Problems and Solutions for Undergraduate Complex Analysis I 9789887607304, 9789887607311, 9789887607328, 9887607312

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Problems and Solutions for Undergraduate Complex Analysis I

by Kit-Wing Yu, PhD [email protected]

c 2022 by Kit-Wing Yu. All rights reserved. No part of this publication may be Copyright reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the author. ISBN: 978-988-76073-0-4 (eBook) ISBN: 978-988-76073-1-1 (Paperback) ISBN: 978-988-76073-2-8 (Hardcover)

ii

About the author

Dr. Kit-Wing Yu received his B.Sc. (1st Hons), M.Phil. and Ph.D. degrees in Math. at the HKUST, PGDE (Mathematics) at the CUHK. After his graduation, he joined United Christian College (UCC) to serve as a mathematics teacher between 2000 and 2020. From 2002 to 2020, he also took the responsibility of the mathematics panel at UCC. Starting from Sept. 2020, Dr. Yu has been promoted to be the Vice Principal (Academic) at Kiangsu-Chekiang College (Kwai Chung). Furthermore, he was appointed as a part-time tutor (2002 – 2005) and then a part-time course coordinator (2006 – 2010) of the Department of Mathematics at the OUHK. Besides teaching, Dr. Yu has been appointed to be a marker of the HKAL Pure Mathematics and HKDSE Mathematics (Core Part) for over thirteen years. Between 2012 and 2014, Dr. Yu was invited to be a Judge Member by the World Olympic Mathematics Competition (China). In the research aspect, he has published research papers in international mathematical journals, including some well-known journals such as J. Reine Angew. Math., Proc. Roy. Soc. Edinburgh Sect. A and Kodai Math. J.. His research interests are inequalities, special functions and Nevanlinna’s value distribution theory. In the area of academic publication, he is the author of the following nine books: • A Complete Solution Guide to Complex Analysis • A Complete Solution Guide to Real and Complex Analysis I • A Complete Solution Guide to Real and Complex Analysis II • A Complete Solution Guide to Real and Complex Analysis • A Complete Solution Guide to Principles of Mathematical Analysis • Problems and Solutions for Undergraduate Real Analysis • Problems and Solutions for Undergraduate Real Analysis I • Problems and Solutions for Undergraduate Real Analysis II • Mock Tests for the ACT Mathematics

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iv

Preface

The book Problems and Solutions for Undergraduate Complex Analysis I, provides a comprehensive problem book for those students and instructors who need to know essential theories and techniques of complex analysis at the undergraduate level. “The only way to learn mathematics is to do mathematics.” – Paul Halmos. My learning and teaching experience has convinced me that this assertion is definitely true. In fact, I believe that “doing mathematics” means a lot to everyone who studies or teaches mathematics. It is not only a way of writing a solution to a mathematical problem, but also a mean of reflecting mathematics deeply, exercising mathematical techniques expertly, exchanging mathematical thoughts with others effectively and searching new mathematical ideas unexpectedly. Thus I hope everyone who is reading this book can experience and acquire the above benefits eventually. The wide variety of problems, which are of varying difficulty, includes the following topics: Complex Numbers, Geometry of Complex Numbers, nth Roots of a Complex Number, Complex Functions and the Analyticity, Power Series, Elementary Theory of Complex Integration, Properties of Analytic and Entire Functions, Further Properties of Analytic Functions and Isolated Singularities of Analytic Functions. Furthermore, the main features of this book are listed as follows: • The book contains 226 problems which cover the topics mentioned above. The solutions are detailed and complete in the sense that every step and every theorem that I applied will be presented. • Each chapter starts with a brief and concise note of introducing the notations, terminologies, basic mathematical concepts or important/famous/frequently used theorems (without proofs) relevant to the topic. • Three levels of difficulty have been assigned to problems: Symbol ⋆ ⋆

⋆

Intermediate

⋆

⋆

Level of difficulty Introductory

⋆

Advanced

Meaning These problems are basic and every student must be familiar with them. The depth and the complexity of the problems increase. Students who targets for higher grades must study them. These problems are very difficult and they may need some specific skills.

v

vi • Different colors are used in appropriate places so as to highlight or explain problems, examples, remarks, main points/formulas involved, or show the steps of manipulation in some complicated proofs. (ebook only) • Different authors may have different setup for the theory of complex integration. To reduce confusions and save your time of cross-checking, Appendix A is made for recording some terminologies of some common complex analysis textbooks, such as Ahlfors [1], Asmar and Grafakos [4], Bak and Newmann [5], Conway [11], Gamelin [14], Rudin [31] as well as Stein and Shakarchi [35]. If you find any typos or mistakes, please feel free to send your valuable comments or opinions to [email protected] Any updated errata of this book or news about my new book will be posted on my new website: https://sites.google.com/view/yukitwing/ Kit Wing Yu January 2022

List of Figures

1.1

The interpretation of a complex number z. . . . . . . . . . . . . . . . . . . . . . .

3

1.2

Geometric interpretations of −z and z. . . . . . . . . . . . . . . . . . . . . . . . .

4

1.3

The parallelogram law. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4

1.4

The triangle formed by 1, z and |z|. . . . . . . . . . . . . . . . . . . . . . . . . .

15

2.1

The complex form of the midpoint M . . . . . . . . . . . . . . . . . . . . . . . . .

19

z2 −z1 z3 −z1 .

2.2

The oriented angle from α = arg

. . . . . . . . . . . . . . . . . . . . . . . .

20

2.3

Stereographic Projection. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

22

2.4

The Midpoint Formula. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

23

2.5

The rhombus ABCD. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

26

2.6

The centroid of ∆ABC. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

27

2.7

The figure of z, z 2 and z 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

32

2.8

The positions of the two points P and Q. . . . . . . . . . . . . . . . . . . . . . .

32

5.1

The circle of convergence of a power series. . . . . . . . . . . . . . . . . . . . . .

80

6.1

The picture for Green’s Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . .

97

6.2

A simply connected region Ω. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

99

Ω′ .

6.3

A multiply connected region

. . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

6.4

A special case for multiply connected region. . . . . . . . . . . . . . . . . . . . . 101

6.5

The rectangle used in the proof of Problem 6.34. . . . . . . . . . . . . . . . . . . 121

7.1

The construction of the angles θ and φ. . . . . . . . . . . . . . . . . . . . . . . . 142

8.1

A simply connected region Ω. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

8.2

The construction of the angles θ and φ. . . . . . . . . . . . . . . . . . . . . . . . 150

vii

viii

List of Figures

Contents

Preface

v

List of Figures

vii

1 Complex Numbers

1

1.1

Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

1.2

Arithmetic of Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

1.3

The Complex Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10

1.4

De Moivre’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17

2 Geometry of Complex Numbers

19

2.1

Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

19

2.2

Applications to Plane Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . .

23

2.3

Locus Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

27

2.4

Transformations of Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . .

31

2.5

Stereographic Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33

3 nth Roots of a Complex Number

37

3.1

Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

37

3.2

nth Roots of a Complex Number . . . . . . . . . . . . . . . . . . . . . . . . . . .

38

3.3

nth Roots of Unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

44

4 Complex Functions and the Analyticity

49

4.1

Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49

4.2

Functions of a Complex Variable . . . . . . . . . . . . . . . . . . . . . . . . . . .

58

4.3

Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

60

4.4

Analytic Functions and the Cauchy-Riemann Equations . . . . . . . . . . . . . .

63

4.5

Elementary Complex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .

69

4.6

The Logarithm Function and Complex Powers

73

ix

. . . . . . . . . . . . . . . . . . .

x

Contents

5 Power Series

77

5.1

Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

77

5.2

Radius of Convergence of Power Series . . . . . . . . . . . . . . . . . . . . . . . .

81

6 Elementary Theory of Complex Integration

95

6.1

Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6.2

Complex Line Integrals and the M -L Formula . . . . . . . . . . . . . . . . . . . . 103

6.3

Cauchy’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

6.4

The Cauchy Integral Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

7 Properties of Analytic and Entire Functions

95

125

7.1

Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

7.2

Zeros of Analytic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

7.3

Properties of Analytic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

7.4

Properties of Entire Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

8 Further Properties of Analytic Functions

145

8.1

Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

8.2

Applications of the Open Mapping Theorem . . . . . . . . . . . . . . . . . . . . . 147

8.3

Applications of Schwarz’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

8.4

Applications of the Schwarz Reflection Principle . . . . . . . . . . . . . . . . . . 161

9 Isolated Singularities of Analytic Functions

165

9.1

Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

9.2

Problems on Removable Singularities . . . . . . . . . . . . . . . . . . . . . . . . . 169

9.3

Problems on Poles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

9.4

Problems on Essential Singularities . . . . . . . . . . . . . . . . . . . . . . . . . . 176

9.5

Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

Appendix

184

A Terminologies

185

A.1 Ahlfors’s Terminologies

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

A.2 Asmar and Grafakos’s Terminologies . . . . . . . . . . . . . . . . . . . . . . . . . 185 A.3 Bak and Newman’s Terminologies . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 A.4 Conway’s Terminologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 A.5 Gamelin’s Terminologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 A.6 Lang’s Terminologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 A.7 Rudin’s Terminologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

Contents A.8 Stein and Sharkarchi’s Terminologies . . . . . . . . . . . . . . . . . . . . . . . . . 187 Index

189

Bibliography

193

CHAPTER

1

Complex Numbers

1.1 Fundamental Concepts Complex analysis is the subject which applies the ideas of calculus to imaginary numbers. But what exactly are imaginary numbers? Usually, we learn about them in high school with introductory remarks from the following situation: “What are the solutions of the equation x2 = −1? We can’t take the square root of a negative number. However, let’s pretend at this moment that we can and since these numbers are really imaginary, it will be convenient to set a number i satisfies i2 = −1.”

In this chapter, the basic theory of complex numbers is studied. Basic references of this chapter are [1, Chap. 1], [5, §1.1, pp. 2 - 4], [21, §1, pp. 3 - 8] and [35, §1.1, pp. 1 - 4]. We assume that you are familiar with various different types of numbers, such as the natural numbers N, the integers Z, the rational numbers Q and the real numbers R.

1.1.1 What is a Complex Number? Definition 1.1. A complex number z is an expression of the form z = x + iy,

(1.1)

where x and y are real numbers and i satisfies i2 = −1. The form (1.1) is called the Cartesian form of z, The real number x is the real part of z and the real number y is the imaginary part of z. We also write x = Re z and y = Im z. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. The set of all complex numbers is denoted by C. A real number x can be thought of as a complex number with zero imaginary part. In this case, x is said to be purely real. If Re z = 0 but Im z 6= 0, then z is said to be purely imaginary. 1

2

Chapter 1. Complex Numbers

1.1.2 Arithmetic of Complex Numbers Definition 1.2. Suppose that z1 = a + bi and z2 = c + di, where a, b, c, d ∈ R and i2 = −1. Then we have (a) z1 + z2 = (a + c) + i(b + d), (b) z1 − z2 = (a − c) + i(b − d), (c) z1 z2 = (ac − bd) + i(ad + bc), (d)

z1 ac + bd bc − ad = 2 +i 2 , where z2 6= 0. 2 z2 c +d c + d2

It is straightforward to see that addition and multiplication of complex numbers are associative and commutative. In fact, these two operations are linked by the distributive law, i.e., z1 × (z2 + z3 ) = z1 × z2 + z1 × z3 . Remark 1.1 (a) Many of the properties of the real number system R hold in the complex number system C, but there are some remarkable differences as well. For example, the concept of order in R does not carry over to C. In other words, we cannot compare two non-real complex numbers. (b) On the other hand, some things that are impossible in real analysis, such as ex = −1 and cos x = 10 if x is a real variable, are perfectly correct, but it is possible in complex analysis when the symbol x is interpreted as a complex variable z. (c) It is a well-known fact that the set C is the smallest field (read [30, pp. 5 - 8] for the definition of a field) containing the field R and the roots of x2 + 1 = 0.

Definition 1.3 (Complex Conjugate). The complex conjugate, denoted by z, of a complex number z = x + iy is given by z = x − iy. Theorem 1.4. Let z, z1 , z2 be complex numbers. Then we have (a) z + z = 2Re z, (b) z − z = 2iIm z, (c) (z) = z, (d) z1 ± z2 = z1 ± z2 , (e) z1 · z2 = z1 · z2 , (f)

z  1

z2

=

z1 . z2

1.1. Fundamental Concepts

3

Since complex numbers satisfy the usual arithmetic properties, all the usual algebraic identities of real numbers are also true for complex numbers. For examples, if z1 , z2 ∈ C, then we have z12 − z22 = (z1 + z2 )(z1 − z2 ) and (z1 ± z2 )2 = z12 ± 2z1 z2 + z22 . Some important identities are collected in the following theorem: Theorem 1.5. Suppose that z, z1 , z2 ∈ C. (a) If n ∈ N, then we have (1 + z)n =

n X

Ckn z k

and

(z1 + z2 )n =

n X

Ckn z1n−k z2k .

k=0

k=0

(b) If n ∈ N, then we have 1−z n = (1−z)(1+z+z 2 +· · ·+z n−1 )

and z1n −z2n = (z1 −z2 )(z1n−1 +z1n−2 z2 +· · ·+z2n−1 ).

The second identity in Theorem 1.5(a) is the Binomial Theorem. Denote z −1 to be the reciprocal 1z . Notice that this is a number such that z · z −1 = z −1 · z = 1. With this notation, if n > 0, then we obtain the formula 1 z −n = n . z

1.1.3 The Complex Plane The Cartesian coordinates (x, y) of a point in R2 can be used to represent the complex number z = x + iy. With this interpretation, we may refer to the plane as the complex plane or the z-plane.a The horizontal axis is called the real axis and the vertical axis is the imaginary axis, see Figure 1.1.

Figure 1.1: The interpretation of a complex number z. a

The complex plane is often called the Argand diagram.

4

Chapter 1. Complex Numbers

Various operations on complex numbers considered in §1.1.2 can be given a geometric interpretation. For instances, −z is obtained by rotating z through the angle π about the origin, while z is the reflection of z with respect to the real axis, see Figure 1.2.

Figure 1.2: Geometric interpretations of −z and z. Addition and multiplication can be given a geometric interpretation too. Since a complex number can be though of as a based vectorb , the sum of z1 and z2 corresponds to the vector sum. This is the so-called parallelogram law (Figure 1.3.).

Figure 1.3: The parallelogram law.

b

The complex number x + iy corresponds to the vector from the point (0, 0) to the point (x, y).

1.1. Fundamental Concepts

5

1.1.4 Polar Form of a Complex Number by

Given z = x + iy, the modulus, or absolute value, of z is denoted by |z| and it is defined |z| = |x + iy| =

p

x2 + y 2 .

√ For example, |3 + 4i| = 32 + 42 = 5. If z1 , z2 ∈ C, then |z1 − z2 | is the distance from 0 to z1 − z2 . Using the parallelogram law, we can deduce that |z1 − z2 | is the distance from z1 to z2 . Some properties of the modulus are listed in the following result: Theorem 1.6 (Properties of the Modulus). (a) |z| ≥ 0, (b) |z| = 0 if and only if z = 0, (c) |z| ≥ |Re z| and |z| ≥ |Im z|, (d) |z| = |z| and | − z| = |z|, (e) |z|2 = z · z, (f) |z1 − z2 | = |z2 − z1 |, z |z | 1 1 provided that z2 6= 0. (g) |z1 z2 | = |z1 | · |z2 | and = z2 |z2 |

Definition 1.7 (Argument). An argument of a nonzeroc complex number z = x + iy, denoted by arg z, satisfies y tan(arg z) = . x It is easy to see that the argument arg z has infinitely many values and all differ by integer multiples of 2π. The one lying in the range (−π, π] is called the principal argument and it is denoted by Arg z.d Once we fix Arg z, other arguments can be found by adding integer multiples of 2π to it. Definition 1.8 (Polar Form of a Complex Number). Suppose that the modulus and an argument of z 6= 0 are r and θ respectively. Then the ordered pair (r, θ) are called polar coordinates of z. The expression z = r(cos θ + i sin θ)

(1.2)

is called the polar form of z. Compare z = x + iy with the expression (1.2), we see immediately that x = r cos θ and y = r sin θ. c d

We remark that no argument is assigned to the point 0. Some textbooks set 0 < Arg z ≤ 2π.

6

Chapter 1. Complex Numbers

1.1.5 Multiplication and Division of Complex Numbers and the Geometric Interpretation Using the polar form of complex numbers, we can derive a geometric interpretation of the multiplication of two complex numbers. To be more precise, we have the following result: Theorem 1.9. Suppose that z1 = r1 (cos θ1 + i sin θ1 ) and z2 = r2 (cos θ2 + i sin θ2 ). Then we see that z1 z2 = r1 r2 [cos(θ1 + θ2 ) + i sin(θ1 + θ2 )]. In other words, the modulus of z1 z2 is the product of the moduli of z1 and z2 and an argument of z1 z2 is the sum of an argument of z1 and an argument of z2 . Geometrically, Theorem 1.9 means that when we multiply z1 by z2 , then z1 is multiplied by the factor |z2 | and rotate it about the origin through the angle Arg z2 . This rotation is clockwise if Arg z2 < 0 and anticlockwise if Arg z2 > 0. However, it is not always true that the principal argument of z1 z2 is the exact sum of the π principal arguments of z1 and z2 . For example, if Arg z1 = 3π 4 and Arg z2 = 2 , then we have Arg z1 + Arg z2 =

5π = Arg (z1 z2 ) + 2π. 4

In general, it is true that Arg (z1 z2 ) = Arg z1 + Arg z2 + 2nπ, where

 −1, if Arg z1 + Arg z2 > π;      0, if −π < Arg z1 + Arg z2 ≤ π; n=      1, if Arg z1 + Arg z2 ≤ −π.

Similar to Theorem 1.9, we have

r1 z1 = [cos(θ1 − θ2 ) + i sin(θ1 − θ2 )]. z2 r2

(1.3)

Geometrically, the expression (1.3) means that when z1 is divided by z2 , then z1 is multiplied by the factor |z12 | and rotate it about the origin through the angle −Arg z2 . This rotation is clockwise if Arg z2 > 0 and anticlockwise if Arg z2 < 0.

1.1.6 De Moivre’s Theorem Applying Theorem 1.9 repeatedly, we can prove the famous Theorem 1.10 (De Moivre’s Theorem). If n is an integer and θ is a real number, then we have (cos θ + i sin θ)n = cos nθ + i sin nθ.

1.2. Arithmetic of Complex Numbers

7

1.2 Arithmetic of Complex Numbers Problem 1.1 ⋆ Express each of the following complex numbers in Cartesian form. (a) (1 + 2i) + (1 + πi). (b) (1 + 2i)(1 + πi). (c) (−1 − 3i)(−1 + 3i). Proof. By Definition 1.2 and the property i2 = −1, we have (a) (1 + 2i) + (1 + πi) = 1 + 2i + 1 + πi = (1 + 1) + (2 + π)i = 2 + (2 + π)i. (b) (1 + 2i)(1 + πi) = 1 + πi + 2i + 2πi2 = 1 − 2π + (π + 2)i. (c) (−1 − 3i)(−1 + 3i) = 1 − 3i + 3i − 3i2 = 1 + 3 = 4. This completes the proof of the problem.



Problem 1.2 ⋆ Write down the real and imaginary parts of z = (3 + i) + 2i(−1 + 2i).

Proof. It is easy to see that z = (3 + i) + 2i(−1 + 2i) = 3 + i − 2i + 4i2 = −1 − i, so we have Re z = −1 and Im z = −1, completing the proof of the problem.  Problem 1.3 ⋆ Express

3 + 4i in Cartesian form. 1 + 2i

Proof. By Definition 1.2 and the property i2 = −1, we have 3 + 4i 3 + 4i 1 − 2i 11 2 11 − 2i 3 − 2i − 8i2 = × = − i. = = 2 1 + 2i 1 + 2i 1 − 2i 1+2 5 5 5 This completes the proof of the problem. Problem 1.4 ⋆ Evaluate i52 and i−26 .



8

Chapter 1. Complex Numbers

Proof. It is clear that i4 = 1, so we have i52 = i4×13 = (i4 )13 = 1 and i−26 = i4×(−7)+2 = (i4 )−7 × i2 = −1. This ends the proof of the problem.



Problem 1.5 ⋆ Evaluate (a) (1 − 3i7 ) + (2 + 4i9 ) − (3 − 5i−5 ). (b) 3(cos π3 + i sin π3 ) + (− 32 +

√

3 2 i).

Proof. (a) Since i7 = i3 = −i, i9 = i and i−5 = −i, we have (1 − 3i7 ) + (2 + 4i9 ) − (3 − 5i−5 ) = (1 + 3i) + (2 + 4i) − (3 + 5i) = 2i. (b) We have √   1 √3   3 √3   √ π  3 π 3 + − + i =3 + i + − + i = 2 3i. 3 cos + i sin 3 3 2 2 2 2 2 2 This ends the proof of the problem.



Problem 1.6 ⋆ Solve

z + 2i z−1 = . z−i z+2

Proof. We get (z − 1)(z + 2) = (z − i)(z + 2i). After simplification, we have (1 − i)z = 4 so that z=

4 4(1 + i) = 2 + 2i, = 1−i (1 − i)(1 + i)

completing the proof of the problem.



Problem 1.7 ⋆ Let (x + iy)2 = 3 − 4i, where x and y are real. Find the values of x and y, and hence the square roots of 3 − 4i. Proof. The equation gives (x2 − y 2 ) + 2xyi = 3 − 4i. By comparing real and imaginary parts, we have x2 − y 2 = 3 and 2xy = −4. Solving the equations, we conclude that x = 2 and y = −1 or x = −2 and y = 1. Now the square roots of 3 − 4i are 2 − i and −2 + i, completing the proof  of the problem.

1.2. Arithmetic of Complex Numbers

9

Problem 1.8 ⋆ Find (1 − i)−1 .

Proof. Since 1 − i = 1 + i and (1 + i)(1 − i) = 2, we have (1 − i)−1 =

1+i 1 i = + . 2 2 2

This completes the proof of the problem.



Problem 1.9 ⋆ Determine the value(s) of the real number x such that (x2 − 3x − 4) + (x2 − 5x − 6)i is purely real.

Proof. Now (x2 − 3x − 4) + (x2 − 5x − 6)i is purely real if and only if x2 − 5x − 6 = 0 if and  only if x = 6 or x = −1. This completes the proof of the problem. Problem 1.10 ⋆ Solve the system of equations for the complex variables z and ω: 

(1 − i)z + (2 + i)ω = 7, (3 + 2i)z − (2 − 3i)ω = 13i.

(1.4)

Proof. Suppose that z = a + bi and ω = c + di, where a, b, c, d ∈ R. Then it follows from the equations (1.4) that (a + b + 2c − d) + (−a + b + 2d + c)i = 7

and (3a − 2b − 2c − 3d) + (3b + 2a − 2d + 3c)i = 13i.

Consequently, we obtain a + b + 2c − d = 7

−a + b + c + 2d = 0

3a − 2b − 2c − 3d = 0

2a + 3b + 3c − 2d = 13.

(1.5) (1.6) (1.7) (1.8)

The sum of the equations (1.5) and (1.6) gives 2b + 3c + d = 7.

(1.9)

Using the equations (1.5) and (1.7), we see that 5b + 8c = 21.

(1.10)

Combining the equations (1.5) and (1.8), we get − b + c = 1.

(1.11)

Finally, by solving the equations (1.9), (1.10) and (1.11), we conclude that c = 2 and then  a = b = 1 and d = −1. Hence z = 1 + i and ω = 2 − i, ending the proof of the problem.

10

Chapter 1. Complex Numbers Problem 1.11

⋆ Suppose that f (z) is a polynomial in z with real coefficients. If a+bi is a root of f (z) = 0, prove that f (a − bi) = 0.

Proof. Write f (z) =

n X

ck z k ,

k=0

where ck are all real. Clearly, ck = ck for all k = 0, 1, . . . , n. Then we have 0=0=

n X k=0

ck (a + bi)k =

n X k=0

ck · (a + bi)k =

n X k=0

ck (a − bi)k

which is exactly the desired result. We complete the proof of the problem.



Problem 1.12 ⋆ ⋆ Suppose that p, q ∈ R and x3 + px + q = 0 has a root a + bi, where a, b ∈ R. Prove that 2a is a root of x3 + px − q = 0.

Proof. Using Problem 1.11, we see that a − bi is also a root of the equation. Furthermore, it must have a real root. Let it be α. Then we have x3 + px + q = [x − (a + bi)] · [x − (a − bi)] · (x − α) = [x2 − 2ax + (a2 + b2 )] · (x − α). By comparing coefficients of the x2 term, we obtain α = −2a so that (−2a)3 + p(−2a) + q = 0 which is equivalent to (2a)3 + p(2a) − q = 0. In other words, 2a is a root of the equation x3 + px − q = 0, completing the proof of the  problem.

1.3 The Complex Plane Problem 1.13 ⋆ Find the modulus and the principal argument of each of the following complex numbers. (a) 1 + 3i. √ (b) −1 − 3i. Proof.

1.3. The Complex Plane

11

(a) Clearly, 1 + 3i lies in the first quadrant, so we follow from Definition 1.7 (Argument) that Arg z = tan−1 3. (b) We have tan−1

√

3 = π3 . Since −1 −

√

3i lies in the third quadrant, we conclude that

 π 2π Arg z = − π − =− . 3 3 We complete the proof of the problem.



Problem 1.14 ⋆ Transform the complex number 2 − 5i in its polar form. √ √ Proof. Let z = 2 − 5i. Then |z| = 22 + 52 = 29 and θ = tan−1 25 . Since z lies in the fourth quadrant, we have Arg z = −θ and thus √ √ z = 29 · [cos(−θ) + i sin(−θ)] = 29(cos θ − i sin θ), completing the proof of the problem.



Problem 1.15 √ √ 2(1 − 3i)(cos θ + i sin θ) ⋆ Simplify . (1 − i)(cos θ − i sin θ) √ √ 5π 7π Proof. Since 1 − 3i = 2(cos 5π 2(cos 7π 3 + i sin 3 ) and 1 − i = 4 + i sin 4 ), we obtain from the expression (1.3) that √

√ √ 5π 2 2(cos 5π 2(1 − 3i)(cos θ + i sin θ) 3 + i sin 3 ) · (cos θ + i sin θ) =√ 7π (1 − i)(cos θ − i sin θ) 2(cos 7π 4 + i sin 4 ) · [cos(−θ) + i sin(−θ)]

5π 2[cos( 5π 3 + θ) + i sin( 3 + θ)] 7π [cos( 7π 4 − θ) + i sin( 4 − θ)]   5π i h  5π 7π 7π +θ− + θ + i sin +θ− +θ = 2 cos 4 3 4  h  3 π i π + i sin 2θ − . = 2 cos 2θ − 12 12

=

This completes the proof of the problem. Problem 1.16 ⋆

⋆ Given that z = r(cos θ + i sin θ) 6= 0. Prove that z −1 =

z . |z|2



12

Chapter 1. Complex Numbers

Proof. Since z = r(cos θ + i sin θ) = r(cos θ − i sin θ), we have z −1 = = = = = =

1 r(cos θ + i sin θ) 1(cos 0 + i sin 0) r(cos θ + i sin θ) 1 [cos(−θ) + i sin(−θ)] r 1 (cos θ − i sin θ) r z r2 z , |z|2

completing the proof of the problem.



Problem 1.17 ⋆

⋆ Let z 6= 0 and −π < Arg z < π. Prove that Arg z = Arg z −1 = −Arg z.

Proof. By Problem 1.16, since

1 |z|2

is real, we see immediately that Arg z −1 = Arg z.

Since z is the reflection of z with respect to the real axis and −π < Arg z < π, we conclude that Arg z = −Arg z which implies the desired result. This completes the proof of the problem.



Problem 1.18 ⋆ Suppose that z1 = x1 +iy1 and z2 = x2 +iy2 are two non-collinear or non-parallel vectors. Let a and b be two real numbers such that az1 + bz2 = 0. Prove that a = b = 0.

Proof. Now az1 + bz2 = 0 implies (ax1 + bx2 ) + i(ay1 + by2 ) = 0. Therefore, we get ax1 + bx2 = ay1 + by2 = 0.

(1.12)

Since z1 and z2 are non-collinear or non-parallel, we have y1 y2 6= x1 x2 so that the equations (1.12) ensure that a = b = 0. This ends the problem of the problem.



1.3. The Complex Plane

13

Problem 1.19 ⋆ Prove that |z1 + z2 |2 = (z1 + z2 ) z1 + z2





and |z1 + z2 |2 + |z1 − z2 |2 = 2 |z1 |2 + |z2 |2 .

Proof. Now we have
(z1 + z2 ) z1 + z2 = (z1 + z2 )(z1 + z2 ) = |z1 + z2 |2 .
This proves the first identity. Next, since |z1 − z2 |2 = (z1 − z2 ) z1 − z2 , we have

|z1 + z2 |2 + |z1 − z2 |2 = (z1 + z2 ) z1 + z2 + (z1 − z2 ) z1 − z2

= |z1 |2 + z1 z2 + z1 z2 + |z2 |2 + |z1 |2 − z1 z2 − z1 z2 + |z2 |2
= 2 |z1 |2 + |z2 |2

which ends the analysis of the problem.



Problem 1.20 ⋆ Let α, β ∈ R \ {0}.

⋆

(a) Prove that αβ is real if and only if β = kα for some k ∈ R. (b) Suppose that a, b, c, d are complex numbers such that a 6= b, ac, bd and ad + bc are all c−d is real. real. Prove that a−b Proof. (a) If β = kα, then αβ = kαα = k|α|2 which is real. Conversely, let α = a + bi and β = c + di, where a, b, c, d ∈ R. Then we have αβ = (a − bi)(c + di) = (ac + bd) + i(ad − bc). Since αβ is real, we know that ad − bc = 0. If a = 0, then b 6= 0 and c = 0 so that d 6= 0. In this case, d d β = di = (bi) = α. (1.13) b b If c = 0, then d 6= 0 and a = 0 so that b 6= 0. This also gives the formula (1.13). Let ac 6= 0. Then we have dc = ab which implies   d  b  c c β = c + di = c 1 + i = c 1 + i = (a + bi) = α. c a a a

(b) We note that c−d a−b c−d = × a−b a−b a−b

14

Chapter 1. Complex Numbers =

ac − ad − bc + bd aa − ab − ab + bb

ac − ad − bc + bd
|a|2 − ab − ab + |b|2
ac − ad + bc + bd = 2 . |a| − 2Re (ab) + |b|2 =

(1.14)

Now it follows from the hypotheses that the expression (1.14) is a real number. This completes the proof of the problem.



Problem 1.21 (The Triangle Inequality) ⋆ For any two complex numbers z1 , z2 , prove that |z1 + z2 | ≤ |z1 | + |z2 |. Prove that the equality holds if and only if z1 = kz2 for some k ≥ 0. Proof. If follows from Problem 1.19 and Theorem 1.6 (Properties of the Modulus) that
|z1 + z2 |2 = (z1 + z2 ) z1 + z2
= |z1 |2 + 2Re z1 z2 + |z2 |2 ≤ |z1 |2 + 2 z1 z2 + |z2 |2 (1.15) = |z1 |2 + 2|z1 | · |z2 | + |z2 |2
2 = |z1 | + |z2 | .

Therefore, we have |z1 + z2 | ≤ |z1 | + |z2 |.

The equality holds trivially if z1
= kz 2 . Suppose that the equality holds. Then it follows from the inequality (1.15) that Re z1 z2 = z1 z2 which means that z1 z2 is a positive real number. Let this number be A. Then we have z1 z2 = A
z1 z2 z2 = Az2 A z1 = · z2 . |z2 |2

This completes the proof of the problem. Problem 1.22 ⋆ Prove that |z1 | − |z2 | ≤ |z1 − z2 |. Proof. Observe from Problem 1.21 (The Triangle Inequality) that |z1 | = |(z1 − z2 ) + z2 | ≤ |z1 − z2 | + |z2 |



1.3. The Complex Plane

15

so that |z1 | − |z2 | ≤ |z1 − z2 |.

(1.16)

− |z1 − z2 | ≤ |z1 | − |z2 |.

(1.17)

Similarly, it can be shown that Hence our result follows from the inequalities (1.16) and (1.17). This ends the proof of the  problem. Problem 1.23 ⋆ Prove that



if |z| = 3.

1 1 ≤ 4 2 z + 5z + 6 48

Proof. Obviously, we have z 4 + 5z 2 + 6 = (z 2 + 2)(z 2 + 3), so Problem 1.22 implies that |z 4 + 5z 2 + 6| = |(z 2 + 2)(z 2 + 3)| ≥ |z 2 | − 2 · |z 2 | − 3 = |9 − 1| · |9 − 3| = 48.

Hence we conclude that



1 1 ≤ 4 2 z + 5z + 6 48

for |z| = 3, completing the proof of the problem. Problem 1.24

⋆ For any z ∈ C, prove that |z − 1| ≤ |z| − 1 + |z| · |Arg z|. Proof. Consider the triangle with vertices 1, z and |z|, see Figure 1.4.

Figure 1.4: The triangle formed by 1, z and |z|.



16

Chapter 1. Complex Numbers It is clear that |z − 1| ≤ |z| − 1 + z − |z| holds. Of course, we also have z − |z| is less than or equal to the arc connecting z and |z|. By the definition, this arc is |z| · |Arg z|, so we glue all these together to get |z − 1| ≤ |z| − 1 + z − |z| ≤ |z| − 1 + |z| · |Arg z|

which completes the proof of the problem.



Problem 1.25 (Lagrange’s Identity) ⋆ Prove that

⋆

n X k=1

|ak |2

n  X k=1

n 2  X ak bk = |bk |2 − k=1

X

1≤k 0, |c|2 > αβ and αzz + cz + cz + β = 0.

Proof. Suppose that
there exist a ∈ C and r > 0 such that |z − a| = r. Then it is easy to see that (z − a) z − a = r 2 which reduces to
zz − az − az + |a|2 − r 2 = 0. In this case, we have α = 1, c = −a and β = |a|2 − r 2 . We observe that they obviously satisfy the condition that |c|2 > αβ. Conversely, we first rewrite the equation as zz +

c β c z + z + = 0. α α α

Further simplification gives  c i |c|2 − αβ  c h = . z+ z+ α α α2

The facts |c|2 > αβ and α > 0 ensure that

Hence it is a circle with radius problem.

√

p |c|2 − αβ c . z + = α α

|c|2 −αβ α

and centered at − αc . We complete the proof of the 

Problem 2.13 ⋆ ⋆ Let α, β, γ ∈ C be such that α+β+γ = 0 and ω be a root of the equation z 2 −z+1 = 0. (a) Prove that β = −γω if and only if αβ + βγ + γα = 0. (b) Prove that the triangle with vertices z1 , z2 , z3 is equilateral if and only if 1 1 1 + + = 0. z1 − z2 z2 − z3 z3 − z1 Proof. We have ω =

√ 1± 3i 2 .

(a) It is clear that β = −γω if and only if β +

γ 2

β 2 + βγ +

=∓

√ 3γ 2 i

which is equivalent to

3γ 2 γ2 =− 4 4

30

Chapter 2. Geometry of Complex Numbers or β 2 + βγ + γ 2 = 0. Since α + β + γ = 0, it gives β(−α − γ) + βγ + γ(−α − β) = 0 αβ + βγ + γα = 0.

(b) ∆z1 z2 z3 is equilateral if and only if |z1 − z2 | = |z1 − z3 | and ∠z2 z! z3 = ± π3 . Therefore, we have  π  π z3 − z1 + i sin ± = cos ± z2 − z1 3 3 z3 − z1 = ω(z2 − z1 ). (2.8) Denote α = z2 − z3 , β = z3 − z1 and γ = z1 − z2 . Clearly, we have α + β + γ = 0 and the equation (2.8) becomes β = −γω. Since z1 , z2 and z3 are distinct, we have αβγ 6= 0. Hence we follow from this and part (a) that αβ + βγ + γα = 0 1 1 1 + + =0 γ β α 1 1 1 + + = 0. z1 − z2 z2 − z3 z3 − z1 We end the analysis of the problem.



Problem 2.14 ⋆ Given that |z − (1 + 2i)| = Re (z − 2 + 3i). Prove that it is a parabola.

Proof. Put z = x + iy into the equation to obtain p (x − 1)2 + (y − 2)2 = x − 2

x2 − 2x + 1 + (y − 2)2 = x2 − 4x + 4

Hence it is a parabola with vertex at

3 2

(y − 2)2 = −2x + 3  1  3 (y − 2)2 = 4 · − x− . 2 2

+ 2i, completing the proof of the problem.



Problem 2.15 ⋆ ⋆ Let P1 , P2 and P3 be three distinct points in C representing the complex numbers z1 , z2 and z3 respectively. Their order is counterclockwise. 2π 2 (a) Define λ = cos 2π 3 + i sin 3 . Assume that the triangle formed by 1, λ and λ is equilateral. Prove that ∆P1 P2 P3 is equilateral if and only if

z1 + λz2 + λ2 z3 = 0. (b) The point (a, b) is called an integral point if both a and b are integers. Show that no triangle with distinct integral points as vertices can be equilateral.

2.4. Transformations of Complex Numbers

31

Proof. (a) It is trivial to check that 1, λ, λ2 are roots of z 3 − 1 = 0. Let E1 , E2 and E3 be the complex numbers corresponding to 1, λ and λ2 respectively. We know that ∆P1 P2 P3 is equilateral if and only if ∆P1 P2 P3 ∼ ∆E1 E2 E3 ∼ ∆E3 E1 E2 . According to the proof of Problem 2.4, we obtain that λ − λ2 z3 − z1 = z2 − z1 1 − λ2 z3 − z1 λ = z2 − z1 1+λ (1 + λ)(z3 − z1 ) = λz2 − λz1

z3 − z1 + λz3 − λz1 = λz2 − λz1

z1 + λz2 − (1 + λ)z3 = 0.

(2.9)

Since 1 + λ + λ2 = 0, we can express the equation (2.9) in the form z1 + λz2 + λ2 z3 = 0. (b) Assume that there was an equilateral triangle ∆P1 P2 P3 with distinct integral points z1 , z2 , z3 as vertices. Denote zk = ak + ibk , where ak , bk ∈ Z for k = 1, 2, 3. Part (a) yields (a1 + ib1 ) + λ(a2 + ib2 ) + λ2 (a3 + ib3 ) = 0 √  1 √3   1 3  i (a2 + ib2 ) + − − i (a3 + ib3 ) = 0 (a1 + ib1 ) + − + 2 2 √ 2 2 √ [(2a1 − a2 − a3 ) + (b3 − b2 ) 3] + i[(2b1 − b2 − b3 ) + (a2 − a3 ) 3] = 0. √ Since ak , bk ∈ Z for k = 1, 2, 3 and 3 is irrational, we have 2a1 − a2 − a3 = b3 − b2 = 2b1 − b2 − b3 = a2 − a3 = 0. Solving the equations to get a2 = a3 and b2 = b3 which mean z2 = z3 , a contradiction. Hence no such equilateral triangle exists. We have completed the proof of the problem.



2.4 Transformations of Complex Numbers Problem 2.16 ⋆ Let a, b ∈ C and a 6= 0. Describe the transformations of f (z) = az + b. Proof. Express a in its polar form |a|eiα . Then we have



f (z) = |a|zeiα + b = f3 f2 f1 (z) ,

where f1 (z) = |a|z, f2 (z) = zeiα and f3 (z) = z + b. Hence it involves three transformations performed successively as follows:

32

Chapter 2. Geometry of Complex Numbers • Enlargement: f1 (z) = |a|z, • Rotation with angle α: f2 (z) = zeiα , • Translation: f3 (z) = z + b.

We end the analysis of the problem.



Problem 2.17 ⋆ Let z = ai, where a is a nonzero real constant. Show that the points corresponding to z, z 2 and z 3 form the vertices of a right-angled triangle.

Proof. The figure of the complex numbers z, z 2 and z 3 is shown in Figure 2.7. Let P and Q

Figure 2.7: The figure of z, z 2 and z 3 . represent the points z − z 2 and z 2 − z 3 respectively, see Figure 2.8 for details. We know that

Figure 2.8: The positions of the two points P and Q. −−→ z 2 − z 3 = ai(z − z 2 ), so we rotate OP by

π 2

−− → and then multiply |OP | by a factor a to obtain

2.5. Stereographic Projection 33 −−→ OQ. In other words, we have ∠P OQ = π2 so that points corresponding to z, z 2 and z 3 form the vertices of a right-angled triangle, ending the proof of the problem.  Problem 2.18 ⋆ Suppose that P1 and P2 represent the nonzero complex numbers z1 and z2 respectively. Show that OP1 ⊥ OP2 if and only if Re (z1 z2 ) = 0. −−→ Proof. that we can enlarge OP1 such that −−→ Observe −−→ that OP1 ⊥ OP2 is equivalent to saying −→ OP1 = OP2 and then rotate the enlarged vector by π to obtain − OP2 . Hence, it means that 2 z2 = iλz1

(2.10)

for some nonzero real number λ. In fact, the equation (2.10) ensures that
z1 z2 = −iλz1 z1 = − λ|z1 |2 i.

By the definition, we conclude that Re (z1 z2 ) = 0 which completes the proof of the problem. 

2.5 Stereographic Projection Problem 2.19 ⋆ In the stereographic projection, express ξ, η and ζ in terms of x and y, where z = x + iy. Proof. We observe from Figure 2.3 that the points (0, 0, 1), (ξ, η, ζ) and (x, y, 0) are collinear. Therefore, we have x y 1 = = ξ η 1−ζ

so that

x= Rewrite the equations (2.11) as

ξ 1−ζ

and y =

η . 1−ζ

ξ = x(1 − ζ) and η = y(1 − ζ).

(2.11)

(2.12)

Recall that the point (ξ, η, ζ) satisfies  1 2 1 ξ 2 + η2 + ζ − = . 2 4

Put the equations (2.12) into the formula (2.13) to get

 1 2 1 x2 (1 − ζ)2 + y 2 (1 − ζ)2 + ζ − = 2 4 2 2 2 2 2 x (1 − 2ζ + ζ ) + y (1 − 2ζ + ζ ) + ζ − ζ = 0

(2.13)

34

Chapter 2. Geometry of Complex Numbers (x2 + y 2 + 1)ζ 2 − (2x2 + 2y 2 + 1)ζ + x2 + y 2 = 0.

Using the quadratic formula, we have p 2x2 + 2y 2 + 1 ± (2x2 + 2y 2 + 1)2 − 4(x2 + y 2 )(x2 + y 2 + 1) ζ= 2(x2 + y 2 + 1) 2x2 + 2y 2 + 1 ± 1 = 2(x2 + y 2 + 1) x2 + y 2 = 2 or 1 (rejected). x + y2 + 1

(2.14)

By substituting the expression (2.14) back into the expressions (2.12), we obtain ξ=

x x2 + y 2 + 1

and η =

y . x2 + y 2 + 1

(2.15)

This completes the proof of the problem.



Remark 2.1 Equivalently, we have the following representation ξ=

z+z , 2(|z|2 + 1)

η=

z−z 2i(|z|2 + 1)

and ζ =

zz . +1

|z|2

Problem 2.20 ⋆ Fix ξ0 ∈ (0, 1). Define S = {(ξ, η, ζ) ∈ Σ | ξ ≥ ξ0 }. Prove that its corresponding set, ζ0 denoted by SC , is the set C \ D(0; 1−ζ ). 0

Proof. It asserts from Problem 2.19 that ξ=

x , 2 x + y2 + 1

η=

Thus we have

Since 0 < ζ0 ≤ ζ < 1, we have

1 1−ζ

y 2 x + y2 + 1

ζ=

x2 + y 2 . x2 + y 2 + 1

ζ = x2 + y 2 . 1−ζ

≥

1 1−ζ0 ,

ζ0 . 1 − ζ0

 SC = C \ D 0;

completing the proof of the problem.

(2.16)

so it follows from the equation (2.16) that

x2 + y 2 ≥ Hence we obtain

and

ζ0  , 1 − ζ0



2.5. Stereographic Projection

35

Problem 2.21 ⋆ Prove that circles on Σ correspond to circles and straight lines in C.

Proof. Express a circle on Σ as the intersection of the sphere Σ and a plane Aξ +Bη+Cζ = D for some complex constants A, B, C and D. Then Problem 2.19 yields the stereographic projection of this circle will consist of points z = x + iy which satisfy Ax By x2 + y 2 + + C · = D. x2 + y 2 + 1 x2 + y 2 + 1 x2 + y 2 + 1 After simplification, we see that (C − D)(x2 + y 2 ) + Ax + By − D = 0.

(2.17)

When C = D, the locus (2.17) is a straight line. Otherwise, we rewrite it as x2 + y 2 +

A B D x+ y− =0 C −D C−D C −D

so that is a circle. This completes the proof of the problem.



Problem 2.22 ⋆ Let z1 , z2 ∈ C and d(z1 , z2 ) be the distance of the chord joining the corresponding points (ξ1 , η1 , ζ1 ) and (ξ2 , η2 , ζ2 ). Prove that

What is d(z, ∞)?

χ(z1 , z2 ) = p

|z1 − z2 | p . 1 + |z1 |2 · 1 + |z2 |2

Proof. Using the formulas (2.14) and (2.15) in Problem 2.19, we see immediately that χ(z1 , z2 )2 = |(ξ1 , η1 , ζ1 ) − (ξ2 , η2 , ζ2 )|2

= (ξ1 − ξ2 )2 + (η1 − η2 )2 + (ζ1 − ζ2 )2

= (ξ12 + η12 + ζ12 ) + (ξ22 + η22 + ζ22 ) − 2(ξ1 ξ1 + η1 η2 + ζ1 ζ2 ).

(2.18)

Since the equation (2.13) reduces to ξ 2 + η 2 + ζ 2 = ζ, put this into the expression (2.18) and then use Remark 2.1 to get χ(z1 , z2 )2 = (ζ1 + ζ2 ) − 2(ξ1 ξ1 + η1 η2 + ζ1 ζ2 )


2 Re z1 · Re z2 + Im z1 · Im z2 + |z1 |2 · |z2 |2 |z2 |2 |z1 |2 + − = 1 + |z1 |2 1 + |z2 |2 (1 + |z1 |2 )(1 + |z2 |2 )
|z1 |2 (1 + |z2 |2 ) + |z2 |2 (1 + |z1 |2 ) − 2 Re z1 · Re z2 + Im z1 · Im z2 + |z1 |2 · |z2 |2 = (1 + |z1 |2 )(1 + |z2 |2 ) |z1 |2 + |z2 |2 − 2(Re z1 · Re z2 + Im z1 · Im z2 ) = (1 + |z1 |2 )(1 + |z2 |2 )

36

Chapter 2. Geometry of Complex Numbers =

|z1 − z2 |2 (1 + |z1 |2 )(1 + |z2 |2 )

(2.19)

which gives our desired result. To evaluate χ(z, ∞), we take z2 → ∞ in the formula (2.19) and obtain χ(z, ∞) = lim d(z, z2 ) z2 →∞

|z − z2 | p = lim p z2 →∞ 1 + |z|2 · 1 + |z2 |2 |1 − zz2 | q = lim p z2 →∞ 1 + |z|2 · 1 + |z12 |2 =p

We complete the proof of the problem.

1

1 + |z|2

.



Remark 2.2 (a) The formula in Problem 2.22 is called the chordal distance and it is easily seen that it is a metric and χ(z1 , z2 ) ≤ 1 because it is the Euclidean distance between two points of a sphere of radius 1. (b) Besides of chordal distance, one can consider the spherical distance ds (z1 , z2 ). This is defined to be the length of the smallest of the two arcs determined by the images (ξ1 , η1 , ζ1 ) and (ξ2 , η2 , ζ2 ) of z1 and z2 on the great circle of the Riemann sphere that connecting those two points. (c) For more derivations and discussions of these distances, please read [17, §1.10 and §1.17].

CHAPTER

3

nth Roots of a Complex Number

3.1 Fundamental Concepts In this chapter, concepts and basic results of nth roots of a complex number ω 6= 0 are introduced. The main references we use here are [2, Chap. 1], [8, Chap. 1] and [20, Chap. 1]. √ 1 If a is a nonnegative real number and n is a positive integer, then the notation n a or a n means the nonnegative nth root of a. This is the unique nonnegative number x such that xn = a. In fact, this can be generalized to complex numbers. Suppose that ω ∈ C \ {0} and n is any positive integer. Each solution of the equation zn = ω

(3.1)

is called an nth root of ω. Fortunately, all solutions of the equation (3.1) can be stated explicitly in the following result: Theorem 3.1 (nth Roots of a Complex Number). Let ω = r(cos θ + i sin θ), where r > 0. Then ω has exactly n distinct nth roots which are given by  1 θ + 2kπ θ + 2kπ  , + i sin zk = r n cos n n where k = 0, 1, 2, . . . , n − 1. Corollary 3.2. The number 1 has exactly n distinct nth roots that are given by zk = cos

2kπ 2kπ + i sin , n n

where k = 0, 1, 2, . . . , n − 1. These are called the nth roots of unity. 37

38

Chapter 3. nth Roots of a Complex Number

3.2 nth Roots of a Complex Number Problem 3.1 ⋆

⋆ Prove Theorem 3.1 (nth Roots of a Complex Number).

Proof. Let z = R(cos φ + i sin φ). Combining the hypotheses and Theorem 1.10 (De Moivre’s Theorem), we get Rn (cos φ + i sin φ)n = r(cos θ + i sin θ) Rn (cos nφ + i sin nφ) = r(cos θ + i sin θ). Now equating the moduli of both sides, and then using the fact that the arguments of two sides differ by an integer multiple of 2π, we obtain immediately that Rn = r

and nφ = θ + 2kπ 1

for some k ∈ Z. The first equation shows that R = r n and the second one implies φ=

θ + 2kπ , n

where k ∈ Z. Therefore, the solutions of the equation (3.1) are of the form

where k ∈ Z.

 1 θ + 2kπ  θ + 2kπ , + i sin zk = r n cos n n

At first sight, it seems that we have found infinitely many solutions. However, not all these solutions are distinct. Actually, if k1 and k2 differ by an integer multiple of n, say k2 = k1 + mn for some m ∈ Z, then we have θ + 2k2 π θ + 2(k1 + mn)π θ + 2k1 π = = 2mπ + . n n n Hence solutions corresponding to k1 and k2 are identical and so all possible solutions of the equation (3.1) must correspond to the integers k = 0, 1, . . . , n − 1. Finally, these n solutions 1 are distinct because they lie on the circle of radius r n centered at the origin, with the angle 2π n between adjacent solutions. This completes the proof of the problem.  Problem 3.2 ⋆ Suppose that z is an nth root of ω 6= 0. Show that z is an nth root of ω.

Proof. It follows from the equation (3.1) that z n = ω so that
n z = ω.

In other words, z is an nth root of ω which ends the proof of the problem.



3.2. nth Roots of a Complex Number

39

Problem 3.3 ⋆ Let z = −1 + i. Find the three cube roots of z. √ √ 3π Proof. It is easy to check that r = 2 and arg z = 3π 2(cos 3π 4 , so we have z = 4 + i sin 4 ). Hence we deduce from Theorem 3.1 (nth Roots of a Complex Number) that   √ √ 3π 3π  13 6 6 2 cos + i sin = 2 cos 4 4

1

z3 =

3π 4

+ 2kπ + i sin 3

3π 4

+ 2kπ  , 3

where k = 0, 1, 2. Writing the cube roots of z explicitly, they are    √ √ π 11π 19π π √ 11π  19π  6 6 6 2 cos + i sin 2 cos 2 cos , and . + i sin + i sin 4 4 12 12 12 12

We have completed the proof of the problem.



Problem 3.4 ⋆ Given that z 6 + z 3 + 1 = 0. (a) Show that z 3 = cos

2π 2π ± i sin . 3 3

(b) Find the roots of the given equation.

Proof. (a) Let ζ = z 3 . Then the equation becomes ζ 2 + ζ + 1 = 0 whose solutions are √ −1 ± 3i . 2 Therefore, we obtain z 3 = ζ = cos

2π 2π ± i sin . 3 3

(b) Now Theorem 3.1 (nth Roots of a Complex Number) implies that zk = cos

2π 3

+ 2kπ ± i sin 3

2π 3

+ 2kπ 3

where k = 0, 1, 2. Hence the roots of the given equation are cos

2π 2π ± i sin , 9 9

cos

8π 8π ± i sin 9 9

This completes the proof of the problem.

and

cos

14π 14π ± i sin . 9 9 

Problem 3.5 ⋆ Solve the equation z n − z n−1 + z n−2 − · · · − z + 1 = 0, where n is a positive even integer.

40

Chapter 3. nth Roots of a Complex Number

Proof. Since n is even, it is apparent that z n+1 + 1 = z n − z n−1 + z n−2 − · · · − z + 1, z+1 so we have z n+1 + 1 = 0 but z + 1 6= 0. Using Theorem 3.1 (nth Roots of a Complex Number), we know that 1

zk = (−1) n+1 1

= (cos π + i sin π) n+1 (2k + 1)π (2k + 1)π = cos + i sin , n+1 n+1 where k = 0, 1, 2, . . . , n. Since z = −1 if and only if k = n2 , the roots are zk = cos

(2k + 1)π (2k + 1)π + i sin , n+1 n+1

where k = 0, 1, 2, . . . , n but k 6= n2 , completing the proof of the problem.



Problem 3.6 ⋆ Let a be a complex constant with |a| = 1 and n be a positive integer. Prove that all the roots of the equation  1 + iz n =a 1 − iz are real.

Proof. Set a = cos θ + i sin θ. Then Theorem 3.1 (nth Roots of a Complex Number) ensures that 1 + izk θ + 2kπ θ + 2kπ = cos + i sin , 1 − izk n n θ+2kπ where k = 0, 1, . . . , n − 1. Denote ωk = cos θ+2kπ for k = 0, 1, . . . , n − 1. Then we 2n + i sin 2n −1 θ+2kπ θ+2kπ have ωk = cos 2n − i sin 2n so that

1 + izk = ωk2 1 − izk 1 + izk = ωk2 − izωk2 zk =

ωk2 − 1 i(ωk2 + 1)

ωk − ωk−1 i(ωk + ωk−1 ) θ + 2kπ = tan 2n

=

which is real for every k = 0, 1, . . . , n − 1. This ends the analysis of the problem.



3.2. nth Roots of a Complex Number

41

Problem 3.7 ⋆

⋆ Let n be a positive integer and n > 1. Prove that

zn + 1 =

 n −1 2 o  Yn  (2k + 1)π  2  + 1 , z − 2z cos   n   k=0

if n is even;

 1  (n−3) n  2 Y o   (2k + 1)π 2   (z + 1) z − 2z cos + 1 , if n is odd.  n r=0

Proof. We consider the equation z n + 1 = 0.

(3.2)

Suppose that n = 2p, where p ≥ 1. Then the roots of the equation (3.2) are given by cos

(2k + 1)π (2k + 1)π + i sin , 2p 2p

where k = 0, 1, . . . , 2p − 1. Since the roots of the equation (3.2) occur in conjugate pairs, we are able to group all the nth roots of −1 as follows: For 0 ≤ k ≤ p − 1, we see that [2(2p − k − 1) + 1]π [2(2p − k − 1) + 1]π + i sin 2p 2p h i h (2k + 1)π (2k + 1)π i = cos 2π − + i sin 2π − 2p 2p (2k + 1)π (2k + 1)π − i sin . = cos 2p 2p cos

Hence we have

(2k + 1)π (2k + 1)π ± i sin , 2p 2p where k = 0, 1, . . . , p − 1. Direct computation gives h (2k + 1)π (2k + 1)π (2k + 1)π ih (2k + 1)π i z − cos z − cos − i sin + i sin 2p 2p 2p 2p h (2k + 1)π (2k + 1)π i2 + sin2 = z − cos 2p 2p (2k + 1)π + 1. = z 2 − 2z cos 2p cos

Hence we obtain z n + 1 = z 2p + 1 =

p−1 Yn

k=0

=

n −1 2

z 2 − 2z cos

Yn

k=0

z 2 − 2z cos

o (2k + 1)π +1 n o (2k + 1)π +1 . n

(3.3)

42

Chapter 3. nth Roots of a Complex Number

Next, let n = 2p + 1, where p > 1. In this case, the roots of the equation (3.2) are given by −1,

cos

(2k + 1)π (2k + 1)π ± i sin , 2p 2p

where k = 0, 1, . . . , p − 1. Therefore, we have z n + 1 = z 2p+1 + 1 = (z + 1)

p−1 Yn

k=0

z 2 − 2z cos

1 (n−3) 2

o (2k + 1)π +1 n

o Y n (2k + 1)π z 2 − 2z cos +1 . n

= (z + 1)

(3.4)

k=0

Consequently, our result follows from the two results (3.3) and (3.4). Thus we have completed  the analysis of the problem. Problem 3.8 ⋆ Let p be a positive integer. Prove that 2p−1

2

·

p−1 Y

sin2

k=0

(2k + 1)π = 1. 4p

Proof. By Problem 3.7 with z = 1 and n = 2p as well as the identity 2 sin2 θ = 1 − cos 2θ, we have p−1 p−1 Y h (2k + 1)π (2k + 1)π i Y 2 1 − cos 4 sin2 2= = 2p 4p k=0

k=0

which reduces to 2p−1

2

p−1 Y

k=0

sin2

(2k + 1)π =1 4p

as required. This completes the proof of the problem.



Problem 3.9 ⋆ Solve the equation (1 + z)2n − (1 − z 2 )n + (1 − z)2n = 0, where n is a positive integer.

Proof. Let u = (1 + z)n and v = (1 − z)n . Then the equation becomes u2 − uv + v 2 = 0.

(3.5)

3.2. nth Roots of a Complex Number

43

As u3 + v 3 = (u + v)(u2 − uv + v 2 ), we have u2 − uv + v 2 =

u3 + v 3 , u+v

where u + v 6= 0. Hence the roots of the equation (3.5) are exactly those roots of u3 + v 3 = 0 excluding those which are also roots of u + v = 0. Now the roots of u3 + v 3 = 0 are exactly those given by (

1 + z 3n ) = −1, 1−z

so Theorem 3.1 (nth Roots of a Complex Number) implies that 1+z (2k + 1)π (2k + 1)π = cos + i sin , 1−z 3n 3n

(3.6)

where k = 0, 1, 2, . . . , 3n − 1. Similarly, the roots of u + v = 0 are those given by (

1+z n ) = −1. 1−z

Again, we obtain from Theorem 3.1 (nth Roots of a Complex Number) that (2p + 1)π (2p + 1)π 1+z = cos + i sin , 1−z n n

(3.7)

where p = 0, 1, 2, . . . , n − 1. Combining the expressions (3.6) and (3.7), the solutions of the required equation are given by 1+z (2k + 1)π (2k + 1)π = cos + i sin , 1−z 3n 3n where k = 0, 1, 2, . . . , 3n − 1 but k 6≡ 1 (mod 3).a Write λk = expression (3.8) that z= = =

(2k+1)π . 3n

(3.8) Then it follows from the

cos λk − 1 + i sin λk 1 + cos λk + i sin λk 2i sin λ2k cos λ2k − 2 sin2

λk 2 2i sin λ2k cos λ2k + 2 cos2 λ2k i sin λ2k (cos λ2k + i sin λ2k ) cos λ2k (cos λ2k + i sin λ2k )

λk 2 (2k + 1)π , = i tan 6n = i tan

where k = 0, 1, 2, . . . , 3n − 1 but k 6≡ 1 (mod 3), ending the proof of the problem. a

The notation a ≡ b (mod n) means that a − b is divisible by n.



44

Chapter 3. nth Roots of a Complex Number

3.3 nth Roots of Unity Problem 3.10 ⋆ Solve the equation z n + z n−1 + · · · + z + 1 = 0, where n is a positive integer. Proof. Multiplying both sides of the equation by (z − 1) to get z n+1 − 1 = 0. It follows from Corollary 3.2 that z = cos

2kπ 2kπ + i sin , n+1 n+1

where k = 0, 1, 2, . . . , n. Since 1 is not a root of the given equation, the required roots are cos

2kπ 2kπ + i sin , n+1 n+1

where k = 1, 2, . . . , n. This completes the proof of the problem.



Problem 3.11 ⋆ Given n ∈ N. Verify that all the roots of the equation (1 + z)2n+1 = (1 − z)2n+1 are given by kπ ±i tan , 2n + 1 where k = 0, 1, 2, . . . , n.

Proof. It is obvious that z = 1 is not a root of the given equation. Thus we can rewrite it as  1 + z 2n+1 1−z

According to Corollary 3.2, we have

= 1.

(3.9)

2kπ 2kπ 1 + zk = cos + i sin , 1 − zk 2n + 1 2n + 1

(3.10)

2kπ where k = 0, 1, 2, . . . , 2n. Denote tk = 2n+1 . Similar to the proof of Problem 3.9, we derive from the equation (3.10) that tk cos tk − 1 + i sin tk = i tan . zk = 1 + cos tk + i sin tk 2

Consequently, the roots of the equation (3.9) are 0, i tan positive integer such that 1 ≤ m ≤ n, then we have tan

kπ 2n+1 ,

where k = 1, 2, . . . , 2n. If m is a

 mπ mπ  2n − (m − 1)π = − tan = tan π − . 2n + 1 2n + 1 2n + 1

3.3. nth Roots of Unity

45

Hence the 2n roots corresponding to k = 1, 2, . . . , 2n can be grouped into pairs of the form ±i tan

kπ , 2n + 1

0 , we get the desired result and complete the proof where k = 1, 2, . . . , n. Since 0 = ±i tan 2n+1 of the problem. 

Problem 3.12 ⋆ Let ω be a non-real cube root of unity and a, b, c ∈ R. Prove that (a) (1 + ω − ω 2 )3 − (1 − ω + ω 2 )3 = 0. (b) (a + b + c)(a + bω + cω 2 )(a + bω 2 + cω) = a3 + b3 + c3 − 3abc.

Proof. Since ω 3 = 1 and ω 6= 1, we have ω 2 + ω + 1 = 0.

(3.11)

(a) Using the result (3.11), we see that (1 + ω − ω 2 )3 − (1 − ω + ω 2 )3 = (1 + ω + ω 2 − 2ω 2 )3 − (1 + ω + ω 2 − 2ω)3 = (0 − 2ω 2 )3 − (0 − 2ω)3

= −8ω 6 + 8ω 3 = −8 + 8 =0

(b) Note that (a + bω + cω 2 )(a + bω 2 + cω) = a2 + b2 ω 3 + c2 ω 3 + ab(ω + ω 2 ) + ac(ω + ω 2 ) + bc(ω 2 + ω 4 ) = a2 + b2 + c2 + ab(1 + ω + ω 2 − 1) + bc(1 + ω + ω 2 − 1) + ac(1 + ω + ω 2 − 1)

= a2 + b2 + c2 − ab − bc − ca.

(3.12)

By the expression (3.12), we obtain that (a + b + c)(a + bω + cω 2 )(a + bω 2 + cω) = (a + b + c)(a2 + b2 + c2 − ab − bc − ca) = a3 + b3 + c3 − 3abc.

Hence we complete the analysis of the problem.



46

Chapter 3. nth Roots of a Complex Number Problem 3.13 ⋆

⋆ Suppose that n is a positive integer and ω = cos

2π 2π + i sin . n n

Find the values of (a) 1 + ω + ω 2 + · · · + ω n−1 . (b) ω × ω 2 × ω 3 × · · · × ω n−1 . (c) (1 − ω)(1 − ω 2 ) · · · (1 − ω n−1 ). (d) 1 + ω m + ω 2m + · · · ω (n−1)m , where m ∈ Z. Proof. Clearly, the distinct roots of the equation z n = 1 are 1, ω, ω 2 , . . . , ω n−1 . (a) Since the sum of roots of the equation z n − 1 = 0 is zero, we have 1 + ω + ω 2 + · · · + ω n−1 = 0. (b) Since the product of roots of the equation z n − 1 = 0 is (−1)n+1 , we have ω × ω 2 × ω 3 × · · · × ω n−1 = 1 × ω × ω 2 × ω 3 × · · · × ω n−1 = (−1)n+1 . (c) By the identity z n − 1 = (z − 1)(z n−1 + z n−2 + · · · + z + 1),

we see that ω, ω 2 , . . . , ω n−1 are roots of the equation z n−1 + z n−2 + · · · + z + 1 = 0. Thus we have (z − ω)(z − ω 2 ) · · · (z − ω n−1 ) = z n−1 + z n−2 + · · · + z + 1. (3.13) We put z = 1 into the equation (3.13) to yield that (1 − ω)(1 − ω 2 ) · · · (1 − ω n−1 ) = 1 + 1 + · · · + 1 = n. (d) By the Division Algorithm, there are integers q and r with 0 ≤ r < n such that m = qn+r. It means that ω m = ω qn+r = (ω n )q · ω r = ω r so that 1 + ω m + ω 2m + · · · + ω (n−1)m = 1 + ω r + ω 2r + · · · + ω (n−1)r . If r = 0, then it is trivial that 1 + ω r + ω 2r + · · · + ω (n−1)r = 1 + 1 + · · · + 1 = n. If r > 0, then the definition of ω guarantees that ω r 6= 1. Thus we have 1 + ω m + ω 2m + · · · + ω (n−1)m =

1 − (ω n )r 1 − (ω r )n = = 0. 1 − ωr 1 − ωr

3.3. nth Roots of Unity

47

In conclusion, we have 1 + ω m + ω 2m + · · · + ω (n−1)m = We complete the proof of the problem.

  n, if m is a multiple of n; 

0,

otherwise.



48

Chapter 3. nth Roots of a Complex Number

CHAPTER

4

Complex Functions and the Analyticity

4.1 Fundamental Concepts Complex function is the part of mathematical analysis that studies functions of complex numbers. It is important in many branches of mathematics including applied mathematics, geometry, number theory as well as in physics (e.g. hydrodynamics and thermodynamics) and also in engineering areas (e.g. nuclear, aerospace, mechanical and electrical engineering). In this section, we are going to introduce some basic concepts, notations and results about functions of one complex variable. We assume that the reader is familiar with the basic theory of topology, such as open sets, closed sets, bounded sets, compact sets, polygonally connected sets, connected sets and etc. We call Ω a region if it is open and connected.a For details of this part, you are encouraged to refer to the references [1, Chap. 1], [4, Chap. 1 and 2], [5, Chap. 2], [14, Chap. I and II], [21, Chap. 1] and [35, Chap. 1]. In the remaining of this book, Ω and U will be reserved for a region and the unit disc D(0; 1) respectively.

4.1.1 Definitions An association which to each element of D ⊆ C associates a complex number ω is called a function of a complex variable. We denote such a function by the symbol f :D→C and ω = f (z). The set D is called the domain of definition of f and f (z) the value of f at z.b We write f (z) = u(z) + iv(z), where u(z) and v(z) are real numbers, so they are real-valued functions. We call u the real part of f and v the imaginary part of f . If we write z = x + iy, where x, y ∈ R, then f can a

Some books use domain instead of region. When the domain of definition is not mentioned, it is agreed that the largest possible subset of C is to be taken. b

49

50

Chapter 4. Complex Functions and the Analyticity

also be written in the form f (z) = f (x, y) = u(x, y) + iv(x, y). Using polar coordinates r and θ, we may write f (z) = f (reiθ ) = u(r, θ) + iv(r, θ). Let n be zero or a positive integer. Let a0 , a1 , . . . , an ∈ C and an 6= 0. Then the function P (z) = a0 + a1 z + a2 z 2 + · · · + an z n is a polynomial of degree n. It is clear that the domain of definition of P is the entire complex P (z) plane. Quotients Q(z) of polynomials are called rational functions and are defined at each point z where Q(z) 6= 0.

Now we have a function f whose domain is D and another function g
with domain E. We suppose that for every z ∈ D, f (z) ∈ E. Thus, for every z ∈ D, g f (z) is well-defined in D and is called the composite function of f and g in that order. Recall that composition of functions is not commutative. For example, take f (z) = 2z + i and g(z) = z 2 + z + 1. Then we have
g f (z) = (2z + i)2 + (2z + i) + 1 = 4z 2 + 4iz − 1 + 2z + i + 1 = 4z 2 + 2(2i + 1)z + i, while


f g(z) = 2(z 2 + z + 1) + i = 2z 2 + 2z + (2 + i).

They imply that g f (1) = 4 + 2(2i + 1) + i = 6 + 5i and f g(1) = 6 + i.

4.1.2 Limits and Continuity

Definition 4.1 (Limits of a Function). The function f : D ⊆ C → C is said to have the limit A as z tends to a ∈ C, lim f (z) = A (4.1) z→a

if and only if for every ǫ > 0, there exists a δ > 0 such that |f (z) − A| < ǫ for all complex numbers of z with |z − a| < δ and z 6= a. The condition (4.1) is equivalent to lim f (z) = A

z→a

so that lim Re f (z) = Re A and

z→a

lim Im f (z) = Im A.

z→a

z gives an example which has no limit at z = 0. In fact, if z is real, then z f (z) = 1, while if z is purely imaginary, then we have f (z) = −1. To say that f has a limit A at ∞ if for every ǫ > 0, there is a large number M > 0 such that

The function f (z) =

|f (z) − A| < ǫ for all complex numbers z satisfying |z| ≥ M .

4.1. Fundamental Concepts

51

Definition 4.2 (Continuity of a Function). The function f : D ⊆ C → C is said to be continuous at a if and only if lim f (z) = f (a). z→a

A continuous function is the function which is continuous at every point of its domain of definition. Write f (z) = u(x, y) + iv(x, y) and z0 = x0 + iy0 . According to Definition 4.2 (Continuity of a Function), f is continuous at z0 if and only if both real functions u and v are continuous at the point (x0 , y0 ). Theorem 4.3 (Properties of Continuous Functions). Suppose that f, g : D ⊆ C → C are continuous functions and c ∈ C. Then (a) Re f (z), Im f (z), cf (z), |f (z)|, f (z) + g(z) and f (z) · g(z) are continuous. (b)

f (z) g(z)

is defined and continuous at every point a such that g(a) 6= 0.

Furthermore, composition of continuous functions is also continuous. More explicitly, if f : D → E and g : E → C are continuous at a and f (a) respectively, then the function g ◦ f : D → C is continuous at a. Theorem 4.4. Suppose that the complex function f is continuous on a closed and bounded region Ω. Then f is bounded on Ω. Remark 4.1 In real analysis, we can visualize a real continuous function f (x) of a real variable as a function whose graph has no holes in it. The discussion in complex analysis is similar. In fact, let f : D → C be defined by f (z) = u(x, y) + iv(x, y) and be continuous on D. Next, the parametric curve γ : [0, 1] → D defined by γ(t) = x(t) + iy(t) is continuous if both the real functions x(t) and y(t) are continuous. Then the image of the parametric curve


ω(t) = f γ(t) = u (x(t), y(t) + iv (x(t), y(t) is continuous on [0, 1].

4.1.3 Differentiability and Analyticity of a Function Definition 4.5 (Differentiability of a Function). A complex-valued function f : D ⊆ C → C is said to be differentiable at z ∈ D if the limit f (z + h) − f (z) h→0 h lim

h∈C

exists. In this case, the limit is denoted by f ′ (z). Here the h is not necessarily real, so it means that the limit must exist irrespective of the way in which h approaches 0 in C. In fact, there is a fundamental difference between the cases of a

52

Chapter 4. Complex Functions and the Analyticity

real and a complex independent variable. For example, the function f (z) = z is not differentiable at any point z because f (z + h) − f (z) h = h h which equals 1 if h is real and −1 if h is purely imaginary.

It is clear from Definition 4.5 (Differentiability of a Function) that if f and g are both differentiable at z, then so are cf,

f ± g,

fg

and

f g

provided that g(z) 6= 0, where c ∈ C. Definition 4.6 (Analyticity of a Function). A complex-valued function f : D ⊆ C → C is said to be analytic at z if f is differentiable in a neighborhood of z. Similarly, f is analytic on S ⊆ D if it is differentiable at all points of some open set containing S.c The class of all analytic functions in the region Ω will be denoted by H(Ω). If Ω = C, then f is called an entire function. Remark 4.2 (a) It should be noted that analyticity is a property that a function is defined on an open set (a neighborhood of a particular point), while differentiability is a property of a function that occurs at a particular point only. (b) A point z at which a complex function f fails to be analytic is called a singular point. z2 For example, i is a singular point of the rational function f (z) = z−i . See Chapter 9 for a more detailed discussion. The sum and the product of two analytic functions are again analytic. The same is true for (z) of two analytic functions, provided that g(z) 6= 0. Besides, if f is differentiable the quotient fg(z) (resp. analytic) at a and g is differentiable (resp. analytic) at f (a), then g ◦ f is differentiable (resp. analytic) at a and
(g ◦ f )′ (a) = g′ f (a) × f ′ (a). (4.2)

4.1.4 The Cauchy-Riemann Equations Theorem 4.7. If f = u + iv is differentiable at z, then it satisfies the Cauchy-Riemann equations ∂u ∂v ∂u ∂v = and =− . (4.3) ∂x ∂y ∂y ∂x Conversely, suppose that u(x, y) and v(x, y) have continuous first partial derivatives with respect to x and y. If u and v satisfy the equations (4.3), then f is differentiable at z. c

Some books use holomorphic functions, regular functions, complex differentiable functions or conformal mapping instead of analytic functions, see [19, p. 16].

4.1. Fundamental Concepts

53

Sometimes, the Cauchy-Riemann equations (4.3) can be expressed in the complex form ∂f ∂f =i . ∂y ∂x In real analysis, it is well-known that if f ′ (x) = 0 on (a, b), then f is constant there. With the aid of Theorem 4.7, this result can be extended to the complex functions. In other words, we obtain Theorem 4.8. If f is analytic in a region Ω and f ′ ≡ 0 there, then f is constant. Furthermore, the following result contains some sufficient conditions for verifying that f a constant function. Theorem 4.9. Let f = u + iv be analytic in a region Ω. If u or |f | is constant in Ω, then f is constant.

4.1.5 Elementary Complex Functions We say f a single-valued function of z if for each value of z in its domain, there corresponds exactly one value of f (z) in its range. If more than one value of f (z) for each value of z, we call f a multi-valued functiond of z. For example, it is easy to see that the function f (z) = z 2 is single-valued, while the equation ω 2 = z defines a multi-valued function of z because there are two values of ω for every z. Before we introduce the study of the elementary complex functions, we first examine briefly the concept of the inverse of a complex function. By definition, we call g(z) an inverse of f (z) if

f g(z) = g f (z) = z.

A basic fact from real analysis is that an inverse of a function is not necessarily a function. However, if f is one-to-one, then its inverse, denoted by f −1 , must be a function. By applying the formula (4.2), we can easily obtain another important result of inverse functions: Theorem 4.10. Suppose that D and E are open in
C. Let f : D → C and g : Ω → C be continuous functions such that f (D) ⊆ E and g f (z) = z in D. If g is differentiable at f (a)
′ and g f (a) 6= 0, then f is differentiable at a and f ′ (a) =

g′

1
. f (a)

Definition 4.11 (The Linear Function). A function of the form f (z) = az + b with z, b ∈ C is called a linear function. Its derivative is given by f ′ (z) = a which is defined at every z. If a = 0, then f is a constant function. If a 6= 0, then f is one-to-one and its inverse function is 1 b z= ω− a a which is also a linear function. In particular, if a = 1 and b = 0, then f becomes the identity function. d

Some authors call it a many-valued function.

54

Chapter 4. Complex Functions and the Analyticity

Definition 4.12 (The Power Function). For every n ∈ N, the function f (z) = z n is called the power function. Obviously, we have f −1 (ω) = {z ∈ C | z n = ω} which is a multi-valued function. In addition, it is clear that f ′ (z) = nz n−1 for every n ∈ N. Definition 4.13 (The Reciprocal Function). The function f : C \ {0} → C \ {0} defined by f (z) =

1 z

is called the reciprocal function. It is clear that f is one-to-one, so the inverse function f −1 (ω) =

1 ω

f ′ (z) = −

1 z2

is also one-to-one. Besides, we know that

for all z ∈ C \ {0}.

4.1.6 The Complex Exponential Function One of the most important functions in all branches of mathematics is the complex exponential function which is defined as follows: Definition 4.14 (The Complex Exponential Function). Let z = x + iy, where x, y ∈ R. The function ez is defined by ez = ex+iy = ex (cos y + i sin y). Therefore, the real and imaginary parts of ez are u(x, y) = ex cos y and v(x, y) = ex sin y respectively. In the case that z is purely imaginary, i.e., x = 0, we get eiy = cos y + i sin y. This is known as Euler’s Formula. Furthermore, it is also well-known that ez 6= 0, e0 = 1, |ez | = |ex+iy | = ex , ez = ez+2πi and it is an entire function such that d z (e ) = ez dz for all z ∈ C.e e

Repeatedly, we have ez = ez+2kπi for every integer k and (ez )(n) = ez for every positive integer n.

4.1. Fundamental Concepts

55

Remark 4.3 With Definition 4.14 (The Complex Exponential Function), we see that it preserves some of the familiar properties of the real exponential function ex , except the injectivity because ez = ez+2πi holds. Thus we cannot define a complex logarithm in the same way as the real logarithm and the full discussion of it will be given in §4.1.8. The trigonometric functions of z are defined in terms of the exponential function. In fact, we have 1 1 (4.4) sin z = (eiz − e−iz ) and cos z = (eiz + e−iz ). 2i 2 Observe that if z is real, then sin z and cos z agree with the usual definitions of sine and cosine of a real variable. The other complex trigonometric functions are defined in terms of the complex sine and cosine functions by tan z =

sin z , cos z

cot z =

cos z , sin z

sec z =

1 cos z

and

csc z =

1 . sin z

Next, the complex hyperbolic functions can be defined in the same way as the real case, namely, ez − e−z ez + e−z sinh z , cosh z = , tanh z = , 2 2 cosh z 1 1 1 coth z = , sech z = , csch z = , tanh z cosh z sinh z where coth z and csch z are not defined at z = kπi, while tanh z and sech z are not defined at z = 21 (2k + 1)πi for k ∈ Z. The hyperbolic functions are closely related to the trigonometric functions. For example, we know from their definitions that sinh z =

sinh(iz) = i sin z.

4.1.7 Branches of Multi-valued Functions Let f be a multi-valued function. A branch of f is a continuous function F on a domain D such that it assigns exactly one of the multiple values of f to each z in D. A branch cut for a branch F of a multi-valued function f is a line or a curve γ that is deleted from the domain D of f so that F is analytic in D \ {γ}. A branch point is any point common to all branch cuts.

It should be noted that it is possible to consider a multi-valued function as being single-valued on a certain surface. Such a surface is called as a Riemann surface which consists of a finite or an infinite number of sheets and these sheets are connected by branch cuts. The end points of a branch cut are called the branch points. The manner in which the sheets are connected depends on the particular multi-valued function. See, for examples, [1, §4.3], [14, Chap. XVI] and [17, pp. 288 - 305].

4.1.8 The Logarithm Function The inverse of the exponential function is the logarithm function. For a nonzero complex number z, we define log z to be any complex number ω such that eω = z.

(4.5)

56

Chapter 4. Complex Functions and the Analyticity

To determine ω in terms of z, we write ω = u + iv and z = reiθ , where r > 0 and θ = arg z. Then the equation (4.5) becomes eu+iv = reiθ and thus eu = r

and eiv = eiθ .

(4.6)

The first part of the equation (4.6) gives u = log r which is our usual natural logarithm of the positive number r. Since ez is not injective, the second part of the equation (4.6) implies that v = θ + 2kπ, where k ∈ Z.

The preceding discussion leads to the fact log z = log |z| + i arg z = log r + i(θ + 2kπ),

(4.7)

where z = reiθ and k ∈ Z. Unlike the real logarithm, this formula (4.7) shows that log z is multi-valued (in fact, infinitely many-valued) because arg z takes multiple values. Now we define a branch of log z to be any single-valued function f that satisfies the formula (4.5). By taking the principal value of the argument Arg z as (−π, π],f we can make the logarithm single-valued. In fact, we have Definition 4.15 (The Principal Branch of the Logarithm). The principal branch of the complex logarithm of z 6= 0 is defined by Log z = log |z| + iArg z,

(4.8)

where −π < Arg z ≤ π. Furthermore, we have log z = Log z + 2kπi for some k ∈ Z. It reduces to the usual real logarithm when z is positive. For examples, log(−1) = (2n + 1)πi and Log (−1) = πi because Arg (−1) = π, where n is an integer. There are several important concepts related to Definition 4.15 (The Principal Branch of the Logarithm): • The expression (4.7) tells us that

elog z = z

(4.9)

if z 6= 0. However, it should be emphasized that the equation log(ez ) = z is not true. To see this, if z = x + iy, then |ez | = ex and arg(ez ) = y + 2nπ with n ∈ Z, so we have log(ez ) = log |ez | + i arg(ez ) = log ex + i(y + 2nπ) = (x + iy) + 2nπi which gives log(ez ) = z + 2nπi. f

See also Definition 1.7 (Argument).

4.1. Fundamental Concepts

57

• Geometrically, the requirement −π < Arg z ≤ π can be treated as a “cut” of the complex plane along the negative real axis. This ray is called a branch cut of log z. Other branch cuts of log z are given by (2k − 1)π < arg z ≤ (2k + 1)π, where k ∈ Z. • The origin evidently is a branch point for all branches of log z. • If we apply Theorem 4.10 to the equation (4.9) with f (z) = log z and g(z) = ez , then we get immediately that d 1 1 (log z) = log z = dz e z for all z ∈ C \ (−∞, 0]. • If x > 0, then Log x = log x. If x < 0, then Log x = log |x| + iπ.

4.1.9 Complex Powers Definition 4.16 (Complex Powers). Let z, α ∈ C and z 6= 0. We define the complex power z α by z α = eα log z , where log z is the logarithm (4.7). If we fix the principal branch (4.8), the principal branch of z α is given by z α = eαLog z . Since log z is multi-valued, the complex power z α is in general multi-valued. To see this, we note that z α = eα log z = eα(Log z+2kπi) = eαLog z · e2kαπi , where k ∈ Z. Since e2kαπi = 1 when α ∈ Z, the function z α is single-valued in this case.

4.1.10 Inverse Trigonometric and Hyperbolic Functions By the following formulas, we know that the inverse trigonometric and hyperbolic functions are infinite-valued functions. i h i h 1 1 sin−1 z = −i log iz + (1 − z 2 ) 2 , cos−1 z = −i log z + (z 2 − 1) 2 , i 1 − iz i iz − 1 log cot−1 z = − log (z 6= ±i), 2 1 + iz 2 iz + 1   1 1 sinh−1 z = log z + (1 + z 2 ) 2 , cosh−1 z = log z + (z 2 − 1) 2 , 1+z 1 z+1 1 , cot−1 z = log (z 6= ±1). tanh−1 z = log 2 1−z 2 z−1 tan−1 z =

58

Chapter 4. Complex Functions and the Analyticity

Their corresponding derivative formulas are given by d d 1 1 (sin−1 z) = (cos−1 z) = − 1 , 1 , 2 dz dz (1 − z ) 2 (1 − z 2 ) 2 1 d 1 d (tan−1 z) = , (cot−1 z) = − , 2 dz 1+z dz 1 + z2 d d 1 1 (sec−1 z) = (csc−1 z) = − 1 , 1 2 2 dz dz z(z − 1) 2 z(z − 1) 2

(Re z > 0),

d d 1 1 (Re z > 0), (sinh−1 z) = (cosh−1 z) = 1 , 1 2 2 dz dz (1 + z ) 2 (z − 1) 2 d 1 d 1 (tanh−1 z) = , (coth−1 z) = , dz 1 − z2 dz 1 − z2 1 1 d d (sech−1 z) = − (csch−1 z) = − (Re z > 0). 1 , 1 dz dz z(1 − z 2 ) 2 z(1 + z 2 ) 2 Formulas connecting trigonometric functions, hyperbolic functions and their inverses can be found in any elementary complex analysis textbooks, so we won’t repeat them here.

4.2 Functions of a Complex Variable Problem 4.1 ⋆ Suppose that f (z) = x2 − y 2 − 2y + i(2x − 2xy) and z = x + iy. Express f in terms of z. Proof. We note that x=

z+z 2

and y =

z−z , 2i

so we have z − z h z + z  z + z  z − z i −2 +i 2 −2 2 2i 2i 2 2 2i   2 2 2 2 2 2 z + 2zz + z z −z z − 2zz + z = + + i(z − z) + i z + z − 4 4 2i 2 2 2 2 z −z z +z + 2iz − = 2 2 2 = z + 2iz.

f (z) =

 z + z 2

−

 z − z 2

This completes the proof of the problem. Problem 4.2 ⋆ Express f (z) = z 4 in the form f (z) = u(x, y) + iv(x, y), where z = x + iy.

Proof. According to the Binomial Theorem, we obtain f (z) = (x + iy)4



4.2. Functions of a Complex Variable

59

= x4 + 4x3 (iy) + 6x2 (iy)2 + 4x(iy)3 + (iy)4 = (x4 − 6x2 y 2 + y 4 ) + i(4x3 y − 4xy 3 ). Thus we have u(x, y) = x4 − 6x2 y 2 + y 4 and v(x, y) = 4x3 y − 4xy 3 , completing the proof of the problem.  Problem 4.3 ⋆ Express f (z) = z 5 + 4z 2 − 6 in the polar coordinate form u(r, θ) + iv(r, θ).

Proof. Let z = r(cos θ + i sin θ). By Theorem 1.10 (De Moivre’s Theorem), we have f (z) = r 5 (cos 5θ + i sin 5θ) + 4r 2 (cos 2θ + i sin 2θ) − 6

= (r 5 cos 5θ + 4r 2 cos 2θ − 6) + i(r 5 sin 5θ + 4r 2 sin 2θ).

Therefore, we see that u(r, θ) = r 5 cos 5θ + 4r 2 cos 2θ − 6 and v(r, θ) = r 5 sin 5θ + 4r 2 sin 2θ. This ends the proof of the problem.  Problem 4.4 real coefficients and z0 = a + ib ⋆ Let P (z) = a0 + a1 z + · · · + an z n be a polynomial with
be a root of P (z) = 0, where a, b ∈ R. Prove that P z0 = 0. Proof. On the one hand, since P (z0 ) = 0, we have P (z0 ) = 0. On the other hand, we know that ak = ak for all 0 ≤ k ≤ n so that P (z0 ) = a0 + a1 z0 + · · · + an z0n

= a0 + a1 · z0 + · · · + an · z0 n

= a0 + a1 z0 + · · · + an z0 n = P (z0 ).

Hence z0 is also a root of P (z), ending the proof of the problem.



Problem 4.5 (Enestr¨ om-Kakeya Theorem) ⋆ ⋆ Suppose that a0 > a1 > a2 > · · · > an−1 > an > 0 and P (z) = a0 + a1 z + · · · + an z n . If ζ is a zero of P , prove that 2na0 . 1 < |ζ| < an

Proof. We first prove that |ζ| > 1. It is easy to see that ζ 6= 1. If |z| ≤ 1 and z 6= 1, then we observe that |(1 − z)P (z)| = |a0 − (a0 − a1 )z − · · · − (an−1 − an )z n − an z n+1 |   > a0 − (a0 − a1 )|z| + · · · + (an−1 − an )|z|n + an |z|n+1

60

Chapter 4. Complex Functions and the Analyticity ≥ a0 − (a0 − a1 ) − · · · − (an−1 − an ) − an

= 0.

This implies that P (z) does not have any zero in U and so |ζ| > 1. 2na0 an .

To see this, if |z| > 1, then we notice that a  a0  n−1 |P (z)| ≥ |an z n | × 1 − + ··· + an z an z n a i h 1  an−1 an−2 0 ≥ |an | · |z|n × 1 − + + ··· + . |z| an an an

Next, we show that |ζ|
1 be large enough so that a  1 1  an−1 an−2 0 + + ··· + ≤ . R an an an 2 If |z| ≥ R, then the inequality (4.11) implies that

1 |P (z)| ≥ |an | · |z|n . 2 In other words, it means that all the zeros of P (z) lie inside D(0; R). Applying this to our situation, it means that a  2na  a 0 n−1 an−2 0 . (4.12) |ζ| ≤ 2 + + ··· + < an an an an

Hence our result follows from combining the inequalities (4.10) and (4.12). This completes the  proof of the problem. Remark 4.4 For the locations of zeros of a polynomial, the reader can read Marden’s book [22].

4.3 Limits and Continuity Problem 4.6 ⋆ Suppose that lim f (z) exists. Prove that this limit is unique. z→z0

Proof. Let lim f (z) = L1 and lim f (z) = L2 . Given ǫ > 0. There exists a δ > 0 such that z→z0

z→z0

ǫ ǫ and |f (z) − L2 | < 2 2 for all z satisfying 0 < |z − z0 | < δ. Therefore, we obtain |f (z) − L1 |
0 such that z→z0

1 |f (z)| > |L| 2 for 0 < |z − z0 | < δ.

Proof. By definition, there corresponds a δ > 0 such that |f (z) − L| < 21 |L| for 0 < |z − z0 | < δ. Since L = L − f (z) + f (z), we see that 1 |L| = |L − f (z) + f (z)| ≤ |L − f (z)| + |f (z)| < |L| + |f (z)| 2 which implies the desired result that |f (z)| > 12 |L|, completing the proof of the problem.



Problem 4.8 ⋆ Evaluate the limit z4 − 1 . z→i z − i lim

4

−1 Proof. Let f (z) = zz−i . Since z 4 −1 = (z−i)(z+i)(z−1)(z+1), we have f (z) = (z+i)(z−1)(z+1) for all z 6= i and then

z4 − 1 = lim(z + i)(z − 1)(z + 1) = 2i(i − 1)(i + 1) = −4i. z→i z − i z→i lim

We complete the proof of the problem.



Problem 4.9 ⋆ Prove that z4 − 1 = 1. z→∞ z 4 + 4z 3 + 2 lim

Proof. By writing 1 − z14 z4 − 1 , = z 4 + 4z 3 + 2 1 + 4z + z24 the result follows immediately because the problem.

1 z

→ 0 and

1 z4

→ 0 as z → ∞, completing the proof of 

62

Chapter 4. Complex Functions and the Analyticity Problem 4.10 ⋆ Let a 6= 0 and n ∈ N. Prove that the function f (z) =

is discontinuous at z = a.

 n  z , if z 6= a; 

0,

otherwise,

Proof. Note that lim f (z) = an . Since f (a) = 0 but a 6= 0, we obtain z→a

lim f (z) 6= f (a).

z→a

By Definition 4.2 (Continuity of a Function), f is discontinuous at z = a, completing the proof  of the problem. Problem 4.11 ⋆ Prove that the function f (z) = z is continuous on C.

Proof. Let z = x + iy, z0 = x0 + iy0 and f (z) = u(x, y) + iv(x, y). Since f (z) = x + iy = x − iy, we have u(x, y) = x and v(x, y) = −y. Now both u(x, y) and v(x, y) are continuous at (x0 , y0 ), so f is continuous at z0 . Since z0 is arbitrary, f is continuous on C, as desired and we complete  the proof of the problem. Problem 4.12 ⋆ Suppose that f : C → C is continuous such that |f (z)| → ∞ as |z| → ∞ and f (C) is open in C. Prove that f (C) = C.

Proof. Assume that Ω = f (C) 6= C. Then Ω is a proper open subset of C so that C \ Ω 6= ∅ and then ∂Ω = Ω ∩ C \ Ω 6= ∅. Fix ω0 ∈ ∂Ω and pick a sequence ωn ∈ Ω such that ωn → ω0

(4.13)

as n → ∞. For each n ∈ N, we consider the set f −1 (ωn ) = {z ∈ C | f (z) = ωn }. Choose the sequence {zn }, where zn ∈ f −1 (ωn ). If {zn } is bounded, then the Bolzano-Weierstrass Theorem ensures that a convergent subsequence {znk } exists. Let its limit be z0 . Now the continuity of f implies that ω0 = lim ωnk = lim f (znk ) = f (z0 ) ∈ Ω, k→∞

k→∞

a contradiction. Consequently, {zn } is unbounded. Without loss of generality, we may assume that zn → ∞ as n → ∞. Therefore, we have |ωn | = |f (zn )| → ∞

(4.14)

as n → ∞. Combining the limits (4.13) and (4.14), we conclude that ω0 = ∞, contradicting to  the fact that ω0 ∈ C. Hence we get f (C) = C and we end the proof of the problem.

4.4. Analytic Functions and the Cauchy-Riemann Equations

63

4.4 Analytic Functions and the CauchyRiemann Equations Problem 4.13 ⋆ ⋆ Assume that the following result is true: If u(x, y) has partial derivatives ux and uy and ux (x, y) = uy (x, y) for every point z = x + iy in a region Ω, then u is constant in Ω. Use this to prove Theorem 4.9. Proof. Let u be constant. Then it is true that ux (x, y) = uy (x, y) = 0 in Ω. By the CauchyRiemann equations (4.3), we get vx (x, y) = vy (x, y) = 0 in Ω. Using the assumption, we know that v is also constant in Ω. Hence f must be constant. Next, suppose that |f | is constant in Ω. If |f | = 0 in Ω, then we are done. Otherwise, we have u2 + v 2 = C 6= 0 in Ω. Taking the partial derivatives with respect to x and y, we have uux + vvx = 0 and uuy + vvy = 0

(4.15)

in Ω. Applying the Cauchy-Riemann equations (4.3) to the equations (4.15), we see that uux − vuy = 0

and uuy + vux = 0

in Ω. Thus we get (u2 + v 2 )ux = 0 in Ω which means that ux = 0 in Ω. By the Cauchy-Riemann equations (4.3) again, we also have vy = 0 in Ω. Similarly, we can prove that uy = vx = 0 in Ω. Now the previous paragraph ensures that f is constant in this case. This completes the  proof of the problem. Problem 4.14 ⋆ Suppose that f (reiθ ) = u(reiθ ) + iv(reiθ ), where ur , uθ , vr and vθ exist in the region Ω. If f is differentiable at Ω, prove that rur = vθ

and uθ = −rvr

in Ω.

Proof. By the Chain Rule, we have ur = ux ·

∂y ∂x + uy · ∂r ∂r

and

uθ = ux ·

∂x ∂y + uy · . ∂θ ∂θ

(4.16)

Since x = r cos θ and y = r sin θ, the equations (4.16) reduce to ur = ux cos θ + uy sin θ

and

uθ = −rux sin θ + ruy cos θ.

(4.17)

64

Chapter 4. Complex Functions and the Analyticity

Similarly, we have vr = vx cos θ + vy sin θ

and

vθ = −rvx sin θ + rvy cos θ.

(4.18)

Since f is differentiable at z, we may combine the Cauchy-Riemann equations (4.3), (4.17) and (4.18) to get vθ 1 ur = vy cos θ − vx sin θ = (vy · r cos θ − vx · r sin θ) = . r r Similarly, we can show that uθ 1 vr = vx cos θ + vy sin θ = − (uy · r cos θ − ux · r sin θ) = − . r r Hence we have completed the analysis of the problem



Problem 4.15 ⋆ Find a real function v to make f (z) = e−x (x sin y − y cos y) analytic.

Proof. Using the Cauchy-Riemann equations (4.3), we obtain vy = ux = e−x sin y − xe−x sin y + ye−x cos y

(4.19)

vx = −uy = e−x cos y − xe−x cos y − ye−x sin y.

(4.20)

and Integrating the equation (4.19) with respect to y, keeping x constant, we obtain v(x, y) = ye−x sin y + xe−x cos y + F (x) for some real function F (x). Putting this into the equation (4.20) and after simplification, we get F ′ (x) = 0 which implies F (x) = c for some constant c. Take c = 0, then one possible function is v(x, y) = ye−x sin y + xe−x cos y. This ends the proof of the problem.



Problem 4.16 ⋆ Let f : Ω → C be analytic in a region Ω. If Im (f ) is constant, prove that f must be constant.

Proof. Write f = u + iv, where Im (f ) = v is constant. Let g = if = −v + iu. Then g is analytic and Re (g) = −v which is constant. It yields from Theorem 4.9 that both g and f are constant,  completing the analysis of the problem.

4.4. Analytic Functions and the Cauchy-Riemann Equations

65

Problem 4.17 ⋆ Suppose that f (z) = y − 2xy + 1 + i(−x + x2 − y 2 ) + z 2 , where z = x + iy. Prove that f is entire.

Proof. We have f (z) = (x2 − 2xy + y − y 2 + 1) + i(−x + 2xy + x2 − y 2 ). Write u(x, y) = x2 − 2xy + y − y 2 + 1 and v(x, y) = −x + 2xy + x2 − y 2 . Then we have ux = 2x − 2y,

uy = −2x + 1 − 2y,

vx = −1 + 2y + 2x

and

vy = 2x − 2y.

Obviously, they have continuous first partial derivatives with respect to x and y. Furthermore, they satisfy the Cauchy-Riemann equations ux = vy

and

vx = −uy

for all x and y. This means that f ′ (z) exists for all complex z, i.e., f is entire which ends the  proof of the problem. Remark 4.5 Alternatively, we obtain the same conclusion of Problem 4.17 if we write f (z) = z 2 + iz 2 − iz + 1. Problem 4.18 ⋆ Prove that f (z) = (z − 1)2 + 5z is nowhere analytic.

Proof. Let z = x + iy. Then we have f (z) = (x − 1 − iy)2 + 5(x − iy)

= (x − 1)2 − 2i(x − 1)y − y 2 + 5x − 5iy

= [(x − 1)2 − y 2 + 5x] − i[2(x − 1)y + 5y]. Define u(x, y) = (x − 1)2 − y 2 + 5x and v(x, y) = −2(x − 1)y − 5y so that ux = 2(x − 1) + 5 = 2x + 3,

uy = −2y,

vx = −2y

and vy = −2(x − 1) − 5 = −2x − 3.

Now ux = vy and uy = −vx if and only if x = − 32 and y = 0. It follows from the CauchyRiemann equations (4.3) that f is differentiable only at (− 32 , 0), so it is nowhere analytic by  Definition 4.6 (Analyticity of a Function). This ends the proof of the problem. Problem 4.19 ⋆ Suppose that z = x + iy and f (z) = cos y − i sin y. Prove that f is nowhere differentiable.

66

Chapter 4. Complex Functions and the Analyticity

Proof. Define u(x, y) = cos y and v(x, y) = − sin y. Then we have ux = 0,

uy = − sin y,

vx = 0 and

vy = − cos y

which have continuous first partial derivatives with respect to x and y. However, if they satisfied the Cauchy-Riemann equations (4.3), then we would have cos y = sin y = 0 which is impossible. Therefore, f does not satisfies the Cauchy-Riemann equations (4.3) at every x and y so that f is nowhere differentiableg and we complete the analysis of the problem.  Problem 4.20 ⋆ Suppose that z = x + iy and define  2 x − y 2 − 2xyi   , if z 6= 0;  x + iy f (z) =    0, otherwise.

Prove that f satisfies the Cauchy-Riemann equations (4.3) at z = 0 but it is not differentiable there.

Proof. We know that

 z2    , if z 6= 0; z f (z) =    0, otherwise.

Let f (z) = u(x, y) + iv(x, y). For z 6= 0, we observe that f (z) =

z3 (x − iy)3 x3 − 3xy 2 y 3 − 3x2 y = = + i · . |z|2 x2 + y 2 x2 + y 2 x2 + y 2

This means that u(x, y) =

x3 − 3xy 2 x2 + y 2

and v(x, y) =

y 3 − 3x2 y x2 + y 2

for z 6= 0. If z = 0, then we have u(0, 0) = v(0, 0) = 0. Thus it is easy to check that 3

h u(h, 0) − u(0, 0) 2 ux (0, 0) = lim = lim h = 1 h→0 h→0 h h

and

3

h v(0, h) − v(0, 0) 2 vy (0, 0) = lim = lim h = 1. h→0 h→0 h h Similarly, we can show that

u(0, h) − u(0, 0) = 0 and h→0 h

uy (0, 0) = lim g

(4.21)

v(h, 0) − v(0, 0) = 0. h→0 h

vx (0, 0) = lim

Of course, it is nowhere analytic by Definition 4.6 (Analyticity of a Function).

(4.22)

(4.23)

4.4. Analytic Functions and the Cauchy-Riemann Equations

67

Consequently, limits (4.21), (4.22) and (4.23) guarantee immediately that ux (0, 0) = vy (0, 0) and uy (0, 0) = −vx (0, 0), i.e., f satisfies the Cauchy-Riemann equations (4.3) at z = 0 as required. To show that f is not differentiable at z = 0, we consider the limits 2

h f (h) − f (0) = lim h = 1 h→0 h h→0 h

lim

and

h2 −h2 −2h2 i

f (h + ih) − f (0) −2h2 i −2i h+ih lim = lim = lim 2 = lim = −1. h→0 h→0 h→0 h (1 + i)2 h→0 2i h + ih h + ih Since the two limits are not the same, we conclude that f is not differentiable at z = 0. Hence we complete the proof of the problem.  Problem 4.21 ⋆ Suppose that f = u + iv is analytic in a region D. Both u and v have continuous second order partial derivatives in D. Prove that ∂2u ∂2u + 2 =0 ∂x2 ∂y holds in D.

Proof. By Theorem 4.7, we see that ∂v ∂u = ∂x ∂y

and

∂u ∂v =− ∂y ∂x

which imply

∂2u ∂2u ∂2v ∂2v and . = =− 2 2 ∂x ∂x∂y ∂y ∂y∂x By Clairaut’s Theorem [39, Theorem 15.12, p. 332], we know that

(4.24)

∂2v ∂2v = ∂x∂y ∂y∂x in Ω, so the equations (4.24) give ∂2u ∂2u = − ∂x2 ∂y 2 in Ω. This is what we want and we end the proof of the problem.



Remark 4.6 The equation in Problem 4.21 is called Laplace’s equation and any function satisfying it is called a harmonic function on Ω. We will see very soon that the assumption “both u and v have continuous second order partial derivatives in Ω” is unnecessary for the truth of Laplace’s equation. Using Problem 4.14, it is easy to show that u and v, when expressed in polar form, satisfy Laplace’s equation in polar form ψrr +

ψr ψθθ + 2 = 0. r r

68

Chapter 4. Complex Functions and the Analyticity Problem 4.22

⋆ Suppose that f is analytic in a region Ω. Define Ω∗ = {z | z ∈ Ω} and f ∗ : Ω∗ → C by f ∗ (z) = f (z). Prove that (f ∗ )′ (z) = f ′ (z) for every z ∈ Ω∗ . Proof. Let z, z0 ∈ Ω∗ with z 6= z0 . Then we have z, z0 ∈ Ω and f ∗ (z) − f ∗ (z0 ) f (z) − f (z0 ) h f (z) − f (z0 ) i = = z − z0 z − z0 z − z0 so that

h f (z) − f (z ) i f ∗ (z) − f ∗ (z0 ) 0 = lim (f ) (z0 ) = lim = f ′ (z0 ). z→z0 z→z0 z − z0 z − z0 ∗ ′

This means that f ∗ ∈ H(Ω∗ ) and we end the proof of the problem.



Problem 4.23 (Rolle’s Theorem (Complex Form)) ⋆ ⋆ Let Ω be an open convex subset of C and f ∈ H(Ω). If a, b ∈ Ω are two distinct points such that f (a) = f (b) = 0, prove that there exist z1 and z2 lying on the segment joining a and b such that Re [f ′ (z1 )] = 0 and Im [f ′ (z2 )] = 0.

Proof. Denote α1 = Re a, α2 = Im a, β1 = Re b, β2 = Im b, u = Re f and v = Im f . Define the real-valued function Φ : [0, 1] → R by

Φ(t) = (β1 − α1 )u a + t(b − a) + (β2 − α2 )v a + t(b − a) . This function is differentiable. Now the hypothesis guarantees that u(a) = u(b) = v(a) = v(b) = 0 which apparently give Φ(0) = (β1 − α1 )u(a) + (β2 − α2 )v(a) = 0 and Φ(1) = (β1 − α1 )u(b) + (β2 − α2 )v(b) = 0. By Rolle’s Theorem, one can find an t1 ∈ (0, 1) such that Φ′ (t1 ) = 0. Let z1 = a + t1 (b − a) which lies on the segment joining a and b. Then we have d Φ(t) dt t=t1

d d + (β2 − α2 ) v a + t(b − a) = (β1 − α1 ) u a + t(b − a) dt dt t=t1 t=t1  ∂v ∂x ∂v ∂y   ∂u ∂x ∂u ∂y  + (β2 − α2 ) . · + · · + · = (β1 − α1 ) ∂x ∂t ∂y ∂t t=t1 ∂x ∂t ∂y ∂t t=t1

Φ′ (t1 ) =

(4.25)

4.5. Elementary Complex Functions

69

Since x = Re a + t · Re (b − a) and y = Im a + t · Im (b − a), we can simplify the expression (4.25) as   0 = (β1 − α1 ) (β1 − α1 )ux (z1 ) + (β2 − α2 )uy (z1 )   + (β2 − α2 ) (β1 − α1 )vx (z1 ) + (β2 − α2 )vy (z1 ) = (β1 − α1 )2 ux (z1 ) + (β1 − α1 )(β2 − α2 )[uy (z1 ) + vx (z1 )] + (β2 − α2 )2 vy (z1 ).

(4.26)

Now the Cauchy-Riemann equations (4.3) ensure that the middle term in the equation (4.26) is zero, so it yields [(β1 − α1 )2 + (β2 − α2 )2 ]ux (z1 ) = 0 and thus ux (z1 ) = 0 which is exactly Re [f ′ (z1 )] = 0. Next, we can apply the previous argument to the analytic function g = −if to obtain that there exists a t2 ∈ (0, 1) such that the point z2 = a + t2 (b − a) satisfies 0 = Re [g′ (z2 )] = vx (z2 ) = −uy (z2 ) = Im [f ′ (z2 )]. Hence we have ended the proof of the problem.



Remark 4.7 The complex version of Rolle’s Theorem as stated in Problem 4.23 was shown by Evard and Jafari [13]. An application of this is the complex version of the Mean Value Theorem: Re [f ′ (z1 )] =

Re [f (b)] − Re [f (a)] Re b − Re a

and Im [f ′ (z2 )] =

Im [f (b)] − Im [f (a)] . Im b − Im a

Besides, we note that one can prove Theorem 4.8 by a combined use of Theorem 7.1 (The Interior Uniqueness Theorem) and this complex version of the Mean Value Theorem, see [13, Corollary 2.3, p. 860] for details.

4.5 Elementary Complex Functions Problem 4.24 ⋆ Prove that cos(z + 2nπ) = cos z for all n ∈ Z and all z ∈ C. Prove also that 2nπ are the only numbers α with cos(z + α) = cos z (4.27) for all z ∈ C. Proof. Using the definitions (4.4), it is clear that cos(z + 2nπ) =

 1
1  i(z+2π) e − e−i(z+2π) = eiz − e−iz = cos z 2 2

70

Chapter 4. Complex Functions and the Analyticity

for all z ∈ C.

Next, we suppose that the equation (4.27) holds for all z ∈ C. Then it follows from the definitions (4.4) that eiz eiα + e−iz e−iα = eiz + e−iz eiz (eiα − 1) = e−iz (1 − e−iα )

= e−iα e−iz (eiα − 1).

(4.28)

If eiα − 1 6= 0, then the equation (4.28) gives eiz = e−iα e−iz for all z. Set z = 0. It gives e−iα = 1 which contradicts the assumption that eiα − 1 6= 0. Hence, we have eiα = 1 and so we conclude that α = 2nπ for some n ∈ Z, completing the proof of the problem.  Problem 4.25 ⋆ Prove that the zeros of cos z are real. What are they?

Proof. Let cos α = 0. It follows from the definitions (4.4) that
1 iα e + e−iα = 0, 2

implying e2iα = −1 = e(2n+1)πi for all n ∈ Z. Hence we have 2iα = (2n + 1)πi and then  1 π α= n+ 2 for all n ∈ Z. This completes the proof of the problem.



Remark 4.8 Similarly, the zeros of sin z are precisely z = nπ for all n ∈ Z. Problem 4.26 ⋆ Prove that sin(z) = sin z.

Proof. If z = x + iy, then we have eiz = ei(x+iy) = e−y · eix = e−y · e−ix = e−i(x−iy) = e−iz and similarly, e−iz = eiz . By the equations (4.4), we see that sin z =



1 iz 1 iz 1 −iz (e − e−iz ) = − − eiz = sin z, e e − e−iz = − 2i 2i 2i

completing the proof of the problem.



4.5. Elementary Complex Functions

71

Problem 4.27 ⋆ Prove that lim tan z =

r→∞ z=reiθ

  i, 

if 0 < θ < π;

−i, if −π < θ < 0.

Proof. Using the equations (4.4), we have tan z =

sin z = cos z

1 iz −iz ) 2i (e − e 1 iz −iz ) 2 (e + e

= −i ·

e2iz − 1 . e2iz + 1

(4.29)

Note that |e2iz | = e−2r sin θ . If 0 < θ < π, then sin θ > 0 so that |e2iz | = e−2r sin θ → 0 as r → ∞. In this case, we conclude that lim tan z = i. r→∞ z=reiθ

For the case −π < θ < 0, we first rewrite the expression (4.29) as tan z = −i ·

1 − e−2iz . 1 + e−2iz

Next, since sin θ < 0, we observe that |e−2iz | = e2r sin θ → 0 as r → ∞. Thus we obtain lim tan z = −i.

r→∞ z=reiθ

Hence we prove the desired result and then we have completed the proof of the problem.



Problem 4.28 ⋆

⋆ Let f : C → C be an entire function such that f (z1 + z2 ) = f (z1 )f (z2 )

(4.30)

f (x) = ex

(4.31)

for all z1 , z2 ∈ C and for all real x. Prove that f (z) = ez .

Proof. Let z = x + iy. According to the equations (4.30) and (4.31), we see that f (z) = f (x + iy) = f (x)f (iy) = ex f (iy). Let f (iy) = U (y) + iV (y), where U and V are real functions. Then we have f (z) = ex U (y) + iex V (y). Since f is entire, it follows from the Cauchy-Riemann equations (4.3) that U (y) = V ′ (y)

and U ′ (y) = −V (y).

72

Chapter 4. Complex Functions and the Analyticity

Thus U ′′ (y) = −U (y) which implies that U (y) = α cos y + β sin y for some α, β ∈ C. By this, we also have V (y) = −β cos y + α sin y. By the condition (4.31), we have U (0) = α = 1 and V (0) = −β = 0 so that f (z) = ex (cos y + i sin y) = ez , as desired and this ends the proof of the problem.



Problem 4.29 ⋆ ⋆ For any n ∈ Z, define zn = ein . Prove that for every ω ∈ C(0; 1), there exists a subsequence {nk } such that znk → ω for k → ∞.

Proof. Let α be irrational. According to Kronecker’s Approximation Theorem [3, §7.4, pp. 148, 149], the set {mα + n | m, n ∈ Z} is dense in R. In particular, we take α = 2π. Let ω = eix for some real x. Then one can find a sequence xk = nk + 2mk π such that xk → x as k → ∞. Consider the sequence znk = eink , where k ∈ N. Therefore, this fact and the continuity of ez together imply that znk = eink = eixk · e−2mk πi = eixk → eix as k → ∞, as required. This completes the analysis of the problem.



Problem 4.30 ⋆ Find all values of θ ∈ [0, 2π) such that the limit iθ

lim ere

r→∞

exists.

iθ

Proof. We note that ere = er cos θ × eir sin θ , so the limit exists if and only if cos θ < 0. In other words, θ must satisfy the inequality 3π π and Re ω > 0. Prove that Log (zω) = Log z + Log ω.

Proof. Let z = reiθ and ω = Reiφ , where − π2 < θ, φ < expression (4.8) imply that Log (zω) = Log [rRei(θ+φ) ]

π 2.

The fact −π < θ + φ < π and the

74

Chapter 4. Complex Functions and the Analyticity = log(rR) + i(θ + φ) = (log r + iθ) + (log R + iφ) = Log z + Log ω,

completing the proof of the problem.



Problem 4.34 ⋆ Let z α take its principal branch. Prove that d α (z ) = αz α−1 . dz

Proof. Since the principal branch z α is taken, the function z α = eα log z is single-valued and analytic in its domain of definition. Hence we observe that d α d αLog z (z ) = e dz dz d = eαLog z (αLog z) dz α = eαLog z × z αLog z e = α Log z e = αe(α−1)Log z = αz α−1 which is the desired result and we end the proof of the problem.



Problem 4.35 1

⋆ Prove that z 2 has no zero at z = 0.

Proof. By the definition, we know that 1

1

1

z 2 = e 2 Log z = e 2 (log |z|+iArg z) . 1

Since log |z| and Arg z are not well-defined at z = 0, z 2 does not have a zero at z = 0, completing the proof of the problem.  Problem 4.36 ⋆

⋆ Let α ∈ C. Prove that |iα | is single-valued if and only if α ∈ R.

4.6. The Logarithm Function and Complex Powers

75

Proof. Let α = a + bi. By the definition, we may write iα = eα log i n h π io = exp (a + bi) · log 1 + i + 2kπ  2π i h π + 2kπ + ia + 2kπ , = exp − b 2 2

where k ∈ Z. Consequently, we obtain

h π i |iα | = exp − b + 2kπ , 2

where k ∈ Z. Now this formula guarantees that |iα | is single-valued if and only if b = 0 if and  only if α ∈ R. This completes the proof of the problem.

76

Chapter 4. Complex Functions and the Analyticity

CHAPTER

5

Power Series

5.1 Fundamental Concepts The most important sequences of functions are those expressible as power series which play an important role in complex analysis because, as we will see very soon in this chapter, every analytic function can be represented by a power series locally. In many circumstances, a power series behaves like a polynomial. The main references we have used in this chapter are [1, §2.2, pp. 33 - 42], [21, Chap. II] and [35, §2.3, pp. 14 - 18].

5.1.1 Basic Facts on Series of Numbers Let’s recall some basic facts about series of numbers. Given a sequence c0 , c1 , c2 , . . . of complex numbers, the infinite series ∞ X cn (5.1) n=0

converges if the limit

lim

N →∞

N X

cn

n=0

exists and finite. Such limit is called the sum of the series, the number cn is called the n-th N X cn is called the N -th partial sum of the series. A series that term and the number sN = n=1

does not converge is said to diverge. The uniqueness of limit of convergent series, bounded series, Cauchy Criterion as well as other necessary and sufficient conditions of convergent series, absolute convergence, tests for convergent series and further properties of the infinite series (5.1) can be found in [39, Chap. 6]. For series of functions

∞ X

fn (x),

n=0

we have two classes of convergence which are pointwise convergence and uniform convergence. Detailed properties of series of functions such as Cauchy Criterion for uniform conver77

78

Chapter 5. Power Series

gence, Weierstrass M -test, uniform convergence and continuity, integrability and differentiability of the limit function, equicontinuity and etc can be found in [39, Chap. 10].

5.1.2 Definitions Definition 5.1 (Power Series). A power series is a series of the form ∞ X

n=0

an z n = a0 + a1 z + a2 z 2 + · · · ,

(5.2)

where a0 , a1 , . . . are constants and they are called coefficients of the series. Power series can be considered as “infinite polynomials” because they can be manipulated in many cases as if they were polynomials. Thus the class of polynomials is contained in the class of power series. The main difference between the two classes is that convergence should be taken into account when we talk about power series and the basic result in this aspect depends on the following number 1 L = lim sup |ak | k . (5.3) k→∞

In fact, we have the following important and useful result: Theorem 5.2 (The Cauchy–Hadamard Theorem). Suppose that we have the power series (5.2) and the number (5.3). (a) If L = 0, then the series (5.2) converges for all z. (b) If L = ∞, then the series (5.2) converges only for z = 0. (c) If 0 < L < ∞, then the series (5.2) converges for all |z|


1 L.

Remark 5.1 Set R =

1 L

and let 0 < δ < R.

(a) This number R is called the radius of convergence of the power series. (b) The series (5.2) converges uniformly and absolutely in any smaller disc {z | |z| ≤ R−δ}. (c) The convergence behavior is not clear on the circle C(0; R) = {z ∈ C | |z| = R}. The radius of convergence can also be found by the formula: a k R = lim sup . ak+1 k→∞

Theorem 5.3 (Abel’s Theorem). Suppose that

∞ X

n=0 ∞ X

we suppose that z0 ∈ C(0; R) is a point such that lim z→z

0

∞ X

z∈D(0;R) n=0

an z n has radius of convergence R. Further, an z0n converges. Then we have

n=0

an z n =

∞ X

n=0

an z0n .

5.1. Fundamental Concepts

79

The most important power series for us will definitely be the geometric series. In fact, we know that 1 − z n+1 Sn = 1 + z + z 2 + · · · + z n = , 1−z so if |z| < 1, then it can be shown easily that ∞

X 1 zn. = 1−z

(5.4)

n=0

5.1.3 Operations on Power Series Theorem 5.4. Suppose that f (z) =

∞ X

an z n and g(z) =

∞ X

bn z n have radii of convergence R1

n=0

n=0

and R2 respectively. Then the functions f (z) + g(z) and f (z)g(z) have power series representations whose radius of convergence R satisfyinga R ≥ min{R1 , R2 }. Theorem 5.5. Suppose that f (z) =

∞ X

n=0

f ′ (z) exists and

an z n converges in D(0; R) = {z ∈ C | |z| < R}. Then

f ′ (z) =

∞ X

nan z n−1

n=1

in D(0; R). Furthermore, f is infinitely differentiable in D(0; R) and for every k ≥ 1, we have f

(k)

(z) =

∞ X

n=k

n(n − 1) · · · (n − k + 1)an z n−k

in D(0; R). Theorem 5.5 is very important because its implication is that every function defined by its power series is analytic inside its radius of convergence. The converse of this result is also true and it will be seen in Chapter 6. As a corollary to Theorem 5.5, we have Corollary 5.6. If the series f (z) =

∞ X

an z n has a positive radius of convergence, then we have

n=0

an =

f (n) (0) n!

for all n = 0, 1, 2, . . .. a

The product of the two power series f and g refers to the Cauchy product. That is, if f (z)g(z) =

∞ X n=0

then we have cn =

n X k=0

for every n ≥ 0.

ak bn−k

cn z n ,

80

Chapter 5. Power Series

Theorem 5.7 (Uniqueness of Power Series). Suppose that f (z) =

∞ X

an z n is zero at all points

n=0

of a nonzero sequence {zk } and zk → 0 as k → ∞. Then f (z) ≡ 0 in C. As an immediate consequence of Theorem 5.7 (Uniqueness of Power Series), we see that if two convergent power series are identical on a set of points with an accumulation point at the origin, then their coefficients are identical too.

5.1.4 Power Series centered at z0 The power series considered in the previous sections is called power series centered at 0. Generally, we may consider power series centered at an arbitrary point z0 : f (z) =

∞ X

n=0

an (z − z0 )n .

In this case, the disc of convergence will be changed from D(0; R) to D(z0 ; R) and the above theorems and corollary remain valid. See Figure 5.1 for the convergence of this general power series.

Figure 5.1: The circle of convergence of a power series.

5.1.5 Power Series centered at Infinity The function f (z) is said to be analytic at z = ∞ if the function g(ω) = f ( ω1 ) is analytic at ω = 0. If f (z) is analytic at ∞, then g(ω) = f ( ω1 ) has a power series expansion centered at ω = 0, i.e., ∞ X an ω n g(ω) = n=0

5.2. Radius of Convergence of Power Series

81

for |ω| < R, where R is its radius of convergence. Thus f (z) has the following series representation ∞ X an f (z) = zn n=0

for |z| >

1 R.

5.2 Radius of Convergence of Power Series Problem 5.1 ⋆ Determine the radii of convergence of the following power series: (a)

∞ X (−2)n z n

n=0

n+1

.

∞ X (2n)! z 2n+1 (b) . · 22n (n!)2 2n + 1 n=0

(c)

∞ X 3n z n . 5n + 7n

n=0

Proof. (a) According to Theorem 5.2 (The Cauchy–Hadamard Theorem), we see that L = lim

k→∞

2 1

(k + 1) k

= 2.

Therefore, the radius of convergence of the power series is 21 . (b) Using the Ratio Test, we see that (2k)! 4(k + 1)2 (2k + 3) 22k+2 [(k + 1)!]2 (2k + 3) · = lim = 1, k→∞ 22k (k!)2 (2k + 1) k→∞ (2k + 1)2 (2k + 2) (2k + 2)!

R = lim

so the radius of convergence of the power series is 1. (c) By the Ratio Test, we know that 5k+1 + 7k+1 5k+1 + 7k+1 1 7 3k lim · = = . k k k+1 k k k→∞ 5 + 7 3 3 k→∞ 5 + 7 3

R = lim

This means that the radius of convergence of the power series is 73 . We complete the proof of the problem.



82

Chapter 5. Power Series Problem 5.2 ⋆ Evaluate the radius of convergence of the power series ∞  X i n3 n z . 1− n

n=1

Proof. It follows from the limit (5.3) that  i k3 k1 L = lim sup 1 − k k→∞  i k2 = lim sup 1 − k k→∞  k 2 + 1  k2 2 = lim sup 2 k k→∞ h 1 k2 i 21 = lim sup 1 + 2 k k→∞ √ = e. Hence we establish from Theorem 5.2 (The Cauchy-Hadamard Theorem) that R=

1 1 =√ . L e

We complete the proof of the problem.



Problem 5.3 ⋆ Construct a power series ∞ X

an z n

n=0

whose radius of convergence is 2, converges at some z0 ∈ C(0; 2) but diverges at some z1 ∈ C(0; 2). Proof. Consider a0 = 0 and an =

1 n2n

for n ∈ N. It is easy to see that

L = lim sup k→∞

so that R = 2. If we take z0 = 2, then we have ∞ X

n=0

1 2×k

1 k

=

1 2

∞ X 1 an · 2 = n n

n=1

which is divergent. Next, if we take z1 = −2, then we have ∞ X

n=0

n

an · (−2) =

∞ X (−1)n

n=1

n

5.2. Radius of Convergence of Power Series

83

which is convergent. Hence this is what we want and we complete the proof of the problem.  Problem 5.4 ⋆ Let the radius of convergence R of the power series

∞ X

an z n is positive. Prove that the

n=0

series

∞ X an z n

n=0

is entire.

nn

Proof. Since R > 0, the series ∞ X

an r n

n=0

converges for all 0 ≤ r < R. Particularly, the sequence {an r n } is bounded, so there is a M > 0 such that |an r n | ≤ M for all n ≥ 0. With a fixed r, we have 1 a 1 Mn n n ≤ →0 n n rn

as n → ∞. In other words, the radius of convergence of the power series is ∞ which means it is entire, completing the proof of the problem.  Problem 5.5 ⋆ Assume that ez has a power series expansion for all values of z. Find the expansion.

Proof. Let f (z) = ez =

∞ X

an z n . According to Corollary 5.6, we obtain

n=0

an =

f (n) (0) 1 = . n! n!

Thus we have ez =

∞ X zn

n=0

n!

.

This representation is unique because of Theorem 5.7 (Uniqueness of Power Series), completing  the proof of the problem. Problem 5.6 ⋆ Can the inequality in Theorem 5.4 be strict?

84

Chapter 5. Power Series

Proof. The answer is affirmative. To see this, let an = 1 and bn = −1 so that R1 = R2 = 1, but f (z) + g(z) =

∞ X

(an + bn )z n = 0

n=0

which means the series converges for all values of z, i.e., the radius of convergence of the sum f (z) + g(z) is ∞. For the product f (z)g(z), we consider a0 = 2, an = 2n for n ≥ 1 and b0 = −1, bn = 1 for n ≥ 1. On the one hand, we have R1 = 12 and R2 = 1. On the other hand, we note that c0 = a0 b0 = −2 and cn =

n X

ak bn−k = a0 bn + an b0 +

k=0

n−1 X k=1

for n ≥ 1. Therefore, we have

ak bn−k = 2 − 2n +

n−1 X

2k = 0

k=1

f (z)g(z) = −2

and it is entire, so the radius of convergence of the product f (z)g(z) is ∞. We end the analysis  of the problem. Problem 5.7 ⋆ Let the power series

∞ X

n=0

an z n converges at z = z0 6= 0. Prove that it converges absolutely

for |z| < |z0 | and uniformly for |z| ≤ r, where r < |z0 |. Proof. Since the series

∞ X

an z0n

n=0

converges, we have an z0n → 0 as n → ∞. Therefore, there exists a positive integer N such that ∞ X z | |n converges and in fact, |an z0n | < 1 for all n ≥ N . For |z| < |z0 |, the geometric series z n=0 0 ∞ X |z0 | z n . = z |z | − |z| 0 0 n=0

Using this, since |z| < |z0 |, we conclude that ∞ X

n=N

∞ X |z0 | z n |an z | < . ≤ z0 |z0 | − |z| n

n=N

Thus, the series converges absolutely for |z| < |z0 |. Next, let Mn = ∞ X Mn converges. For |z| ≤ r and n ≥ N , we have the series

rn |z0 |n ,

where r < |z0 |. Then

n=0

|an z n | = |an z0n | ·

rn |z|n < = Mn . |z0 |n |z0 |n

By the Weierstrass M -test, our power series converges uniformly for |z| ≤ r. We end the proof  here.

5.2. Radius of Convergence of Power Series

85

Problem 5.8 ⋆ Suppose that a0 = 1,

a1 = −1 and 3an + 12an−1 − an−2 = 0

for all n = 2, 3, . . .. By considering the polynomial P (z) = 3 + 12z − z 2 , find the radius of convergence of the power series ∞ X an z n . f (z) = n=0

Proof. Note that 2

P (z)f (z) = (3 + 12z − z )

∞ X

an z n

n=0

= 3a0 + (3a1 + 12a0 )z +

∞ X

(3an + 12an−1 − an−2 )z n

n=2

= 3 + 9z, so we get

3 + 9z . (5.5) 3 + 12z − z 2 Since the disc of convergence is the largest disc centered at 0 in which the expression (5.5) is well-defined, its radius of convergence is the √ shortest distance from the origin to the singularities √ of f . Obviously, the roots of P (z) are 6 ± 39, so the radius of convergence of f is 39 − 6,  completing the proof of the problem. f (z) =

Problem 5.9 ⋆ Find the set of complex numbers z for which the power series ∞ X zn nlog n

(5.6)

n=1

and its term by term derivatives of all orders converge absolutely.

Proof. Suppose that R is the radius of convergence of the power series. Then we obtain from Theorem 5.2 (The Cauchy-Hadamard Theorem) that 1

R = lim sup |klog k | k k→∞   1 = lim sup exp log k k log k k→∞

= lim sup exp k→∞ 0

=e

h1 k

(log k)2

i

86

Chapter 5. Power Series = 1.

Using Theorem 5.5 and Theorem 5.2 (The Cauchy-Hadamard Theorem), the series and all its derivatives converge absolutely inside U and diverge in |z| > 1. Let |z| = 1. Then the k-th derivative of the power series is given by ∞ X

n=k

n(n − 1) · · · (n − k + 1)

z n−k nlog n

which implies ∞ ∞ X 1 z n−k X n(n − 1) · · · (n − k + 1) log n n(n − 1) · · · (n − k + 1) log n ≤ n n n=k

≤

=

n=k ∞ X

n=k ∞ X

n=k

nk nlog n 1

nlog n−k

.

For large enough n, we have log n − k > 2, we deduce from the convergence of the power series ∞ X 1 and the comparison test that the series (5.6) converges absolutely on |z| = 1. Hence the 2 n n=1

desired set is U and we complete the proof of the problem.



Problem 5.10 ⋆ Consider the power series

∞ X zn

n=0

n2

. Let R be its radius of convergence.

(a) Prove that the series converges in U . (b) Is the same conclusion true for the series of derivatives?

Proof. (a) It is clear that R = lim sup k→∞

 k 2 = 1. k+1

Therefore, the power series converges for |z| < 1 and diverges for |z| > 1. If |z| = 1, then n we know that | zn2 | = n12 so that the series converges absolutely. Hence the series converges in U . (b) The series of the first derivative is ∞ X z n−1

n=1

n

.

5.2. Radius of Convergence of Power Series

87

Using similar argument as in part (a), this series converges for |z| < 1. However, on |z| = 1, ∞ X 1 diverges. In other words, the same result does not hold for the series of the series n n=1 derivatives. This completes the proof of the problem.



Problem 5.11 ⋆ Suppose that f (z) =

∞ X

an z n and g(z) =

∞ X

bn z n are analytic in a neighborhood

n=0

n=0

containing U . Prove that the power series h(z) =

∞ X

an bn z n

n=0

defines an analytic function in a neighborhood containing U .

Proof. The hypothesis ensures that f and g must have radius of convergence greater than 1. Thus there exists a number d > 1 such that bn an → 0 and →0 n d dn as n → ∞. This gives a n bn lim 2n = 0 n→∞ d which means that the series h(z) has radius of convergence at least d2 > 1. Hence we conclude that the power series defines an analytic function in a neighborhood containing U . We complete  the proof of the problem. Problem 5.12 ⋆ Suppose that f (z) =

∞ X

n=0

an z n has a positive radius of convergence R and f ′ (0) = a1 6= 0.

Prove that there exists an r ∈ (0, R) such that f is injective in D(0; r). Proof. Fix an r with r ∈ (0, R) at this moment. Set z, ω ∈ D(0; r). Thus we have f (z) − f (ω) = a1 (z − ω) +

∞ X

n=2

an (z n − ω n )

= a1 (z − ω) + (z − ω)

∞ X

an

n=2

n X

z n−k ω k−1 .

k=1

Repeated applications of the triangle inequality asserts that ∞ n X X an |f (z) − f (ω)| ≥ |a1 | · |z − ω| − |z − ω| · z n−k ω k−1 n=2

k=1

88

Chapter 5. Power Series ≥ |a1 | · |z − ω| − |z − ω| · 

≥ |a1 | − 

= |a1 | −

∞ X

n=2 ∞ X

n=2

|an | ·

is small enough, then we have

n=0

k=1

n=2

|an |



n X k=1

|z|n−k |ω|k−1

r n−1 · |z − ω|

 n|an |r n−1 · |z − ω|.

According to Theorem 5.5, the power series ∞ X

n X

∞ X

∞ X

n=0

n|an |z n−1 has radius of convergence R, so if r

|a1 | and this says that 2

n|an |r n−1


|a1 | · |z − ω|. 2

Hence f is injective in D(0; r) and we end the proof of the problem. Problem 5.13 ⋆ Use Problem 5.5 to verify that |ez − 1| ≤ e|z| − 1 ≤ |z| · e|z| . Proof. By Problem 5.5 and the triangle inequality, on the one hand, we have ∞ X zn − 1 |e − 1| = n! n=1 z2 z3 = z + + + ··· 2! 3! |z|2 |z|3 + + ··· ≤ |z| + 2! 3! ∞ X |z|n = − 1 n! n=1 z

= |e|z| − 1|.

On the other hand, we see that |z|2 |z|3 + + ··· 2! 3! ∞ X |z|n−1 = |z| · n!

|e|z| − 1| = |z| +

= |z| ·

n=1 ∞ X

n=0

|z|n (n + 1)!



5.2. Radius of Convergence of Power Series ≤ |z| ·

89

∞ X zn

n=0 |z|

n!

= |z| · e . Hence we complete the proof of the problem.



Remark 5.2 The result in Problem 5.13 can be generalized to n n z X z k |z| X |z|k e − ≤ e − ≤ |z|n+1 · e|z| , k! k! k=0

where n ≥ 0.

k=0

Problem 5.14 (Tauber’s Theorem for Power Series) ⋆ Suppose that the power series f (z) =

∞ X

ak z k

k=0

converges in U and kak → 0 as k → ∞. Prove that

lim

n→∞

n X k=1

k|ak | = 0.

n

Prove also that if f (x) → L as x → 1 and x ∈ R, then we have ∞ X

an = L.

n=0

Proof. Given ǫ > 0. Then there is a positive integer N such that n|an | < gives n X k=1

=

N X


0 such that |bk | > cr k for infinitely many positive integers k. Proof of Lemma 5.1. The definition gives 1 = (a0 + a1 z + · · · + an z n )(b0 + b1 z + · · · ). If we compare the constant term on both sides, then we have a0 b0 = 1. Similarly, if we compare the coefficients of z k+n on both sides, we get a0 bk+n + a1 bk+n−1 + · · · + an bk = 0

(5.10)

for all k ≥ 0. The condition a0 b0 = 1 implies that a0 6= 0 and b0 6= 0. Thus there is a positive number c such that 0 < c < |b0 |. Since an 6= 0, one can pick a positive integer N such that |a0 | + |a1 |R + · · · + |an−1 |Rn−1 ≤ |an |Rn (5.11) for all R ≥ N . Let r =

1 R.

Then the inequality (5.11) becomes

|a0 |r n + |a1 |r n−1 + · · · + |an−1 |r ≤ |an | for all sufficiently small r > 0. We know that |b0 | > c = cr 0 . Suppose that |bk | > cr k for some nonnegative integer k. We claim that there exists an p ∈ {1, 2, . . . , n} such that |bk+p | > cr k+p . Assume that it is not the case. In other words, we have |bk+p | ≤ cr k+p

(5.12)

for all p = 1, 2, . . . , n. It yields from the equation (5.10) and the inequality (5.12) that |a0 | · |bk+n | + |a1 | · |bk+n−1 | + · · · + |an−1 | · |bk+1 | |an | k+n k+n−1 |a0 | · cr + |a1 | · cr + · · · + |an−1 | · cr k+1 ≤ |an |  |a |r n + |a |r n−1 + · · · + |a  0 1 n−1 |r = cr k |an |

|bk | ≤

≤ cr k

which contradicts the fact |bk | > cr k . This proves Lemma 5.1.



5.2. Radius of Convergence of Power Series We return to the proof of the problem. Now we put z = f (r −1 ) = b0 +

93 1 r

into the power series (5.9) to get

bp b1 + ··· + p + ··· . r r

(5.13)

By Lemma 5.1, there are infinitely many k such that b |b | k k k = k > c > 0 r r

which forces the series (5.13) diverges. However, this contradicts the fact that the radius of convergence of f is infinite. Therefore, we have completed the proof of the problem.  Remark 5.3 The argument above is actually due to Velleman [37]. A few more different proofs of the Fundamental Theorem of Algebra will be seen in Chap. 5 to 8. In fact, if you are interested to know more about this famous theorem, you are advised to read the book by Benjamin [6].

94

Chapter 5. Power Series

CHAPTER

6

Elementary Theory of Complex Integration

6.1 Fundamental Concepts In Chapter 5, we see that every function defined by its power series is in fact analytic inside its radius of convergence. For the converse of this statement, we have to study the complex integration. We assume that the reader is familiar with the definition and basic properties of the Riemann integral Z b

f (x) dx

a

of a real-valued continuous function f defined on the interval [a, b]. The extension to complexvalued continuous function f : [a, b] → C is Z b Z b Z b Im f (x) dx. Re f (x) dx + i f (x) dx = a

a

a

Integrals are important in the study of functions of a complex variable. The integral we frequently use in complex function theory is a type of line integral around a closed curve in the complex plane. In this chapter, such line integral is introduced. Then Cauchy’s Theorem and the Cauchy’s Integral Formula, which are the central theme of complex function theory, as well as their applications are studied. The main references of this chapter are [1, Chap. 4], [4, Chap. 3], [5, Chap. 4], [19, Chap. 2], [31, Chap. 10] and [35, Chap. 2]. Recall from Chapter 4 that Ω, Ω and H(Ω) denote a region, the unit disc D(0; 1) and the space of all analytic functions in the region Ω.

6.1.1 Terminologies We introduce some terminologies about a curve in the complex plane. Any continuous mapping γ : [a, b] → C is called a curve. Suppose that the curve γ : [a, b] → C has the parametrization γ(t) = x(t) + iy(t), where a ≤ t ≤ b. The curve is piecewise differentiable and we set γ ′ (t) = x′ (t) + iy ′ (t) if x and y are continuous on [a, b] and continuously differentiable on [a, b] except at finitely many points. 95

96

Chapter 6. Elementary Theory of Complex Integration

We call the curve smooth if γ ′ (t) 6= 0 for all t ∈ [a, b], except at finitely many points. We say that γ is a closed curve if γ(a) = γ(b). In particular, γ is a simple closed curve if no other points coincide, except the end points. Remark 6.1 (a) It seems that different authors may have different setup for the theory of complex integration. As a result, they may use different meanings for terms like arcs, curves and paths and you will probably find confusions when you read these books. To save your time of cross-checking, Appendix A is made for recording some terminologies of some common complex analysis textbooks, such as Ahlfors [1], Asmar and Grafakos [4], Bak and Newmann [5], Conway [11], Gamelin [14], Rudin [31] as well as Stein and Shakarchi [35]. (b) The terminologies we employ in this book follow those used by Bak and Newmann.

Definition 6.1 (Line Integrals). Suppose that γ : [a, b] → C is a smooth curve and f is continuous at all points γ(t), where a ≤ t ≤ b. Then we define the (line) integral of f over the curve γ as Z b Z Z
f γ(t) · γ ′ (t) dt. f = f (z) dz = γ

a

γ

If γ is a closed curve, then we may write

I

f (z) dz γ

for this integral. We note that orientation of the integral should be taken into account. More explicitly, the orientation of γ : [a, b] → C is the image of γ(t) as t travels from a to b. Then the reverse of γ is the curve −γ : [a, b] → C defined by (−γ)(t) = γ(b + a − t).a For example, the positive orientation (counterclockwise) of the circle C(z0 ; r) is the one given by the standard parametrization γ(t) = z0 + reit , where −π ≤ t ≤ π. Thus the negative orientation (clockwise) of C(z0 ; r) is assumed to be γ(t) = z0 + re−it ,

where −π ≤ t ≤ π. In the following text, we assume that all circles are positively oriented, If γ is smooth, then −γ is also smooth and Z Z f (z) dz = − f (z) dz. −γ

γ

A curve may have more than one parameterization. Two parameterizations γ1 : [a, b] → C and γ2 : [c, d] → C are said to be equivalent if there exists a continuously differentiable bijection s : [c, d] → [a, b] such that s′ (t) > 0 and
γ2 (t) = γ1 s(t) . a

Intuitively, −γ is the points of γ traced in the opposite direction.

6.1. Fundamental Concepts

97

Thus, if γ1 and γ2 are equivalent smooth curves, then we must have Z Z f (z) dz. f (z) dz = γ1

γ2

Using Definition 6.1 (Line Integrals), it is very easy to see the following additive property holds: If γ is subdivided into a finite number of curves, say γ = γ1 + γ2 + · · · + γn , then we have Z

f (z) dz =

γ1 +γ2 +···+γn

n Z X

f (z) dz.

k=1 γk

Theorem 6.2. Suppose that f and g are continuous functions on the smooth curve γ and α, β ∈ C. Then we have Z Z Z   αf (z) + βg(z) dz = α f (z) dz + β g(z) dz. γ

γ

γ

Let f, g ∈ H(Ω) and γ : [a, b] → Ω be a smooth curve. Then we have the complex analogue of the integration by parts: Z

γ





f (z)g (z) dz = f γ(b) g γ(b) − f γ(a) g γ(a) − ′

Z

g(z)f ′ (z) dz. γ

6.1.2 Green’s Theorem We recall a basic result from advanced calculus that enables us to express line integrals as double integrals.b Theorem 6.3 (Green’s Theorem). Let Ω be a bounded region whose boundary ∂Ω be a simple closed curve with positive orientation. Let P (x, y) and Q(x, y) be continuously differentiable functions on Ω ∪ ∂Ω. Then we have ZZ  Z ∂Q ∂P  P dx + Q dy = dx dy. − ∂y Ω ∂x ∂Ω

Figure 6.1: The picture for Green’s Theorem. b

See also [21, p. 252] or [30, Theorem 10.45, pp. 282, 283].

98

Chapter 6. Elementary Theory of Complex Integration

It should be emphasized that this theorem is valid for both simply- and multiply-connected regions. In addition, Green’s Theorem can take the complex form: Z ZZ P (z, z) dz + Q(z, z) dz = 2i (Pz − Qz ) dx dy Ω

∂Ω

or equivalently,

Z

f (z, z) dz = 2i

∂Ω

where f (z, z) = P (z, z) + iQ(z, z).

ZZ

Ω

∂f dx dy, ∂z

6.1.3 The Fundamental Theorem of Calculus Let f be a function on the region Ω. A primitive for f on Ω is a function F analytic in Ω and such that F ′ = f on Ω. Theorem 6.4 (The Fundamental Theorem of Calculus). Suppose that Ω is a region and that f : Ω → C is continuous and has a primitive F in Ω. If γ : [a, b] → Ω is a curve, then we have Z

f (z) dz = F γ(b) − F γ(a) . γ

In the case that γ is closed, we have

Z

f (z) dz = 0. γ

6.1.4 An Estimate of the Integral: The M-L Formula Theorem 6.5 (The M -L Formula). Suppose that γ : [a, b] → C is a smooth
curve of length L and f is a continuous complex-valued function on γ. Let M = max{|f γ(t) | | a ≤ t ≤ b}. Then we have the estimate Z f (z) dz ≤ M L. γ

This estimate can be applied to prove the following result:

Theorem 6.6. Let γ be a smooth curve and {fn } be a sequence of continuous complex-valued functions on γ such that fn → f uniformly on γ. Then we have Z Z fn (z) dz = f (z) dz. lim n→∞ γ

γ

Corollary 6.7. Suppose that {fn } is a sequence of continuous complex-valued functions and ∞ X fn (z) converges uniformly on a smooth curve γ. Then we have that n=1

∞ Z X

n=1 γ

fn (z) dz =

Z X ∞

γ n=1

fn (z) dz.

6.1. Fundamental Concepts

99

6.1.5 Cauchy’s Theorems Cauchy’s theorem (or called Cauchy’s Integral Theorem) is a type of results which investigate conditions so that the integral of an analytic function over a simple closed curve vanishes. In fact, there are a variety of conditions available and we are going to list some of them here. Theorem 6.8 (Cauchy’s Theorem for a Triangle). Suppose that ∆ is a triangle in C and f is analytic in a region Ω containing ∆ and its interior. Then we must have Z f (z) dz = 0. ∆

It is sometimes called Goursat’s Theorem. We notice that the above result is also true in the
case that ∆ replaced by a rectangle R or in the case that f is continuous on Ω and f ∈ H Ω\{p} for some p ∈ Ω, see [1, Theorem 2, p. 109] or [31, Theorem 10.13, pp. 205, 206]. Theorem 6.9 (Cauchy’s Theorem in a Convex Region). Suppose that Ω is a convex region and f ∈ H(Ω). Then it is true that f = F ′ for some F ∈ H(Ω) and Z f (z) dz = 0 γ

for every closed path γ in Ω. Theorem 6.10 (Cauchy’s Theorem in a Disc). Given that Ω is an open disc. If f ∈ H(Ω), then we have Z f (z) dz = 0 γ

for every closed curve γ in Ω. A region Ω is said to be simply connected if Ω is a connected set such that the interior of any simple closed curve contained in Ω is also contained in Ω. Intuitively, it means that Ω has no “holes” in it, see Figure 6.2 for an illustration.

Figure 6.2: A simply connected region Ω.

100

Chapter 6. Elementary Theory of Complex Integration

For examples, the disc D(z0 ; R), the horizontal strip {z ∈ C | a < Im z < b} and every convex region are simply connected. However, neither the annulus {z ∈ C | R1 < |z − z0 | < R2 } nor the punctured disc {z ∈ C | 0 < |z − z0 | < R} is simply connected.c

A region Ω′ that is not simply connected is said to be multiply connected. That is, a multiply connected region has a “hole” in it, see Figure 6.3 for an example.

Figure 6.3: A multiply connected region Ω′ . Theorem 6.11 (Cauchy’s Theorem in a Simply Connected Region). Given that Ω is a simply connected region. If f ∈ H(Ω), then there exists an F ∈ H(Ω) such that F ′ = f throughout Ω so that Z f (z) dz = 0

γ

for every smooth closed curve γ in Ω. Our Theorem 6.11 (Cauchy’s Theorem in a Simply Connected Region) can be treated as a generalization of the other forms of Cauchy’s Theorem because any triangle, any convex region or any disc is simply connected, see [27, Examples 12.3, p. 143]. As a corollary to it, we have Corollary 6.12 (Independence of Paths). Given that Ω is a simply connected region. Let f ∈ H(Ω) and z1 , z2 ∈ Ω, If Γ and γ are two paths in Ω joining z1 and z2 , then we obtain Z Z f (z) dz = f (z) dz. Γ

γ

Theorem 6.13 (Cauchy’s Theorem in a Multiply Connected Region). Let Ω be the region bounded by the non-overlapping simple closed curves γ, γ1 , γ2 , . . . , γn all traversed counterclockwise, where γ1 , γ2 , . . . , γn lie inside the interior of γ. If f ∈ H(Ω ∪ γ ∪ γ1 ∪ · · · ∪ γn ), then we have Z n Z X f (z) dz. f (z) dz = γ

c

k=1 γk

Topologically, a simply connected region can be “continuously” shrunk to a point.

6.1. Fundamental Concepts

101

Corollary 6.14. If γ1 and γ2 denote simple closed curves traversed counterclockwise, γ1 lies inside γ2 and f is analytic in a set consisting of the curves γ1 , γ2 and all points between them, then we obtain Z Z f (z) dz.

f (z) dz =

γ2

γ1

See Figure 6.4 below.

Figure 6.4: A special case for multiply connected region. Remark 6.2 (a) The general version of Cauchy’s Theorem can be found in [31, Theorem 10.35, pp. 218 - 220]. Besides, the converse of Cauchy’s Theorem, which is called Morera’s Theorem will be stated in Chapter 7. (b) Corollary 6.12 (Independence of Path) is equivalent to the fact that there exists an F ∈ H(Ω) such that F ′ = f . (c) For a proof of Theorem 6.13 (Cauchy’s Theorem in a Multiply Connected Region), please read [4, Theorem 3.7.2, pp. 198 - 200].

6.1.6 Cauchy’s Integral Formula and its Consequences As an application of Cauchy’s Theorem, Cauchy’s Integral Formula is derived. It gives the value of an analytic function in a disk in terms of the values on the boundary. As a result, an analytic function may be represented by a power series. Theorem 6.15 (The Cauchy Integral Formula). Suppose that Ω is an open set containing the closed disc D(a; r). If f ∈ H(Ω), then for all z0 ∈ D(a; r), we have 1 f (z0 ) = 2πi

Z

C(a;r)

f (z) dz . z − z0

102

Chapter 6. Elementary Theory of Complex Integration

We note that Cauchy’s Integral Formula) has no analog for functions of a real variable. The functions fn (x) = xn share the same boundary values fn (0) = 0 and fn (1) = 1 for all n ∈ N, but they are all different at all points of (0, 1). Cauchy’s Integral Formula leads to many important properties of analytic functions. The first remarkable fact of this kind is that every analytic function is infinitely differentiable. Explicitly, we have Corollary 6.16 (The Cauchy Integral Formula for Derivatives). Suppose that Ω is an open set containing the closed disc D(a; r). If f ∈ H(Ω), then for all z0 ∈ D(a; r) and every positive integer n, we have Z n! f (z) dz f (n) (z0 ) = . 2πi C(a;r) (z − z0 )n+1 Another striking consequence of Theorem 6.15 (The Cauchy’s Integral Formula) is the power series representation of an analytic function. This is the content of Taylor’s Theorem: Corollary 6.17 (Taylor’s Theorem). Suppose that Ω is an open set containing the closed disc D(a; r). Then f has a power series expansion at a: f (z) =

∞ X

n=0

cn (z − a)n

for all z ∈ D(a; r) and the coefficients cn are given by f (n) (a) 1 cn = = n! 2πi

Z

C(a;r)

f (z) dz (z − a)n+1

for all n ≥ 0. The series in Corollary 6.17 (Taylor’s Theorem) is called the Taylor series of f at a. If a = 0, then the series becomes the Maclaurin series of f .

6.1.7 Cauchy’s Type Integrals Definition 6.18 (Cauchy’s Type Integrals). Integrals of the form Z f (ζ) dζ γ ζ −z

(6.1)

are called Cauchy’s type integrals, where γ : [a, b] → C is a rectifiable curved (not necessarily closed or simple) intersecting itself only a finite number of times, z ∈ C \ γ([a, b]) and f is continuous on γ. It is well-known that γ divides the plane into a finite number of disjoint regions Ω0 , Ω1 , . . . , Ωn , where Ω0 denotes the unbounded region. Since the integrand f (ζ) ζ −z d

Check Appendix A.7 for its definition.

6.2. Complex Line Integrals and the M -L Formula

103

is continuous for all z ∈ C \ γ([a, b]), the integral (6.1) indeed defines a function F (z) on C \ γ([a, b]). It is a fact that F is analytic in every Ωk and Z f (ζ) dζ F ′ (z) = 2 γ (ζ − z) in Ωk . However, F may represent different analytic functions in different Ωk . For further discussion of this topic, please read [17, §7.20, pp. 486 - 490].

6.2 Complex Line Integrals and the M -L Formula Problem 6.1 ⋆ If f (z) is continuous in a neighborhood of z = 0, prove that Z

lim

r→0 C(0;r)

f (z) dz = 2πif (0). z

Proof. Given ǫ > 0. There exists a δ > 0 such that ǫ 2π for all 0 < r < δ and −π ≤ t ≤ π. Thus, for all 0 < r < δ, it follows from Definition 6.1 (Line Integrals) that Z π Z π f (reit ) Z f (z) it f (0) dt dz − 2πif (0) = × ire dt − i it z re −π −π C(0;r) Z π [f (reit ) − f (0)] dt = Z π−π |f (reit ) − f (0)| dt ≤ |f (reit ) − f (0)|
0 and −π ≤ t ≤ π. Prove that Z

f (z) dz =

γ

  0, 


Proof. Note that γ ′ (t) = iReit and f γ(t) = (Line Integrals) that Z

γ

Z

if n 6= 1;

2πi, if n = 1.

e−int Rn .

Then it can be checked from Definition 6.1

π

iReit · e−int dt Rn −π Z π i ei(1−n)t dt = n−1 R −π  if n 6= 1;  0, =  2πi, if n = 1.

f (z) dz =

We have completed the proof of the problem.



6.2. Complex Line Integrals and the M -L Formula

105

Problem 6.5 ⋆ Let Ω be the region bounded by the smooth closed curve γ and A(Ω) be its area. Prove that Z 1 A(Ω) = z dz. 2i γ

Proof. Put f (z) = z into the complex form of Theorem 6.3 (Green’s Theorem), we get Z ZZ dx dy = 2iA(Ω) z dz = 2i Ω

γ

which gives the desired result. We complete the proof of the problem.



Remark 6.3 If γ is the unit circle C(0; 1), then we see from Problem 6.4 and Problem 6.5 that A(U ) = π.

Problem 6.6 ⋆ Prove the complex form of Theorem 6.3 (Green’s Theorem).

Proof. Write f (z, z) = P (x, y) + iQ(x, y). Then we have Z Z f (z, z) dz = (P + iQ)( dx + i dy) ∂Ω ∂Ω Z Z (Q dx + P dy) (P dx − Q dy) + i = ∂Ω ∂Ω ZZ  ZZ  ∂P ∂Q ∂P  ∂Q  dx dy + i dx dy + − =− ∂x ∂y ∂y Ω ∂x Z Z Ωh  ∂Q ∂P i ∂P ∂Q  =i +i dx dy − + ∂x ∂y ∂x ∂y Z ZΩ ∂f = 2i dx dy, Ω ∂z completing the analysis of the problem.



Problem 6.7 ⋆ Show that

Z

C(0;1)

1 e z dz ≤ 2πe.

Proof. Since the length of C(0; 1) is 2π, Theorem 6.5 (The M -L Formula) states that Z 1 e z dz ≤ 2πM, C(0;1)

106

Chapter 6. Elementary Theory of Complex Integration 1

where M denotes an upper bound of |e z | on C(0; 1). To prove that M = e, let z = eit , where −π ≤ t ≤ π. Then we have z1 = e−it so that 1

|e z | = |ecos t−i sin t | = ecos t ≤ e. Hence we see that M = e and then we conclude that Z 1 e z dz ≤ 2πe. C(0;1)

This completes the proof of the problem.



Problem 6.8 ⋆ Show that

Z

C

dz 2π , ≤ z 2 + 10 3

where C is the circle γ(t) = 2eit for −π ≤ t ≤ π.

Proof. Since |z 2 + 10| ≥ 10 − |z|2 = 10 − 4 = 6 if z ∈ C, it follows from Theorem 6.5 (The M -L Formula) that Z dz 1 2π . ≤ × 4π = 2 6 3 C z + 10

This completes the proof of the problem.



Problem 6.9 ⋆ Prove that

Z

γ

π dz ≤ , 2 9 3z + z + 1

where γ is the arc of the circle C(0; 2) from z = 2 to z = 2i in the first quadrant.

Proof. On C(0; 2), we have |z| = 2 so that

which implies

2 3z + z + 1 ≥ 3 z 2 − |z| − 1 = 3|z|2 − |z| − 1 = 9

1 1 ≤ 9 3z 2 + z + 1

on C(0; 2). Since the length of the arc γ is π, we establish from Theorem 6.5 (The M -L Formula) that π Z dz ≤ 2 9 γ z +z+1 as required. We complete the proof of the problem.



6.3. Cauchy’s Theorem

107

Problem 6.10 ⋆ Let f (z) = z n + an−1 z n−1 + · · · + a1 z + a0 . Prove that max |f (z)| ≥ 1.

z∈C(0;1)

Proof. Let ak = |ak |eitk , where 0 ≤ k ≤ n − 1. Let z = eit , where −π < t ≤ π. Then we have a1 a0 an−1 + · · · + n−1 + n |f (z)| = |z|n · 1 + z z z i(tn−1 −t) = 1 + |an−1 |e + · · · + |a1 |ei[t1 −(n−1)t] + |a0 |ei(t0 −nt) ≥ 1 + |an−1 | cos(tn−1 − t) + · · · + |a1 | cos[t1 − (n − 1)t] + |a0 | cos(t0 − nt).

For 0 ≤ k ≤ n − 1, since

Z

π

−π

cos[tk − (n − k)t] dt = 0,

it follows from the inequality (6.4) that Z Z π Z π it |f (e )| dt ≤ max |f (z)| dt ≤ 2π = −π

(6.4)

z∈C(0;1)

−π

π −π

dt ≤ 2π max |f (z)|. z∈C(0;1)

Thus we conclude that max |f (z)| ≥ 1

z∈C(0;1)

which ends the proof of the problem.



6.3 Cauchy’s Theorem Problem 6.11 ⋆ Let f (z) =

z3 . z 2 +7z+12

Show that Z

f (z) dz = 0

C(0;1)

without computing the integral explicitly. Proof. Since the roots of z 2 + 7z + 12 = 0 are −3 and −4, f is analytic in C \ {−3, −4}. Obviously, C(0; 1) ⊆ C \ {−3, −4}, so we may apply Theorem 6.10 (Cauchy’s Theorem in a Disc) to conclude immediately that Z f (z) dz = 0. C(0;1)

This completes the proof of the problem.



108

Chapter 6. Elementary Theory of Complex Integration

Problem 6.12 ⋆ Suppose that γ is the unit square and

f (z) = Evaluate

 3  z , 

if z ∈ D(0; 4);

|z|2 , otherwise. Z

f (z) dz. γ


Proof. It is clear that f ∈ H D(0; 4) and D(0; 4) contains the unit square γ, so Theorem 6.10 (Cauchy’s Theorem in a Disc) implies that Z f (z) dz = 0. γ

We end the proof of the problem.



Problem 6.13 ⋆ Suppose that ∆ denotes the boundary of the triangle with vertices at the points 0, 3i and −4 oriented counterclockwise. Compute the integral Z (ez − z) dz. ∆

Proof. Let Z1 , Z2 and Z3 correspond to the complex numbers 0, 3i and −4 respectively. Thus Z1 , Z2 and Z3 form the vertices of the triangle ∆. Since ∆ is a simple closed curve and ez is entire, it yields from Theorem 6.8 (Cauchy’s Theorem for a Triangle) that Z ez dz = 0. ∆

Therefore, we get Z Z (ez − z) dz = − z dz ∆ ∆ Z Z Z = − −−−→ z dz − −−−→ z dz − −−−→ z dz Z1 Z2 1

=− −

Z

Z

(−3it) · 3i dt −

0

0

Z2 Z3

1

Z

Z3 Z1

1

0

[−3i(1 − t) − 4t] · (−3i − 4) dt

−4(1 − t) · 4 dt

9 7 = − − − 12i + 8 2 2 = −12i. This ends the proof of the problem.



6.3. Cauchy’s Theorem

109

Problem 6.14 ⋆

⋆ Prove the Fundamental Theorem of Algebra by Cauchy’s Theorem.

Proof. We adapt the proof of Boas [7]. Let P (z) be a nonconstant polynomial of degree n. Without loss of generality, we may assume that P (z) is real for real z. Otherwise, we consider the nonconstant polynomial P (z)P (z). Assume that P (z) 6= 0 in C. Now the continuity of P and the hypothesis imply that P does not change sign for real z. Suppose that P (x) > 0 for all x ∈ R. Then we have Z π dt > 0. (6.5) −π P (2 cos t)

Note that cos t = 12 (z + z1 ), so we have Z π Z Z dt dz z n−1 dz 1 1 = , = 1 i C(0;1) zP (z + z ) i C(0;1) Q(z) −π P (2 cos t)

where Q(z) = z n P (z + z1 ). This Q is also a polynomial and Q(z) 6= 0 for z 6= 0. If an is the leading coefficient of P , then it is easy to check that Q(0) = an . This means that Q(z) 6= 0 in C and the integrand z n−1 f (z) = Q(z) is entire and Theorem 6.10 (Cauchy’s Theorem in a Disc) yields the result Z π dt =0 −π P (2 cos t) which is contrary to the inequality (6.5). Hence P (z) has a zero in the complex plane. This completes the proof of the problem.  Problem 6.15 ⋆

⋆ Given that Ω is a simply connected region. If f ∈ H(Ω), prove that Z z F (z) = f (ζ) dζ z0

is a primitive of f in Ω, where z0 ∈ Ω and the path of integration is any smooth curve in Ω joining z0 and z.

Proof. Since f ∈ H(Ω), it follows from Corollary 6.12 (Independence of Paths) that F is welldefined. Let z ∈ Ω. In order to prove the problem, it suffices to establish lim

h→0

F (z + h) − F (z) = f (z). h

(6.6)

To this end, we note that 1 F (z + h) − F (z) = h h

Z

z+h

z0

f (ζ) dζ −

1 h

Z

z

f (ζ) dζ z0

110

Chapter 6. Elementary Theory of Complex Integration Z 1 z+h f (ζ) dζ = h z Z 1 z+h [f (ζ) − f (z) + f (z)] dζ = h z Z 1 z+h = [f (ζ) − f (z)] dζ + f (z) h z

which gives 1 F (z + h) − F (z) − f (z) = h h

Z

z+h

z

[f (ζ) − f (z)] dζ.

(6.7)

Since f ∈ H(Ω), f is continuous on Ω. Given ǫ > 0, there exists a δ > 0 such that |f (z)−f (ζ)| < ǫ for every |z − ζ| < δ. By taking the path as the line segment joining z and z + h in the integral (6.7) and choosing |h| < δ, we see from Theorem 6.5 (The M -L Formula) that Z z+h F (z + h) − F (z) 1 1 [f (ζ) − f (z)] dζ ≤ − f (z) = × × |h| × ǫ = ǫ. h |h| |h| z

As ǫ is arbitrary, we obtain the desired result (6.6) and we complete the proof of the problem.  Problem 6.16 / Ω. Pick z0 ∈ Ω, fix a value of log z0 ⋆ Suppose that Ω is a simply connected region and 0 ∈ and define Z z dζ + log z0 . f (z) = z0 ζ Prove that f is an analytic branch of log z in Ω.

Proof. Since Ω does not contain the origin, primitive of z1 in Ω, i.e.,

1 ζ

∈ H(Ω). Thus Problem 6.15 ensures that f is a

f ′ (z) =

1 z

for every z ∈ Ω. To show that f satisfies the equation (4.5), we consider the function F (z) = ze−f (z) . It is obvious that F ′ (z) = e−f (z) − zf ′ (z)e−f (z) = 0 in Ω, it deduces from Theorem 4.8 that F is a constant function. Since F (z0 ) = 1, we must have ef (z) = z, i.e., f satisfies the equation (4.5). Hence f is an analytic branch of log z, completing the proof  of the problem.

6.3. Cauchy’s Theorem

111

Remark 6.4 Suppose that Ω is a simply connected region, f ∈ H(Ω) and f (z) 6= 0 in Ω. Fix z0 ∈ Ω and define Z z ′ f (ζ) dζ, F (z) = z0 f (ζ) where the integral is taken over a smooth curve in Ω connecting z0 and z. Using similar argument as in the proof of Problem 6.16, it can be shown that F (z) is an analytic branch of log f (z) and f ′ (z) F ′ (z) = f (z) for every z ∈ Ω. Problem 6.17 Find the power series representation of Log z for |z − 1| < 1. 1 Proof. Recall that the geometric series (5.4) converges uniformly to 1−z in U , so its power series can be integrated term by term. In other words, we have Z zX Z z ∞ Z z ∞ ∞ X X z n+1 dζ n ζ dζ = ζ n dζ = = , n+1 0 0 0 1−ζ n=0

n=0

n=0

where z ∈ U and the path of integration is the segment from 0 to z. By Remark 6.4, we see immediately that ∞ X z n+1 −Log (1 − z) = . n+1 n=0

By replacing 1 − z by z, we obtain

Log z =

∞ X (−1)n+1

n=1

n

for |z − 1| < 1, completing the proof of the problem.

(z − 1)n 

Problem 6.18 ⋆ Given that γ is a simple closed curve. If f is analytic and f ′ is continuous on and inside γ, prove that Z f (z) dz = 0. γ

Proof. We note that f (z) is independent of z, so ∂f ∂z = 0. Since our f satisfies the hypotheses of Theorem 6.3 (Green’s Theorem), we apply its complex form to establish immediately that Z f (z) dz = 0, γ

ending the proof of the problem.



112

Chapter 6. Elementary Theory of Complex Integration

6.4 The Cauchy Integral Formula Problem 6.19 ⋆ Compute the integral

Z

Proof. Rewrite the integral as

Z

ez dz. 1+z

C(0;2)

C(0;2)

ez dz. z − (−1)

Since −1 is inside the circle C(0; 2), we can apply Theorem 6.15 (The Cauchy Integral Formula) with f (z) = ez to obtain Z 2πi ez dz = 1 , e C(0;2) z − (−1) completing the proof of the problem.



Problem 6.20 ⋆ Compute the integral

Z

Proof. Rewrite the integral as

Z

dz . z(z − πi)

C(3i;2)

1 z C(3i;2)

z − πi

dz.

Let f (z) = 1z . Clearly, f is analytic everywhere except the origin. Of course, we have 0 ∈ / D(3i; 2) but πi ∈ D(3i; 2). Hence we may employ Theorem 6.15 (The Cauchy Integral Formula) to conclude that Z

C(3i;2)

dz = z(z − πi)

Z

1 z

C(3i;2)

z − πi

ending the proof of the problem.



Problem 6.21 ⋆ Compute the integral

dz = 2πif (πi) = 2,

Z

2π 0

it

ee dt.

6.4. The Cauchy Integral Formula

113

Proof. Set f (z) = ez and z0 = 0. Since z0 = 0 lies inside the circle C(0; 1). According to Theorem 6.15 (The Cauchy Integral Formula), we see that 1 2π

Z

2π

eit

e

0

1 dt = 2πi

which implies

Z

2π

Z

ez dz = e0 = 1 z−0

C(0;1)

it

ee dt = 2π.

0

This completes the analysis of the problem.



Problem 6.22 ⋆ Compute the integral

Z

C(0;8)

z 4 + ez dz. (z + 2πi)3

Proof. Put f (z) = z 4 + ez and z0 = −2πi. Since f is entire and −2πi ∈ D(0; 8), we deduce from Corollary 6.16 (The Cauchy Integral Formula for Derivatives) that Z

C(0;8)

z 4 + ez dz = (z + 2πi)3

Z

C(0;8)

f (z) dz [z − (−2πi)]3

2πi ′′ f (−2πi) 2! = πi[12(−2πi)2 + e−2πi ]

=

= π(1 − 48π 2 )i. We end the proof of the problem.



Problem 6.23 ⋆ Suppose that P (z) = a0 + a1 z + · · · + an z n . Prove that Z z n−1 |P (z)|2 dz = 2πian an . C(0;1)

Proof. We have Z

z C(0;1)

n−1

2

|P (z)| dz = =

Z

z

n−1

C(0;1)

Z

C(0;1)

=i

Z

π −π

·

z n−1 ·

eint

n X k=0

n X

n X

k,p=0

k=0

n  X  ak z · ak z k dz k

k=0

n  X  ak z k · ak z k dz k=0

ak ap ei(k−p)t dt

114

Chapter 6. Elementary Theory of Complex Integration Z π X n =i (6.8) ak ap ei(n+k−p)t dt. −π k,p=0

By definition, we know that Z

π

ei(m−n)t dt =

−π

  2π, if m = n; 

0,

otherwise.

Putting this into the equation (6.8), we see immediately that Z π Z n−1 2 a0 an dt = 2πia0 an . z |P (z)| dz = i −π

C(0;1)

This completes the proof of the problem.



Problem 6.24 ⋆ Suppose that f ∈ H(U ) and f (z) = u(z) + iv(z), where u and v are real functions such that either u(0) = v(0) or u(0) = −v(0). Prove that Z π Z π

2 it v 2 reit dt u re dt = −π

−π

for all r ∈ (0, 1).

Proof. Since f ∈ H(U ), we have f 2 ∈ H(U ). Since r ∈ (0, 1), we deduce from Theorem 6.15 (The Cauchy Integral Formula) that Z Z π Z π 1 f 2 (z) 1 1 2 2 it it dt f (0) = dz = = f (re ) · rie f 2 (reit ) dt. (6.9) 2πi C(0;r) z − 0 2πi −π reit 2π −π



Since f 2 (0) = u2 (0) + 2iu(0)v(0) − v 2 (0) and f 2 (reit ) = u2 reit + 2iu reit v reit − v 2 reit , we compare the real parts of both sides of the equation (6.9) and get Z π  2 it

 1 u re − v 2 reit dt = u2 (0) − v 2 (0). 2π −π

Using the hypotheses, we see that u2 (0) − v 2 (0) = 0 and thus Z π Z π

2 it v 2 reit dt u re dt = −π

−π

holds for all r ∈ (0, 1). We have completed the analysis of the problem.



Problem 6.25 (Cauchy’s Estimate or Cauchy’s Inequality) ⋆ Suppose that Ω is a region containing D(a; r) and |f (z)| ≤ M for all z ∈ C(a; r). Prove Cauchy’s Estimate: n! · M |f (n) (a)| ≤ . rn

6.4. The Cauchy Integral Formula

115

Proof. From the hypotheses, we know that

M f (z) |f (z)| = n+1 ≤ n+1 n+1 (z − a) r r

whenever z ∈ C(a; r). Combining Theorem 6.5 (The M -L Formula) and Corollary 6.16 (The Cauchy Integral Formula for Derivatives), we establish |f

(n)

n! Z f (z) n! M n! · M (a)| = × n+1 × 2πr = dz . ≤ n+1 2πi C(a;r) (z − a) 2π r rn

This completes the proof of the problem.



Problem 6.26 ⋆ Suppose that the radius of convergence of the power series f (z) =

∞ X

an z n is R, where

n=0

a0 6= 0. For every 0 < λ < R, we define Mλ = max |f (z)|. Prove that if |z| < z∈C(0;λ)

|a0 | |a0 |+Mλ λ,

then f (z) 6= 0.

Proof. Combining Problem 6.25 and Corollary 6.17 (Taylor’s Theorem), we see that, on C(0; λ), |an | =

Mλ |f (n) (0)| ≤ n n! λ

for every n ≥ 0. Next, using this bound (6.10), if |z| < |f (z) − a0 | ≤

∞ X

n=1

n

|an | · |z| ≤ Mλ

(6.10)

|a0 | |a0 |+Mλ λ,

∞ X |z|n

n=1

λn

then we have

= Mλ ·

|z| . λ − |z|

Therefore, we obtain |f (z) − a0 | < Mλ ·

λ

|a0 | |a0 |+Mλ λ 0| − |a0|a |+Mλ λ

= |a0 |.

Finally, the triangle inequality implies that |f (z)| = |a0 − (f (z) − a0 )| ≥ |a0 | − |f (z) − a0 | > 0 which completes the proof of the problem.



Problem 6.27 ⋆

⋆ Compute the integral

Z

C(0;2)

z 2021 dz. z 2022 + z + 1

116

Chapter 6. Elementary Theory of Complex Integration

Proof. Assume that z0 ∈ D(0; 2) was a zero of z 2022 + z + 1. Clearly, z0 6= 0. On the one hand, we have 1 1 1 + 2021 + 2022 = 0. (6.11) z0 z0 On the other hand, if |z| ≥ 2, then we know that 1 1 1 1 1 1 1 + + ≥ 1 − 2021 − 2022 ≥ 1 − 2021 − 2022 > 0 z 2021 z 2022 |z| |z| 2 2

which contradicts the result (6.11). Thus all zeros of z 2022 + z + 1 must lie inside C(0; 2) and then the function z 2021 f (z) = 2022 z +z+1 is analytic in |z| ≥ 2. Applying Corollary 6.14, we can write Z Z z 2021 z 2021 dz = dz, (6.12) 2022 + z + 1 2022 + z + 1 C(0;r) z C(0;2) z where r > 2. Note that

1 z+1 z 2021 − =− . z 2022 + z + 1 z z(z 2022 + z + 1)

(6.13)

Next, we observe that if |z| = r > 2, then r+1 z+1 , ≤ 2022 2022 z(z + z + 1) r(r − r − 1)

so an application of Theorem 6.5 (The M -L Formula) gives Z z+1 r+1 2π(r + 1) dz × 2πr = 2022 . ≤ 2022 2022 +z+1 r(r − r − 1) r −r−1 C(0;r) z

This implies immediately that

lim

Z

r→∞ C(0;r)

z+1 dz = 0. +z+1

z 2022

Hence we conclude from the expressions (6.12) and (6.13) and Problem 6.4 that Z Z h1 i z 2021 z+1 dz dz = lim − 2022 + z + 1 r→∞ C(0;r) z z 2022 + z + 1 C(0;2) z Z dz = lim r→∞ C(0;r) z = 2πi.

This completes the proof of the problem.



Problem 6.28 ⋆ Prove that

Z

C(2i;4)

z dz = πi. z2 + 9

6.4. The Cauchy Integral Formula

117

Proof. We first factorize the denominator to get z 2 + 9 = (z − 3i)(z + 3i). It is easy to see that 3i is the only point lying inside C(2i; 4). Therefore, we may write z f (z) = , z2 + 9 z − 3i z . Since f is analytic in a region containing C(2i; 4), we apply Theorem 6.15 where f (z) = z+3i (The Cauchy Integral Formula) to obtain Z Z f (z) z dz = dz = 2πif (3i) = πi, 2+9 z z C(2i;4) − 3i C(2i;4)

completing the proof of the problem.



Problem 6.29 ⋆

⋆ Let f be entire and satisfy the equation f (z + ω) = f (z) + f (ω)

for all z, ω ∈ C. Prove that f ′′ (z) = 0 in the complex plane. Proof. Let z1 , z2 , . . . , zn ∈ C. Then repeated use of the hypothesis implies that n X k=1

f (zk ) = f

n X k=1

 zk .

Consequently, by taking z1 = z2 = · · · = zn = nz , the formula z  f (z) = nf n

(6.14)

holds for all z ∈ C. Since f is entire, |f | is continuous on C. Since C(0; 1) is compact, the Extreme Value Theorem ensures that |f | attains its maximum on C(0; 1). Let it be M and we follow from the expression (6.14) that  z  |f (z)| = n f ≤ nM n on C(0; n).

Let n ≥ 2. If |ζ| < n and |z| = n, then the triangle inequality |z| ≤ |z − ζ| + |ζ| implies that 0 < n − |ζ| ≤ |z − ζ|. Combining this, Theorem 6.5 (The M -L Formula) and Corollary 6.16 (The Cauchy Integral Formula for Derivative) to establish 2! Z f (z) ′′ |f (ζ)| = dz 3 2πi C(0;n) (z − ζ) nM 1 ≤ × 2πn × π (n − |ζ|)3

118

Chapter 6. Elementary Theory of Complex Integration 2n2 M

≤


3 . n − |ζ|

Since lim

n→∞

2n2 M


3 = 0, n − |ζ|

we see immediately that f ′′ (ζ) = 0 for all ζ ∈ C, as required. We complete the analysis of the problem.  Problem 6.30 ⋆ Given that f (z) =

∞ X

an z n and g(z) =

∞ X

bn z n have radii of convergence R1 and R2

n=0

n=0

respectively. Let h(z) = f (z)g(z). Express the Taylor series of h in terms of an and bn .


Proof. By Theorem 5.5, we have f ∈ H D(0; R1 ) and g ∈ H D(0; R2 ) so that h ∈ H D(0; R) , where R ≥ min{R1 , R2 }. Therefore, it asserts from Corollary 6.17 (Taylor’s Theorem) that h(z) =

∞ X

cn z n

and cn =

n=0

h(n) (0) . n!

Now the product rule for differentiation shows that h(n) (0) =

n X

Ckn f (k) (0)g(n−k) (0).

(6.15)

k=0

Recall from Corollary 5.6 that

f (n) (0) and n! Put these into the expression (6.15) to derive an =

h(n) (0) =

n X k=0

which yields

bn =

g (n) (0) . n!

Ckn × k!ak × (n − k)!bn−k = n!

cn =

n X

n X

ak bn−k

k=0

ak bn−k .

k=0

This completes the proof of the problem. Problem 6.31

⋆ Suppose that f is entire and there exist positive constants M and α such that |f (z)e−α|z| | ≤ M for all z ∈ C. Prove that holds in C.

|f ′ (z)e−α|z| | ≤ αeM



6.4. The Cauchy Integral Formula

119

Proof. For all z ∈ C and r > 0, it follows from Corollary 6.16 (The Cauchy Integral Formula for Derivatives) that Z 1 f (ζ) ′ f (z) = dζ 2πi C(z;r) (ζ − z)2 which gives |f ′ (z)| ≤ r · max

ζ∈C(z;r)

|f (ζ)| 1 1 1 = · max |f (ζ)| ≤ · max M eα|ζ| ≤ · M eα(|z|+r) . |ζ − z|2 r ζ∈C(z;r) r ζ∈C(z;r) r

Consequently, we get

αr ′ f (z)e−α|z| ≤ M · e . r By elementary differentiation, we see that the function g(r) = eαr r −1 attains its absolute minimum αe at the point r = α1 . Hence we conclude that ′ f (z)e−α|z| ≤ αeM.

This completes the proof of the problem.



Problem 6.32
⋆ Suppose that f ∈ H D(0; R) , where R > 1. Prove that Z π π t f (eit ) cos2 dt = [2f (0) + f ′ (0)]. 2 2 −π

Proof. By Theorem 6.10 (Cauchy’s Theorem in a Disc), we have Z π Z 1 1 it it f (z) dz = 0. f (e )e dt = 2π −π 2πi C(0;1) Combining Theorem 6.15 (The Cauchy tegral Formula for Derivatives), we get Z 1 f (0) = 2πi C(0;1) Z 1 f ′ (0) = 2πi C(0;1)

(6.16)

Integral Formula) and Corollary 6.16 (The Cauchy Inf (z) 1 dz = z−0 2π

Z

π

−π

f (z) 1 dz = 2 (z − 0) 2π

f (eit ) dt,

Z

π

f (eit )e−it dt.

(6.17) (6.18)

−π

Hence we apply the identity (4.4) to the sum of the expressions (6.16), (6.17) and (6.18) to conclude that Z π 1 ′ f (eit )(2 + eit + e−it ) dt 2f (0) + f (0) = 2π −π Z π it it
2 1 = f (eit ) e 2 + e− 2 dt 2π −π Z t 2 π f (eit ) cos2 dt. = π −π 2 This certainly implies our desired result and we complete the proof of the problem.



120

Chapter 6. Elementary Theory of Complex Integration

Problem 6.33 ⋆ ⋆ Suppose that γ : [a, b] → C is a closed curve in U such that 0 lies inside the interior of γ and γ(a) = γ(b) = i. Let f ∈ H(U ). For every k ∈ N, we let logk z = Log z + 2kπi. Prove that Z 1 f ′ (z) logk z dz = f (i) − f (0). 2πi γ

Proof. As z traces from i along the curve γ and ends at i, the value of logk z will change continuously from logk i to logk+1 i = logk i + 2πi. According to integration by parts in (see §6.1.1) and Theorem 6.15 (The Cauchy Integral Formula), we see that Z γ(b) Z − f (z)(logk z)′ dz f ′ (z)Log k z dz = f (z) logk z γ(a) γ γ Z

f (z) dz = f γ(b) logk+1 γ(b) − f γ(a) logk γ(a) − γ z
= f (i) logk i + 2πi − f (i) logk i − 2πif (0)   = 2πi f (i) − f (0) .

This gives the desired formula and we complete the proof of the problem.



Problem 6.34 (Cauchy’s Integral Formula in a Half Plane) ⋆ ⋆ Let α be real and f be analytic in {z ∈ C | Re z > α}. If there are positive constants M and p such that M |f (z)| ≤ p |z|

for large enough z, then for Re z > a > α, we have Z a+iρ f (ζ) 1 lim dζ. f (z) = − 2πi ρ→∞ a−iρ ζ − z

Proof. Fix z with Re z > a > α. Let ρ > max{|Re z|, |Im z|}. Thus the rectangle R = {(x, y) | a ≤ x ≤ ρ, −ρ ≤ y ≤ ρ} contains this z, see Figure 6.5 for details. Define ℓ = {z | z = a − it, −ρ ≤ t ≤ ρ} and γ = ∂R − ℓ. Now Theorem 6.15 (The Cauchy Integral Formula) asserts that Z 1 f (ζ) f (z) = dζ 2πi ∂R ζ − z Z Z 1 f (ζ) f (ζ) 1 = dζ + dζ 2πi ℓ ζ − z 2πi γ ζ − z Z a+iρ Z f (ζ) 1 f (ζ) 1 =− dζ + dζ. (6.19) 2πi a−iρ ζ − z 2πi γ ζ − z

6.4. The Cauchy Integral Formula

121

For sufficiently large enough ρ and ζ ∈ γ, we have |ζ| ≥ ρ and | zζ |
e−n and + 2 + · · · | < . 2 n n 2

122

Chapter 6. Elementary Theory of Complex Integration

However, we observe from the triangle inequality that ∞ X  1  ak |a0 | > e−n ≥ |a0 | − > f n nk 2 k=1

for all n ≥ N , a contradiction. Therefore, a0 = 0. Next, let ak be the first nonzero coefficient in {a1 , a2 , . . .}. Then we have f

1 n

=

ak ak+1 + k+1 + · · · . k n n

(6.21)

For sufficiently large enough n, the expression (6.21) gives  1  k 2nk |ak | ≤ 2 f n < n . n e

As a result, it implies ak = 0 if we take n → ∞ which contradicts the assumption that ak is nonzero. Thus we have a1 = a2 = · · · = 0, but this forces that f ≡ 0 which contradicts our hypothesis that |f ( n1 )| > 0. Hence we conclude  that no such function f exists and this completes the proof of the problem. Problem 6.36 ⋆ Let f ∈ H(Ω) and U ⊆ Ω. Prove that f

(n)

n! (0) = π

Z

π −π

where n ≥ 1.



 Re f (eit ) e−int dt,

Proof. Write f (z) =

∞ X

am z m .

m=0

For any n ≥ 1, on the one hand, we know from Corollary 6.7 that it is legitimate to interchange the integral and the summation in the following computation: n! 2π

Z

π

f (eit )e−int dt

−π

n! = 2π n! = 2π

Z

π

−π

∞ X

m=0

∞ X

m=0

am

hZ

 am e−imt e−int dt π

−π

i e−i(n+m)t dt .

(6.22)

Since n + m ≥ 1, the integral in the expression (6.22) is always zero, we obtain n! 2π

Z

π −π

f (eit )e−int dt = 0.

(6.23)

6.4. The Cauchy Integral Formula

123

On the other hand, we follow from Corollary 6.16 (The Cauchy Integral Formula for Derivatives) that Z Z n! f (ζ) n! π (n) f (0) = dζ = f (eit )e−int dt. (6.24) 2πi C(0;1) ζ n+1 2π −π Finally, the results (6.23) and (6.24) reveal that Z Z n! π n! π it −int (n) f (e )e dt + f (0) = f (eit )e−int dt 2π −π 2π −π Z  n! π  = Re f (eit ) e−int dt, π −π completing the proof of the problem.



124

Chapter 6. Elementary Theory of Complex Integration

CHAPTER

7

Properties of Analytic and Entire Functions

7.1 Fundamental Concepts In Chapter 6, we know the importance and usefulness of Cauchy’s Theorems and Cauchy’s Integral Formula. In fact, some of their applications have already been seen there. For examples, every analytic function is infinitely differentiable (Corollary 6.16 (The Cauchy Integral Formula for Derivatives)), the power series representation of an analytic function (Corollary 6.17 (Taylor’s Theorem)) and Cauchy’s Estimate (Problem 6.25). In this chapter, we are going to study other important properties of analytic and entire functions. The main references are [1, Chap. 4], [4, Chap. 3], [5, Chap. 6], [9, Chap. III and V], [11, Chap. IV and VI], [14, Chap. III and IV], [16, Chap. 3 and 5] and [31, Chap. 10].

7.1.1 Zeros of Analytic Functions Zeros of analytic functions can be treated in more or less the same way as zeros of polynomials. Suppose that Ω is a region and f ∈ H(Ω). Let a be a point of Ω such that f (a) = 0. Now a is called a zero of order p,a where p ∈ N, if f ′ (a) = f ′′ (a) = · · · = f (p−1) (a) = 0

and f (p) (a) 6= 0.

Using Corollary 6.17 (Taylor’s Theorem), we see that f (z) =

∞ X

n=p

an (z − a)n = (z − a)p g(z),

where g ∈ H(Ω) and g(a) 6= 0. Furthermore, the fact g(a) 6= 0 guarantees that g is nonzero in a neighborhood of a which means Z(f ) = {a ∈ C | f (a) = 0} has no limit point in Ω.b Theorem 7.1 (The Interior Uniqueness Theorem). Let Ω be a region and f ∈ H(Ω). Suppose that {zn } is a sequence of Ω such that zn → z0 ∈ Ω and f (zn ) = 0 for all n ∈ N. Then we have f ≡0 a b

Or we say it is of multiplicity p. In fact, Z(f ) is at most countable, see [31, Theorem 10.18, pp. 208, 209].

125

126

Chapter 7. Properties of Analytic and Entire Functions

in Ω. In particular, if f, g ∈ H(Ω) agree at a set of points with an accumulation point in Ω, then it must be true that f ≡ g in Ω. By this theorem, it is easy to show that the following trigonometric identities are valid for all complex numbers z, z1 and z2 : sin(−z) = − sin z,

cos(−z) = cos z, sin z , cos2 z + sin2 z = 1, tan z = cos z cos(z1 + z2 ) = cos z1 cos z2 − sin z1 sin z2 ,

sin(z1 + z2 ) = sin z1 cos z2 + cos z1 sin z2 , 1 + cos 2z 1 − cos 2z cos2 z = , sin2 z = . 2 2

If a ∈ Ω is a zero of f of infinite order, then we have f (n) (a) = 0 for every n ≥ 0. By Corollary 6.17 (Taylor’s Theorem), we have f (z) ≡ 0 in D(a; r) for some r > 0. By Theorem 7.1 (The Interior Uniqueness Theorem), f must be the zero function in Ω. Remark 7.1 We notice that it is possible to construct a real-valued nonconstant infinitely differentiable function on R having a zero of infinite order. In fact,  −1  e x2 , if x 6= 0; f (x) =  0, otherwise is an example of such kind, see [30, Exercise 8.1, p. 196].

7.1.2 Properties of Analytic Functions Theorem 7.2 (Morera’s Theorem). Let f be a continuous function on a region Ω. If the integral Z f (z) dz = 0 γ

holds for every simple closed curve γ in Ω, then f ∈ H(Ω).c Theorem 7.3 (The Mean Value Property for Analytic Functions). Let Ω be a region, α ∈ Ω and f ∈ H(Ω). Then we have Z π 1 f (α) = f (α + reit ) dt, 2π −π where D(α; r) ⊆ Ω. Theorem 7.4 (The Maximum Modulus Theorem). Suppose that Ω is a region and f ∈ H(Ω). If there exists an α ∈ Ω at which |f | has a local maximum, then f is constant. c

The result remains true when the simple closed curve is replaced by every triangle, every square, every rectangle or every circle. Some generalizations of Morera’s Theorem can be found in [26], [32, p. 196] and [34].

7.2. Zeros of Analytic Functions

127

Theorem 7.5 (The Minimum Modulus Theorem). Suppose that Ω is a region, f ∈ H(Ω) and f never vanishes on Ω. If there exists an α ∈ Ω such that |f (α)| ≤ |f (z)| for all z ∈ Ω, then f is constant. Remark 7.2 In the above two theorems, if the region Ω is bounded and f is continuous on Ω, then we also have max |f (z)| = max |f (z)| and min |f (z)| = min |f (z)|, z∈∂Ω

z∈Ω

z∈Ω

z∈∂Ω

where ∂Ω denotes the boundary of the set Ω.

7.1.3 Properties of Entire Functions Theorem 7.6 (Liouville’s Theorem). A bounded entire function is constant. Corollary 7.7. If f is entire and f (z) → ∞ as z → ∞, then f has at least one zero in the complex plane. Corollary 7.8 (The Fundamental Theorem of Algebra). Every polynomial P (z) of degree n ≥ 1 has a zero in C.

7.2 Zeros of Analytic Functions Problem 7.1 ⋆ Let f be entire and define

Prove that F is entire.

 f (z) − f (a)   , if z 6= a;  z−a F (z) =    f ′ (a), otherwise.

Proof. Since f is entire, we get from Corollary 6.17 (Taylor’s Theorem) that, if z 6= a, then F (z) =

f ′′ (a) f (3) (a) f (z) − f (a) = f ′ (a) + (z − a) + (z − a)2 + · · · . z−a 2! 3!

(7.1)

The definition of F tells us that this formula (7.1) remains valid for z = a. Since the power series is convergent everywhere, it follows from Theorem 5.5 that F is actually entire as required. We  end the analysis of the problem.

128

Chapter 7. Properties of Analytic and Entire Functions

Remark 7.3 With the aid of Problem 7.1, it can be shown similarly that if f is entire and a1 , a2 , . . . , aN are zeros of f for some N ∈ N, then the function g : C \ {a1 , a2 , . . . , aN } → C defined by g(z) =

f (z) (z − a1 )(z − a2 ) · · · (z − aN )

has limits at a1 , a2 , . . . , aN . If we define g(an ), where n = 1, 2, . . . , N , by these limits, then the (extended) function g must be entire.

Problem 7.2 ⋆ Determine all entire functions with zeros set { n1 | n = 1, 2, . . .}.

Proof. Let f be such a function. It is clear that f is continuous on C which gives  1 1 f (0) = f lim = lim f = 0. n→∞ n n→∞ n

In other words, f has a zero at the origin. For any ǫ > 0, since D(0; ǫ) contains at least one n1 for sufficiently large n. This means that the origin is not isolated. Hence Theorem 7.1 (The Interior  Uniqueness Theorem) reveals that f ≡ 0 in C. This completes the proof of the problem. Problem 7.3 ⋆ ⋆ Let Ω be a region and f ∈ H(Ω). For each z ∈ Ω, there exists a nonnegative integer nz such that f (nz ) (z) = 0. Prove that f is a polynomial.

Proof. Let z ∈ Ω and choose r > 0 such that D(z; r) ⊆ Ω. For each n = 0, 1, 2, . . ., let

Z f (n) = {ζ ∈ Ω | f (n) (ζ) = 0} and An = Z f (n) ∩ D(z; r). Then the hypothesis allows us to write

D(z; r) =

∞ [

An .

n=0

Of course, D(z; r) is uncountable, so at least one of An is uncountable. Let any one of them be Anz . By the definition, every An is bounded. Hence the Bolzano-Weierstrass Theorem [39, Problem 5.25, p. 68] implies that this Anz has a limit point in D(z; r) and then also in Ω. Now we apply Theorem 7.1 (The Interior Uniqueness Theorem) to conclude that f (nz ) (z) = 0 holds in Ω. By Theorem 4.8, we get f (nz −1) (z) = c0 for some constant c0 . If we define f1 = f −

c0 z nz −1 , (nz − 1)!

7.2. Zeros of Analytic Functions (nz −1)

then we have f1

129

= f (nz −1) − c0 ≡ 0 in Ω. By Theorem 4.8 again, we have (nz −2)

f1

(z) ≡ c1

for some constant c1 . Next, we define f2 = f1 − (n −2)

c0 c1 c1 z nz −2 = f − z nz −1 − z nz −2 (nz − 2)! (nz − 1)! (nz − 2)!

(n −2)

so that f2 z = f1 z − c1 ≡ 0 in Ω. Therefore, by repeated applications of Theorem 4.8, we see that f is a polynomial of degree at most nz − 1. This completes the proof of the problem.  Problem 7.4 ⋆ Suppose that an ∈ R are distinct for n ∈ N, an → 0 as n → ∞ and f
(an ) is real for all large n. Suppose, further, that there is an r > 0 such that f ∈ H D(0; r) . Prove that f (z) = f (z)

(7.2)

for all z ∈ D(0; r). Proof. Let F (z) = f (z).
Since D ∗ (0; r) = {z | z ∈ D(0; r)} = D(0; r), it follows from Problem 4.22 that F ∈ H D(0; r) . Next,
we consider the function g : D(0; r) → C defined by g = F − f . It is clear that g ∈ H D(0; r) and g(an ) = F (an ) − f (an ) = f (an ) − f (an ) = f (an ) − f (an ) = 0

for all large enough n. Hence Theorem 7.1 (The Interior Uniqueness Theorem) implies that  F ≡ f which is exactly the requirement (7.2), completing the analysis of the problem. Problem 7.5 ⋆ ⋆ Let Ω be a region and f ∈ H(Ω). If f ′′ (z) + f (z) = 0 in Ω and f (a) = f ′ (a) = 0 for some a ∈ Ω, prove that f ≡ 0 in Ω. Proof. Consider F = (f ′ )2 + f 2 . Then F ∈ H(Ω) and it is easy to see that F ′ = 2f ′ f ′′ + 2f ′ f = 2f ′ (f ′′ + f ) = 0 in Ω, so Theorem 4.8 shows that F is a constant function. In fact, F (a) = [f ′ (a)]2 + f (a)2 = 0 so that F (z) ≡ 0 in Ω. Notice that F = (f ′ + if )(f ′ − if ). We claim that either f ′ + if ≡ 0 or f ′ − if ≡ 0. Assume that f ′ + if 6≡ 0.d It means that f ′ (α) + if (α) 6= 0 for some α ∈ Ω. By the continuity of f ′ + if , there exists a neighborhood d

See Problem 7.7 for the general case.

130

Chapter 7. Properties of Analytic and Entire Functions

D(α; δ) ⊆ Ω such that f ′ + if 6= 0 in D(α; δ). Combining this and the fact F ≡ 0, we conclude that f ′ − if ≡ 0

in D(α; δ). Using Theorem 7.1 (The Interior Uniqueness Theorem), we have f ′ − if ≡ 0 in Ω and the claim follows. If f ′ + if ≡ 0, then we have f ′ ≡ −if in Ω. Consider the function g(z) = eiz f (z). Since g ′ (z) = ieiz f (z) + eiz f ′ (z) = eiz [f ′ (z) + if (z)] ≡ 0 in Ω, we see from Theorem 4.8 that g is constant throughout Ω, i.e., eiz f (z) = A for some constant A. Since f (a) = 0, we have A = 0 and hence f ≡ 0 in Ω. Now the case for f ′ − if ≡ 0 can be treated similarly. Hence we end the proof of the problem.  Problem 7.6 ⋆ ⋆ Let f ∈ H(U ). Show that there exists a complex constant ℓ and a sequence {zn } ⊆ U such that |zn | → 1 and f (zn ) → ℓ as n → ∞. Proof. There is nothing to prove if f ≡ 0. Without loss of generality, we may assume that f has finitely many zeros. Otherwise, Z(f ) is an infinite subset of U . Let {zn } be any countable subset of Z(f ). By the Bolzano-Weierstrass Theorem, we may assume that {zn } is convergent. If the limit belongs to U , then Theorem 7.1 (The Interior Uniqueness Theorem) yields that f ≡ 0, a contradiction. Thus we must have |zn | → 1 and f (zn ) → 0 as n → ∞. Suppose that 0 is a zero of f of order p ≥ 1 and the other zeros of f in U are a1 , a2 , . . . , ak for some positive integer k. Then we consider the function F (z) =

z p (z − a1 )(z − a2 ) · · · (z − ak ) f (z)

and it satisfies F ∈ H(U ). For each n = 1, 2, . . ., we let Mn =

max

1 ) z∈C(0;1− n

|F (z)|.

Let N be a positive integer such that 1 − N1 > max{|a1 |, |a2 |, . . . , |ak |}. By Theorem 7.4 (The Maximum Modulus Theorem), we know that MN ≤ Mn for all n ≥ N . Since there exists a point αn ∈ C(0; 1 − n1 ) such that |αn

|p

|f (αn )| 1 = × |αn − a1 | × |αn − a2 | × · · · × |αn − ak | |F (αn )| 1 = Mn 1 ≤ MN 1 = |F (αN )| =

|αN

|p

|f (αN )| × |αN − a1 | × |αN − a2 | × · · · × |αN − ak |

7.3. Properties of Analytic Functions

131

for all n ≥ N , we have |f (αN )| · |αn − a1 | × |αn − a2 | × · · · × |αn − ak | |αN |p × |αN − a1 | × |αN − a2 | × · · · × |αN − ak | |αn |p (1 + |a1 |)(1 + |a2 |) · · · (1 + |ak |) × |f (αN )| ≤ |αN − a1 | × |αN − a2 | × · · · × |αN − ak | 0 such that f (z) 6= 0 for all z ∈ D(p; δ) ⊆ Ω. Therefore, we must have g(z) = 0 for all z ∈ D(p; δ). Using Theorem 7.1 (The Interior Uniqueness Theorem), we have g ≡ 0 in Ω, completing the proof of the problem.  Remark 7.4 By contrast with Problem 7.7, one can find real functions f, g such that f 6≡ 0 and g 6≡ 0 in R, but f g ≡ 0 in R. Therefore, this is a difference between real analysis and complex analysis.

7.3 Properties of Analytic Functions Problem 7.8 ⋆ Use Cauchy’s Estimate to prove Theorem 7.6 (Liouville’s Theorem). Proof. Let f be a bounded entire function and M be a bound of f in the plane. Put n = 1 in Problem 6.25 (Cauchy’s Estimate) and replace a by arbitrary z, we get |f ′ (z)| ≤

M . r

Note that M is independent of r, so if r → ∞, then we have |f ′ (z)| = 0

132

Chapter 7. Properties of Analytic and Entire Functions

for all z ∈ C. This means that f ′ (z) = 0 for all z ∈ C. By Theorem 4.8, we conclude that f is constant. This ends the proof of the problem.  Problem 7.9 ⋆ Use Theorem 7.4 (The Maximum Modulus Theorem) to prove Corollary 7.8 (The Fundamental Theorem of Algebra)

Proof. Let P (z) = a0 + a1 z + · · · + an z n be nonconstant, where n ≥ 1. Assume that P had no zeros in C. Then P1 is entire. Obviously, we have P (z) → ∞ as |z| → ∞ (see Remark 7.5), so Theorem 7.4 (The Maximum Modulus Theorem) implies that max z∈C

1 1 = lim max = 0. |P (z)| R→∞ C(0;R) |P (z)|

Consequently, we have P (z) ≡ ∞ in the complex plane which is a contradiction. Hence P must  have a zero in C which ends the proof of the problem. Problem 7.10 ⋆ Prove that there is no polynomial in the form P (z) = z n + an−1 z n−1 + · · · + a1 z + a0 such that |P (z)| < 1 when |z| = 1.

Proof. Assume that P (z) was a polynomial with the properties stated in the problem. By Problem 6.25 (Cauchy’s Estimate) and the hypothesis, we see that |P (n) (0)|
0 for all z ∈ Ω and F = hypothesis states that 1 |F (z)| ≤ |f (a)|

1 f

∈ H(Ω). The

for all z ∈ Ω. By Theorem 7.4 (The Maximum Modulus Theorem), we can say that F is constant. In other words, it must be true that f is a constant as required and we complete the  analysis of the problem. Problem 7.14 ⋆ Let f ∈ H(U ) and there correspond a positive constant M such that |f (z)| ≤ M for all z ∈ U . Prove that M |f ′ (z)| ≤ 1 − |z| in U .

Proof. Fix a ∈ U and set δ > 0 such that |a| < δ < 1. If z ∈ D(a; δ − |a|), then we have |z| ≤ |z − a| + |a| ≤ δ − |a| + |a| = δ < 1 which implies D(a; δ − |a|) ⊆ U. Consequently, we have |f | ≤ M in D(a; δ − |a|). Now we apply Problem 6.25 (Cauchy’s Estimate) to obtain M |f ′ (a)| ≤ . (7.3) δ − |a| Since δ is arbitrary, we deduce from the inequality (7.3) that |f ′ (a)| ≤ lim

δ→1

M M = . δ − |a| 1 − |a|

Since a is any point of U , we get the desired inequality in the question and we end the analysis  of the problem. Problem 7.15 ⋆ Prove Theorem 7.2 (Morera’s Theorem).

7.3. Properties of Analytic Functions

135

Proof. Let α ∈ Ω and consider D(α : r) ⊆ Ω for some r > 0. We define F : Ω → C by Z F (z) = f (ζ) dζ, γ

where γ is any smooth curve in Ω connecting α and z. Since Ω is connected, this curve must lie in Ω. Furthermore, the function F is well-defined. To see this, if τ is another smooth curve in Ω connecting α and z, then γ ◦ τ −1 is a smooth closed curve in Ω and the hypothesis asserts that Z Z Z f (ζ) dζ = 0 f (ζ) dζ = f (ζ) dζ + which means

γ◦τ −1

τ −1

γ

Z

f (ζ) dζ = γ

Z

f (ζ) dζ. τ

Suppose that z ∈ Ω. It is clear that

Z F (z + h) − F (z) 1 z+h − f (z) = f (ζ) dζ − f (z) h h z Z 1 z+h [f (ζ) − f (z)] dζ. = h z

(7.4)

Given ǫ > 0, there exists a δ > 0 such that |f (ζ) − f (z)| < ǫ whenever |ζ − z| < δ. If h is chosen so that |h| < δ, then the segment joining z and z + h lies inside D(z; δ). Applying Theorem 6.5 (The M -L Formula) to the integral (7.4), we establish that Z 1 F (z + h) − F (z) 1 z+h [f (ζ) − f (z)] dζ < · hǫ = ǫ − f (z) = · h |h| h z

for all |h| < δ. Consequently, we have F ′ (z) = f (z) in Ω and our result follows. This completes the analysis of the problem.  Problem 7.16 ⋆ Find a nonanalytic function f satisfying Theorem 7.3 (The Mean Value Property for Analytic Functions)

Proof. Consider f (z) = 1 + iy. For any z = x + iy ∈ C, we have Z π Z π

1 1 it f z + re dt = f x + r cos t + i(y + r sin t) dt 2π −π 2π −π Z π 1 [1 + i(y + r sin t)] dt = 2π −π Z π Z 1 ir π = (1 + iy) dt + sin t dt 2π −π 2π −π = 1 + iy + 0 = f (z). However, since ux = uy = vx = 0 but vy = 1, Theorem 4.7 reveals that f is not analytic at any point in the complex plane. This f is an example that we want, so we have completed the  analysis of the problem.

136

Chapter 7. Properties of Analytic and Entire Functions

Problem 7.17
⋆ Let z0 ∈ C and f ∈ H D(z0 ; R) for some R > 0. Prove that Z π n! (n) f (z0 + reit )e−int dt f (z0 ) = 2πr n −π for every n ≥ 0 and 0 < r < R. Use this to prove Theorem 7.3 (The Mean Value Property for Analytic Fuunction).

Proof. According to Corollary 6.16 (The Cauchy Integral Formula for Derivatives), we get Z f (ζ) n! dζ. (7.5) f (n) (z0 ) = 2πi C(z0 ;r) (ζ − z0 )n+1 Put ζ = z0 + reit into the formula (7.5), we obtain Z π Z π n! f (z0 + reit ) n! (n) it f (z0 ) = · ire dt = f (z0 + reit )e−int dt. 2πi −π (z0 + reit − z0 )n+1 2πr n −π

(7.6)

Take n = 0 into the formula (7.6), it forces that Z π 1 f (z0 + reit ) dt f (z0 ) = 2π −π which is what we need. Thus we end the proof of the problem.



Problem 7.18 ⋆ Suppose that f, g ∈ H(Ω) and f, g are continuous on Ω. Prove that |f (z)| + |g(z)| attains its maximum on ∂Ω.

Proof. Fix p ∈ Ω. Without loss of generality, we may assume that f (p)g(p) 6= 0. Otherwise, f (z)g(z) = 0 for all z ∈ Ω and then Problem 7.7 asserts that f ≡ 0 or g ≡ 0 in Ω and our result follows easily: If f ≡ 0 in Ω, then |f (z)| + |g(z)| = |g(z)| and Remark 7.2 guarantees that max |f (z)| + |g(z)| = max |g(z)| = max |g(z)|. Ω

Ω

∂Ω

The case for g ≡ 0 is similar.

Now we set |f (p)| = f (p)eiθ and |g(p)| = g(p)eiφ for some real θ and φ. Define F (z) = f (z)eiθ + g(z)eiφ .

By the hypotheses, we know that F ∈ H(Ω) and F is continuous on Ω. Observe from Remark 7.2 again that   0 < |f (p)| + |g(p)| = F (p) = |F (p)| ≤ max |F (z)| ≤ max |f (z)| + |g(z)| . z∈∂Ω

This ends the proof of the problem.

z∈∂Ω



7.3. Properties of Analytic Functions

137

Problem 7.19 ⋆ Let f ∈ H(U ) and |f (z 2 )| ≥ |f (z)| for all z ∈ U . Prove that f is constant.

Proof. For each n = 1, 2, . . ., we set 1
1
and Cn = C 0; . n n Then Theorem 7.4 (The Maximum Modulus Theorem) and Remark 7.2 reveal that there exists a sequence {zn } such that |zn | = n1 and Dn = D 0;

max |f (z)| = max |f (z)| = |f (zn )|. z∈Cn

z∈Dn

Clearly, this implies that |f (zn2 )| = max |f (z)| = max |f (z)| = max |f (z 2 )| ≥ max |f (z)| = |f (zn )|, z∈Cn2

z∈Dn2

z∈Dn

z∈Dn

but |zn2 | < |zn | and this contradicts Theorem 7.4 (The Maximum Modulus Theorem). Hence we complete the proof of the problem.  Problem 7.20
⋆ Given that R > 0 and z ∈ C. If f ∈ H D(z; R) and 0 < r < R, show that Z π Z r 1 f (z + teiθ )t dt dθ. f (z) = 2 πr −π 0

Proof. According to Theorem 7.3 (The Mean Value Property for Analytic Functions), we have Z π 1 f (z + teiθ ) dθ. f (z) = (7.7) 2π −π We notice that

Z

r

f (z)t dt =

0

r2 f (z). 2

By combining the formulas (7.7) and (7.8), we get Z rh Z π Z πZ r i 1 1 r2 iθ f (z) = f (z + te ) dθ t dt = f (z + teiθ )t dt dθ 2 2π 2π 0 −π −π 0 which gives the desired formula. This ends the analysis of the problem.

(7.8)



Problem 7.21 ⋆ Find all α ∈ C such that there exists an f ∈ H(U ) such that f for n = 2, 3, . . ..

1 n

=

1 n+α

(7.9)

138

Chapter 7. Properties of Analytic and Entire Functions

Proof. Since f ∈ H(U ), f is continuous at the origin. Thus the expression (7.9) implies f (0) = 0. Let zn = n1 . Furthermore, the expression (7.9) can be expressed as f (zn ) =

zn . 1 + αzn

By Theorem 7.1 (The Interior Uniquenss Theorem), f have the form f (z) =

z 1 + αz

in U . We require that 1 + αz 6= 0 because f ∈ H(U ). Now 1 + αz = 0 if and only if α = − 1z . Therefore, 1 + αz 6= 0 if and only if |α| ≤ 1. We conclude that α ∈ U and we complete the proof of the problem. 

7.4 Properties of Entire Functions Problem 7.22 ⋆ Let f and g be entire. If there exists an λ ∈ R such that Im f (z) ≤ λ × Im g(z) for all z ∈ C, prove that we have f (z) = Ag(z) + B for some constants A and B. Proof. Define h(z) = i[f (z) − λg(z)]. Then h is entire and Re h(z) = Im f (z) − λIm g(z) ≤ 0 for all z ∈ C. This implies that |eh(z) | = eRe h(z) ≤ 1

for all z ∈ C. Thus eh(z) is bounded entire. By Theorem 7.6 (Liouville’s Theorem), eh(z) and then h(z) are constants. The definition of h gives immediately that f (z) = Ag(z) + B for some constants A and B. We complete the proof of the problem.



Problem 7.23 ⋆ Let f be nonconstant entire. Suppose that there exist a, b ∈ C such that f (az + b) = f (z) for all z ∈ C. Prove that an = 1 for some n ∈ N. Proof. Without loss of generality, we assume that a 6= 1. It is clear that the equation az + b = z b b has the unique solution z = − a−1 . Define α = − a−1 and F (z) = f (z + α). Then F is also entire and we have
F (az) = f (az + α) = f (az + aα + b) = f a(z + α) + b = f (z + α) = F (z) (7.10)

7.4. Properties of Entire Functions

139

for every z ∈ C. By Theorem 6.17 (Taylor’s Theorem), we write F (z) =

∞ X

ck z k .

k=0

Then it follows from the equation (7.10) that ∞ X

ck ak z k =

k=0

∞ X

ck z k

k=0

for every z ∈ C. Now we conclude from Theorem 7.1 (The Interior Uniqueness Theorem) that ak ck = ck

(7.11)

for every k ≥ 0. Since f is nonconstant, F is also nonconstant and then cn 6= 0 for some nonnegative integer n. In other words, we have the desired result from the formula (7.11),  completing the proof of the problem. Problem 7.24 (Extended Liouville’s Theorem I) ⋆ If f is entire and for some nonnegative integer N , there exists a positive constant A such that |f (z)| ≤ A|z|N for all z ∈ C, then prove that f is a polynomial of degree at most N . Proof. Since f is entire, we may write f (z) =

∞ X

an z n

n=0

in the complex plane. By Corollary 6.17 (Taylor’s Theorem) and Theorem 6.5 (The M -L Formula), we know that 1 Z f (z) dz Ar N A |an | = ≤ r · = n−N n+1 n+1 2πi C(0;r) z r r

for r > 0 and for all n = 1, 2, . . .. If n = N + 1, N + 2, . . ., then we have n − N ≥ 1 so that r → ∞ implies that |an | = 0,

so f is a polynomial of degree less than or equal to N . We complete the analysis of the  problem. Problem 7.25 (Extended Liouville’s Theorem II) ⋆ Let f be entire and N be a nonnegative integer. Suppose that there exist constants A ≥ 0 and B > 0 such that |f (z)| ≤ A + B|z|N for all sufficiently large |z|. Prove that f is a polynomial of degree at most N .

140

Chapter 7. Properties of Analytic and Entire Functions

Proof. The case N = 0 is Theorem 7.6 (Liouville’s Theorem). Assume that the statement is true for some nonnegative integer N = k. For N = k + 1, there exist constants A ≥ 0 and B > 0 such that |f (z)| ≤ A + B|z|k+1 (7.12) for all |z| ≥ M , where M is a large positive number. We consider  f (z) − f (0)   , if z 6= 0;  z−0 F (z) =    f ′ (0), otherwise.

By Problem 7.1, this F is entire and the assumption step (7.12) implies that |F (z)| =

|f (z) − f (0)| |f (z)| + |f (0)| A + B|z|k+1 + A 2A 2A ≤ ≤ = + B|z|k ≤ + B|z|k |z| |z| |z| |z| M

for all |z| ≥ M . Therefore, F is a polynomial of degree at most k and consequently, f is a polynomial of degree at most k + 1. By induction, the statement is true for all nonnegative integers N . This completes the analysis of the problem.  Problem 7.26 ⋆

⋆ Let f be entire and f (z) → ∞ as z → ∞. Prove that f is a polynomial.

Proof. The hypothesis ensures that there exists a positive constant M such that |f (z)| > 1 in C \ D(0; M ). Assume that f had infinitely many distinct zeros in D(0; M ). Let this collection be {an }. Since {an } is bounded, the Bolzano-Weierstrass Theorem [39, Problem 5.25, pp. 68, 69] implies that it has a convergent subsequence in D(0; M ) ⊆ D(0; M + 1). Now we know from Theorem 7.1 (The Interior Uniqueness Theorem) that f ≡0 in D(0; M + 1), contradicting the fact that |f (z)| > 1 in C \ D(0; M ). Hence f has only finitely many zeros in D(0; M ), say a1 , a2 , . . . , ap . Define F (z) =

f (z) . (z − a1 )(z − a2 ) · · · (z − ap )

It follows from Remark 7.3 that F is entire and F (z) 6= 0 for every z ∈ C. Therefore, the function (z − a1 )(z − a2 ) · · · (z − ap ) 1 = (7.13) g(z) = F (z) f (z) 1 is also entire and g(z) 6= 0 for every z ∈ C. Since f (z) → ∞ as z → ∞, we have f (z) → 0 as z → ∞. Therefore, it follows from the expression (7.13) that there exist positive constants A and B such that |g(z)| ≤ A + B|z|p .

7.4. Properties of Entire Functions

141

By Problem 7.25 (Extended Liouville’s Theorem II), we know that g is a polynomial of degree at most p. Recall that g(z) 6= 0 in C, so Corollary 7.8 (The Fundamental Theorem of Algebra) asserts that g must be a nonzero constant. Let it be α. Then we have f (z) =

1 (z − a1 )(z − a2 ) · · · (z − ap ) α

This ends the proof of the problem.



Remark 7.6 (a) Of course, Problem 7.26 is a generalization of Corollary 7.7. (b) In addition, if P (z) = an z n + an−1 z n−1 + · · · + a0 with an 6= 0, then, for sufficiently large z, we have a0 |a0 | |an−1 | an−1 + · · · + n ≥ |z|n · |an | − − ··· − n |P (z)| = |z|n · an + z z |z| |z|

so that |P (z)| → ∞ as |z| → ∞. This and Problem 7.26 combines to imply that an entire function f is a polynomial if and only if f (z) → ∞ as z → ∞. Problem 7.27 ⋆ Let f be entire and |f (z 2 )| ≤ 2|f (z)| for all z ∈ C. Prove that f is constant. n

Proof. We claim that for every z ∈ C and n = 0, 1, 2, . . ., we have |f (z 2 )| ≤ 2n |f (z)|. When n = 0, it is in fact our hypothesis. Assume that it is also true for n = k, i.e., k

|f (z 2 )| ≤ 2k |f (z)| for all z ∈ C. If n = k + 1, then we have

f z 2k+1 = f (z 2k )2 ≤ 2|f (z 2k )| ≤ 2k+1 |f (z)|.

Our claim follows from induction. Let

M = max |f (z)| and C(0;2)

n

rn = 22

n

for n = 0, 1, 2, . . .. If |ωn | = rn , then we have ωn = zn2 for some zn ∈ C(0; 2) so that n

|f (ωn )| = |f (zn2 )| ≤ 2n |f (zn )| ≤ 2n M.

(7.14)

Fix p ∈ N and consider ωn ∈ C(0; rn ), we deduce from Problem 6.25 (Cauchy’s Estimate) and the estimate (7.14) that |f (p) (0)| ≤

p! · 2n M p! · M = p·2n −n . p rn 2

(7.15)

Since p is fixed and M is independent of n, if we take n → ∞, then the inequality (7.15) implies that f (p) (0) = 0. Finally, Corollary 6.17 (Taylor’s Theorem) yields that f is constant,  completing the proof of the problem.

142

Chapter 7. Properties of Analytic and Entire Functions

Problem 7.28 ⋆ Suppose that f is entire and [Re f (z)]2 ≥ [Im f (z)]2 for all z ∈ C. Prove that f is a constant.

Proof. Define F : C → C by F (z) = e−f

2 (z)

. It is clear that F is also entire and

|F (z)| = e[Im f (z)]

2 −[Re f (z)]2

≤1

for all z ∈ C. By Theorem 7.6 (Liouville’s Theorem), F is constant so that f is also constant. This completes the analysis of the problem.  Problem 7.29 ⋆ Let f be entire and satisfy |f (z)| ≤ for all z ∈ C. Prove that f ≡ 0.

1 |Re z|

(7.16)

Proof. Let R > 0 and define F (z) = (z 2 + R2 )f (z). Suppose that |z| = R and Im z ≥ 0. Let θ be the acute angle between the line passing through z and iR and the line Lθ passing z and perpendicular to the imaginary axis. In this case, we have 0 ≤ θ ≤ π4 and so z − iR √ = sec θ ≤ 2. (7.17) Re z See Figure 7.1 for an illustration.

Figure 7.1: The construction of the angles θ and φ.

7.4. Properties of Entire Functions

143

Similarly, suppose that |z| = R and Im z < 0. Let φ be the angle between the line passing through z and −iR and the line Lφ passing z and perpendicular to the imaginary axis. In this case, we have 0 ≤ φ < π4 and so z + iR √ = sec φ < 2. Re z

(7.18)

Combining the hypothesis, the estimates (7.17) and (7.18), we get (z − Ri)(z + Ri) √ |F (z)| = |(z − Ri)(z + Ri)f (z)| ≤ ≤ 2 2R. Re z

For any fixed z ∈ D(0; R), it deduces from Theorem 7.4 (The Maximum Modulus Theorem) that √ 2 R F (z) . (7.19) |f (z)| = ≤ 2 (z − Ri)(z + Ri) R − |z|2

By taking R → ∞ in the inequality (7.19), we conclude that f (z) = 0. Since this holds for all z in D(0; R) and R can be taken arbitrary large, we obtain the desired result and we complete  the analysis of the problem. Remark 7.7 Of course, the conclusion of Problem 7.29 remains true if the condition (7.16) is replaced by 1 for all z ∈ C. |f (z)| ≤ |Im z| Problem 7.30 ⋆ Let f be nonconstant and entire. Prove that f (C) is dense in C.

Proof. Assume that f (C) was not dense in C. Then there exist p ∈ C and ǫ > 0 such that D(p; ǫ) ∩ f (C) = ∅, i.e., |f (z) − p| ≥ ǫ > 0 for all z ∈ C. Then the function F (z) =

1 f (z) − p

is also entire and bounded by 1ǫ . Now Theorem 7.6 (Liouville’s Theorem) comes into play that F is constant. This implies that f is also constant, a contradiction to the hypothesis. Hence we  have completed the proof of the problem.

144

Chapter 7. Properties of Analytic and Entire Functions

CHAPTER

8

Further Properties of Analytic Functions

8.1 Fundamental Concepts In Chapter 7, we have studied many basic and important properties of functions analytic in a region. In this chapter, we continue to study deeper properties of analytic functions and we just state the properties without going into greater detail. For proofs of these theorems in this chapter, please read [4, Chap. 4], [5, Chap. 7], [9, Chap. VI], [11, Chap. VI and VII], [14, Chap. VIII, IX and X], [16, Chap. 5] and [35, §2.5]. Of course, we recall that Ω and U always denote a region and the unit disc D(0; 1) respectively.

8.1.1 The Open Mapping Theorem and Schwarz’s Lemma Theorem 8.1 (The Open Mapping Theorem). If f ∈ H(Ω) and f is nonconstant, then f (Ω) is open in C. In particular, if f ∈ H(Ω), then either f (Ω) is an open set or f (Ω) is a single point. With Theorem 8.1 (The Open Mapping Theorem), Theorem 4.9 can be shown easily. We say that f is an open map if f maps open sets to open sets. Thus Theorem 8.1 (The Open Mapping Theorem) can be rephrased as “a nonconstant analytic function f is an open map”. Theorem 8.2 (Schwarz’s Lemma). Let f ∈ H(U ), |f (z)| ≤ 1 in U and f (0) = 0. Then we have |f (z)| ≤ |z| and

|f ′ (0)| ≤ 1.

If either |f (z)| = |z| for some z 6= 0 or if |f ′ (0)| = 1, then f (z) = eiθ z for some real θ. We notice that both theorems above can be proved by the applications of Theorem 7.4 (The Maximum Modulus Theorem), see [5, pp. 93, 94].

8.1.2 The Schwarz Reflection Principle By the upper half plane Π+ we mean the set of all z = x + iy with y > 0. Similarly, the lower half plane Π− consists of those z with y < 0. Let Ω be an open set that is symmetric with respect to R, i.e., z ∈ Ω and z ∈ Ω. 145

146

Chapter 8. Further Properties of Analytic Functions

Denote Ω+ = Ω ∩ Π+ , Ω− = Ω ∩ Π− and ℓ = Ω ∩ R. See Figure 8.1 for an illustration. As an application of Theorem 7.2 (Morera’s Theorem) (see also [11, pp. 211, 212]) and [35, pp. 58 60]), we obtain

Figure 8.1: A simply connected region Ω. Theorem 8.3 (The Schwarz Reflection Principle). Suppose that f ∈ H(Ω+ ), f can be extended continuously to ℓ and f (x) ∈ R for all x ∈ ℓ. Then there exists an F ∈ H(Ω) such that F =f on Ω+ and this F satisfies the relation F (z) = f (z) = F (z) for every z ∈ Ω− . Theorem 8.3 (The Schwarz Reflection Principle) can be applied to other situations with respect to analytic arc. See the discussion in [5, pp. 102, 103], [14, pp. 284 - 286], [19, p. 96] and [31, Exercise 2, p. 293]. For convenience, we quote the Schwarz Reflection Principle for U from [40, Lemma 14.2, pp. 415, 416]: Theorem 8.4 (The Schwarz Reflection Principle for U ). Every eit ∈ L ⊆ C(0; 1) is the center of an open disc Dt such that Dt ∩ U lies in Ω. Denote Ω∗ to be the reflection of Ω, i.e., o n 1 Ω∗ = z ∈ C z ∗ = ∈ Ω . z Suppose that f ∈ H(Ω) and Im f (zn ) → 0 for every sequence {zn } in Ω which converges to a point of L. Then there exists a function F , analytic in the set Ω ∪ L ∪ Ω∗ , such that   f (z), if z ∈ Ω ∪ L;   F (z) = 1    f , if z ∈ Ω∗ . z

8.2. Applications of the Open Mapping Theorem

147

8.2 Applications of the Open Mapping Theorem Problem 8.1 ⋆ Let f ∈ H(Ω) and f 2 (z) = f (z) for all z ∈ Ω. Prove that f is constant on Ω. Proof. The hypothesis ensures that F = f 3 ∈ H(Ω). Thus for every z ∈ Ω, we have F (z) = f 2 (z) × f (z) = f (z) × f (z) = |f (z)|2 on Ω. Consequently, F (Ω) ⊆ R and Theorem 8.1 (The Open Mapping Theorem) asserts that F (Ω) must be a single point, i.e., f 3 (z) = F (z) = α for some α ∈ C and all z ∈ C. Denote ω = e

2πi 3

. Then we have 1

1

1

f (Ω) ⊆ {|α| 3 , ω|α| 3 , ω 2 |α| 3 }, so f (Ω) is a finite set. By Theorem 8.1 (The Open Mapping Theorem) again, f (Ω) must be infinite, a contradiction. This forces that f is a constant, as required. This completes the proof  of the problem. Problem 8.2 ⋆ Let Ω ⊆ C be an open set, f ∈ H(Ω) and f be nonconstant. Prove that f (Ω) is an open set.

Proof. Obviously, Ω is a union of open discs, i.e., [ Ω= D(z; δz ), z∈Ω

where δz > 0. Since each D(z; δz ) is
a region, we apply Theorem 8.1 (The Open Mapping Theorem) to conclude that f D(z; δz ) is open in C. Using [39, Problem 1.4, p. 5], we obtain f (Ω) =

[

z∈Ω


f D(z; δz )

which implies f (Ω) is open in C, completing the proof of the problem.



Remark 8.1 More can be said in Problem 8.2. In fact, if f is analytic in the region Ω, then it is continuous on Ω. By [23, Theorem 23.5, p. 150], the set f (Ω) is connected. Hence f (Ω) is in fact a region.

148

Chapter 8. Further Properties of Analytic Functions

Problem 8.3 ⋆ Suppose that Ω is an open set containing U , f ∈ H(Ω) and f is nonconstant. If |f (z)| = 1 on C(0; 1), prove that f has a zero in U .

Proof. Assume that f (z) 6= 0 for all z ∈ U . Then F = f1 ∈ H(Ω) and |F (z)| = 1 on C(0; 1). According to Theorem 7.4 (The Maximum Modulus Theorem), we know that |f (z)| ≤ 1 and |F (z)| ≤ 1 for all z ∈ U . Consequently, we have |f (z)| = 1 in U , i.e., f (U ) is a subset of C(0; 1). By Theorem 8.1 (The Open Mapping Theorem), f (U ) is a single point which contradicts our hypothesis that f is nonconstant. Hence there exists a z0 ∈ U such that f (z0 ) = 0, completing the proof of the problem.  Problem 8.4 ⋆ Suppose that f ∈ H(Ω) and there exist fixed m, n ∈ N such that f m (z) = [f (z)]n for every z ∈ Ω. Prove that f is constant. Proof. Since we have f m+n (z) = |f (z)|n in Ω, it means that the analytic function f m+n is real on Ω. Now Theorem 8.1 (The Open Mapping Theorem) states that f m+n is constant. Thus f is also constant, ending the proof of the problem.  Problem 8.5 ⋆ Prove Theorem 7.4 (The Maximum Modulus Theorem) by Theorem 8.1 (The Open Mapping Theorem).

Proof. Let f ∈ H(Ω) and f have a local maximum at α ∈ Ω. Assume that f was not constant.
Then there exists a δ > 0 such that f D(α, δ) is open in C by Theorem 8.1 (The Open Mapping Theorem). In other words, there exists a ρ > 0 such that
D(f (α); ρ) ⊆ f D(α, δ) . Set f (α) = reiθ0 for some real θ0 and r > 0. Pick a sufficiently small ǫ > 0 such that f (ω) = (r + ǫ)eiθ0 ∈ D(f (α); ρ). Then we have |f (ω)| = r + ǫ > |f (α)| so that α is not a maximum of |f (z)|. This completes the proof of the problem.  Problem 8.6 (The Inverse Function Theorem (Complex Form)) ⋆ Suppose that f ∈ H(Ω) and injective. Prove that f −1 : f (Ω) → C is analytic and (f −1 )′ (ω) = where ω = f (z).

1 , f ′ (z)

8.2. Applications of the Open Mapping Theorem

149

Proof. Recall that f −1 (ω) = {z ∈ Ω | f (z) = ω}. Since f is injective, f −1 is well-defined. For each open set V of C, the set (f −1 )−1 (V ) = f (V ) is open in f (Ω) by Theorem 8.1 (The Open Mapping Theorem). This implies that f −1 is continuous ([23, p. 102]) on f (Ω) and since z = f −1 (f (z)) for every z ∈ Ω, it follows from Theorem 4.10 that (f −1 )′ (ω) =

1 , f (z)

completing the proof of the problem.



Remark 8.2 In fact, the Inverse Function Theorem is a local property. See [14, p. 234] and [31, Theorem 10.30, pp. 215, 216].

Problem 8.7 ⋆ Let f ∈ H(U ) and f be one-to-one in U \ {0}. Prove that f is one-to-one in U .

Proof. Assume that f was not one-to-one in U . Then there exists an α ∈ U \ {0} such that f (α) = f (0) = ω. Let V and W be neighborhoods of 0 and α such that V ∩ W = ∅. By Theorem 8.1 (The Open Mapping Theorem), f (V ) and f (W ) are open sets containing ω. Thus their intersection is a nonempty open set. Pick ζ ∈ f (V ) ∩ f (W ) with ζ 6= ω. Then one can find a ∈ V and b ∈ W such that f (a) = f (b) = ζ. (8.1) Clearly, a 6= 0 and b = 6 0 because ζ 6= ω. Now the result (8.1) contradicts the fact that f is  one-to-one in U \ {0} and we complete the proof of the problem. Problem 8.8 ⋆ ⋆ ⋆ Let r > 0. Suppose that f is nonconstant and analytic in a neighborhood containing D(0; r). Suppose, further, that for some k ∈ N, we have k

f (z) = f (0) + ck z +

∞ X

cn z

n

and

n=k+1


for every z ∈ D(0; r). Denote E = f C(0; r) and

∞ X

n=k+1

|cn |r n−k < |ck |

δf (0) = inf{|f (0) − ζ| | ζ ∈ E}.

(a) Prove that δf (0) > 0.
(b) Prove that D(f (0); 2−1 · δf (0) ) ⊆ f D(0; r) .

(c) Use (b) to prove Theorem 8.1 (The Open Mapping Theorem).

(8.2)

(8.3)

150

Chapter 8. Further Properties of Analytic Functions

Proof. Basically, we follow Cater’s argument in [10] to show the results of this problem. We note that the equation (8.3) is the distance from the point f (0) to the set E and it can be regarded as a uniformly continuous function on E, see [30, Exercise 20, p. 101]. (a) If z ∈ C(0; r), then we get from the inequality in (8.2) that ∞ ∞ ∞ X X X n n c z ≤ |c | · |z| = |cn |r n < |ck z k | n n n=k+1

so that

n=k+1

n=k+1

∞ ∞ X X |f (z) − f (0)| = ck z k + cn z n ≥ |ck z k | − cn z n > 0. n=k+1

n=k+1

In other words, it means that f (0) ∈ / E. Since f is continuous on the compact set C(0; r), E is also compact. By [30, Exercise 21, p. 101], it must be true that δf (0) > 0. (b) Let

 δf (0)  . (8.4) ω ∈ D f (0); 2

It suffices to prove that ω ∈ f D(0; r) . Assume, on the contrary, that ω ∈ / f D(0; r) .
As f D(0; r) is compact, there exists a point p ∈ D(0; r) such thata  |ω − f (p)| = inf |ω − f (z)| z ∈ D(0; r) .

By the definitions (8.3) and (8.4), we see that ω is closer to f (0) than to any point in E, see Figure 8.2 below:

Figure 8.2: The construction of the angles θ and φ. 
Recall that δω = inf |ω−f (z)| z ∈ D(0; r) is a uniformly continuous function on the compact set f D(0; r) , so the existence of the point p is guaranteed by the Extreme Value Theorem [39, p. 100]. a

8.2. Applications of the Open Mapping Theorem

151

It follows that f (p) ∈ / E and thus p ∈ D(0; r). Using Corollary 6.17 (Taylor’s Theorem), we have ∞ X m an αn f (p + α) = f (p) + am α + n=m+1

for some α in a neighborhood of 0, where am 6= 0. Select a real a as small as we want such that ∞ X

n=m+1

0 < a < r − |p|,

(8.5)

|an |an−m < |am |,

(8.6)

|am |am < |ω − f (p)|.

(8.7)

Let θ1 and θ2 be such that am = |am |eiθ1

and

ω − f (p) = |ω − f (p)|eiθ2 .

Next, we define θ3 and α by mθ3 + θ1 = θ2

and

α = aeiθ3

(8.8)

respectively. Then it follows from the definitions (8.8) that am αm = |am |am ei(mθ3 +θ1 ) = |am |am eiθ2 . Therefore, both am αm and ω − f (p) lie on the same straight line and we deduce from the inequality (8.7) that |am αm | = |am |am < |ω − f (p)| and it implies |ω − f (p)| − |am αm | = |ω − f (p) − am αm |.

(8.9)

By the inequality (8.6), we obtain ∞ X an αn < |am αm |.

(8.10)

n=m+1

Combining the expression (8.9) and the estimate (8.10), we see immediately that ∞ X n m an α |ω − f (p + α)| = ω − f (p) − am α − n=m+1 ∞ X

≤ |ω − f (p) − am αm | +

n=m+1 ∞ X

= |ω − f (p)| − |am αm | + < |ω − f (p)|.

an αn

n=m+1

Finally, we observe from the inequality (8.5) that |p + α| ≤ |p| + |α| = |p| + a < r

an αn

152

Chapter 8. Further Properties of Analytic Functions and this means that p + α ∈
D(0; r). However, this contradicts the choice of p and we conclude that ω ∈ f D(0; r) . Since ω is arbitrary, we actually have 
δE  ⊆ f D(0; r) . D f (0); 2

(c) Let ζ ∈ Ω. By translation, we may assume that ζ = 0.b We choose r > 0 sufficiently small such that the inequality (8.2) and D(0; 2r) ⊆ Ω hold. Now part (b) guarantees that 
δE  ⊆ f D(0; r) ⊂ f (Ω). D f (0); 2

Consequently, f (0) is an interior point of f (Ω) which gives Theorem 8.1 (The Open Mapping Theorem). We complete the analysis of the problem.



Remark 8.3 Classically, proofs of Theorem 8.1 (The Open Mapping Theorem) based on ideas of Rouch´e’s Theorem ([4, pp. 359, 360] or [25, p. 344]), or the Minimum Modulus Theorem ([5, pp. 93, 94] or [29, pp. 256, 257]). The present proof of Problem 8.8 depends solely on power series and do not use integration of any kind. Similar proof can be found in [21, pp. 76 - 82] and read also Reem’s manuscript [28] for more references.

Problem 8.9 ⋆ ⋆ ⋆ Prove that Theorem 8.1 (The Open Mapping Theorem) and Corollary 7.8 (The Fundamental Theorem of Algebra) are equivalent in the case of polynomials.

Proof. The proof we present here mainly follows the argument used by Thompson in [36].c We assume that Corollary 7.8 (The Fundamental Theorem of Algebra) is true. Suppose that f (z) = z n + an−1 z n−1 + · · · + a0 , where a0 , a1 , . . . , an−1 are complex numbers. We want to prove that f is an open map. Assume that f was not open. Then there exists an open set V and a point p ∈ V such thatd f (p) ∈ / f (V )◦ .

(8.11)

Otherwise, we obtain f (V ) ⊆ f (V )◦ and this implies that f (V ) = f (V )◦ b

Openness is invariant under translation. As Thompson pointed out that with the α obtained in (8.17) below, an elementary proof of the Fundamental Theorem of Algebra can easily be given without assuming that f is an open map. Read [36, Theroem 2] for details. d Here E ◦ denotes the set of all interior points of E. This is always an open set. c

8.2. Applications of the Open Mapping Theorem

153

for any open set V . In other words, f is an open map, a contradiction. Since V is open, there exists an r ∈ (0, 1) such that D(p; r) ⊆ V so that
◦ (8.12) f D(p; r) ⊆ f (V )◦ . Now the relations (8.11) and (8.12) together imply that one can find a δ ∈ (0, r) such that
◦ f (p) ∈ / f D(p; δ) .
Therefore, we can pick an α ∈ / f D(p; δ) such that |f (p) − α|
0 such that |z| > R implies a0 1 n an−1 n + · · · + n > |z| > |a0 | > 0. (8.15) |f (z)| = |z| · 1 + z z 2

Denote K = D(0; R) and W = C \ K. Then we have the fact C=K ∪W

and

K ∩ W = ∅.

Clearly, the inequality (8.15) means that if z ∈ W , then |f (z)| > |a0 | so that the distance from 0 to the set f (W ) is at least |a0 |, i.e., inf |f (z)| > |a0 |.

z∈W

(8.16)

Since f (0) = a0 , we have a0 ∈ f (K). As f (C) = f (K) ∪ f (W ). Combining this fact and the inequality (8.16), we get inf |f (z)| = inf |f (z)|. z∈C

z∈K

Recall that f (K) is compact because f is continuous on the compact set K. Thus there exists an α ∈ K such that inf |f (z)| = |f (α)|. (8.17) z∈C

By the assumption, we know that f (α) 6= 0, so the result (8.17) implies that no points of f (C) lie on the line segment connecting 0 and f (α); otherwise, we get inf |f (z)| < |f (α)|

z∈C

154

Chapter 8. Further Properties of Analytic Functions
which contradicts the result (8.17). Therefore, D f (α); ǫ contains points not belong to f (C) for every ǫ > 0. However, this contradicts the fact that f (C) is open. Hence it forces that f (α) = 0. This completes the proof of the problem.



8.3 Applications of Schwarz’s Lemma Problem 8.10 ⋆ Suppose that f ∈ H(U ), f (U ) ⊆ U and f (0) = 0. Show that |f (z) + f (−z)| ≤ 2|z|2 for every z ∈ U . Proof. Our f satisfies the hypotheses of Theorem 8.2 (Schwarz’s Lemma), so we have |f (z)| ≤ |z| in U . Let f (z) = a1 z + a2 z 2 + · · · and F (z) =

f (z) + f (−z) = a2 z + a4 z 3 + · · · . 2z

By the definition, F ∈ H(U ) and F (0) = 0. Since |F (z)| ≤ 1, we have F (U ) ⊆ U . Therefore, Theorem 8.2 (Schwarz’s Lemma) implies that |F (z)| ≤ |z| and this is equivalent to |f (z) + f (−z)| ≤ 2|z|2

(8.18)

for every z ∈ U . This completes the proof of the problem.



Problem 8.11 ⋆

⋆ Let z 6= 0. Prove that the inequality in Problem 8.10 is strict unless f (z) = eiθ z 2

for some real θ.

Proof. If the equality in (8.18) holds at some point a ∈ U \ {0}, then we have |F (a)| = |a| so that Theorem 8.2 (Schwarz’s Lemma) ensures that F (z) = eiθ z for some real θ. This implies that ∞ X a2n z 2n = f (z) + f (−z) = 2zF (z) = 2eiθ z 2 . 2 n=1

If we compare the coefficients of both sides, then we get a2 = eiθ

and

a2n = 0

8.3. Applications of Schwarz’s Lemma

155

for every n ≥ 2. Consequently, f can be expressed as f (z) = eiθ z 2 +

∞ X

a2n+1 z 2n+1 = eiθ z 2 + g(z),

n=0

where g is odd in U . We claim that g ≡ 0 in U . To see this, we note that if z ∈ U , then we have |eiθ z 2 + g(z)|2 = |f (z)|2 ≤ 1.

(8.19)

|eiθ z 2 − g(z)|2 = |eiθ (−z)2 + g(−z)|2 = |f (−z)|2 ≤ 1

(8.20)

As g is odd, we have

for every z ∈ U . We apply Problem 1.19 to the sum of the two inequalities (8.19) and (8.20) to get 2 ≥ |eiθ z 2 + g(z)|2 + |eiθ z 2 − g(z)|2 = 2 |z|4 + |g(z)|2 in U . Thus it means that




|g(z)|2 ≤ 1 − |z|4 in U .



√ √ Given ǫ > 0. Denote D(ǫ) = D 0; 4 1 − ǫ and C(ǫ) = C 0; 4 1 − ǫ . Recall that g ∈ H(U ), so Theorem 7.4 (The Maximum Modulus Theorem) implies that max |g(z)|2 ≤ max |g(z)|2 ≤ max (1 − |z|4 ) = 1 − (1 − ǫ) = ǫ.

z∈D(ǫ)

z∈C(ǫ)

z∈C(ǫ)

In other words, we know that |g(z)| ≤

√

ǫ

for every z ∈ D(ǫ). Since ǫ is arbitrary, our claim g(z) = 0 for all z ∈ U is true so that our definition shows f (z) = eiθ z 2 in U , completing the proof of the problem.



Problem 8.12
⋆ Prove the following version of Schwarz’s Lemma: If f ∈ H D(0; R) , f (0) = 0 and |f (z)| ≤ M in D(0; R) for some M > 0 and R > 0, then we have |f ′ (0)| ≤

M R

and |f (z)| ≤

M |z| R

for every z ∈ D(0; R). If any one of the above equality holds, then f (z) = real θ.

M iθ Re

for some

156

Chapter 8. Further Properties of Analytic Functions

Proof. Since f (0) = 0, we may write f (z) = a1 z + a2 z 2 + · · · . Define F : D(0; R) → C by  f (z) − f (0)   , if z ∈ D(0; R) \ {0};  z−0 F (z) =    f ′ (0), if z = 0.


Similar as Problem 7.1, it is easily seen that F ∈ H D(0; R) . Let 0 < r < R. Then we have max |F (z)| ≤

z∈C(0;r)

M . r

It follows from Theorem 7.4 (The Maximum Modulus Theorem) that |F (z)| ≤ z ∈ D(0; r). In fact, this implies that M M = r→R r R

|F (z)| ≤ lim

M r

for every

(8.21)

for all z ∈ D(0; R). Equivalently, we obtain |f (z)| = |z| · |F (z)| ≤

M |z| R

in D(0; R). By the definition of F and the inequality (8.21), we see that |f ′ (0)| = F (0) ≤ This is exactly our first assertion. If |f ′ (0)| =

M R

or |f (ω)| =

M R |ω|

M R.

for some ω ∈ D(0; R) \ {0}, then we get

|F (0)| =

M R

or

|F (ω)| =

|f (ω)| M = . |ω| R

Thus |F | attains a local maximum in D(0; R) and it equals Modulus Theorem). Consequently, we obtain F (z) =

M R

by Theorem 7.4 (The Maximum

M iθ e R

in D(0; R) for some real θ and this gives the second assertion. Hence we have completed the  proof of the problem. Problem 8.13
⋆ Suppose that f ∈ H D(0; R) and |f (z)| ≤ M in D(0; R) for some M > 0 and R > 0. Prove that 2M |z| |f (z) − f (0)| ≤ R for every z ∈ D(0; R).
Proof. Set F (z) = f (z) − f (0). Then F ∈ H D(0; R) , F (0) = 0 and |F (z)| ≤ 2M in D(0; R). Using Problem 8.12, we get |f (z) − f (0)| = |F (z)| ≤

2M |z| R

for every z ∈ D(0; R) which ends the proof of the problem.



8.3. Applications of Schwarz’s Lemma

157

Problem 8.14 ⋆ Use Problem 8.13 to prove Theorem 7.6 (Liouville’s Theorem).

Proof. Let |f (z)| ≤ M for every z ∈ C. Fix r > 0 and choose ω ∈ C(0; r). Given ǫ > 0. Let R > 2rM ǫ . Then it follows from Problem 8.13 that |f (ω) − f (0)| ≤

2|ω|M 2rM = < ǫ. R R

As ǫ and ω are arbitrary, we have f (z) = f (0) on C(0; r). Now Theorem 7.1 (The Interior Uniqueness Theorem) guarantees that f is constant  in C. Hence we complete the analysis of the problem. Problem 8.15 (The Schwarz-Pick Lemma) ⋆

⋆ Let f ∈ H(U ) and |f (z)| < 1 for every z ∈ U . For any z, ω ∈ U , prove that f (z) − f (ω) z − ω ≤ and 1 − ωz 1 − f (ω)f (z)

|f ′ (z)| ≤

1 − |f (z)|2 . 1 − |z|2

Proof. Let α ∈ U and define ϕα : U → C bye ϕα (z) =

z−α . 1 − αz

(8.22)

Since the inverse of ϕ−α , it is straight forward to check that ϕα is injective. Since |α|−1 > 1, we know that ϕα ∈ H(U ). On |z| = 1, we see that |z|2 − αz − αz + |α|2 z−α z−α · = =1 1 − αz 1 − αz 1 − αz − αz + |α|2 · |z|2
so that ϕα (U ) ⊆ U and ϕα C(0; 1) ⊆ C(0; 1). Since the same is true for its inverse ϕ−α , we actually have
ϕα (U ) = U and ϕα C(0; 1) = C(0; 1). |ϕα (z)|2 =

Let ω ∈ U . The hypotheses imply that f (ω) ∈ U so that the functions Φ and Ψ defined by Φ(z) =

z−ω 1 − ωz

and

Ψ(ζ) =

ζ − f (ω)

1 − f (ω)ζ

satisfy Φ, Ψ ∈ H(U ) and Φ(U ) = Ψ(U ) = U . Next, we define F = Ψ ◦ f ◦ Φ−1 : U → C. It is clear that F ∈ H(U ) and


F (0) = Ψ f Φ−1 (0) = Ψ f (ω) = 0. e

This is an example of the so-called linear fractional transformations. See [31, Theorem 12.4, pp. 254, 255] for some basic properties of the mapping ϕα .

158

Chapter 8. Further Properties of Analytic Functions

Furthermore, the properties of ϕα ensure that |F (z)| ≤ 1 for every z ∈ U . Hence it follows from Theorem 8.2 (Schwarz’s Lemma) that |F (z)| ≤ |z|

−1 Ψ f Φ (z) ≤ |z| ζ − f Φ−1 (z)

≤ |z|. 1 − f Φ−1 (z) ζ

(8.23)

Replacing Φ−1 (ω) and ζ by z ∈ U and f (z) in the inequality (8.23) respectively, we see that f (z) − f (ω) ζ − f (ω) z−ω = ≤ |Φ(z)| = . 1 − ωz 1 − f (ω)f (z) 1 − f (ω)ζ

(8.24)

Finally, we follow from the inequality (8.24) that

1 f (z) − f (ω) 1 · ≤ . z−ω 1 − ωz 1 − f (ω)f (z)

(8.25)

Taking ω → z in the inequality (8.25), we conclude that

|f ′ (z)| 1 1 − |f (z)|2 ≤ |1 − |z|2 | .

Since |z| < 1 and |f (z)| < 1, we actually have

|f ′ (z)| 1 ≤ 1 − |f (z)|2 1 − |z|2 which gives the second inequality. This completes the proof of the problem.



Remark 8.4 The second inequality in Problem 8.15 (The Schwarz-Pick Lemma) answers the question: What is max |f ′ (α)| among all f ∈ H(U ) such that |f (z)| < 1 in U and f (α) = β? In fact, since 1 − |β|2 |f ′ (α)| ≤ , 1 − |α|2 we have

max |f ′ (α)| = at f (α) = 0.

1 1 − |α|2

Problem 8.16 ⋆ Suppose that f ∈ H(U ) is bounded. Prove that sup(1 − |z|2 )|f ′ (z)| ≤ sup |f (z)|. z∈U

z∈U

8.3. Applications of Schwarz’s Lemma

159

Proof. Since f is bounded, we write sup |f (z)| = M . If M = 0, then there is nothing to prove. z∈U

(z) Let M 6= 0. Define F (z) = fM . Then we have F ∈ H(U ) and |F (z)| ≤ 1, so we may apply Problem 8.15 (The Schwarz-Pick Lemma) to yield

1 − |F (z)|2 1 |f ′ (z)| = |F ′ (z)| ≤ ≤ . M 1 − |z|2 1 − |z|2

Hence for every z ∈ U , we obtain (1 − |z|2 )|f ′ (z)| ≤ M and then sup(1 − |z|2 )|f ′ (z)| ≤ sup |f (z)|.

z∈U

z∈U

We end the proof of the problem.



Problem 8.17 ⋆ Suppose that f ∈ H(U ), f (U ) ⊆ U and f (0) = f ′ (0) = · · · = f (n−1) (0) = 0 for some n ∈ N. Prove that |f (z)| ≤ |z|n for every z ∈ U . Proof. By Corollary 6.17 (Taylor’s Theorem), we see that f (z) =

f (n) (0) n f (n+1) (0) n+1 z + z + ··· . n! (n + 1)!

If we define F (z) = then F ∈ H(U ), F (z) =

f (z) zn

f (n) (0) f (n+1) (0) + z + ··· , n! (n + 1)!

for all z ∈ U \ {0} and F (0) =

f (n) (0) n! .

Fix z ∈ U and choose r such that |z| < r < 1 so that z ∈ D(0; r). It follows from Theorem 7.4 (The Maximum Modulus Theorem) that |F (ζ)| attains its maximum on D(0; r) at some point on the boundary C(0; r). Since |f (z)| ≤ 1 for all z ∈ U , we obtain |f (ζ)| 1 ≤ n. n r ζ∈C(0;r) |ζ|

|F (z)| ≤ max |F (ζ)| = max ζ∈C(0;r)

(8.26)

Notice that the left-hand side of the inequality (8.26) is independent of r, so we take r → 1 to get |F (z)| ≤ 1 for every z ∈ U . Hence we have

|f (z)| ≤ |z|n

(8.27)

for every z ∈ U \ {0}. However, the inequality (8.27) also holds for z = 0 because f (0) = 0.  Consequently, we have the conclusion which completes the proof of the problem. Problem 8.18 ⋆ Let f ∈ H(U ) and f (0) = a 6= 0. Prove that f does not have a zero in D(0; |a|).

160

Chapter 8. Further Properties of Analytic Functions

Proof. Assume that f (α) = 0 for some α ∈ D(0; |a|) which means |α| < |a|. Define F (z) =
f ϕ−α (z) , where ϕα is given by the formula (8.22). It is obvious that F ∈ H(U ), |F (z)| ≤ 1 in
U and F (0) = f ϕ−α (0) = 0, so we observe from Theorem 8.2 (Schwarz’s Lemma) that |F (z)| ≤ |z|

for every z ∈ U . Substituting z = −α into the inequality to obtain |a| = |f (0)| = |F (−α)| ≤ |α| which contradicts the assumption that |α| < |a|. Hence f does not have a zero in D(0; |a|) and we complete the proof of the problem.  Problem 8.19 ⋆ Let f ∈ H(U ), |f (z)| ≤ 1 in U and f ( 34 ) = 0. Prove that  7  20 ≤ . f 16 43

Proof. The first inequality in Problem 8.15 (The Schwarz-Pick Lemma) remains true if |f (z)| ≤ 1 7 and ω = 34 to get for every z ∈ U , so we apply it directly with z = 16  7  7 − 3 20 16 4 ≤ f 7 = 43 . 16 1 − 43 × 16

This ends the proof of the problem.



Problem 8.20 ⋆ Let f : U → U be analytic and f have more than one fixed point. Prove that f is the identity map.

Proof. Assume that f (α) = α and f (β) = β, where α 6= β. Consider the mapping φα : U → C given by α−z φα (z) = . 1 − αz Using similar argument as in the proof of Problem 8.15 (The Schwarz-Pick Lemma), we see that φα is bounded by 1 and bijective in U . It is clear that φα (0) = α. Define the mapping F = φ−1 α ◦ f ◦ φα . The properties of f and φα ensure that F ∈ H(U ) and |F (z)| < 1 in U . Next, direct computation gives F (0) = 0. Furthermore, we have
−1 −1 F φ−1 (8.28) α (β) = φα (f (β)) = φα (β).

−1 Since φ−1 α is injective and α 6= β, we have φα (β) 6= 0. By Theorem 8.2 (Schwarz’s Lemma), we obtain F (z) = eiθ z (8.29)

8.4. Applications of the Schwarz Reflection Principle

161

for some real θ. Combining the two results (8.28) and (8.29), we conclude that eiθ = 1 and F is the identity map. This implies that f is also the identity map which completes the  proof of the problem.

8.4 Applications of the Schwarz Reflection Principle Problem 8.21 ⋆ Let f be entire and f (R) ⊆ R as well as f (iR) ⊆ iR. Prove that f is odd. Proof. By Theorem 8.3 (The Schwarz Reflection Principle), since f (R) ⊆ R, we know that symmetric points with respect to R will be mapped to symmetric points (under f ) with respect to R. In other words, we have (8.30) f (z) = f (z) for every z ∈ C. Next, consider g(z) = if (iz). Then g is entire and g(R) ⊆ R. By Theorem 8.3 (The Schwarz Reflection Principle) again, we have g(z) = g(z) which gives if (iz) = if iz




f (iz) = −f iz




f (iz) = −f (−iz).

(8.31)

By replacing iz by z in the equation (8.31), we get f (z) = −f (−z)

(8.32)

for every z ∈ C. Combining the two equations (8.30) and (8.32), we conclude that f (z) = −f (−z) = −f (−z) = −f (−z) for every z ∈ C. This means that f is odd and we complete the proof of the problem. Problem 8.22 ⋆ Suppose that f ∈ H(Π+ ∪ R), f (R) ⊆ R and f is bounded. Prove that f is constant.



162

Chapter 8. Further Properties of Analytic Functions

Proof. According to Theorem 8.3 (The Schwarz Reflection Principle), f can be extended to an entire function F such that F (z) = f (z) on Π+ . Since f is bounded in Π+ ∪R, F is bounded in C. By Theorem 7.6 (Liouville’s Theorem),  F and then f are both constant which completes the proof of the problem. Problem 8.23 ⋆ Prove Theorem 8.4 (The Schwarz Reflection Theorem for U ).

Proof. Here we repeat the proof in [40, pp. 415, 416]. Define the function ψ : Π+ → U by ψ(z) =

z−i . z+i

(8.33)

Then it is easy to see that ψ is a bijection, ψ ∈ H(Π+ ) and ψ(R) ⊆ C(0; 1). Furthermore, its inverse is given by i(1 + ζ) . ψ −1 (ζ) = 1−ζ b = ψ −1 (L) and Ω b = ψ −1 (Ω) are a segment of R and a region in Π+ respectively. Now if Thus L b then we have z ∈ Ω, b so it follows from the definition (8.33) that b − = {z ∈ C | z ∈ Ω}, z∈Ω ψ(z) =

1

ψ(z)

.

(8.34)

b and z → L” b is equivalent to saying that “ψ(z) ∈ Ω and ψ(z) → L”. Next, we see that “z ∈ Ω b ⊆ Π+ → C satisfies This property implies that the function fb given by fb = f ◦ ψΩb : Ω
Im fb(z) = Im f ψ(z) → 0

b and z → L. According to [31, Theorem 11.14], there is a function Fb, analytic in as z ∈ Ω b b b − , such that Ω∪L∪Ω 
b ∪ L; b   f ψ(z) , if z ∈ Ω b (8.35) F (z) =   f ψ(z
), if z ∈ Ω b −. Using the formula (8.34), the function (8.35) can be expressed in the following form:  f (ζ), if ζ ∈ Ω ∪ L;    F (ζ) = 1    f , if ζ ∈ Ω∗ . ζ

This completes the proof of problem.



8.4. Applications of the Schwarz Reflection Principle

163

Problem 8.24
⋆ Suppose that f ∈ H(C \ {0}), f (z) = f ( 1z ) for every z ∈ C \ {0} and f C(0; 1) ⊆ R. Prove that f (x) ∈ R for all nonzero real x. Proof. By Theorem 8.4 (The Schwarz Reflection Principle for U ), we have f (z) = f

1 z

for every z ∈ C \ {0}. Take x ∈ R \ {0} and apply the hypothesis, we have f (x) = f

1 x

= f (x),

so f (x) ∈ R. This completes the proof of the problem.



164

Chapter 8. Further Properties of Analytic Functions

CHAPTER

9

Isolated Singularities of Analytic Functions

9.1 Fundamental Concepts In this chapter, we focus on the behavior of a function at points where it fails to be analytic. Thus the notion of isolated and non-isolated singularities are introduced as well as their characteristics are also given. Next, the concept of Laurent series is given and it provides a powerful way to determine the singularity properties of an analytic function f . Further references for this chapter are [1, §3.1 & 3.2], [4, Chap. 4], [5, Chap. 9], [8, Chap. 5], [16, Cchap. 4] and [25, Chap. 9].

9.1.1 Classification of Singularities Let z0 ∈ C. Denote D ′ (z0 ; r) to be the punctured disc {z ∈ C | 0 < |z − z0 | < r} = D(z0 ; r) \ {z0 }. Sometimes, the term deleted neighborhood is also used for D ′ (z0 ; r). There are two types of singularities: • Isolated singularity, • Non-isolated singularity. A point at which a single-valued function f is analytic is called a regular point of f . A point z0 is called an isolated singularity of f if f is analytic in D ′ (z0 ; r) for some r > 0. The singularity z0 is non-isolated if every neighborhood of z0 contains at least another singularity of f . For examples, the singularities of the following functions are all isolated.

• f1 (z) = • f2 (z) =

  z + 2, if z 6= 0; 

10,

if z = 0.

1 if z 6= 0. z2 165

166

Chapter 9. Isolated Singularities of Analytic Functions 1

• f3 (z) = e z if z 6= 0. By the definition of Log z, it is easy to see that every point of (−∞, 0] is a non-isolated singularity of it because every neighborhood of a point of (−∞, 0] contains another point on the negative real axis at which Log z is not analytic. As the title suggests, we only concentrated on the discussion of isolated singularities. In fact, the behavior of a function f near an isolated singularity z0 can be further classified as follows:
(i) Removable singularity: There exists an F ∈ H D(z0 ; r) for some r > 0 such that F (z) = f (z) on D ′ (z0 ; r).
(ii) Poles: There exist g, h ∈ H D(z0 ; r) for some r > 0 such that g(z0 ) 6= 0, h(z0 ) = 0 and f (z) =

g(z) h(z)

on D ′ (z0 ; r). The order of a pole of f at z0 is defined as the order of a zero of h at z0 . A pole of order one is called a simple pole, and a pole of order two is called a double pole. (iii) Essential singularity: Neither (i) nor (ii). In the case of a removable singularity, we can simply replace f by the corresponding analytic function F . A function f on an open set
Ω is meromorphic if there exists a subset A that has no limit points in Ω, f ∈ H Ω \ A and f has a pole at each point of A. At an essential singularity, the function oscillates and grows faster than any power z n and the behavior of f at such a point is sometimes not easy to figure out.

9.1.2 Singularities at Infinity In the extended complex plane, the infinity can be a singularity of a function f . To be precise, a function f has an isolated singularity at z = ∞ if f is analytic in a deleted neighborhood of ∞. By this, we mean that there is a positive constant R such that f is analytic in the
region 1 {z ∈ C | R < |z| < ∞}. Clearly, this is equivalent to saying that f (ζ) ∈ H D(0; R ) , where ζ = z1 . Consequently, ∞ is said to be a removable singularity, a pole or an essential singularity of f according as the same type of the singularity of the corresponding function F (ζ) = f

1 ζ

at the point ζ = 0.

9.1.3 Characteristics of Isolated Singularities Theorem 9.1 (Riemann’s Theorem on Removable Singularities). Let z0 be an isolated singularity of f . If we have lim (z − z0 )f (z) = 0, z→z0

then the singularity is removable.

9.1. Fundamental Concepts

167
Corollary 9.2. Let f ∈ H D ′ (z0 ; r) for some r > 0. Then f is bounded in D ′ (z0 ; r) if and only if z0 is a removable singularity of f . Theorem 9.3. If f is analytic in D ′ (z0 ; r) and if there exists a positive integer k such that lim (z − z0 )k f (z) 6= 0 and

lim (z − z0 )k+1 f (z) = 0,

z→z0

z→z0

then f has a pole of order k at z0 . Theorem 9.4 (The
Casorati-Weierstrass Theorem). If f has an essential singularity at z0 , then the set f D ′ (z0 ; r) is dense in C for every r > 0. Remark 9.1

Theorem 9.4 (The Casorati-Weierstrass Theorem) has a generalization, namely, the Picard Great Theorem: If an analytic function f has an essential singularity at a point z0 , then on any D ′ (z0 ; r), f (z) takes on every complex value, with at most one exception, infinitely often.

9.1.4 Existence and Uniqueness of the Laurent Series In Chapter 6, we know that any function analytic in a disc can be expanded there by a power series. If f has a removable singularity at z0 , then the definition ensures that f has a power series representation at z0 . However, it is not true for poles and essential singularities. Fortunately, a tool similar to power series can be developed when we study behavior near poles and essential singularities. This tool is called Laurent Series. ∞ X

We say that the infinite series

αn converges if both the series

n=−∞

αn and

n=0

converge in the usual sense. In this case, we can write ∞ X

∞ X

αn =

n=−∞

∞ X

αn +

−1 X

−1 X

αn

n=−∞

αn .

n=−∞

n=0

Now a Laurent series about the point z = z0 is a formal expression of the form ∞ X

n=−∞

an (z − z0 )n .

(9.1)

Its domain of convergence is determined by the following result: Theorem 9.5. Denote the series (9.1) by f (z). Let R1 =

1 lim sup |an |

1 n

and

1

R2 = lim sup |a−n | n . n→∞

n→∞

Then the series converges in the domain {z ∈ C | R1 < |z − z0 | and |z − z0 | < R2 }.

168

Chapter 9. Isolated Singularities of Analytic Functions

If R1 < R2 , then the series (9.1) converges uniformly in any compact subset of the annulus ann (z0 ; R1 ; R2 ) = {z ∈ C | R1 < |z − z0 | < R2 } and f is analytic in ann (z0 ; R1 ; R2 ). Particularly, we have f (z) = f1 (z) + f2 (z), where f1 (z) =

∞ X

n=0

an (z − z0 )n

and

f2 (z) =

−1 X

n=−∞

an (z − z0 )n

are analytic in D(z0 ; R2 ) and C \ D(z0 ; R1 ) respectively. For simplicity, when z0 = 0, we just write ann (R1 , R2 ) = ann (0; R1 ; R2 ). The above theorem says that any Laurent series (9.1) must be analytic in the annulus ann (z0 ; R1 ; R2 ). Conversely,
the existence and the uniqueness of the Laurent series of an f ∈ H ann (z0 ; R1 ; R2 ) are given by the following two theorems: Theorem 9.6 (The
Cauchy Integral Formula for an Annulus). Let 0 ≤ R1 < R2 ≤ +∞. If f ∈ H ann (z0 ; R1 ; R2 ) , then for each r1 , r2 with R1 < r1 < r2 < R2 and every z ∈ ann (z0 ; r1 ; r2 ), we have Z Z f (ζ) f (ζ) 1 1 dζ − dζ. f (z) = 2πi C(z0 ;r2 ) ζ − z 2πi C(z0 ;r1 ) ζ − z
Theorem 9.7 (Laurent Series). Suppose that 0 ≤ R1 < R2 ≤ +∞. If f ∈ H ann (z0 ; R1 , R2 ) , then f has the Laurent series representation (9.1), where 1 an = 2πi

Z

f (ζ) dζ (ζ − z0 )n+1

C(z0 ;R)

(9.2)

for any R1 < R < R2 . The Laurent series, like the power series for analytic functions, is unique because the coefficient (9.2) is uniquely determined. Here we call −1 X

n=−∞

an (z − z0 )n

and

∞ X

n=0

an (z − z0 )n

the principal part and analytic part of the Laurent series respectively. Since each term of the Laurent series (9.1) is analytic in the annulus ann (z0 ; R1 ; R2 ) and the series converges uniformly on any compact subsets of ann (z0 ; R1 ; R2 ), the whole series can be integrated or differentiated term by term in the annulus. That is ′

f (z) =

∞ X

n=−∞

n−1

nan (z − z0 )

and

Z

f (z) dz =

γ

where γ is any smooth curve contained in ann (z0 ; R1 ; R2 ).

∞ X

n=−∞

an

Z

γ

(z − z0 )n dz,

9.2. Problems on Removable Singularities

169

9.1.5 Singularities and the Laurent Series In Theorem 9.7 (Laurent Series), if R1 = 0, then z0 becomes an isolated singularity and this allows us to classify the nature of the singularity at z = z0 by examining its Laurent series (9.1). In fact, we have (i) Removable singularity: f has a removable singularity at z = z0 if and only if an = 0 for all n < 0. In other words, its principal part vanishes. (ii) Pole: f has a pole of order m ≥ 1 at z = z0 if and only if a−m 6= 0 and an = 0 for all n < −m. Equivalently, it means that the ‘tail’ after the term a−m (z − z0 )−m vanishes in the principal part. (iii) Essential singularity: f has an essential singularity at z = z0 if and only if an 6= 0 for infinitely many negative integers n, i.e., the ‘tail’ won’t terminate.

9.2 Problems on Removable Singularities Problem 9.1 ⋆ Let f be entire and

f (z) z

→ 0 as z → ∞. Prove that f is a constant,

Proof. Consider f1 (z) = f (z) − f (0) and f2 (z) =

f1 (z) z .

Then we have f1 (0) = 0. Since

f1 (z) = lim f1 (z) = 0, z→0 z→0 z→0 z we get from Theorem 9.1 (Riemann’s Theorem on Removable Singularity) that the origin is a removable singularity of f2 . Therefore, we may assume that f2 is actually entire. lim zf2 (z) = lim z ·

Next, the hypothesis implies that f (z) − f (0) = 0. z→∞ z→∞ z In other words, given ǫ > 0, there exists an R > 0 such that lim f2 (z) = lim

|f2 (z)| < ǫ

(9.3)

for all z with |z| > R. By Theorem 7.4 (The Maximum Modulus Theorem), we see that the inequality (9.3) holds in the complex plane. The arbitrariness of ǫ implies that f2 ≡ 0 so that f  is a constant. This completes the proof of the problem. Problem 9.2 ⋆ Show that there is no function f ∈ H(C \ {0}) such that

in C \ {0}.

1 |f (z)| ≥ p |z|

170

Chapter 9. Isolated Singularities of Analytic Functions

Proof. Assume that f was such a function. The hypotheses ensure that |f (z)| > 0 for all z ∈ C \ {0} so that F = f1 ∈ H(C \ {0}) and satisfies |F (z)| ≤

p

|z|

(9.4)

in C \ {0}. Obviously, the inequality (9.4) implies that F is bounded in D ′ (0; 1), so Corollary 9.2 implies that F has a removable singularity at 0 and we may assume that F is entire by the definition. Fix z0 ∈ D ′ (0; 1) and select R > |z0 |. Combining the inequality (9.4), Corollary 6.16 (The Cauchy Integral Formula for Derivatives) and Theorem 6.5 (The M -L Formula), we establish that √ √ 1 Z F (ζ) 1 R R R ′ |F (z0 )| = · dζ ≤ · 2πR = →0 2πi C(0;R) (ζ − z0 )2 2π (R − |z0 |)2 (R − |z0 |)2

as R → ∞. It means that F ′ ≡ 0 in D ′ (0; 1). Now we employ Theorem 7.1 (The Interior Uniqueness Theorem) to conclude that F′ ≡ 0

in C. Finally, we deduce from Theorem 4.8 that F and hence f is constant in C. However, this contradicts the fact that |f (z)| → ∞ as z → 0. Thus no such function exists and we complete  the proof of the problem. Problem 9.3 ⋆ ⋆ Suppose that 0 < r < R < ∞ and f : D ′ (0; 1) → ann (r; R) is analytic. Prove that f is not bijective.

Proof. For simplicity, we denote A = ann (r; R). Assume that f was bijective in D ′ (0; 1). It means that f is bounded in D ′ (0; 1), so we apply Corollary 9.2 to conclude that the origin is a removable singularity of f . We extend f to an F ∈ H(U ). As 0 is a limit point of D
′ (0; 1),
the continuity of F ensures that F (0) is a limit point of the set F D ′ (0; 1) = f D ′ (0; 1) = A. Thus we obtain F (U ) ⊆ A. Furthermore, since f is one-to-one, F is not constant. We see from Theorem 8.1 (The Open Mapping Theorem) that
◦ F (U ) ⊆ A = A.
Let p = F (0) ∈ A. Since f D ′ (0; 1) = A, there exists an z ∈ D ′ (0; 1) such that f (z) = p.

On the one hand, since z 6= 0, we can find neighborhoods V ⊆ U and W ⊆ D ′ (0; 1) of 0 and z such that V ∩ W = ∅. Using Theorem 8.1 (The Open Mapping Theorem) again, both F (V ) and F (W ) are open subsets of A so that F (V ) ∩ F (W ) is open in A.a On the other hand, since f is a bijection and a

Recall that, in general, f (V ∩ W ) ⊆ f (V ) ∩ f (W ) for any function f .

9.2. Problems on Removable Singularities

171

V \ {0} ⊆ D ′ (0; 1), we obtain

  F (V ) ∩ F (W ) = {F (0)} ∪ F (V \ {0}) ∩ F (W )   = {f (0)} ∪ f (V \ {0}) ∩ f (W )   = {p} ∪ f (V \ {0}) ∩ f (W )     = {p} ∩ f (W ) ∪ f (V \ {0}) ∩ f (W ) .

(9.5)

Since (V ∩ {0}) ∩ W = ∅ and f is a bijection, f (V \ {0}) ∩ f (W ) = ∅. Therefore, we deduce from the expression (9.5) that F (V ) ∩ F (W ) = {p} which contradicts the fact that F (V ) ∩ F (W ) is open in A. In conclusion, f is not bijective and we complete the analysis of the problem.  Problem 9.4 P (z) be rational, where P and Q are polynomials. Prove that f has a Q(z) removable singularity at ∞ if and only if deg P ≤ deg Q. ⋆ Let f (z) =

Proof. Obviously, f has a removable singularity at ∞ if and only if F (ζ) = f ( ζ1 ) has a removable singularity at ζ = 0. According to Corollary 9.2, there exist constants M > 0 and R > 0 such that |f ( 1ζ )| ≤ M for 0 < |ζ| < R. Since z = 1ζ , the last condition is equivalent to |f (z)| ≤ M

(9.6)

for |z| > R1 . Hence the estimate (9.6) and Remark 7.6 guarantee that deg P ≤ deg Q, ending the proof of the problem.  Problem 9.5 ⋆ Suppose that f has an isolated singularity at ∞. Prove that if ∞ is removable, then lim f ′ (z) = 0.

z→∞

Proof. Suppose that ∞ is a removable singularity of f . Then ζ = 0 is a removable singularity of F (ζ) = f ( ζ1 ), where ζ ∈ D(0; R) for some R > 0. By the definition, F (ζ) can be assumed to be analytic at ζ = 0. Thus, F ′ (ζ) is also analytic at ζ = 0 and then it is bounded at ζ = 0. Since F ′ (ζ) = − ζ12 · f ′ ( 1ζ ), we have 1 1 (9.7) f ′ (z) = − 2 F ′ z z for

1 R

< |z| < ∞. Taking z → ∞ in the expression (9.7), we obtain 1 ′ 1  =0 F z→∞ z 2 z

lim f ′ (z) = − lim

z→∞

because F ′ (z) is bounded when z → ∞. This completes the proof of the problem.



172

Chapter 9. Isolated Singularities of Analytic Functions

9.3 Problems on Poles Problem 9.6 ⋆ Let f have a pole of order n at z = z0 . Prove that f ′ has a pole of order n + 1 at z = z0 .
Proof. By the definition, there exist g, h ∈ H D(z0 ; r) for some r > 0 such that g(z0 ) 6= 0, h(z0 ) = 0 and g(z) (9.8) f (z) = h(z) on D ′ (z0 ; r). Since the order of zero of h at z = z0 is n, we write h(z) = (z − z0 )n h1 (z)


for some h1 ∈ H D(z0 ; r) and h1 (z0 ) 6= 0. Now we rewrite the formula (9.8) as f (z) = where g1 =

g h1

g1 (z) , (z − z0 )n

and g1 (z0 ) 6= 0. Then we have f ′ (z) =

1 [−ng1 (z) − (z − z0 )g1′ (z)]. (z − z0 )n+1

(9.9)

Since −ng1 (z0 ) 6= 0, the formula (9.9) tells usthe fact that f ′ has a pole of order n + 1 at z = z0 , completing the proof of the problem.  Remark 9.2 Similarly, it can be shown that if f has a pole of order n ≥ 1 at z = z0 , then for m ≥ 1, the function f m has a pole of order mn at z = z0 .

Problem 9.7 ⋆ Let f be meromorphic in the region Ω. Prove that poles of f are also poles of f ′ .

Proof. Let zk ∈ Ω be a pole of f . Using Problem 9.6, we know that f (z) =

gk (z) , (z − zk )pk

where gk ∈ H(D(zk ; rk )) for some rk > 0 with gk (zk ) 6= 0 and pk ∈ N. We follow from Problem 9.6 that f ′ has a pole (of order pk + 1) at z = zk . Since this is true for every pole of f , we obtain  the desired result and we end the proof of the problem.

9.3. Problems on Poles

173

Problem 9.8 ⋆ ⋆ Let z0 be an isolated singularity of f and f (z) → ∞ as z → z0 . Prove that f has a pole at z = z0 .

Proof. Given R > 0, there exists a δ > 0 such that f is analytic and |f (z)| > R in D ′ (z0 ; δ). Particularly, since f (z) 6= 0 in D ′ (z0 ; δ), the function g = f1 is analytic and bounded by R1 in D ′ (z0 ; δ). By Theorem 9.1 (Riemann’s Theorem on Removable Singularities), g has a removable
singularity at z = z0 so that we may assume that g ∈ H D(z0 ; δ) . Since f (z) → ∞ as z → z0 , we have 1 = 0. lim g(z) = lim z→z0 z→z0 f (z) Thus it follows from this and Corollary 6.17 (Taylor’s Theorem) that g(z) =

∞ X

n=1

an (z − z0 )n

in D ′ (z0 ; δ). Since g 6≡ 0 in D ′ (z0 ; δ), not all an ’s are zero. Let ap be the first nonzero coefficient in the Taylor’s series of g(z), i.e., g(z) =

∞ X

n=p

an (z − z0 )n

(9.10)

in D ′ (z0 ; δ). Now the series (9.10) implies that 1 = ap + ap+1 (z − z0 ) + · · · → ap 6= 0 (z − z0 )p f (z) as z → z0 . This means that lim (z − z0 )p f (z) =

z→z0

1 ap

lim (z − z0 )p+1 f (z) = 0.

and

z→z0

Hence we observe from Theorem 9.3 that f has a pole of order p at z = z0 . We have completed  the proof of the problem. Problem 9.9 ⋆ If f is entire with a pole of order k at ∞, prove that f is a polynomial of deg f = k.

Proof. Since f is entire, it has the series expansion f (z) =

∞ X

an z n

n=0

for all z ∈ C. We set

∞

1
X an = F (ζ) = f ζ ζn n=0

174

Chapter 9. Isolated Singularities of Analytic Functions

which has a pole of order k at ζ = 0 so that ak 6= 0 and an = 0 for all n > k. Consequently, f (z) = a0 + a1 z + · · · + ak z k is a polynomial of degree k. This ends the proof of the problem.



Problem 9.10 ⋆ Let f be a meromorphic function in C and have a simple pole at α. Prove that the function F defined by f ′′′ (z) 3 h f ′′ (z) i2 − F (z) = ′ f (z) 2 f ′ (z) is entire.

Proof. We express f in the form f (z) =

g(z) , z−α

where g is entire and g(α) 6= 0. Observe that f ′ (z) =

g1 (z) , (z − α)2

f ′′ (z) =

g2 (z) (z − α)3

and f ′′′ (z) =

g3 (z) , (z − α)4

where g1 (z) = (z −α)g′ (z)−g(z), g2 (z) = (z −α)g1′ (z)−2g1 (z) and g3 (z) = (z −α)g2′ (z)−3g2 (z). We note that all g1 (α), g2 (α) and g3 (α) are nonzero, so we obtain f ′′′ (z) 3 h f ′′ (z) i2 − f ′ (z) 2 f ′ (z) g3 (z) (z − α)2 3 h g2 (z) (z − α)2 i2 = · − · · (z − α)4 g1 (z) 2 (z − α)3 g1 (z) 2 3 g2 (z) g3 (z) − · = 2 (z − α) g1 (z) 2 (z − α)2 g12 (z)

F (z) =

=

2g3 (z)g1 (z) − 3g22 (z) 1 × . (z − α)2 2g12 (z)

(9.11)

Denote G(z) = 2g3 (z)g1 (z) − 3g22 (z). Direct computation gives G(α) = G′ (α) = 0. It means that we can express G(z) = (z−α)2 H(z) for some entire H. Put this into the expression (9.11) to get H(z) F (z) = 2 2g1 (z) which is clearly entire. This completes the proof of the problem.



9.3. Problems on Poles

175

Problem 9.11 ⋆ Suppose that f and g are meromorphic functions in C and |f (z)| ≤ |g(z)| for all z except their poles. Prove that there exists a constant α such that f (z) = αg(z) for all z except the poles of f and g.

Proof. There is nothing to prove if g ≡ 0. Thus we may assume that g 6≡ 0. Since | fg | ≤ 1 for all z except the poles of f and g, all its singularities must be removable by Corollary 9.2. Hence fg can be extended to a bounded entire function. Now Theorem 7.6 (Liouville’s Theorem) establishes that f (z) =α g(z) for some constant α. This implies the expected result and we complete the proof of the problem.  Problem 9.12 ⋆ Let f be meromorphic in C. Suppose that there exist R, A > 0 and nonnegative integer N such that |f (z)| ≤ A|z|N (9.12) for all |z| > R. Prove that f is rational.

Proof. The condition (9.12) implies that f does not have any pole in |z| > R. Let E be the set of poles of f inside D(0; R). If E is infinite, then the Bolzano-Weierstrass Theorem implies that E has a convergent sequence {zn }. Let zn → ω as n → ∞. Obviously, every zn is a zero of f1 inside D(0; R). If ω ∈ D(0; R), then Theorem 7.1 (The Interior Uniqueness Theorem) implies that f1 ≡ 0 in D(0; R) and also in C, but this contradicts the hypothesis that |f (z)| ≤ A|z|N for all |z| > R. Hence we have E = {α1 , α2 , . . . , αn } for some n ∈ N. Set

F (z) = (z − α1 )(z − α2 ) · · · (z − αn )f (z).

Then F must be entire and satisfy |F (z)| = |(z − α1 )(z − α2 ) · · · (z − αn )f (z)| ≤ A′ |z|N +n for sufficiently large enough z, where A′ is a positive constant. Using Problem 7.25, we see that F is a polynomial with deg F ≤ N + n. Rewrite f (z) =

F (z) (z − α1 )(z − α2 ) · · · (z − αn )

which shows the f is a rational function, ending the proof of the problem.



176

Chapter 9. Isolated Singularities of Analytic Functions

9.4 Problems on Essential Singularities Problem 9.13 ⋆ Let f have an essential singuarlity at z = z0 . Prove that, for every ω ∈ C, there exists a sequence {zn } converging to z0 such that lim f (zn ) = ω.

n→∞

Proof. By Theorem 9.4 (Casorati-Weierstrass Theorem), there exists a sequence {zn } such that |f (zn ) − ω|
0. Thus g(z) is bounded for |z| > R1 . By Theorem 7.4 (The Maximum Modulus Theorem), we see that g is entire bounded, so Theorem 7.6 (Liouville’s Theorem) implies that g is constant. By this, h is also constant, a contradiction. Next, if g(z) has an essential singularity at ∞, then G(ζ) has an essential singularity at ζ = 0. By Problem 9.13, there exists a sequence {ζn } such that ζn → 0 and G(ζn ) → 0 as n → ∞. Therefore, there exists a sequence {zn } such that zn → ∞ and g(zn ) → 0 as n → ∞. It yields that
lim h(zn ) = lim f g(zn ) = f (0). (9.13) n→∞

n→∞

However, since h is a nonconstant polynomial, we know from Remark 7.6 that lim h(z) = ∞

z→∞

which contradicts the limit (9.13). Hence g(z) has a pole at ∞ and then Remark 7.6 or Problem 9.9 says that g is a polynomial. Finally, given an unbounded sequence {ζn }. Since g is a polynomial, Theorem 6.4 (The Fundamental Theorem of Algebra) guarantees that g is surjective. Thus there exists an unbounded sequence {zn } such that ζn = g(zn ) → ∞ as n → ∞. Since h is also a polynomial, we obtain
f (ζn ) = f g(zn ) = h(zn ) → ∞ as n → ∞. Now Remark 7.6 or Problem 9.9 implies that f is also a polynomial. We end the  analysis of the problem.

9.4. Problems on Essential Singularities

177

Problem 9.15 ⋆

⋆ Let f be an injective entire function. Prove that f (z) = az + b

for some a, b ∈ C and a 6= 0.

Proof. Since f is injective, it is not constant. Let g(z) = f ( z1 ). Then g ∈ H(C \ {0}). If g has a removable singularity at 0, then g can be extended to an entire function G. Since G is bounded in D(0; R) for some R > 0, f is also bounded outside D(0; R1 ). We deduce from Theorem 7.4 (The Maximum Modulus Theorem) that f is also bounded in D(0; R1 ). Thus f is a bounded entire function and then f is constant by Theorem 7.6 (Liouville’s Theorem), but this contradicts the injectivity of f . If g has an essential
singularity at 0, then Theorem 9.4 (The Casorati-Weierstrass1
Theorem) ′ shows that g D (0; r) is dense in C for some r > 0. Thus the set f {z ∈ C | |z| >
r must be dense in C. Using Theorem 8.1 (The Open Mapping Theorem), the set f D(0; 1r ) is open in C. Consequently, we get   1  n 1 o f D 0; ∩ f z ∈ C |z| > 6= ∅, r r

but this again contradicts the injectivity of f .

By the analysis in the previous paragraphs, we see that 0 is actually a pole of g. Hence there are only finitely many terms in the principal part of the Laurent series of g. As the principal part (resp. analytic part) of g is the same as the analytic part (resp. principal part) of f , f is a polynomial. If deg f ≥ 2, then Theorem 6.4 (The Fundamental Theorem of Algebra) implies that f is not injective, a contradiction. Hence we must have deg f = 1 so that f (z) = az + b for some a, b ∈ C with a 6= 0. This completes the proof of the problem.



Problem 9.16 ⋆ Let f be entire and map every unbounded sequence to another unbounded sequence. Prove that f is a polynomial.

Proof. The hypothesis implies that f does not have a removable singularity at ∞. Assume that f had an essential singularity at ∞. By Problem 9.13, for every ω ∈ C, there exists an unbounded {zn } such that f (zn ) → ω as n → ∞, but this is a contradiction to our hypothesis. Hence f has a pole at ∞ and we conclude from Remark 7.6 or Problem 9.9 that it is a polynomial. We  complete the analysis of the problem.

178

Chapter 9. Isolated Singularities of Analytic Functions

9.5 Miscellaneous Problems Problem 9.17 ⋆ Prove that the coefficients (9.2) are uniquely determined. Proof. Suppose that f (z) =

∞ X

n=−∞

an (z − z0 )n

and f (z) =

∞ X

m=−∞

bm (z − z0 )m

both are valid in ann (z0 ; R1 ; R2 ). Then each series converges uniformly on any circle C(z0 ; R), where R1 < R < R2 . This fact yields Z Z ∞ ∞ X X an bm n+k (z − z0 ) dz = (z − z0 )m+k dz, 2πi 2πi C(z ;R) C(z ;R) 0 0 n=−∞ m=−∞ where k ∈ Z. Recall the basic fact that 1 2πi

Z

C(z0 ;R)

(z − z0 )n dz =

we obtain from the expression (9.14) that

(9.14)

  1, if n = −1; 

0, otherwise,

a−k−1 = b−k−1 for all k ∈ Z. In other words, we have an = bn for every n ∈ Z and this shows the uniqueness of  the Laurent series of f . We complete the proof of the problem. Problem 9.18 ⋆ ⋆ Let f have a pole of order n ≥ 1 at z0 . Prove that g = ef has an essential singularity at z0 .

Proof. We first show that z0 is not a pole of g. To this end, assume that g had a pole of order m ≥ 1 at z0 . By Problem 9.6, z0 is a pole of order n + 1 and m + 1 of f ′ and g′ respectively. Since g′ = f ′ · ef = f ′ g, g′ has a pole of order m + n + 1 at z0 . The equality m + n + 1 = m + 1 implies n = 0, a contradiction. Hence z0 is not a pole of g. To complete the proof, we need to show that z0 is not a removable singularity of g. To this end, write A(z) , f (z) = (z − z0 )n

9.5. Miscellaneous Problems

179

where A is analytic at z0 and A(z0 ) 6= 0. Let A(z0 ) = reiθ for some r > 0 and real θ. Set 1 iθ zk = z0 + e n k for k = 1, 2, . . .. Then it is easy to see that zk → z0 as and A(zk ) → A(z0 ) = reiθ as k → ∞. Combining this and the fact that f (zk ) = we get

A(zk ) = e−iθ kn A(zk ), (zk − z0 )n


n lim g(zk ) = lim ef (zk ) = lim exp e−iθ kn A(zk ) = lim er·k = ∞.

k→∞

k→∞

k→∞

k→∞

This result indicates that z0 is not a removable singularity of g. By the definition, it is an essential singularity of g, completing the proof of the problem.  Problem 9.19
⋆ Let f ∈ H D ′ (0; 1) and

|f (z)| ≤ log

1 |z|

in D ′ (0; 1). Prove that f has a removable singularity at 0.

Proof. Let the Laurent series of f in D ′ (0; 1) be f (z) =

∞ X

an z n .

n=−∞

By Theorem 9.7 (Laurent Series), we know that Z f (ζ) 1 dζ, an = 2πi C(0;r) ζ n+1

(9.15)

where 0 < r < 1. Thus the hypothesis and the formula (9.15) together imply that |an | ≤ If n < 0, then r → 0 gives

1 1 1 1 1 × 2πr × n+1 × log = n log . 2π r r r r a−1 = a−2 = · · · = 0.

In other words, 0 is a removable singularity of f , completing the proof of the problem.



Problem 9.20 ⋆ Consider the function f (z) =

1 . (z − 1)(z − 2)

Determine the Laurent series in the regions D ′ (1; 1) and D ′ (2; 1).

(9.16)

180

Chapter 9. Isolated Singularities of Analytic Functions
Proof. It is clear from the definition (9.16) that f ∈ H C \ {1, 2} . By Theorem 9.7 (Laurent Series), we can expand f in (different) Laurent series valid in D ′ (1; 1) and D ′ (2; 1) respectively. To find the explicit expansions, we apply the geometric series (5.4) that is valid in U : ∞

X 1 zn. = 1 − z n=0 In D ′ (1; 1), we have ∞ ∞ X X 1 1 n f (z) = − (z − 1)n . (z − 1) = − =− · (z − 1)[1 − (z − 1)] z−1 n=−1

n=0

Similarly, in D ′ (2; 1), we see that f (z) =

∞ X 1 (−1)n+1 (z − 2)n . = (z − 2)[1 + (z − 2)] n=−1

We end the proof of the problem.



Problem 9.21 ⋆ Repeat Problem 9.20 for the regions Ω1 = U , Ω2 = ann (0; 1; 2) and Ω3 = ann (0; 2; ∞). Determine the principal part and analytic part in each of the regions.

Proof. Observe that f (z) =

1 1 1 =− + . (z − 1)(z − 2) z−1 z−2

In Ω1 = U , we have |z| < 1 and | z2 | < 1 so that f (z) =

1 1 1 − · 1−z 2 1−

z 2

=

∞ X

n=0

zn −

∞ ∞  X X 1  n zn 1 − = z . 2n+1 n=0 2n+1 n=0

(9.17)

The expression (9.17) shows that f has no principal part in Ω1 and its analytic part there is given by ∞  X 1  1 − n+1 z n . 2 n=0

Next, in Ω2 = ann (0; 1; 2), we know that | z1 | < 1 and | 2z | < 1 so that 1 1 f (z) = − · z 1−

1 z

−

1 1 · 2 1−

z 2

=−

∞ X

n=0

1 z n+1

−

∞ ∞ ∞ X X X zn zn 1 = − − . n+1 n+1 2 2 zn

n=0

n=0

n=1

By the expression (9.18), the principal part and the analytic part of f in Ω2 are ∞ X 1 − n z n=1

respectively.

and

∞ X zn − 2n+1 n=0

(9.18)

9.5. Miscellaneous Problems

181

Finally, in Ω3 = ann (0; 2; ∞), the facts | 2z | < 1 and | z1 | < 1 imply that 1 1 1 1 f (z) = − · + · 1 z 1− z z 1 − 2z ∞ ∞ X X 2n 1 + =− z n+1 n=0 z n+1 n=0 =− =−

∞ X 2n − 1

n=0 ∞ X

n=1

z n+1

2n−1 − 1 . zn

(9.19)

In Ω3 , we see from the expression (9.19) that f has no analytic part there and its principal part is exactly ∞ X 2n−1 − 1 . − zn n=1

We end the analysis of the proof.



Problem 9.22 ⋆ Find the isolated singularities in the extended complex plane of each of the following functions and the corresponding Laurent series at its singularlities. (a) f (z) =

z6 (z − 1)2 1

(b) g(z) = ez− z . (c) h(z) =

cot z . z

Proof. (a) Since (z − 1)2 f (z) → 1 and (z − 1)3 f (z) → 0 as z → 1, it follows from Theorem 9.3 that f (z) z = 1 is a pole of order 2. Similarly, since fz(z) 4 → 1 and z 5 → 0, f has a pole of order 4 at ∞. By the Binomial Theorem, we have z 6 = [(z − 1) + 1]6

= (z − 1)6 + 6(z − 1)5 + 15(z − 1)4 + 20(z − 1)3 + 15(z − 1)2 + 6(z − 1) + 1,

so the Laurent series of f at z = 1 is given by f (z) = (z − 1)4 + 6(z − 1)3 + 15(z − 1)2 + 20(z − 1) + 15 + in D ′ (1; ∞).

1 6 + z − 1 (z − 1)2

182

Chapter 9. Isolated Singularities of Analytic Functions 1 |z|

For the singularity at ∞, if |z| > 1, then

< 1 and we have ∞

X 1 1 = z − 1 n=0 z n+1 so we apply Cauchy product to get X 1  X 1  X n+1 1 = · = (z − 1)2 z n+1 z n+1 z n+2 ∞

∞

∞

n=0

n=0

n=0

in |z| > 1. Thus the Laurent series of f at ∞ is given by 6

f (z) = z ·

∞ X n+1 n=0

z n+2

=

∞ X n+1

n=0

z n−4

4

3

2

= z + 2z + 3z + 4z +

∞ X n+5 n=0

zn

.

1

(b) The Laurent series of ez and e z are given by ∞ X zn

n=0

and

n!

∞ X 1 n!z n n=0 1

respectively. From these, we know that ez has an essential singularity at ∞ and e z has 1 a removable singularity at ∞. Similarly, ez has a removable singularity at 0 and e z has an essential singularity at 0. Thus the function g must have essential singularities at the 1 origin and ∞.b As g(z) = ez · e− z , Cauchy product gives g(z) =

∞ X zn  n=0

n!

×

∞ X (−1)n  n=0

n!z n

=

∞ ∞ hX X

n=−∞

k=0

(−1)k i n z . k!(n + k)!

(9.20)

Classically, the summation inside the brackets of the expression (9.20) is Bessel function of the first kind Jn (2). The general formula of this function is given by Jν (z) =

∞ X

n=0

 z ν+2n (−1)n , · n!Γ(ν + n + 1) 2

where Γ(z) is the Gamma function and ν any complex number, read Watson’s book [38, Chap. II & III] for details. Hence the expression (9.20) can be represented as ∞ X

g(z) =

Jn (2)z n .

n=−∞

(c) We write

cos z 1 · . z sin z Since z 2 h(z) = cos z · sinz z → 1 and z 3 h(z) = z cos z · sinz z → 0 as z → 0, h has a pole of order 2 at z = 0 by Theorem 9.3. Similarly, given n ∈ Z \ {0}, since h(z) =

lim (z − nπ)h(z) = lim

z→nπ b

z→nπ

Or we may use Problem 9.18 directly because z −

1 z

z − nπ 1 cos z · = n z (−1) sin(z − nπ) nπ have a pole at 0 and ∞.

9.5. Miscellaneous Problems

183

and

z − nπ cos z · = 0, n z→nπ z→nπ z (−1) sin(z − nπ) it follows from Theorem 9.3 again that h has simple poles at every nπ, where n ∈ Z \ {0}. As ∞ is the limit point of the sequence {. . . , −3π, −2π, −π, π, 2π, 3π, . . .}, ∞ is not an isolated singularity of h. lim (z − nπ)2 h(z) = lim (z − nπ)

Using the formula from [15, Eqn. (7), p. 42] directly, we have h(z) =

∞

X (−1)n B2n 22n 1 cot z = 2+ z 2n−2 z z (2n)! n=1

(9.21)

in D ′ (0; π). Here Bn denotes the Bernoulli’s number and it is defined by the coefficients of the following series ∞ X Bn n z = z ez − 1 n=0 n! for |z| < 2π. Hence the formula (9.21) is the Laurent series of h in D ′ (0; π).

Let k ∈ Z \ {0}. On the one hand, the fact cot(z − kπ) = cot z implies that cot z =

∞ X (−1)n B2n 22n n=0

(2n)!

(z − kπ)2n−1

(9.22)

holds in D ′ (kπ, π). On the other hand, we know that ∞ 1 1 1 X (−1)n = · (z − kπ)n = z−kπ n z kπ (kπ) kπ(1 + kπ ) n=0

(9.23)

holds in D ′ (kπ, π). We combine the expressions (9.22) and (9.23) to get cot z z ∞ ∞ i i hX 1 h X (−1)n (−1)n B2n 22n 1 2n n · (z − kπ) (z − kπ) × = kπ z − kπ n=0 (kπ)n (2n)! n=0

h(z) =

where

i i hX 1 1 h X (−1)n n n = × b (z − kπ) · (z − kπ) n kπ z − kπ (kπ)n ∞

∞

n=0

n=0

bm Hence we conclude that

 (−1)n B2n 22n   , if m = 2n;  (2n)! =    0, otherwise.

h(z) =

∞ X 1 1 cn (z − kπ)n , · · kπ z − kπ n=0

where

cn =

n X (−1)p p=0

We end the analysis of the problem.

(kπ)p

bn−p 

APPENDIX

A

Terminologies

A.1 Ahlfors’s Terminologies Definition A.1 (Arcs, Differentiable Arcs and Regular Arcs). An arc is a continuous function γ : [a, b] → C. The arc γ is said to be differentiable if γ ′ exists and is continuous on [a, b]. In addition, if γ ′ (t) 6= 0 for all t ∈ [a, b], then it is regular. An arc is said to be piecewise differentiable or piecewise regular if the same conditions hold except for a finite number of values t. Definition A.2 (Simple Arcs and Closed Curves). The function γ is called a simple arc or Jordan arc if γ(t1 ) = γ(t2 ) only for t1 = t2 . An arc is a closed curve if the end points coincide, i.e., γ(a) = γ(b). If the boundary of a set R is a closed curve, then we call it the boundary curve or contour of R and we denote it by ∂R. The concepts of chains and cycles are studied in [1, pp. 137, 138].

A.2 Asmar and Grafakos’s Terminologies Definition A.3 (Curves). A curve is a continuous function γ : [a, b] → C. Definition A.4 (Paths or Contours). A path or a contour is a curve γ : [a, b] → C if γ ′ is continuous on [a, b] or if there exist points a = a0 < a1 < · · · < an = b such that γ ′ is continuous on [ak , ak+1 ], where k = 0, 1, . . . , n − 1. A curve or a path γ is closed if γ(a) = γ(b).

A.3 Bak and Newman’s Terminologies 185

186

Appendix A. Terminologies

Definition A.5 (Curves, Piecewise Differentiable Curves and Smooth Curves). A curve is a function γ : [a, b] → C. It is called piecewise differentiable if γ is continuous on [a, b] and if there exist points a = a0 < a1 < · · · < an = b such that γ is continuously differentiable on [ak , ak+1 ], where k = 0, 1, . . . , n − 1. In addition, if γ ′ (t) 6= 0 for all t ∈ [a, b] except at a finite number of points, then γ is called smooth. A smooth curve is closed if γ(a) = γ(b). It is simple closed if no other points coincide, i.e., if γ(t1 ) = γ(t2 ) with t1 < t2 , then t1 = a and t2 = b. In the textbook [5, p. 46], Bak and Newmann assume all curves are smooth.

A.4 Conway’s Terminologies Definition A.6 (Paths, Smooth Paths and Piecewise Smooth Paths). A path is a continuous function γ : [a, b] → C. If γ ′ exists in [a, b] and is continuous on [a, b], then γ is called smooth. It is called piecewise smooth if there exist points a = a0 < a1 < · · · < an = b such that γ ′ is smooth on each [ak , ak+1 ], where k = 0, 1, . . .. We call γ a rectifiable path if it is a function of bounded variation, i.e., there is a constant M > 0 such that for any partition P = {a = t0 < t1 < · · · < tn = b} of [a, b], we have υ(γ; P ) =

n X k=1

|γ(tk ) − γ(tk−1 )| ≤ M.

A curve γ is an equivalence class of paths. It is called smooth (resp. piecewise smooth) if and only if at least one representative of the class is smooth (resp. piecewise smooth).

A.5 Gamelin’s Terminologies Definition A.7 (Paths, Smooth Paths and Piecewise Smooth Paths). A path is a continuous function γ : [a, b] → C. The path γ is called smooth if γ has as many derivatives as is necessary. It is called piecewise smooth if there exist points a = a0 < a1 < · · · < an = b such that the curve γ : [ak , ak+1 ] → C

is smooth, where k = 0, 1, . . . , n − 1.

A path is simple if γ(x) 6= γ(y) for x 6= y. It is closed if γ(a) = γ(b). A simple closed path is a closed path γ such that γ(x) 6= γ(y) for a ≤ x < y ≤ b. In [14, p. 71], Gamelin defines a curve as a smooth or piecewise smooth path.

A.6 Lang’s Terminologies Definition A.8 (Curves and Paths). A curve γ : [a, b] → C is a continuously differentiable function. A path γ : [a, b] → C is a sequence of curves γ1 , γ2 , . . . , γn such that the end point of γk is equal to the beginning point of γk+1 .

A.7. Rudin’s Terminologies

187

By a closed path, we mean a path γ such that γ(a) = γ(b).

A.7 Rudin’s Terminologies Definition A.9 (Curves). A curve is a continuous function γ : [a, b] → C. If γ(a) = γ(b), then γ is called a closed curve. Definition A.10 (Paths). A path is a continuous function γ : [a, b] → C such that there exist a = a0 < a1 < · · · < an = b and the curve γ : [ak , ak+1 ] → C has a continuous derivative, where k = 0, 1, . . . , n − 1. The left- and right-derivatives of γ at points a1 , a2 , . . . , an−1 may differ. A closed path is a closed curve which is also a path. The concepts of chains and cycles are studied in [31, §10.34, pp. 217, 218]. In [30, pp. 136, 137], a curve is an arc if γ is one-to-one and γ is said to be rectifiable if γ ′ is continuous on [a, b]. In this case, the length of γ, denoted by Λ(γ) is given by Z b |γ ′ (t)| dt. Λ(γ) = a

A.8 Stein and Sharkarchi’s Terminologies Definition A.11 (Curves and Smooth Curves). A curve is a function γ : [a, b] → C. It is called smooth if γ ′ exists and is continuous on [a, b] and γ ′ (t) 6= 0 for all t ∈ [a, b]. Definition A.12 (Piecewise-smooth Curves). The curve γ : [a, b] → C is called piecewisesmooth if γ is continuous on [a, b] and if there exist points a = a0 < a1 < · · · < an = b such that the curve γ : [ak , ak+1 ] → C is smooth, where k = 0, 1, . . . , n − 1. A smooth or piecewise-smooth curve is closed if γ(a) = γ(b). A smooth or piecewise-smooth curve is simple if it is not self-intersecting, i.e., γ(x) 6= γ(y) unless x = y. Stein and Sharkarchi call any piecewise-smooth curve a curve in [35].

188

Appendix A. Terminologies

Index

Symbols nth roots of ω, 37 nth roots of unity, 37

for an Annulus, 168 Cauchy product, 79 Cauchy’s Estimate, 114 Cauchy’s Inequality, 114 Cauchy’s Integral Formula, 101 in a Half Plane, 120 Cauchy’s Integral Theorem, 99 Cauchy’s Theorem, 99 for a Triangle, 99 in a Convex Region, 99 in a Disc, 99 in a Multiply Connected Region, 100 in a Simply Connected Region, 100 Cauchy’s type integral, 102 Cauchy-Hadamard Theorem, 78 Cauchy-Riemann equations, 52 centroid, 27 chordal distance, 36 Clairaut’s Theorem, 67 closed curve, 96 complex conjugate, 2 complex hyperbolic function, 55 complex number, 1 conformal mapping, 52 constant function, 53 curve, 95

A absolute value, 5 analytic arc, 146 analytic at z, 52 analytic part, 168 annulus, 100 arc, 185 differentiable, 185 Jordan, 185 regular, 185 simple, 185 Argand diagram, 3 argument, 5 B Bernoulli’s number, 183 Bessel function of the first kind, 182 Binomial Theorem, 3 boundary curve, 185 bounded variation, 186 branch, 55 of log z, 56 branch cut, 55 of log z, 57 branch point, 55

D De Moivre’s Theorem, 6 deleted neighborhood of z0 , 165 domain, 49 domain of definition of f , 49

C Cartesian form, 1 Casorati-Weierstrass Theorem, 167 Cauchy Criterion, 77 Cauchy Criterion for uniform convergence, 78 Cauchy Integral Formula

E Enestr¨om-Kakeya Theorem, 59 entire function, 52 equicontinuity, 78 189

190 essential singularity, 166 Euler’s Formula, 22, 54 exponential function, 54 Extended complex plane, 22 F Fundamental Theorem of Algebra, 127 Fundamental Theorem of Calculus, 98 G Gamma function, 182 geometric series, 79 Goursat’s Theorem, 99 Green’s Theorem, 97 H harmonic function, 67 holomorphic function, 52 I imaginary axis, 3 imaginary part, 1 imaginary part of f , 49 inner division point, 20 integers, 1 integral point, 30 integration by parts, 97 Interior Uniqueness Theorem, 125 Inverse Function Theorem (Complex Form), 148 inverse of a complex function, 53 isolated singularity, 165 L Lagrange’s Identity, 16 Laplace’s equation, 67 Laurent Series, 167 line integral, 96 linear fractional transformation, 157 linear function, 53 Liouville’s Theorem, 127 logarithm function, 55 lower half plane, 145 M Maclaurin series, 102 many-valued function, 53 Maximum Modulus Theorem, 126 Mean Value Property for Analytic Functions, 126

Index median point, 27 meromorphic function, 166 Midpoint Formula, 23 Minimum Modulus Theorem, 127 modulus, 5 Morera’s Theorem, 101, 126 multi-valued function, 53 multiply connected, 100 N natural numbers, 1 non-isolated singularity, 165 O one-point compactification, 23 open map, 145 Open Mapping Theorem, 145 order of pole, 166 orientation, 96 P parallelogram law, 4 parametrization, 95 Picard Great Theorem, 167 piecewise differentiable, 95 pointwise convergence, 77 pole, 166 double, 166 simple, 166 power function, 54 primitive, 98 principal argument, 5 principal part, 168 punctured disc, 100 purely imaginary, 1 purely real, 1 R radius of convergence of the power series, 78 rational numbers, 1 real axis, 3 real numbers, 1 real part, 1 real part of f , 49 reciprocal function, 54 rectifiable curve, 187 rectifiable path, 186 region, 49 regular function, 52

Index regular point, 165 removable singularity, 166 Riemann sphere, 23 Riemann surface, 55 Riemann’s Theorem on Removable singularities, 166 Rolle’s Theorem (Complex Form), 68 Rouch´e’s Theorem, 152 S Schur function, 132 Schwarz Reflection Principle, 145 Schwarz’s Lemma, 133, 145 Schwarz-Pick Lemma, 157 Section Formula, 19 simple closed curve, 96 simply connected, 99 single-valued function, 53 singular point, 52 smooth, 96 spherical distance, 36 Stereographic Projection, 22, 33

191 T Tauber’s Theorem for Power Series, 89 Taylor series, 102 Taylor’s Theorem, 102 The M -L Formula, 98 The Cauchy Integral Formula, 101 The Cauchy Integral Formula for Derivatives, 102 U uniform convergence, 77 upper half plane, 145 V value, 49 W Weierstrass M -test, 78 Z zero of order, 125

192

Index

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