ln ⎢ i z + 1− z 2 ⎣

!

(E.4-9)

(

⎡ + 1− z 2 ⎛ ⎞ 1 ⎢1 −1 −1 sec z = cos ⎜ ⎟ = − i ln ⎢ z ⎝ z⎠ ⎢⎣

z ≠ 0 ! (E.5-2)

(E.5-5)

(E.5-3)

694

d 1 cot −1 z = − ! dz 1+ z 2

!

E.6! !

!

!

!

(E.5-12)

INVERSE COMPLEX HYPERBOLIC FUNCTIONS

(

⎛ sinh z = ln ⎜ z + z 2 + 1 ⎝ −1

)

1 2⎞

⎟⎠ !

(

(

)

)

1 2

⎤ ⎥! ⎥ ⎥⎦

1 2⎞

⎟⎠ !

(

d sinh −1 z = dz

!

d cosh −1z = dz

!

d 1 ! tanh −1z = dz 1− z 2

(E.6-1)

⎡ + 1+ z 2 ⎛ 1⎞ ⎢1 cosech −1z = sinh −1 ⎜ ⎟ = ln ⎢ z ⎝ z⎠ ⎢⎣ ⎛ cosh −1z = ln ⎜ z + z 2 − 1 ⎝

!

1 2

z +1 1 2

z −1

!

(E.6-7)

!

(E.6-8)

(E.6-9)

z ≠ 0 ! (E.6-2)

(E.6-3)

)

⎡ + z2 − 1 1 ⎢1 −1 −1 ⎛ 1 ⎞ sech z = cosh ⎜ ⎟ = ln ⎢ z ⎝ z⎠ ⎢⎣

2

1 1+ z ! ln 2 1− z

!

tanh −1 z =

!

⎛ 1 ⎞ 1 ⎡ z + 1⎤ ! coth −1 z = tanh −1 ⎜ ⎟ = ln ⎢ ⎝ z ⎠ 2 ⎣ z − 1 ⎥⎦

⎤ ⎥! ⎥ ⎥⎦

z ≠ 0 ! (E.6-4)

(E.6-5)

z ≠ ± i ! (E.6-6) 695

Andersson, M., 1997, Topics in complex analysis, SpringerVerlag, New York.

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