The Elements of Complex Analysis [3 ed.] 9788122457117

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PUBLISHING GLOBALLY

NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS 7/30A, Daryaganj, New Delhi - 110002 Visit us at www.newagepublishers.com

To My Guides and Teachers: Late Professor L. S. Bosanquet Late Professor B. Kuttner Late Dr. P. Vermes

PREFACE TO THE THIRD EDITION

This book is meant as a textbook for a course in Complex Analysis at graduate level of universities in India and abroad. It is also intended to be useful for scientists and engineers. The second edition of this book was reviewed by learned mathematicians in many reputed international journals of mathematics and the response was very encouraging. In this third edition we have incorporated reviewer’s suggestions and comments. We welcome any further comments and suggestions from readers. Finally, I thank the publisher for his cooperation in bringing out the book. B. Choudhary

PREFACE TO THE SECOND EDITION

Complex Analysis was originally developed for the sake of its application. Keeping in view the applied nature of the subject I have included in this edition topics like some physical applications of conformal mapping, extension of the maximum modulus principle, some consequences of Jensen’s theorem and Hadmard’s gap theorem. The justifiable criticism was that the chapters on conformal mapping and analytic continuation were not given more attention than most other books. In this edition, both these chapters have been strengthened by new results and examples. Solutions to some more selected exercises which involve lot of new ideas and theoretical considerations have been provided at the end. I hope this will help the students to get actively involved in the subject. Finally, I thank the publisher for his cooperation in bringing out the book. Readers are urged to send their comments and suggestions so that I can improve the usefulness of the book. B. Choudhary

PREFACE TO THE FIRST EDITION

This book is an introductory course in Complex Analysis of one Complex Variable. It contains sufficient material for one semester course. The prerequisite for this book is a course in advanced calculus and a course in elementary modern algebra. The book consists of fourteen chapters and two appendices. Chapter 1 is very elementary and deals with the field of complex numbers. Assuming that the reader is familiar with metric spaces, continutity, convergence, compactness, connectedness, etc., we briefly introduce these concepts in Chapter 2. Chapter 3 to Chapter 10, includes the essentials of Complex Analysis, which cannot be left out in one semester course. Since the proof of the Riemann Mapping Theorem is somewhat more difficult than the study of the specific cases considered in Chapter 9, it has been presented in the Appendix. Chapters 12-14 deal with further analytic aspects of functions in many directions, which may lead to some other branches of mathematics. Chapter 11 initiates the reader in the consideration of functions as points in a metric space and presents Weierstrass Factorization Theorem and its application. The homological version of Cauchy’s Theorem is presented in the Appendix. Exercises are given at the end of each chapter. Solutions to some selected problems are special feature of the book. This book grew out of a regular course given by me at M.Sc. level of Indian Institute of Technology, Delhi. I have been influenced by many books on the same subject, especially by L. Ahlfors, H. Cartan, J.B. Conway and Serge Lang. I have also referred, in various places, to those books which have been of particular assistance to me in preparing lecture notes for this book. This book covers the syllabus on Complex Analysis at graduate level of the universities in India and undergraduate level of the universities in North America and Europe. I have resisted the temptation to include everything of the subject in this text, consequently there are many facets of the subject which have been omitted.

xii

PREFACE TO THE FIRST EDITION

I have chosen the topics with greatest care and have tried to present them systematically with diagrams and illustrations. In each chapter the level grows gradually and accumulations of lengthy theorems at one place have been avoided. My thanks are due to many people who have read drafts of the text. I am afraid that despite all my efforts, if some of the mistakes are still survived, I will be grateful to the readers who are kind enough to point them out. B. Choudhary

CONTENTS

Preface to the Third Edition Preface to the Second Edition Preface to the First Edition 1.

Sets, Functions and Complex Numbers 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9

2.

3.

19

Definition ......................................................................................... 19 Convergence, Completeness ............................................................ 23 Continuous Functions ...................................................................... 25 Compactness .................................................................................... 26 Connectedness .................................................................................. 29 Exercises .......................................................................................... 33

Elementary Properties of Analytic Functions 3.1 3.2 3.3

1

Sets ..................................................................................................... 1 Functions ............................................................................................ 3 Countable Sets ................................................................................... 5 Fields .................................................................................................. 7 Complex Numbers ............................................................................. 8 The Complex Plane .......................................................................... 10 The Roots of Complex Numbers ..................................................... 11 Stereographic Projection .................................................................. 14 Spherical Representation.................................................................. 14 Exercises .......................................................................................... 15

Metric Spaces 2.1 2.2 2.3 2.4 2.5

vii ix xi

36

Limits and Continuity ...................................................................... 36 Complex Differentiability ................................................................ 37 The Cauchy-Riemann Equations ..................................................... 40

xiv

CONTENTS

3.4 3.5 3.6 3.7 4.

Line Integral and Cauchy’s Theorem 4.1 4.2 4.3 4.4 4.5 4.6 4.7

5.

7.

79

Infinite Series in C ........................................................................... 79 Series of Functions and Uniform Convergence ............................... 82 Power Series ..................................................................................... 84 Exercises .......................................................................................... 88

Laurent Series, Singularities 7.1 7.2 7.3 7.4 7.5 7.6 7.7

66

The Winding Number ...................................................................... 66 Statement of Cauchy’s Theorem ...................................................... 69 Some Consequences of Cauchy’s Theorem ..................................... 70 Application of Cauchy’s Integral Formula....................................... 72 Exercises .......................................................................................... 76

Power Series 6.1 6.2 6.3

50

Definitions........................................................................................ 50 Riemann-Stieltjes Integral ................................................................ 51 Line-Integral .................................................................................... 52 Modulus of Continuity ..................................................................... 53 Local Primitive ................................................................................. 55 Cauchy’s Theorem (Homotopy Form) ............................................. 59 Global Primitives ............................................................................. 62 Exercises .......................................................................................... 63

Application of Cauchy’s Theorem 5.1 5.2 5.3 5.4

6.

Exponential Function ....................................................................... 42 Trigonometric and Hyperbolic Functions ........................................ 44 Logarithm ......................................................................................... 45 Inverse Trigonometric and Hyperbolic Functions ........................... 46 Exercises .......................................................................................... 47

90

Power Series Representation of Analytic Functions ........................ 90 Taylor Series .................................................................................... 91 Zeros of Analytic Function .............................................................. 92 Laurent Series .................................................................................. 93 Isolated Singularities ........................................................................ 96 Limit Points of Zeros and Poles ....................................................... 97 Meromorphic Functions ................................................................... 98 Exercises ........................................................................................ 100

xv

CONTENTS

8.

Residue Theorem and Its Applications 8.1 8.2 8.3

9.

Definition ....................................................................................... 102 Applications of the Residue Theorem ............................................ 104 The Logarithmic Residue ............................................................... 113 Exercises ........................................................................................ 116

Conformal Mapping 9.1 9.2 9.3 9.4 9.5 9.6 9.7

119

Definition ....................................................................................... 120 Linear Fractional Transformation .................................................. 122 Definition ....................................................................................... 125 Symmetry ....................................................................................... 126 The Schwarz-Christoffel Transformation ...................................... 130 The Transformations w = sin z and w = cos z ................................ 131 Riemann Surfaces .......................................................................... 132 Exercises ........................................................................................ 135

10. Harmonic Functions 10.1 10.2 10.3 10.4 10.5 10.6

102

137

Definitions of Harmonic Functions ............................................... 137 Harmonic Functions and Analytic Functions ................................. 138 Harmonic Functions on a Disk ...................................................... 140 Construction of Harmonic Functions on a Disk ............................ 143 Some Physical Applications of Conformal Mapping ..................... 147 Some Other Physical Interpretations.............................................. 151 Exercises ........................................................................................ 153

11. Weierstrass Factorization Theorem

157

Part I 11.1 11.2 11.3

Metric on C(G, Ω) .......................................................................... 157 Spaces of Analytic Functions ......................................................... 160 Weierstrass Factorization Theorem ................................................ 161 Exercises ........................................................................................ 170 Part II

Extension of the Maximum Modulus Principle 11.4 11.5 11.6 11.7 11.8

173

Hadamard’s Product Representation .............................................. 173 The Effect of Zeros, Jensen’s Formula .......................................... 174 Some Consequences of Jensen’s Theorem ..................................... 176 Phragmen-Lindelöf Theorem ......................................................... 179 The Gamma Function .................................................................... 182 Exercises ........................................................................................ 186

xvi

CONTENTS

12. Elliptic Functions 12.1 12.2 12.3 12.4 12.5 12.6

187

Groups ............................................................................................ 187 Elliptic Functions ........................................................................... 188 Weierstrass’ Elliptic Functions ....................................................... 191 The Addition Theorems ................................................................. 194 The Weierstrass’ Zeta Function ...................................................... 196 The Weierstrass’ Sigma Function ................................................... 198 Exercises ........................................................................................ 200

13. Analytic Continuation, Differential Equations

202

13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8

Analytic Continuation .................................................................... 202 Continuation along a Path .............................................................. 203 Continuation by Reflection ............................................................ 204 Nowhere-Continuable Power Series .............................................. 209 Differential Equations .................................................................... 213 Solutions at Infinity........................................................................ 216 The Hypergeometric Differential Equation ................................... 217 Some Simple Consequences of the Function F(a, b, c, z) in (33) ......................................................................... 218 13.9 Bessel’s Differential Equation ....................................................... 220 13.10 Legendre’s Differential Equation ................................................... 221 Exercises ........................................................................................ 222 14. Approximation by Rational Functions and Polynomials

225

14.1 Uniform Approximation................................................................. 225 14.2 Locally Analytic Functions ............................................................ 230 Exercises ........................................................................................ 235 APPENDIX 1: Riemann Mapping Theorem

237

APPENDIX 2: Homological Version of Cauchy’s Theorem

243

Solutions to Some Selected Exercises

249

Bibliography

293

List of Symbols

294

Index

297

1 1.1

SETS, FUNCTIONS AND COMPLEX NUMBERS

SETS

Let A be a given set. We write x ∈ A if x is an element of A. The negation of x ∈ A is written in the form x ∉ A. If each element of E is also an element of F, then we say that E is a subset of F, or that E is contained in F, or that F contains E and write E ⊂ F or F ⊃ E. In particular, E ⊂ E for every set E. The set which contains no element is called the empty set. We denote the empty set by φ. Note that the empty set is a subset of every set. Two sets E and F are said to be equal if they contain the same elements. In order to show that the sets E and F are equal we must show that E ⊂ F and F ⊂ E. If E ⊂ F and F ⊂ E, we write E = F. If E ⊂ F and E ≠ F, E is called a proper subset of F. Let P denotes a property for a collection of elements. We use the symbol {x : P} to denote the set of all elements x which have the property P. Let A and B be two sets. The union of two sets A and B is defined to be the set of all elements which belong either to A or to B or to both A and B. In symbols, A ∪ B = {x : x ∈ A or x ∈ B}. The intersection of two sets A and B is defined to be the set of all elements which belong to both A and B. In symbols, A ∩ B = {x : x ∈ A and x ∈ B}. Let A and B be two sets. If A ∩ B = φ, then we say that A and B are disjoint. We now list some of the algebraic properties of the operations on sets that we have just defined. The proofs of these assertions are left to the reader. 1

2

THE ELEMENTS OF COMPLEX ANALYSIS

A

B

A∪B

Fig. 1.I

A

B

A∩B

Fig. 1.II

Properties A ∪ A = A, A ∪ φ = A, A ∩ A = A, A ∩ φ = φ; A ∪ B = B ∪ A, A ∩ B = B ∩ A; (A ∪ B) ∪ C = A ∪ (B ∪ C), A ∩ (B ∩ C) = (A ∩ B) ∩ C; A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C); A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). These properties are called the idempotent, the commutative, the associative, and the distributive properties, respectively, of the operations of union and intersection of sets. Let A and B be two sets. The complement of B relative to A is the set of all elements of A which are not in B. In symbols, A – B = {x : x ∈ A, x ∉ B}. (a) (b) (c) (d)

A

B

A

–

B

Fig. 1.III

We now state the De Morgan’s laws for three sets. De Morgan’s Laws (a) A – (B ∪ C) = (A – B) ∩ (A – C); (b) A – (B ∩ C) = (A – B) ∪ (A – C). We verify the first of the two equalities. If x ∈ A – (B ∪ C), then x ∈A, x ∉ B and x ∉ C. Hence x ∈ A – B and x ∈ A – C. This shows that x ∈ (A – B) ∩ (A – C).

3

SETS, FUNCTIONS AND COMPLEX NUMBERS

Conversely, if x ∈ (A – B) ∩ (A – C), then x ∈ (A – B) and x ∈ (A – C). Thus x ∈ A, x ∉ B and x ∉ C. This shows that x ∈ A and x ∉ B ∪ C, i.e. x ∈ A – (B ∪ C). The verification of the second equalities is left to the reader. Let A and B be two non-empty sets. The Cartesian product A × B of the sets A and B is defined to be the set of all ordered pairs (x, y) where x ∈ A and y ∈ B. In symbols, A × B = {(x, y) : x ∈ A and y ∈ B}. Let A = {a, b, c, d} and B = {1, 2, 3}. Example : A × B = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3), (c, 1), (c, 2), (c, 3), (d, 1), (d, 2), (d, 3)}. Note that (a, 1) ∈ A × B, but (1, a) ∉ A × B. We shall use the following notations for designating intervals. If a and b are real numbers such that a < b, then we define [a, b] = {x : a ≤ x ≤ b}; (a, b) = {x : a < x < b}; [a, b) = {x : a ≤ x < b}; (a, b] = {x : a < x ≤ b}. 1.2

FUNCTIONS

Let A and B be arbitrary non-empty sets. A function from A to B is defined to be a set f of ordered pairs in A × B such that if (x, y) and (x, y′) belong to f, then y = y′. In other words, a function from A to B is some rule whereby to each element x ∈ A corresponds a uniquely determined element y ∈ B. The y which corresponds in this way to a given x is denoted by f (x), and is called the value of f. The set A is called the domain of definition of f and the set of values of f is called the range of f. The terms “function” or “mapping” are synonymous and we denote them by f : A → B, with domain A, and range contained in B. Let A and B be two sets and let f : A → B be a map. Suppose that D is a subset of A. The image f (D) is the subset of B defined by f (D) = {f (x) : x ∈ D}. f B

A D

f(D)

Fig. 1.IV. The image f(D)

4

THE ELEMENTS OF COMPLEX ANALYSIS

It is clear that f (A) ⊂ B. If f (A) = B, then we call f a mapping of A onto B. The mapping f of A into B is called one-to-one mapping if two different elements in A have different images under f. In other words, f is one-to-one mapping of A into B if whenever x1 ≠ x2, x1 ∈ A, x2 ∈ A implies that f (x1) ≠ f (x2). If f : A → B is both onto and one-to-one, then we can define the inverse mapping. f –1 : B → A in the following way : f –1 (y) = x if and only if f (x) = y for every y ∈ B and for every x ∈ A. Let f : A → B be a map. Suppose now that G is a subset of B. The inverse image f –1 (G) is the subset of A defined by f –1 (G) = {x : f (x) ∈ G}. f

A

B G

f –1(G) f –1

Fig. 1.V. The inverse image f –1 (G)

We list below certain properties of sets that are preserved under the direct image and inverse image, respectively. The reader should convince himself of the validity of these properties. Properties: Let f : A → B be a map. Suppose that D and E are subsets of A. (a) If D ⊂ E, then f (D) ⊂ f (E); (b) f (D ∩ E) ⊂ f (D) ∩ f (E); (c) f (D ∪ E) = f (D) ∪ f (E); (d) f (D \ E) ⊂ f (D). Properties: Let f : A → B be a map. Suppose that G and H are subsets of B. (a) If G ⊂ H, then f –1 (G) ⊂ f –1 (H); (b) f –1 (G ∩ H) = f –1 (G) ∩ f –1 (H); (c) f –1 (G ∪ H) = f –1 (G) ∪ f –1 (H); (d) f –1 (G – H) = f –1 (G) – f –1 (H). Observe that the inverse image is better behaved than the direct image. Let f : A → B and g : B → C be any two functions. The composition g o f is the function from A → C defined as follows: (x, z) ∈ g o f if and only if for some y, (x, y) ∈ f and (y, z) ∈ g.

5

SETS, FUNCTIONS AND COMPLEX NUMBERS A

B g

f x

C (g o f)x

f(x)

gof

Fig. 1.VI. Composition of functions

Note that if f and g are one-to-one, then g o f is also one-to-one and its inverse function is (g o f) –1 = f –1 o g– 1. Let f : A → B be a function and let E be a subset of A. We call the set of elements (x, f (x)) with x ∈ E the restriction of f to E and write f | E : E → B. Clearly, if f is one-to-one, so is f | E. 1.3

COUNTABLE SETS

Definition: Let A and B be any two sets. If the function f : A → B is one-to-one and onto, then we say that f is a one-to-one correspondence between A and B. Example: Let A denotes the set of positive odd integers and B the set of positive even integers. Define f : A → B by f (n) = n + 1. Then f is one-to-one correspondence between A and B. Whenever there exists a one-to-one correspondence from set A to set B, we say that A and B have the same cardinal number, or briefly, that A and B are equivalent, and write A ≈ B. This relation has the following properties and can be easily verified. (a) reflexive : A ≈ A for any set A. (b) symmetric : if A ≈ B, then B ≈ A. (c) transitive : if A ≈ B and B ≈ C, then A ≈ C. A set A is called finite if it is empty or if there is a natural number n such that A is equivalent to the set {1, 2, 3, ..., n}. A set A is called infinite if A is not finite. The most basic infinite set is the set of all positive integers: I = {1, 2, 3, .., n, n + 1, ...}. Definition: We say that a set A is countable if A ≈ I. Example: Let A be the set of all integers.

6

THE ELEMENTS OF COMPLEX ANALYSIS

Define f : I → A by

R| n f (n) = S n2− 1 |T− 2

(n even ); (n odd ).

The function is one-to-one correspondence from A to I. The sets A and I are arranged as follows: A: 0, 1, –1, 2, –2, ...

?

?

?

?

?

I: 1, 2, 3, 4, 5, ... We say that a set A is uncountable if A is neither finite nor countable. Definition: Let I be the set of all positive integers. A sequence in a set A is a function f : I → A. In other words, a sequence assigns to each n ∈ I, a uniquely determined element of A. If f (n) = xn, for n ∈ I, we usually denote the sequence f by the symbol {xn} or x1, x2, x3, ..., xn, ... . If xn ∈ A for all n ∈ I, then we say that xn is a sequence in A. We need to define unions and intersections of large class of sets. Let {Ai} be a collection of sets, where i runs through some index set I. We define their union and intersection as follows: ∪ Ai = {x : x ∈ Ai for at least one i ∈ I}

i ∈I

and

∩ A = {x : x ∈ A for every i ∈ I} i i

i ∈I

If I is the set of all positive integers, then their union and intersection are often written in the form ∞

∪ Ai

i =1

and

∞

∩ Ai .

i =1

Note that De Morgan’s laws can be generalized for an arbitrary collection of sets in the following way: (a) A − ∪ Ai = ∩ ( A − Ai ); i ∈I

i ∈I

(b) A − ∩ Ai = ∪ ( A − Ai ). i ∈I

i ∈I

We shall now establish two useful properties of countable sets. Theorem 1. Every infinite subset F of a countable set E is countable. Proof: Let the element x ∈ E be arranged in a sequence x1, x2, x3, ..., xn, ... of distinct elements. We construct a sequence n1, n2, n3, ... in the following way.

7

SETS, FUNCTIONS AND COMPLEX NUMBERS

Let n1 be the smallest positive integer such that xn1 ∈ F. Having chosen n1, let n2 be the smallest integer greater than n1 such that xn2 ∈ F. In this way, having chosen n1, n2, ..., nk – 1 (k = 2, 3, 4, ...), let nk be the smallest integer greater than nk –1 such that xnk ∈ F. f (k) = xnk (k = 1, 2, 3, ...).

Set

We thus see that there exists a one-to-one correspondence between F and the set of all positive integers. Theorem 2. The union of a countable collection of countable sets is countable. Proof: Let {Ai} be a sequence of countable sets, and let ∞

E = ∪ Ai . i =1

Then we have to prove that E is countable. Let {A1, A2, A3, ...} be arranged in a sequence {aij}, i, j = 1, 2, 3, ... . Let all the elements aij be written in the following order. Note that we have listed the elements of A1 in the first row, the elements of A2 in the second row, and so on. Write all these elements (as indicated by the arrows) in the form of a sequence a11, a12, a21, a13, a22, a31, ... a11

a12

a13

a14

a15

a21

a22

a23

a24

a25

a31

a32

a33

a34

a35

a41

a42

a43

a44

a45

a51

a52

a53

a54

a55

Observe that apq precedes amn if p + q < m + n, or, in case p + q = m + n, if p < m. Hence, it follows from Theorem 1 that E is countable. 1.4

FIELDS

In order to introduce the notion of a “field” we shall follow a convention that is familiar to the reader from elementary courses in modern algebra. By a binary operation in a set F, we mean a function

8

THE ELEMENTS OF COMPLEX ANALYSIS

h : F × F → F. In general, h(a, b) denotes the value of the binary operation h at the point (a, b) in F × F, but we shall use symbols such as a + b or a · b. Definition: A set F is called a field if there are two binary operations + and · such that the following properties are satisfied. If a, b, c belong to F, then (i) a + b = b + a; (ii) (a + b) + c = a + (b + c); (iii) a⋅b = b⋅a; (iv) (a⋅b)⋅c = a·(b⋅c); (v) a⋅(b + c) = (a⋅b) + (a⋅c) and (b + c)⋅a = (b⋅a) + (c⋅a); (vi) there exists a unique element θ in F such that θ + a = a and a + θ = a; (vii) for each element a in F there is an element a in F such that a + a = θ and a + a = θ. (viii) there exists a unique element e ≠ θ in F such that e⋅a = a, a⋅e = a; (ix) for each element a ≠ θ in F there is an element a′ in F such that a ⋅ a′ = e, and a′ ⋅ a = e. The element θ is called the zero element of F and the element e is called the identity or unit element of F. We assume that the reader is familiar with the “algebraic” structure of the real number system. Before we discuss the “algebraic” structure of the complex number system, we shall give two examples of fields. Examples: (i) Consider the system R of real numbers. Two of the binary operations in R are addition and multiplication. Note that the familiar operations subtraction and division are defined in terms of addition and multiplication respectively. Here θ = 0 and e = 1 are the additive and multiplicative identities of R. For every a ∈ R, there exists a = (– 1) a in R such that a + (– 1) a = 0 and (–1) a + a = 0. For every a ≠ 0 in R, there exists a number a′ = 1/a such that a′a = 1 and aa′ = 1. We conclude that the set R of real numbers forms a field. (ii) Let Q denotes the set of rational numbers; that is, real numbers of the form p/q where p and q are integers and q ≠ 0. In this case also, θ = 0 and e = 1. It is readily checked that the set Q of all rational numbers forms a field. 1.5

COMPLEX NUMBERS

Let R be the set of real numbers. A complex number z is defined as an ordered pair of real numbers: z = (a, b) where a ∈ R, b ∈ R.

9

SETS, FUNCTIONS AND COMPLEX NUMBERS

The real numbers a and b are called the real and the imaginary part of z, respectively. In symbols, a = Re z, b = Im z. Two complex numbers z1 = (a1, b1) and z2 = (a2, b2) are said to be equal if and only if a1 = a2 and b1 = b2. We denote by C the set of all complex numbers. The operations of addition and multiplication in C are defined as follows: z1 + z2 = (a1 + a2, b1 + b2); z1z2 = (a1a2 – b1b2, a1b2 + a2b1); λz = λ (a, b) = (λa, λb) where λ ∈ R. It can be easily checked that with these definitions C satisfies the associative, commutative and distributive laws for addition and multiplication. The complex numbers (0, 0) and (1, 0) are the additive and multiplicative identities of C. The operations of subtraction and division in C are defined as z1 – z2 = (a1 – a2, b1 – b2);

FG H

z1 a a + b1b2 a2 b1 − a1b2 = 1 22 , z2 a2 + b22 a22 + b22

IJ if K

z2 = (a2, b2) ≠ (0, 0). It is readily seen that the set C of all complex numbers with respect to the operations described above is a field. Note that for a, b ∈ R, the following properties hold. (i) (a, 0) + (b, 0) = (a + b, 0); (ii) (a, 0) (b, 0) = (ab, 0);

F I H K

( a, 0) a = , 0 if b ≠ 0. b ( b, 0) This shows that we can identify (a, 0) with a and hence may consider R as a subset of C. It is now desirable to show that the notation (a, b) is equivalent to the more customary a + ib. If we introduce i = (0, 1) and identify (a, 0) and (b, 0) with a and b, respectively, then a + ib = (a, 0) + (0, 1) (b, 0) = (a, 0) + (0, b) = (a, b). Note that i2 = (0, 1) (0, 1) = (– 1, 0) = – 1. It may be remarked here that the equation x2 + 1 = 0, which has no solution in R, becomes solvable in C. The roots of the equation are ± i. Before we conclude this section, it is important to mention that the field of complex numbers is not an ordered field like that of real numbers and rational numbers. We shall not discuss it here and readers interested in this topic may consult Birkhoff and Maclane [5], or any other good book on modern algebra. (iii)

10

1.6

THE ELEMENTS OF COMPLEX ANALYSIS

THE COMPLEX PLANE

It is customary to use the letter R to denote the set of all real numbers as well as the real line, and identify the real numbers with the points on the real line. For the geometric representation of the complex numbers, it is necessary to introduce first the coordinate plane. Take two mutually perpendicular axes, the x-axis and the y-axis, as shown in Fig. 1.VII. Let P be a point in the plane. Project P perpendicularly onto points Px and Py on the x and y axes, respectively. Let x and y be the coordinates of Px and Py on the x and y axes, respectively. Then it can be easily seen that there is a one-to-one correspondence between the ordered pairs (x, y) of real numbers and point P in the plane. The set of ordered pairs (x, y) with x ∈ R and y ∈ R is called the coordinate plane and is denoted by R × R, or R2. It follows from the definition of complex numbers that each z = x + iy in C can be identified with the unique point (x, y) in the plane R2. Thus with each complex number z = x + iy, we associate the point (x, y) in the plane, and vice-versa, each point of that plane corresponds to one and only one complex number. We call this plane the complex plane and denote by C itself. In the complex plane C, the x-axis of the coordinate system is called the real axis and the y-axis is called the imaginary axis. y-axis Py = y

P = (x, y)

Px = x

x-axis

Fig. 1.VII

Let z = x + iy be a complex number. We introduce the polar coordinate (r, θ): x = r cos θ, y = r sin θ. The positive quantity r is called the absolute value of z and is denoted by r = | z |. The angle θ is the angle between the positive real axis and the line segment from 0 to z. We illustrate this in Fig. 1.VIII. Y r X¢

q

O

Y¢

Fig. 1.VIII

x

P(x, y) y X

SETS, FUNCTIONS AND COMPLEX NUMBERS

11

The angle θ is called the argument of z and is denoted by θ = arg z. If θ is one of the value of arg z, then arg z = θ + 2kπ, k = 0, ± 1, ± 2, ... The value of θ which lies in the interval –π 0.

(iv) | z | = | z |.

Note also that | z1 – z2 | is exactly the distance between z1 and z2. The fundamental properties of the distance function are: If z, z1, z2 ∈ C, then (a) | z | ≥ 0, | z | = 0 if and only if z = 0 ; (b) | z1z2 | = | z1 | | z2 | ; (c) | z1 + z2 | ≤ | z1 | + | z2 | ; (d) | z1 – z2 | ≥ | z1 | – | z2 |. We prove only the statement (c), which is called the triangle inequality.

13

SETS, FUNCTIONS AND COMPLEX NUMBERS

Observe that for any z1 ∈ C, – | z1 | ≤ Re z1 ≤ | z1 | ; – | z1 | ≤ Im z1 ≤ | z1 |. Hence Re ( z1 z2 ) ≤ | z1 z 2 | = | z1 | | z2 |. Thus | z1 + z2 |2 = | z1 |2 + 2 Re ( z1 z2 ) + | z2 |2 ≤ | z1 |2 + 2 | z1 | | z2 | + | z2 |2 = (| z1 | + | z2 |)2, from which (c) follows. The triangle inequality can easily be generalized i.e. (e)

n

n

i =1

i =1

| ∑ zi | ≤ ∑ | zi |

The equality in (e) holds if and only if arg z1 = arg z2 = ... = arg zn. A complex number z = x + iy can be represented by the directed line segment OP joining the origin O to the point P (x, y). We often call OP the position vector of P. Let OP1 and OP2 represent the complex numbers z1 = x1 + iy1 and z2 = x2 + iy2, respectively. We complete the parallelogram OP1 PP2 whose side OP1 and OP2 correspond to z1 and z2. It is readily verified that addition of complex numbers corresponds, in the complex plane, to a parallelogram law for addition of vectors. This fact is illustrated in Fig. 1.IX. y 2) x 2, P 2(

P(x1 + x2, y1 + y2)

z2 z1 O

Fig. 1.IX

P1(x1, y1)

14

1.8

THE ELEMENTS OF COMPLEX ANALYSIS

STEREOGRAPHIC PROJECTION

Let C be the complex plane. Consider a unit sphere S which is centered at z = 0 of the complex plane C. Suppose that the sphere S is cut by the complex plane C along its equator, which coincides with the unit circle centred at z = 0, as indicated in Fig. 1.X. The diameter NN ′ is perpendicular to C. We call points N and N ′ the north and south poles of the sphere S. A point P in C is mapped onto the point P′ on the sphere S by joining N to P by a line. The line NP intersects the sphere S at P′. Points inside the unit circle are mapped onto the lower hemisphere, and points outside the unit circle are mapped onto the upper hemisphere. The origin z = 0, is mapped onto the south pole N ′ of the sphere S. We now put the north pole N in correspondence with the “point at infinity” of the complex plane C. N P′

O P

N′

Fig. 1.X

The method described above for mapping the plane onto the sphere is called the Stereographic Projection. The set of all points of the complex plane including the point at infinity is called the extended complex plane. The sphere is often called the Riemann Sphere. This is the manner in which a one-to-one correspondence is established between the sphere and the extended complex plane. 1.9

SPHERICAL REPRESENTATION

We denote the extended plane by C∞ = C ∪ {∞}. Let P = x + iy and let P′ = (x1, x2, x3) be the corresponding point on S. The line in R3 through P and N is given by (5) [{(1 – u) x, (1 – u) y, u} : – ∞ < u < ∞] In order to find the coordinates of P′ we have to find the value of u at which this line intersects S. If u is the value, then 1 = (1 – u)2 x2 + (1 – u)2 y2 + u2 = (1 – u)2 | P |2 + u2. It follows that 1 – u2 = (1 – u)2 | P |2

15

SETS, FUNCTIONS AND COMPLEX NUMBERS

Since u ≠ 1 (P ≠ ∞) we get | P |2 − 1 | P |2 + 1 | P |2 − 1 2x 2y , x2 = , x3 = Thus x1 = 2 2 | P |2 + 1 | P| +1 | P| +1 We can write the above expression as u=

| P |2 − 1 P−P P+P , , = = x x 2 3 | P |2 + 1 | P |2 + 1 | P |2 + 1 Now we have to find P when the point P′ (P′ ≠ N) is given. Setting u = x3 and using (5) we get x + ix2 P= 1 1 − x3 Let P, Q be in C∞. Define the distance from P to Q, d (P, Q) to be the distance between the corresponding points P′, Q′ in R3. If P′ = (x1, x2, x3) and Q′ = (x′1 , x′2 , x′3), then (7) d (P, Q) = {(x1 – x′1)2 + (x2 – x′2)2 + (x3 – x′3)2}1/2 Since P′ and Q′ are on S, it follows from (7) that [d (P, Q)]2 = 2 – 2 {x1 x′1 + x2 x′2 + x3 x′3}. Using (6) we find that

(6)

x1 =

d (P, Q) =

2| P −Q| , (P, Q ∈ C) {(1 + | P |2 ) (1 + | Q |2 )}1/ 2

Proceeding in a similar way we can obtain 2 d (P, ∞) = , (P ∈ C) (1 + | P |2 )1/ 2 Thus, the correspondence between points of S and C∞ gives the spherical representation of the extended plane.

EXERCISES 1. The symmetric difference between two sets A and B, denoted by A Δ B, is defined by AΔB = (A – B) ∪ (B – A). (i) Show that Δ satisfies associative law and commutative law. (ii) A ∩ (BΔC) = (A ∩ B) Δ (A ∩ C). (iii) A Δ φ = A, A ΔA = φ. 2. Let f : A → B ; let S and T be subsets of A, and D and E be subsets of B, Prove the following; (i) f (f –1 (D)) ⊂ D, and give an example in which f (f –1 (D)) ≠ D.

16

THE ELEMENTS OF COMPLEX ANALYSIS

(ii) f (S) – f (T) ⊂ f (S – T), and give an example in which f (S) – f (T) ≠ f (S – T). (iii) f

–1

(D) – f

–1

(E) = f –1 (D – E).

3. Let X be any set. The mapping f defined on X by f (x) = x for every x ∈ X, is called the identity mapping on X. We will denote this mapping by IX . Note that such a mapping keeps every element of X fixed. Let f : X → Y be one-to-one and onto, and let f –1 be its inverse. Prove that f o f –1 = IY . Explain how does IY differ from IX . 4. Prove that there exists a one-to-one correspondence between the set of positive integers and the set of all positive rational numbers. 5. Prove that the set of all real numbers between zero and one is not countable. 6. Prove that the set of all rational numbers is countable. 7. Prove that the set of all points whose coordinates are both rational is countable. 8. If A and B are countable, then prove that A × B is countable. 9. Express the following complex numbers in the form a + ib where a, b are real numbers (i)

1 − 1 + 2i

(iii) (1 + πi) (i + π)

(ii) i (2 + 3i)4 (iv)

1+ i 1− i − 1− i 1+ i

(1 + i ) 3 (2 + i ) (1 + 2i ) 10. Express the following complex numbers in polar form

(v)

(i) 1 + i 3 (iii) – 4i

(ii) – 2 3 − 2i (iv) – 1 – i

(v) 1 − i 2

(vi) 2 2 + i 2 2

(vii)

2i

(viii) – 4

3 i 2 (3 + 4i) (12 − 5i) 11. If z = , compute 1+ i | z |, Re z and Im z.

(ix)

3 /2 −

12. Compute all the roots of the equation z3 = 1. If ω is one of the complex cube roots of unity show that the other is ω2. Verify that 1 + ω + ω2 = 0. 13. Compute all roots and plot them in the complex plane: (i) (– 1)1/4 (ii) (– 1 + i)1/3 (iii) i2/3 (iv) (– 16i)1/4

17

SETS, FUNCTIONS AND COMPLEX NUMBERS

(v) (– 27i)1/6

(vi) (1 – i)–1/2 (viii) (– 1 – i)4/5

(vii) 81/3

(ix) ( 4 2 + i 4 2 )1/ 2 14. If a + ib is a root of a0zn + a1zn – 1 + ... + an = 0 where an ≠ 0, a1, a2, ..., an, a and b are real, prove that a – ib is also a root. 15. If x1 + iy1 is one complex root of the equation z5 = 1 show that 4x1 ( y14 − x14 ) = 1. 16. If z is a complex number such that z z = 1, compute | 1 + z |2 + | 1 – z |2. 17. If w =

z1 z2 , show that

z1 + z 2 z + z2 +w|+| 1 – w |. 2 2 18. If z1, z2 are complex numbers, show that

| z1 | + | z2 | = |

(a) | z1 – z2 |2 + | z1 + z2|2 = 2 | z1 |2 + 2 | z2 |2 ; (b) | | z1 | – | z2 | | ≤ | z1 – z2 |. 19. Show that three points z1, z2, z3 in the complex plane are collinear iff there exist real numbers a, b, c not all zero, such that a + b + c = 0 and az1 + bz2 + cz3 = 0. 20. Show that triangles whose vertices are the points z1, z2, z3 and z4, z5, z6, respectively, are similar iff z1 z2 z3

z4 1 z5 1 = 0. z6 1

21. Show that if the points z1, z2, z3 are the vertices of an equilateral triangle, then z12 + z 22 + z32 = z1z2 + z2z3 + z3z1. Is the converse true ?

22. If z1, z2, z3, z4 are the position vectors of the vertices for quadrilateral ABCD, show that ABCD is a parallelogram iff z1 – z2 + z3 – z4 = 0. 23. Show that the locus of z defined by the equation | z – w1 | = | z – w2 | is the perpendicular bisector of the line AB, where A and B are the points in the complex plane which represent the complex numbers w1 and w2 respectively. Deduce that 2 z − w1 − w2 arg = ± π/2. w1 − w2 24. If z1, z2, z3 and z4 are four distinct points of the complex plane, show that they lie on a circle or a straight line iff ( z3 − z1 ) ( z 4 − z2 ) ( z3 − z2 ) ( z 4 − z1 ) is real.

18

THE ELEMENTS OF COMPLEX ANALYSIS

25. Describe geometrically the set of points z satisfying the following conditions. (a) | z – i | = 2 (b) | z + 2 i | + | z – 2 i | = 6 (c) | z – 3 | – | z + 3 | = 4 (d) z ( z + 2) = 3 (e) | z + 3 i | ≤ 1 (f) Re z ≥ 0 (g) Im z ≥ 0 (h) Re z + Im z = 0 (i) Im z2 = 4 (j) | Re z | + | Im z | = 1. 26. Determine the locus of the points z which satisfy arg

FG z − a IJ = π/2. H z − bK

2 2.1

METRIC SPACES

DEFINITION

A metric space is a pair (X, d) where X is a set and d is a mapping from X × X into R which satisfies the following conditions: (i) d (x, y) ≥ 0; (ii) d (x, y) = 0 iff x = y; (iii) d (x, y) = d ( y, x); (iv) d (x, z) ≤ d (x, y) + d (y, z) for x, y, z ∈ X. Examples: (a) Let X = R or C and define d by d(z1, z2) = | z1 – z2 |. n (b) Let X = R and define d by

L O d (x, y) = | x – y | = M ∑ ( x − y ) P N Q n

i =1

i

i

2

1/ 2

,

where x = (x1, x2 , ..., xn) and y = (y1, y2, ..., yn) ; xi, yi ∈ R, i = 1, 2, ..., n. (c) Let X be a non-empty set and define d by d (x, y) =

RS0 if x = y; T1 if x ≠ y.

This metric space is called a discrete metric space. Open Ball: Let (X, d) be a metric space. If a ∈ X and r > 0, then the set {x : x ∈ X, d (x, a) < r}, denoted by Br (a), is called the open ball with centre a and radius r. The open ball Br (a) on R is the bounded open interval (a – r, a + r) with mid-point a and total length 2r. The open ball Br (a) on C is the set {z ∈ C : | z – a | < r}. 19

20

THE ELEMENTS OF COMPLEX ANALYSIS

Figure 2.I illustrates an open ball in the complex plane C. The open ball on C is also called the open disc. Y r a |z a| < r O

X

Fig. 2.I

Open Sets: Let (X, d) be a metric space. A set G ⊂ X is open if for each x ∈ G there is an r > 0 such that Br (x) ⊂ G. Examples: (a) The set G = {z ∈ C : a < Re z < b} is open. (b) The set S = {z ∈ C : Re z < 0} ∪ {0} is not open. Note that the empty set φ and the full space X are open sets. Observe that in any metric space (X, d), each open ball is an open set. Closed Sets: Let (X, d) be a metric space. The set G ⊂ X is said to be closed if the complement X – G is open. We need the following characterization of open sets in terms of open balls. Lemma 2.1 Let (X, d) be a metric space. A subset E of X is open if and only if it is a union of open balls. Proof: Suppose that E is the union of a collection F of open balls. If F is empty, then E is also empty and hence it is open. Suppose that F is non-empty. E is also non-empty. Let x ∈ E. Since E is the union of the open balls in F, x ∈ Br (x0) which is contained in F. Since each open ball is an open set, there exist an open ball Br (x) ⊂ Br (x0). Hence, 1

Br1 (x) ⊂ E, and so, E is open. To prove the converse, assume that E is open. Since E is open, each of its points is the centre of an open ball contained in E. Hence E is the union of all the open balls contained in E. Theorem 2.1. Let (X, d) be a metric space. Then (i) the sets X and φ are open; (ii) the union of any number of open sets in X is open; (iii) the intersection of a finite number of open sets in X is open.

21

METRIC SPACES

Proof: (i) Since every open ball centred on each of its points is contained in X, hence X is open. φ is clearly open, since the requirement of the definition of open set is automatically satisfied. To prove (ii), let {Hi : i ∈ I}, where I is any index set, be a class of open sets in X. Let H = ∪ Hi . If {Hi} is empty, then H is empty and hence H is open. i ∈I

Assume that {Hi} is non-empty. Since each Hi is a union, of open balls, H is the union of all the open balls. Hence, by Lemma 2.1, H is open. To prove (iii), let {Gi : i = 1, 2, ..., n} be a finite number of open sets in X. Let G = ∩ Gi . If {Gi} is empty, then G becomes the whole space X and hence G is i ∈I

open. Assume that G is non-empty. Let x ∈ G. Since x is in each Gi, and each Gi is open. Thus for each i there is a positive real number ri such that Bri (x) ⊂ Gi. Set r = min {r1, r2, ..., rn} and note that n

Br (x) ⊂ ∩ Gi . Hence i =1

n

∩ Gi is open.

i =1

Theorem 2.2. Let (X, d) be a metric space. Then (i) the sets X and φ are closed; (ii) the intersection of any number of closed sets in X is closed; (iii) the union of a finite number of closed sets in X is closed. Proof: (i) The result follows from Theorem 2.1(i) and the fact that Xc = φ, φc = X, where “c” stands for the complement. To prove (ii), let {Fi : i ∈ I} be a class of closed sets. Then by De Morgan’s law and Theorem 2.1 we see that

F ∩ FI H K i ∈I

Thus

i

c

= ∪ Fi c is open. i ∈I

∩ Fi is closed.

i ∈I

The proof of (iii) is similar to that of (ii). Note that a set which is not closed is not necessarily open, and vice-versa. Interior: Let (X, d) be a metric space and S a subset of X. A point x ∈ S is called an interior point of S if there exist an open ball Br (x) such that Br (x) ⊂ S. The interior of S, denoted by S 0, is the set of all its interior points. Observe that S 0 ⊂ S. It is easy to verify that S is open iff S = S 0. Closure: Let (X, d) be a metric space and S a subset of X. A point x ∈ X is called a closure point of S if every open ball centred on x contains at least one point of S. In other words, a point x ∈ X is a closure point of S if

22

THE ELEMENTS OF COMPLEX ANALYSIS

Br (x) ∩ S ≠ φ for all r > 0. The closure of S, denoted by S , is the set of all its closure points. Observe that S ⊂ S . Closed Ball: Let (X, d) be a metric space. Let a ∈ X and let r > 0. Then the set {x ∈ X : d (x, a) ≤ r} is called the closed ball with centre a and radius r. It can be easily proved that in a metric space (X, d), each closed ball is a closed set. For, let d(x, a) > r, and let r1 = d(x, a) – r > 0. If d (y, x) < r1, then d (a, y) ≥ d(a, x) – d(y, x) > d(a, x) – r1 = d(a, x) – [d(a, x) – r] = r. This shows the S c is open, and thus, S is closed. In general, it is not true that in every metric space the closure of every open ball of radius r is the closed ball of radius r. Here is an example. Let (X, d) be a discrete metric space (i.e., d(x, y) = 1 if x ≠ y, d (x, x) = 0). Then B1 (a) = {x ∈ X : d (x, a) < 1} = {a}; {x ∈ X : d (x, a) ≤ 1} = X; B1 ( a) = {a}. Limit Point: Let (X, d) be a metric space and S a subset of X. A point x ∈ X is called a limit point (an accumulation point) of S if each open ball Br(x) contains at least one point of S different from x. In other words, a point x ∈ X is a limit point of S if Br (x) ∩ (S – {x}) ≠ φ for each r > 0. It is clear that every limit point of a set must be a closure point of that set. The set of all limit points of S is called the derived set of S, and is denoted by S′.

Note that S = S ∪ S′. Also, note that S is closed iff it contains all its limit points.

RS T

UV W

1 1 Examples: (a) Let X = R and S = 1, , , ... . 2 3 0 is a–the only limit point of S. (b) Let X = C and S = [0, 1) ∪ {i}. The set of all points of [0, 1] is a limit point of S but i is not. Boundary: Let (X, d) be a metric space and S a subset of X. A point x ∈ X is called a boundary point of S if every open ball Br (x) intersects both S and Sc. In other words, a point x ∈ X is a boundary point of S if Br (x) ∩ S ≠ φ and Br (x) ∩ Sc ≠ φ for all r > 0. The boundary of S, denoted by δS, is the set of all its boundary points.

23

METRIC SPACES

Note that δS = δS c. We list below some useful properties which can be easily verified. Properties: Let (X, d) be a metric space and S a subset of X. Then (i) S 0 = X – ( X − S) , S = X – (X – S)0, δS = S – S 0; (ii) S 0 is the largest open set contained in S; (iii) S is the smallest closed set that contains S. 2.2

CONVERGENCE, COMPLETENESS

Let (X, d) be a metric space. Let {xn} be a sequence in X. The sequence {xn} is said to be convergent to x in X if for every ε > 0 there is an integer n0 such that d(xn, x) < ε for n ≥ n0. In symbols, we write lim xn = x. Note that a sequence in a metric space can have at most one limit. The following theorem characterizes the closure points of a set in terms of sequences. Theorem 2.3. Let (X, d) be a metric space and S a subset of X. Then a point x ∈ X is a point of S iff there exists a sequence {xn} of S such that lim xn = x. In particular, if x is a limit point of S, then there exists a sequence of distinct points of S that converges to x. Proof: Assume that x ∈ S . For each n choose xn ∈ S such that d(xn, x)
n0. Thus,

Br(x ) ∩ S ≠ φ for each r > 0. Hence x ∈ S . Now for the particular case, assume that x is a limit point of S. Then choose x1 ∈ S such that x1 ≠ x and d(x, x1) < 1. Having chosen x1 ∈ (S – {x}) such that d (x, x1) < 1, choose x2 ∈ S such that x2 ≠ x and d (x, x2) < 1/2. In this way, having chosen x1, x2 , ..., xn ∈ (S – {x}), choose xn + 1 ∈ S – {x} such that d(xn + 1, x) < min

FG 1 , d( x , x)IJ . H n +1 K n

Then {xn} is a sequence of S where xn ≠ xm (n ≠ m) and lim xn = x. Dense Set: Let (X, d) be a metric space and S a subset of X. The set S is called dense in X if S = X. It follows from Theorem 2.3 that a set S is dense in X iff for every x ∈ X there exists a sequence {xn} of S such that lim xn = x.

24

THE ELEMENTS OF COMPLEX ANALYSIS

Example: The set Q of rational numbers is dense in R. Cauchy Sequence: Let {xn} be a sequence in X. We say that {xn} is a Cauchy sequence if for every ε > 0 there is an integer n0 such that d (xm , xn) < ε for m, n ≥ n0. Note that a convergent sequence in X is a Cauchy sequence in X but the converse is not true. As an example, consider X = (0, ∞) with d (x, y) = | x – y |, and define the sequence by xn = 1/ n. Then the sequence {xn} is a Cauchy sequence in X that does not converge in X. A metric space (X, d) is said to be complete if every Cauchy sequence in X converges in X. The simple examples of complete metric spaces are: (a) R or C with the usual metric d (x, y) = | x – y |; (b) Rn with the metric

L O d (x, y) = | x – y | = M ∑ ( x − y ) P N Q n

i =1

i

i

2

1/ 2

.

Observe that the set Q of rational numbers is not complete. Theorem 2.4. Let (X, d) be a complete metric space and let S a subset of X. Then S is closed iff S is a complete metric space with metric d. Proof: Assume that S is closed. Let {yn} be a Cauchy sequence in S. Then {yn} is a Cauchy sequence in X. Since X is complete, {yn} converges in X to a point x in X. But since S is closed, x ∈ S. This shows that (S, d) is a complete metric space. Conversely, assume that (S, d) is a complete metric space. If there is a sequence {yn} of S such that line yn = x in X, then {yn} is a Cauchy sequence in X. It is clear that {yn} is a Cauchy sequence in S, and since S is complete, it must converge to a unique point in S. This point must be x. Thus, x ∈ S, and S is therefore closed. Diameter: Let (X, d) be a metric space and S a subset of X. The diameter of S, denoted by diam S, is defined as diam S = sup {d (x, y) : x, y ∈ S}. The set S is called bounded if diam S < + ∞. Theorem 2.5 (Cantor). Let (X, d) be a complete metric space and let {Fn} be a sequence of closed, non-empty subsets of X such that Fn ⊃ Fn +1 for each n. Further, let lim (diam Fn) = 0 Then

∞

∩ Fn is a single point.

n =1

∞

Proof: Let x1, x2 ∈ ∩ Fn . Then x1, x2 ∈ Fn for each n. Hence n =1

d (x1, x2) ≤ diam (Fn) for each n.

25

METRIC SPACES

Thus,

d(x1, x2) = 0 implies x1 = x2. ∞

This shows that ∩ Fn cannot contain more than one point. n =1

∞

It remains to prove that ∩ Fn is non-empty. For each n, choose xn ∈ Fn. n =1

Since lim (diam Fn) = 0, then for ε > 0, there is n0 such that diam ( Fn ) < ε. Now if 0

m, n ≥ n0, then Fm ⊂ Fn , Fn ⊂ Fn , and this implies that xm and xn are in Fn . 0 0 0 Therefore, d (xm, xn) < ε for m, n ≥ n0 which shows that {xn} is a Cauchy sequence in X. Since X is complete, it follows that there is a point x ∈ X such that lim xn = x. ∞

The proof will be complete if we show that x ∈ ∩ Fn . Since xm ∈ Fn for n =1

m ≥ n, it follows that x ∈ Fn for each n. But by assumption, each Fn is closed, hence Fn = Fn. Thus, x ∈ Fn for each n, and therefore, ∞

x ∈ ∩ Fn . n =1

This important theorem is due to G. Cantor. Before we conclude this section, we shall define “nowhere dense” set. Let (X, d) be a metric space and S a subset of X. The set S is said to be nowhere dense if its closure has an empty interior. In other words, the set S is nowhere dense if ( S )° = φ. A classical example of nowhere dense subset of the real line is the Cantor Set. We shall not describe it and readers may consult Rudin [13], Principles of Mathematical Analysis, or Bartle, [3]. The Elements of Real Analysis. 2.3

CONTINUOUS FUNCTIONS

Let (X1, d1) and (X2, d2) be metric spaces. The function f : X1 → X2 is said to be continuous at a ∈ X1 if for each ε > 0 there exists δ > 0 such that d2 (f (x), f(a)) < ε whenever d1 (x, a) < δ. Note that δ depends on ε as well as on a. The function f : X1 → X2 is said to be continuous if it is continuous at each point of X1. The following theorem characterizes continuous functions in terms of open sets. Theorem 2.6. Let (X1, d1 ) and (X2, d2) be metric spaces and let f : X1 → X2 be a function. Then f is continuous iff f –1 (G) is open in X1 whenever G is open in X2. Proof: Let f be continuous. If f –1 (G) is empty, then the proof is trivial. Assume that it is non-empty. Let x ∈ f –1 (G). Then f (x) ∈ G. Since G is open, there

26

THE ELEMENTS OF COMPLEX ANALYSIS

exists some r > 0 such that Br (f (x)) ⊂ G. Now by the continuity of f, there exists an open ball Bδ(x) such that f(Bδ(x)) ⊂ Br (f (x)), and since Br(f (x)) ⊂ G, it follows that Bδ (x) ⊂ f –1 (G). Hence, f –1 (G) is open. Conversely, assume that f –1 (G) is an open subset of X1 whenever G is an open subset of X2. Let ε > 0. Consider the open set G = Bε (f (x)). Since x ∈ f –1 (G), there exists some δ > 0 such that Bδ(x) ⊂ f –1 (G). Thus f (Bδ (x)) ⊂ Bε (f (x)). Hence f is continuous at x. This proves the theorem. Uniform Continuity: Let (X 1, d1) and (X 2, d 2) be metric spaces and let f : X 1 → X2 be a function. We say that f is uniformly continuous if for every ε > 0 there exists δ > 0 (depending only on ε) such that d2 ( f (x1)), ( f (x2)) < ε whenever d1 (x1, x2) < δ. Note that every uniformly continuous function is continuous but the converse is not true. As an example, let X1 = (0, 1) and X2 = R both with d (x, y) = | x – y |. Then f (x) = 1/x is continuous but not uniformly continuous. 2.4

COMPACTNESS

Let (X, d) be a metric space. A collection {Gi} of open subsets of X is said to be an open cover of X if X ⊂ ∪ Gi . Let S be a subset of X. The set S is said to be i ∈I

compact if every open cover of S has finite subcover. In other words, there is a finite number of sets G1, G2, ..., Gn in the collection {Gi} such that S ⊂ G1 ∪ G2 ∪ ... ... ∪ Gn. Note that the empty set and all finite sets are compact. The set S = {z ∈ C : | z | < 1} is not compact. A metric space (X, d) is said to be sequentially compact if every sequence in it has a convergent sub-sequence. The Heine-Borel theorem states that every closed and bounded set in Rn is compact. We prove here the converse result. Theorem 2.7. Let E be a subset of Rn . Then the following statements are equivalent. (i) E is compact. (ii) E is closed and bounded. Proof: As noted above, (ii) ⇒ (i). If we prove that (i) ⇒ (ii), this will establish the equivalence of the statements. Assume (i) holds. We shall prove first that E is bounded. For each k ∈ I, let Bk be the open ball defined by Bk = {x ∈ Rn : | x | < k}. m

Clearly, E ⊂ ∪ Bk . Since E is compact, E ⊂ ∪ Bi . This shows that there k ∈I

i =1

exists m ∈ I such that E ⊂ Bm, and hence E is bounded.

27

METRIC SPACES

Next we prove that E is closed. Suppose E is not closed. Then there is a limit point y of E such that y ∉ E. Now for x ∈ E and for rx = | x – y |/2, construct the open ball Brx (x), with centre at x and radius rx. The collection of open balls { Br (x) : x ∈ E} x

is an open covering of E. By compactness of E, there exists a finite subcover, say Br1 (x1), Br2 (x2), ..., Brn (xn) such that n

E ⊂ ∪ Bri ( xi ) . i =1

Let r = min {r1, r2, ..., rn}. Then it is easy to show that the open ball Br (y) has no points in common with any of the collection Brk (xk). In fact, if x ∈ Br(y), then | x – y | < r ≤ rk | x – xk | = | y – xk + x – y | ≥ | y – xk | – | x – y | = 2rk – | x – y | > rk ,

and

so that x ∉ Brk (xk). Therefore, Br (y) ∩ E is empty. This contradicts that y is a limit point of E. Henec E is closed. Totally Bounded Set: Let (X, d) be a metric space and E a subset of X. The set E is said to be totally bounded if for every r > 0 there exists a finite number of points x1, x2, ..., xn such that n

E ⊂ ∪ Br ( xi ) . i =1

Note that a compact metric space is totally bounded but the converse is not necessarily true. As an example, take E = (0, 1), X = R, with d (x, y) = | x – y |. Theorem 2.8. Let (X1 , d1 ) and (X2, d2 ) be metric spaces. Let f : X1 → X2 be a continuous function, and let E be a compact subset of X1. Then f(E) is a compact subset of X2. Proof: Let f (E) ⊂ ∪ Gi be an open cover. Then E ⊂ ∪ f –1 (Gi). Since f is i ∈I

i ∈I

continuous, it follows from Theorem 2.6 that each of the sets f –1 (Gi) is open. Since E is compact, there exist indices i1, i2, ..., in such that E ⊂ f –1 (Gi ) ∪ f –1 (Gi2 ) ∪ ... ∪ f –1 (Gin ) . Thus

F∪ H

1

n

f (E) ⊂ f

k =1

n

I K

f −1 (Gik )

n

−1 = ∪ f ( f (Gik )) = k∪= 1 Gik , k =1

which shows that f (E) is compact. Theorem 2.9. Let (X1, d1) and (X2 , d2) be metric spaces. Let f : X1 → X2 be a continuous function, and let X1 be compact. Then f is uniformly continuous. Proof: Let ε < 0. Since f is continuous, there exist δx > 0 such that (1) d2 (f (x), f (y)) < ε/2 whenever d1 (x, y) < δx .

28

THE ELEMENTS OF COMPLEX ANALYSIS

Let Gx be the set defined by

RS T

UV W

1 δx 2 Let X1 ⊂ ∪ (Gxi ) be an open cover. Since X1 is compact, there exist finite

Gx = y ∈ X1 : d1 ( x, y)
0. Note that δ does not depend on x. Now if x and y are any two points in X1 such that d1(x, y) < δ, then x ∈ Gx m

1 δ x . Also, 2 m d1 (xm, y) ≤ d1 (xm, x) + d1(x, y) 1 < δ xm + δ ≤ δ x m . 2 It now follows from (1) that

for at least one Gx m . This shows that d1(x, xm)
. 2 2 It is immediately verified that the pair G, H form a disconnected of I. (b) The set E of all positive rational numbers is disconnected in R.

Let G = x ∈ R : x
2 }. The pair G, H form a disconnection of E. (c) Consider the sets G = {x ∈ R : – 1 < x ≤ u}, H = {x ∈ R : u < x < 2} where 0 < u < 1. Then G and H split the interval [0, 1] into two disjoint non-empty subsets whose union is [0, 1]. But since G is not open it does not follows that [0, 1] is disconnected. We shall prove that [0, 1] is a connected set in R. Note that in order to show that a set is connected we need to show that no disconnection can exist. The following theorem establishes a simple characterization of connected sets in R. Theorem 2.11. A subset S of the real line R is connected if and only if S has the following property. If a ∈ S, b ∈ S, and a < c < b, then c ∈ S. Proof: The proof is by contradiction. Assume that a ∈ S, b ∈ S, a < c < b, but c ∉ S. Let G be the set of all x < c and H be the set of all y > c, then by the definition of connectedness it follows that S is not connected. Let a ∈ S, b ∈ S and a < b, and suppose that S is disconnected. Then there exist disjoint open sets G, H in R such that a ∈ G, b ∈ H, S ⊂ G ∪ H. Let c = sup (G ∩ [a, b]) Since b ∈ H and H is open, therefore c < b. If c ∈ G, then c < b; since G is open, there are points in G ∩ [a, b] which exceed c, contrary to its definition. Hence c ∉ G.

30

THE ELEMENTS OF COMPLEX ANALYSIS

Since a ∈ G and G is open, therefore a < c. If c ∈ H, then since H is open there is a point c1 < c such that the interval [c1, c] is contained in H ∩ [a, b]. Consequently, [c1, c] ∩ G = φ. This also contradicts the definition of c as sup (G ∩ [a, b]). Hence c ∉ H. Since S ⊂ G ∪ H, it follows that c ∉ S, which proves the theorem. Corollary: A subset S of R is connected if and only if S is an interval. In particular, R is connected. There is no such simple characterization of connected sets in the complex plane. Connectedness in Rn Theorem 2.12. The space Rn is connected. Proof: Assume that Rn is disconnected. Then there exist two disjoint non-empty open sets A, B such that Rn ⊂ A ∪ B. Let a ∈ A and b ∈ B. Define the line segment E joining a and b by E = {a + t(b – a) : t ∈ [0, 1]}. Let G = {t ∈ R : a + t (b – a) ∈ A} and H = {t ∈ R : a + t (b – a) ∈ B}. It can be easily verified that G and H are disjoint non-empty open subsets of R and the pair G, H form a disconnection for [0, 1]. This contradicts that [0, 1] is connected. Corollary: The only subsets of Rn which are both open and closed are the empty set φ and the set Rn itself. The next theorem characterizes connectedness in Rn. Definition: Let x and y be two points in Rn. A polygonal curve in Rn from a point x to a point y is an ordered set of a finite number of line segments {L1, L2, ..., Ln} such that (a) x is the beginning point of L1, (b) y is the end point of Ln, and (c) the end point of Li is the beginning point of Li + 1. This definition is illustrated in Fig. 2.II. x2

x

Rn x1

y

xn–1

Fig. 2.II

Theorem 2.l3. Let E be an open set in Rn. Then E is connected iff any pair of points x, y in E can be joined by a polygonal curve which lies entirely in E.

31

METRIC SPACES

Proof: Assume that E is disconnected. Then there exist two open sets A and B which form a disconnection for E. Let x ∈ A ∩ E and y ∈ B ∩ E. Let x and y be joined by a polygonal curve (L1, L2, ..., Ln) which lies entirely in E. Let i ∈ I be the smallest number such that the end point xi – 1 of Li belongs to A ∩ E and the end point xi belongs to B ∩ E. Define G = {t ∈ R : xk – 1 + t(xk – xk – 1) ∈ A ∩ E} and H = {t ∈ R : xk – 1 + t(xk – xk – 1) ∈ B ∩ E}. Then it can be easily verified that G and H are disjoint non-empty open subsets of R and the pair G, H form a disconnection for [0, 1]. This contradicts the fact that [0, 1] is connected. Thus, if E is disconnected, then any two points in E cannot be joined by a polygonal curve lying entirely in E. Conversely, assume that E is connected open set in R n. Let x ∈ E. Let S be the set of all those points of E which can be joined to x by a polygonal curve lying entirely in E. Let T be the set of all those points of E which cannot be joined to x by a polygonal curve lying entirely in E. Clearly, S ∩ T = φ. The set S is non-empty since it contains the point x. We now prove that S is open in Rn. Let y ∈ S. Since E is open there exist some positive number r such that | z – y | < r, and so, z ∈ E. It now follows from the definition of S that if a segment from y to z is added to a polygonal curve from y to x, then z ∈ S. This shows that S is open in Rn. Similarly, we can prove that T is open in Rn. If T is not empty, then the pair S, T form a disconnection of E and this contradicts that E is connected. Therefore, T = φ and every point of E can be joined to x by a polygonal curve which lies entirely in E. This completes the proof of the theorem. Examples: (a) Consider the set G = {(x, y) ∈ R2 : 0 < y ≤ x2, x ≠ 0} ∪ {(0, 0)}. The set G is connected in R2, but it is not true that any two points in G can be joined by a polygonal curve lying entirely in G. (b) Consider the set

RS T

H = ( x , y ) ∈ R 2 : y = sin

FG 1 IJ , x ≠ 0UV ∪ {(0, y) : – 1 ≤ y ≤ 1}. H xK W

The set H is connected in R2, but it is not true that every pair of points in H can be joined by a polygonal curve lying entirely in H. Theorem 2.14. Let (X1 , d1) and (X2, d2) be metric spaces: Let f : X1 → X2 be a continuous function. If X1 is connected, then f (x1) is a connected subset of X2. Proof: Assume that f (x 1) is not connected. Then there exist disjoint non-empty open sets A and B in X2, both of which intersect f (X1) and f (X1) ⊂ A ∪ B.

32

THE ELEMENTS OF COMPLEX ANALYSIS

Since f is continuous, it follows from Theorem 2.6 that f –1 (A) and f –1 (B) are open in X1. It is immediately verified that f –1 (A) and f –1 (B) are disjoint and non-empty and f –1(A) ∪ f –1 (B) = X1. This shows that X1 is disconnected, which is a contradiction. Note that if E ⊂ X2 is connected and f is continuous, then it is not necessarily true that f –1 (E) is connected in X1. But if we take disjoint segments in the complex plane C and project them on the real axis, then the projection is connected. The following figure illustrates this fact. y E F

x

O

Fig. 2.III

Convex Set: Let S be a subset of R n. The set S is said to be convex if λx + (l – λ)y ∈ S whenever x ∈ S, y ∈ S, and 0 < λ < 1. Example: Open and closed balls in Rn are convex sets. Region: Let S be a subset of the complex plane C. If S is open and connected, then S is called a region. Examples: (a) An open convex set is a region. (b) An open disc, an open ellipse, an open triangle and an open square are convex regions. (c) We say that an open set S ⊂ C is starlike if there exists a point z0 ∈ S such that for any z ∈ S the segment [z0, z] belongs to S. The point z0 is referred to as the centre of the star.

z2

z1

Convex region

Fig. 2.IV

z2

z1 Non-convex region

Fig. 2.V

It is immediately verified that an open starlike set is connected, and so, is a region.

33

METRIC SPACES

We illustrate these examples in the following figure.

z2

z1

Starlike region

Fig. 2.VI

EXERCISES 1. Let (X, d) be a metric space. Prove that every open ball is open set and every closed ball is a closed set. 2. Prove that a set E ⊂ X is open iff X – E is closed. 3. Let (X, d) be a metric space and Y ⊂ X. Suppose that E ⊂ X is open. Prove that E ∩ Y is open in (Y, d). Conversely, show that if E1 ⊂ Y is open in (Y, d), there is an open set E ⊂ X such that E1 = E ∩ Y.

FG H

IJ K

d is also a metric space. 1+ d 5. Let (X, d) be a metric space. Prove that any two distinct points of X can be separated by open balls in the following sense: if x and y are distinct points in X, then there exists a disjoint pair of open balls each of which is centred on one of the points.

4. Let (X, d) be a metric space. Prove that X ,

6. Find the interior of each of the following subsets of the real line; the set of all integers; the set of all rationals; the set of all irrationals (0, 1); [0, 1]; [0, 1) ∪ {1, 2}. Find also the interior of each of the following subsets of the complex plane: {z : | z | < 1}; {z : | z | ≤ | 1}; {z : I (z) = 0}; {z : R (z) is rational}. 7. Show that a subset of a metric space is bounded iff it is non-empty and is contained is some closed ball. 8. Find the boundary of each of the following subsets of the real line: the integers; the rationals; [0, 1]; (0, 1). Find also the boundary of each of the following subsets of the complex plane: {z : | z | < 1}; {z : | z | ≤ | 1} ; {z : I (z) > 0}. 9. Prove that a Cauchy sequence is convergent iff it has a convergent subsequence. 10. Give three examples of metric spaces which are not complete. 11. Let X and Y be metric spaces and let f be a mapping of X into Y. If f is a constant mapping, prove that f is continuous. Use this to show that a continuous mapping need not have the property that the image of every open set is open. 12. Let X and Y be metric spaces and f a mapping of X into Y. Show that f is continuous iff f –1 (F) is closed in X whenever F is closed in Y iff f ( B ) ⊆ f ( B) for every subset B of X.

34

THE ELEMENTS OF COMPLEX ANALYSIS

13. Examine the uniform continuity of the following functions on the open unit interval (0, 1) 1 (ii) sin x (i) 1− x 1 (iv) x1/2 (iii) sin x (v) x3 Examine also which are uniformly continuous in the open interval (0, + ∞). 14. Let (X, d1) and (Y, d2) be metric spaces. Suppose that f : X → Y is uniformly continuous. Show that if {xn} is a Cauchy sequence in X, then {f (xn)} is a Cauchy sequence in Y. Does the result hold good if we only assume that f is continuous. 15. Suppose that (Y, d2) is a complete metric space and suppose that f : (Z, d3) → (Y, d2) is uniformly continuous, where Z is dense in (X, d1). By using exercise 14 show that there is uniformly continuous function g : X → Y with g (x) = f (x) for every x ∈ Z. 16. Let E be a compact subset of the metric space (X, d) and let x ∈ X. Define the distance dist. (x, E) = inf {d (x, y) : y ∈ E}. Show that if x ∉ E, then dist. (x, E) > 0. 17. Let E and F be two compact subsets of the metric space (X, d). Define the distance dist. (E, F) by dist. (E, F) = inf {d (x, y) : x ∈ E, y ∈ F}. Show that if E ∩ F = φ, then dist. (E, F) > 0. 18. Prove that the union of a finite number of compact sets is compact. 19. Let (X, d) be a metric space and let E ⊂ X. Prove that if E is connected, then its

F I H K

closure E is also connected. 20. Let X be a vector space. A real valued function || ⋅ || defined on X is called a norm if it satisfies the following properties: (i) || x || ≥ 0 for each x ∈ X, and || x || = 0 iff x = 0; (ii) || αx || = | α | || x || for all x ∈ X and α ∈ R; (iii) || x + y || ≤ || x || + || y || for all x, y ∈ X. Property (iii) is called the triangle inequality. Note that the absolute value function yields a norm for R or C. A vector space together with a norm for X is called a normed space. On a normed space X a metric is defined by d (x, y) = || x – y ||. (a) Verify that d (x, y) is indeed a metric on X. (b) Show that | || x || – || y || | ≤ || x – y ||. (c) Show that the vector space Rn with the norm || x || =

FG ∑ x IJ H K n

i =1

1/ 2

2 i

for each x = (x1, x2, ..., xn) ∈ Rn is a normed space. (d) Let B (X) be the vector space of all bounded real valued functions defined on a non-empty set X. Define a norm on B (X) by

35

METRIC SPACES

|| f ||∞ = sup {| f(x) | : x ∈ X} for each f ∈ B (X). This norm is referred to as the sup norm. Prove that the vector space B (X) with the sup norm is a normed space. (e) Let X be a vector space. We say that two norms || ⋅ ||1 and || ⋅ ||2 on X are equivalent if there exist constants K > 0 and M > 0 such that K || x ||1 ≤ || x ||2 ≤ M || x ||1 for every x ∈ X. Let v = (x, y) ∈ R2 Define the following norms on R2 by (i) || v ||1 = | x | + | y |; (ii) || v ||2 = (x2 + y2)1/2; (iii) || v ||∞ = max {| x |, | y |}; 1/ 2

⎛ x2 y2 ⎞ (iv) || v || = ⎜ 2 + 2 ⎟ , b ⎠ ⎝a where a and b are two fixed positive numbers.

Prove that the above norms are all equivalent. Illustrate geometrically the closed unit ball for each norm described above.

3 3.1

ELEMENTARY PROPERTIES OF ANALYTIC FUNCTIONS

LIMITS AND CONTINUITY

Let Ω be an open subset of the complex plane C. Let f be a function on Ω. Let A be a complex constant. We say that lim f (z) = A z → z0

if the following condition is satisfied. Given ε > 0 there exists a number δ > 0 such that if z ∈ Ω and | z – z0 | < δ, then | f (z) – A | < ε. Note that this definition of a limit implies that z may approach z0 from any direction in the complex plane C. For example, let us take f (z) = It can be easily seen that

LM N

( x + y)2 . x 2 + y2

OP Q

LM N

OP Q

lim lim f ( z ) = 1 = lim lim f ( z ) . y→0 y→0 x→0

x→0

But along the path y = mx, we have

(1 + m) 2 . z→0 1 + m2 The limiting value here depends on m and hence lim f (z) does not exist. lim f ( z ) =

z→0

Let f be a function on Ω. Let z0 ∈ Ω. We say that f is continuous at z0 if lim f (z) = f (z0). z → z0

If f is continuous at every point of Ω, we say that f is continuous on Ω. Examples: (i) The function f (z) = z, z ∈ C is continuous on C. (ii) The functions Re z and Im z are continuous on C. 36

37

ELEMENTARY PROPERTIES OF ANALYTIC FUNCTIONS

1 is coutinuous on C – 0 . z | z |, z ≠ 0 (iv) The function f (z) = is not continuous at the origin. 1, z = 0 The sum and product of two continuous functions are continuous; the quotient f/g is continuous at z0 provided g(z0) ≠ 0. If r > 0 and z1 is a complex number. We denote D (z1, r) = {z : | z – z1 | < r }. D (z1, r) is called the open circular disc † with centre at z1 and radius r. The closure of D (z1, r) is

kp

(iii) The function f (z) =

RS T

D (z1, r) = {z : | z – z1 | ≤ r }.

We also denote D′ (z1, r) = {z : 0 < | z – z1 | < r } which is called the punctured disc with centre at z1 and radius r. 3.2

COMPLEX DIFFERENTIABILITY

Let Ω be an open set in the complex plane C. Suppose f is a function on Ω. The function f is said to be differentiable at a point z0 of Ω if (1) zlim →z

0

f ( z ) − f ( z0 ) z − z0

exists. The derivative is denoted by f′ ( z0). Note that the limit is independent of the path along which z → z0 in the complex plane. Writing z – z0 = h in the above definition, we get the equivalent form

f ( z0 + h) − f ( z 0 ) = f ′ ( z0). h→0 h It is important to note that h is a complex number. Replacing z0 by z in (1′) we have (1′) lim

f ( z + h) − f ( z ) = f ′ ( z). h→0 h Denote h by Δz. Then (1′′) takes the form (1′′) lim

(1′′′) lim

Δz → 0

f ( z + Δz ) − f ( z ) = f ′ ( z). Δz

Suppose w = f (z). We sometimes define Δw = f (z + Δz) – f ( z) † The open circular disc is also called a neighbourhood of the point z1.

38

THE ELEMENTS OF COMPLEX ANALYSIS

and write the derivative as

dw Δw = Δzlim . →0 dz Δz By the definition of a limit, (1) means that f ′ (z0) exists if to every ε > 0 there corresponds a δ > 0 such that f ( z ) − f ( z0 ) − f ′( z 0 ) < ε z − z0

for all z ∈D′ (z0, δ). Analytic Functions: A function f is said to be analytic at a point z0, if it is differentiable throughout some ε – neighbourhood of z0. A function f is analytic in a region if it is analytic at every point of the region. Observe that a function which is differentiable at a point, need not necessarily be analytic at that point. Example (v) below illustrates this fact. Examples (i) If n is a positive integer, then f (z) = zn is differentiable in the entire complex plane. (ii) f (z) = Re z and f (z) = Im z are not differentiable. 1 (iii) f (z) = is differentiable in C – {0}. 2 (iv) f (z) = z is not differentiable. (v) f (z) = | z |2 is differentiable only at z = 0. Theorem 1. If f : Ω → C is differentiable at a point z0 in Ω, then f is continuous at z0. Proof :

LM N

lim | f (z) – f (z0) | = lim

z → z0

z → z0

| f ( z ) − f ( z0 ) | | z − z0 |

= f ′ (z0) ⋅ 0 = 0.

OP L lim | z − z |O PQ Q MN z → z0

0

Hence f is continuous at z0. Let f and g be functions defined on the open set Ω. Suppose that f and g are differentiable at z. Then (i) the sum f + g is differentiable at z and (f + g)′ (z) = f ′ (z) + g′ (z); (ii) the product fg is differentiable at z, and (fg)′ (z) = f ′ (z) g (z) + f (z) g′ (z); (iii) the quotient f /g is differentiable at z provided g (z) ≠ 0, and

g ( z ) f ′ ( z ) − f ( z ) g′ ( z ) ( g ( z ))2 The proof is left to the reader which is well known in real analysis. (f /g)′ (z) =

ELEMENTARY PROPERTIES OF ANALYTIC FUNCTIONS

39

Chain Rule. Let f and g be differentiable on Ω1 and Ω2 respectively and let f (Ω1) ⊂ Ω2. Then g o f is differentiable on Ω1 and (g o f)′ (z) = g′ (f (z)) f ′ (z). Proof: In order to prove the chain rule we need the following relations. Let f be differentiable on Ω, z ∈Ω, and h ≠ 0. Consider f ( z + h) − f ( z ) – f ′ (z). η (z, h) = h Then f (z + h) – f (z) = h f ′ (z) + hη (z, h) lim η ( z, h ) = 0.

and

h→0

Conversely, suppose that f (z + h) = f (z) + h f1 (z) + h ζ (z, h) where

lim ζ ( z, h) = 0.

h→0

Then f is differentiable at z with f1 (z) = f ′ (z). We apply the above relations to a proof of the chain rule. Let D (z, r) ⊂ Ω1 and | h | < r. Then, since f is differentiable, we have k = f (z + h) – f (z), where k = hf ′ (z) + h η (z, h) and

lim η (z, h) = 0.

h→0

Hence g (f (z + h)) = g (f (z) + k) = g (f (z)) + kg′ (f (z)) + kζ (f(z), k), where

lim ζ (f(z), k) = 0.

k→0

Substituting for k its expression in terms of h, we have η (z, h) g′ (f (z)) + (f ′ (z) + η (z, h)) ζ (f (z), k) = ξ (z, h) (say). Then g (f (z + h)) = g (f (z)) + hg′ (f (z)) f ′ (z) + h ξ (z, h). It can be easily seen that lim ξ (z, h) = 0.

h→0

Therefore

lim

h→0

i.e.

g ( f ( z + h)) − g ( f ( z )) = g′ (f (z)) f ′ (z) h (g o f)′ (z) = g′ (f (z)) f ′ (z).

40

3.3

THE ELEMENTS OF COMPLEX ANALYSIS

THE CAUCHY-RIEMANN EQUATIONS

Any complex-valued function may be decomposed into its real part, Re f (z) = u (x, y) and its imaginary part, Im f (z) = v (x, y) such that f (z) = u (x, y) + iv (x, y). Here u and v are real valued functions of the two real variables x and y. We shall now derive a simple basic criterion for analyticity of a complex valued function. Let f : Ω → C be analytic. By definition, f ( z + Δz ) − f ( z ) (1) f ′ (z) = lim Δz → 0 Δz (2) {u ( x + Δx, y + Δy) + iv ( x + Δx, y + Δy)} − {u ( x, y) + iv ( x, y)} = lim Δx → 0 Δx + i Δy Δy → 0

Δy = 0, we obtain

If

LM{u ( x + Δx, y) + iv ( x + Δx, y)} − {u ( x, y) + iv ( x, y)}OP Δx N Q {v ( x + Δx, y) − v ( x, y)} O u ( x + Δx , y ) − u ( x , y ) L lim M +i PQ Δx Δx N

f ′(z) = lim

Δx → 0

=

Δx → 0

∂u ∂v +i . ∂x ∂x On the other hand, if Δx = 0, we get from (2)

(3)

=

LM{u ( x, y + Δy) + iv ( x, y + Δy)} − {u ( x, y) + iv ( x, y)}OP iΔy N Q L{u ( x, y + Δy) − u ( x, y)} + i {v ( x, y + Δy) − v ( x, y)}OP = lim M iΔy iΔy N Q

f ′ (z) = lim

Δy → 0

Δy → 0

1 ∂u ∂v + . i ∂y ∂y ∂v ∂u (4) = −i . ∂y ∂y Hence, from (3) and (4) =

∂u ∂v ∂v ∂u +i = −i . ∂x ∂x ∂y ∂y Thus

∂u ∂v ∂u ∂v = and =− . ∂x ∂y ∂y ∂x These basic relations are called Cauchy-Riemann equations. (5)

41

ELEMENTARY PROPERTIES OF ANALYTIC FUNCTIONS

Suppose that u and v have continuous second partial derivatives. Differentiating the C – R equations again we get ∂2u ∂2v ∂2u ∂2v . = and = − ∂x 2 ∂x ∂y ∂y 2 ∂y ∂x Hence, ∂2u ∂2u + = 0. ∂x 2 ∂y 2 Any function satisfying (6) is said to be harmonic. We shall discuss harmonic functions in Chapter 10. Let Ω be a region in the plane and let u and v be functions defined on Ω with continuous partial derivatives. Suppose also that u and v satisfy the C – R equations. If f (z) = u (x, y) + iv (x, y), then f can be shown to be analytic in Ω. To see this, let z = x + iy ∈ Ω and let Br (z) ⊂ Ω. If Δz = Δx + iΔy ∈ Br (0), then u (x + Δx, y + Δy) – u (x, y) (7) = [u (x + Δx, y + Δy) – u (x, y + Δy)] + [u (x, y + Δy) – u (x, y)]. By the mean value theorem for the derivative of a function of one variable, we have (6)

(8)

RSu ( x + Δx, y + Δy) − u ( x, y + Δy) = u ( x + Δx , y + Δy) Δx Tu ( x, y + Δy) − u ( x, y) = u ( x, y + Δy ) Δy x

y

1

1

where | Δx1 | < | Δx | and | Δy1 | < | Δy |. Denote φ (Δx, Δy) = [u (x + Δx, y + Δy) – u (x, y)] – [ux (x, y) Δx + uy (x, y) Δy]. Now from (8) we obtain φ( Δx, Δy) Δx = [ux (x + Δx, y + Δy) – ux (x, y)] Δx + iΔy Δx + iΔy Δy [uy (x, y + Δy) – uy (x, y)] + Δx + iΔy But | Δx | ≤ | Δx + iΔy |, | Δy | ≤ | Δx + i Δy |, | Δx1 | < | Δx |, | Δy1 | < | Δy | and the fact that ux and uy are continuous gives that φ ( Δx, Δy) lim = 0. (9) Δz → 0 Δz Hence u (x + Δx, y + Δy) – u (x, y) = ux (x, y) Δx + uy (x, y) Δy + φ (Δx, Δy) where φ satisfies (9).

42

THE ELEMENTS OF COMPLEX ANALYSIS

Similarly v (x + Δx, y + Δy) – v (x, y) = vx (x, y) Δx + vy (x, y) Δy + ψ (Δx, Δy) where ψ satisfies ψ ( Δx, Δy) lim =0 (10) Δz → 0 Δz Since u (x, y) and v (x, y) satisfy the C – R equations, it can be easily verified that φ ( Δx, Δy) + ψ ( Δx, Δy) f ( z + Δx + i Δy ) − f ( z ) = ux (x, y) + ivx (x, y) + . Δ x + i Δy Δ x + i Δy By (9) and (10), we see that f is differentiable and f ′ (z) = ux (x, y) + ivx (x, y). Since ux and vx are continuous, f ′ is continuous and thus f is analytic. We summarise these results in the following theorem. Theorem 2. Let u (x, y) and v (x, y) be real valued functions defined on a region Ω and suppose that u (x, y) and v (x, y) have continuous partial derivatives. Then f : Ω → C defined by f (z) = u (x, y) + iv (x, y) is analytic iff u and v satisfy the C – R equations. 3.4

EXPONENTIAL FUNCTION

The exponential function ez is of basic importance in complex analysis. In fact, it serves as a basis for defining all other elementary transcendental functions. In calculus we have seen that the real exponential function ex (also written exp x) has the properties d x (e ) = ex dx and e x1 + x2 = e x1 e x2 . While defining ez we will preserve as many of the familiar properties of the real exponential function ex. We wish that (a) e2 shall be single-valued and analytic; d z (e ) = ez; (b) dz (c) ez = ex when Im z = 0. Let (11) ez = u + iv.

ELEMENTARY PROPERTIES OF ANALYTIC FUNCTIONS

43

In order that ez should satisfy condition (b), we have ∂u ∂v +i = u + iv. ∂x ∂x

Hence ∂u = u. ∂x ∂v (13) = v. ∂x Now (12) will be satisfied if (14) u = ex φ(y) where φ (y) is any function of y. Furthermore, since ez is to be analytic, u and v must satisfy the C – R equations. Hence

(12)

∂u = v. ∂y Differentiating this with respect to y; we obtain

(15) –

∂2u ∂v =− 2 ∂y ∂y 2 ∂ u ∂u =− . 2 ∂y ∂x

i.e. Using (12), this becomes

∂ 2u =–u ∂y 2 which, on substituting u = ex φ(y) from (14), reduces to ex φ″ (y) = – ex φ (y) i.e. φ″ (y) = – φ (y). This is a simple linear differential equation whose solution is φ (y) = A cos y + B sin y. Hence, from (14), u = ex φ (y) = ex (A cos y + B sin y)

and from (15),

v=–

∂u = – ex (– A sin y + B cos y). ∂y

Therefore, from (11) ez = u + iv = ex [(A cos y + B sin y) + i (A sin y – B cos y)] This reduces to ex when y = 0, as required by condition (c). Hence ex = ex (A – iB) which holds iff A = 1 and B = 0.

44

THE ELEMENTS OF COMPLEX ANALYSIS

Thus we arrive at the definition that if there is a function of z satisfying the conditions (a), (b) and (c), then it must be (16) ez = ex + iy = ex (cos y + i sin y). Note that | ez | = ex and arg ez = y. The possibility of writing any complex number in exponential form is now apparent. Applying (16) with x = 0 and y = θ, we have (17) eiθ = cos θ + i sin θ. e iθ + e − iθ e iθ − e − iθ , sin θ = . 2 2i A function f is periodic with period c if f (z + c) = f (z) for all z ∈ C. If c is a period of ez then ez = ez + c = ez ec implies that ec = 1. Since 1 = | e c | = eRe (c), Re (c) = 0. Thus c = iθ for some θ in R. c iθ But 1 = e = e = cos θ + i sin θ gives that the periods of ez are the integral multiples of 2πi. Thus, if we divide the plane into infinitely many horizontal strips by the lines z = 2πik, k being any integer, the exponential function behaves the same in each of these strips. This property of periodicity is one which is not present in the real exponential function.

Thus

3.5

cos θ =

TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS

On the basis of equation (17) we extend the definitions of cos z and sin z 1 cos z = (eiz + e–iz); 2 1 sin z = (eiz – e–iz). 2i From these definitions one can establish the familiar formulas: cos2 z + sin2 z = 1. cos (z1 ± z2) = cos z1 cos z2 ∓ sin z1 sin z2. sin (z1 ± z2) = sin z1 cos z2 ± cos z1 sin z2. d (cos z) = – sin z. dz d (sin z) = cos z. dz Expanding cos z in exponentials, we have e i ( x + iy ) + e − i ( x + iy ) cos z = 2

ELEMENTARY PROPERTIES OF ANALYTIC FUNCTIONS

45

e − y (cos x + i sin x ) + e y (cos x − i sin x ) 2 e y + e−y e y − e−y = cos x . − i sin x 2 2 Using the usual definitions of the hyperbolic functions of real variables, we

=

get (18) cos z = cos (x + iy) = cos x cosh y – i sin x sinh y. Similarly (19) sin z = sin (x + iy) = sin x cosh y + i cos x sinh y. In particular, taking x = 0 in (18) and (19), we obtain (20) cos iy = cosh y; (21) sin iy = i sinh y. The remaining trigonometric functions of z are defined in terms of cos z and sin z by means of the usual identities. We define the hyperbolic functions of z by ez + e− z ; 2 ez − e− z . (23) sinh z = 2 By expanding the exponentials and simplifying we get (24) cosh z = cosh x cos y + i sinh x sin y; (25) sinh z = sinh x cos y + i cosh x sin y. Setting x = 0, we obtain (26) cosh iy = cos y; (27) sinh iy = i sin y.

(22) cosh z =

3.6

LOGARITHM

Let Ω be a region of C. We define a branch of the logarithm in Ω to be a continuous function f : Ω → C such that (28) exp (f (z)) = z for all z ∈ Ω. Setting z = reiθ and w = f (z) = u + iv in (28), we have eu = r, i.e. u = ln r and v = θ. Thus w = u + iv = ln r + iθ. (29) w = ln | z | + i arg z. If we let θ1 be the principal argument of z, (29) can be written (30) ln z = ln | z | + i(θ1 + 2nπ), (n = 0, ± 1, ± 2, ...). This shows that the logarithmic function is infinitely many valued. For any particular value of n a unique branch of the function is determined, and the logarithm

46

THE ELEMENTS OF COMPLEX ANALYSIS

becomes effectively single valued. If n = 0, the resulting branch of the logarithm function is the principal value. At all points except the points on the non-positive real-axis, each branch of ln z is continuous and analytic. In fact, from the definition 1 y ln z = ln | z | + i arg z = ln (x2 + y2) + i tan–1 . 2 z One can verify that the C – R equations are satisfied everywhere except at the origin. y 1 Also, u = ln (x2 + y2) and v = tan–1 x 2 are continuous except on the non-positive real axis. Hence d ∂u ∂v (ln z ) = +i dz ∂x ∂x y x −i 2 = 2 2 x +y x + y2 z 1 = = zz z We now define the general power of a complex number z = r (cos θ + i sin θ) by (31) zα = eα ln z, (α complex, z ≠ 0). Since ln z is infinitely many valued, it follows that zα is also infinitely many valued. We can write equation (31) as zα = exp {α [ln | z | + i (θ + 2nπ)]} = exp {α (ln | z |)eiαθ e2nαπi where θ is the principal value of the arg z.

3.7

INVERSE TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS

The function w = sin–1z is defined as the function which satisfies the relation sin w = z. Similarly, we can define cos–1z, tan–1z, cot–1z, sec–1z, cosec–1z, sinh–1z, cosh–1 z, tanh–1 z, coth–1 z, sech–1 z, and cosech–1 z. Note that the inverse trigonometric and hyperbolic functions can be expressed in terms of natural logarithms. We derive the following formula: (32) sin–1 z = – i ln (iz ± 1 − z 2 ) Let

w = sin–1z. Then

Hence,

e iw − e − iw 2i 2iw iw e – 2ize – 1 = 0

z = sin w =

47

ELEMENTARY PROPERTIES OF ANALYTIC FUNCTIONS

Solving this equation as a quadratic in eiw, we obtain eiw = iz ± 1 − z 2 Taking logarithms and solving for w, we have w = – i ln (iz ± 1 − z 2 ). We now simply write the other formulas (33) cos–1 z = – i ln (z ± (34) tan–1 z =

z2 − 1)

i i+z ln 2 i−z

(35) cosh–1 z = ln (z ±

z2 − 1)

(36) sinh–1 z = ln (z ± z 2 + 1 ) 1 1+ z . (37) tanh–1 z = ln 2 1− z EXERCISES 1. Let f be the function defined by f (z) = lim

n→∞

Fz GH z

n n

I JK

−1 . +1

Show that the limit exists for | z | ≠ 1. Is it possible to define f (z) when | z | = 1 in such a way to make f continuous? 2. Let f be the function defined by

zn . n → ∞ 1 + zn For what values of z the limit exists? 3. For what values of z are each of the following functions continuous? 1 1 (i) f (z) = + 2 z z +1 f (z) = lim

(ii) f (z) = (x + y2) + ixy

R| z (iii) f (z) = S |T Rz (iv) f (z) = S T

2

+ 3iz − 2 , z≠−i z+i i, z=−i

2

+ iz + 2, z ≠ i i, z=i

4. Show that the function f (x, y) = is not continuous at the origin.

x + iy x − iy

48

THE ELEMENTS OF COMPLEX ANALYSIS

5. Show that the following functions are not differentiable (i) f (z) = | z | (ii) f (z) = Im z (iii) f (z) = z . 6. Use the Cauchy-Riemann equations and examine which of the following functions are differentiable 1 (ii) f (z) = (i) f (z) = x2 + y2 x + iy (iii) f (z) = ey (cos x + i sin x). At which points do the C–R conditions hold in each case? 7. Prove that the function

R| x f (z) = S x |T

3 2

− y3 x 3 + y3 + , z≠0 i + y2 x 2 + y2 0, z=0

satisfies the C – R equations at the origin but does not have a derivative. 8. Show that the C – R equations in polar coordinates are given by

r

∂u ∂v ∂u ∂v = , =−r ∂r ∂φ ∂φ ∂r

where f (z) = f (r, φ) = u (r, φ) + iv (r, φ). 9. If f (z) and g (w) are analytic functions, prove that g (f(z)) is analytic. 10. Prove that an analytic function cannot have a constant absolute value without reducing to a constant. 11. Let G be a region and define G1 = {z : z ∈ G}. If f : G → C is analytic, prove that f1 : G1 → C defined by f1 (z) = f ( z ) is analytic. 12. If f (z) is an analytic function, prove that

LM ∂ | f (z) |OP + LM ∂ | f (z) |OP = | f ′ (z) | N ∂x Q N ∂y Q F ∂ + ∂ I | f (z) | = 4 | f ′ (z) | . GH ∂x ∂y JK 2

2

(i)

2

(ii)

2

2

2

2

2

2

13. Show that the following functions are harmonic and find a corresponding analytic function f (z) = u (x, y) + iv (x, y). (i) u = arg z (ii) u = ex cos y (iii) u = x2 – y2. 14. If u and v are harmonic in a region Ω, prove that

FG ∂u − ∂v IJ + i FG ∂u + ∂v IJ is analytic in Ω. H ∂y ∂x K H ∂x ∂y K

15. Prove that a harmonic function satisfies the differential equation ∂2u = 0. ∂z ∂z 16. Let Ω be a region which is a symmetric with respect to the real axis; that is, when z ∈ Ω. Also z ∈ Ω. Show that if f is analytic on Ω1 = Ω ∩ {z : Im z > 0}, then the

function φ with values φ(z) = f ( z ) is analytic on Ω2 = Ω ∩ {z : Im z < 0}.

ELEMENTARY PROPERTIES OF ANALYTIC FUNCTIONS

49

17. Let U (x, y) = x/(x2 + y2), (x, y) ∈ R2 ~ {(0, 0)}. By using the C – R equations, find a function V (x, y) such that if u (z) = u (x + iy) = U (x, y) and v (z) = v (x + iy) = V (x, y), the function f = u + iv is analytic on C ~ {0}. 18. Prove that there cannot exist a function analytic on an open set E ⊂ C with real part x – 2y2. 19. Find the principal value of the following functions (i) (1 + i)2 – i (ii) (1 – i)2 – 3i (iii) tan i (iv) ln (– 3 + 4i) i (v) 2 . 20. Show that there is no branch of the logarithm defined on G = C ~ {0}.

4 4.1

LINE INTEGRAL AND CAUCHY’S THEOREM

DEFINITIONS

Let [a, b] be a closed interval. A set of points P = {t0, t1, t2, ..., tn} satisfying a = t0 < t1 < ... < tn – 1 < tn = b is called a partition of [a, b]. The interval [tk – 1, tk] is called the kth sub-interval of P so that n

∑ (t k – tk – 1 ) = b – a.

k =1

Let Ω ⊂ C be a region. Let [a, b] be some interval. A path in Ω is defined to be a continuous function γ : [a, b] → Ω. We call γ(a) the initial point, and γ(b) the terminal point. We call γ a differentiable path if γ′(u) exists for each u ∈ [a, b] and γ′ : [a, b]→ C is continuous. γ is said to be piecewise differentiable if there is a partition of [a, b] such that γ is differentiable on each sub-interval [tt – 1, ti], 1 ≤ i ≤ n. Note that a function γ : [a, b] → C has a derivative γ′(u) for each u ∈ [a, b] implies that γ (u + h ) – γ (u ) lim = γ′(u) h→0 h exists for a < u < b and the right and left limits exist for u = a and u = b. Note also that in the definition of differentiable path the continuity of γ′ is included. Many authors call this “continuously differentiable” path. Let γ : [a, b] → C. Let P = {t0, t1, t2, ..., tn} be a partition of [a, b]. Write Δγk = γ(tk) – γ(tk – 1), k = 1, 2, ..., n.

50

51

LINE INTEGRAL AND CAUCHY’S THEOREM

If there is constant M > 0 such that n

ν(γ, P) = ∑ | Δγ k | ≤ M k =1

for all partitions of [a, b], then γ is said to be of bounded variation of [a, b]. By total variation of γ, denoted by V(γ), we mean V(γ) = sup {ν(γ, P) : P is a partition of [a, b]}. Note that V(γ) ≤ M < ∞. A partition Q of [a, b] is called a refinement of P if Q ⊃ P. We state one useful formula regarding the variation of g when g′ is continuous on [a, b]. (1) V(g) =

z

b a

| g ′| dx.

Properties (i) γ is of bounded variation iff Re γ and Im γ are of bounded variation. (ii) If γ is real valued and non-decreasing, then γ is of bounded variation and V(γ) = γ(b) – γ(a). (iii) Let γ : [a, b] → C be of bounded variation. If P and Q are partitions of [a, b] and P ⊂ Q, then ν(γ, P) ≤ ν(γ, Q). We leave it to the reader to verify the validity of these properties. 4.2

RIEMANN-STIELTJES INTEGRAL

In order to define the integral of a function along a path in C we need the notion of Riemann-Stieltjes integral. For detailed discussion on Riemann-Stieltjes integral we refer to books on Real Analysis by Rudin, Apostol and Bartle. Let f, g : [a, b] → C be bounded functions. Corresponding to every partition P = {x0, x1, ..., xn} of [a, b] we choose points u0, u1, ..., un – 1 such that xk ≤ uk ≤ xk + 1, k = 0, 1, 2, ..., n – 1. Introduce n –1

S (P, f, g) = ∑ f ( uk ) [g(xk + 1) – g(xk)] k =0

and consider the behaviour of S(P, f, g) when the norm of P is small. We say that f is Riemann-Stieltjes integrable with respect to g provided that there exists a number A ∈ C such that for each positive η there exists a positive δ such that | S (P, f, g) – A | < η when || P || < δ. [(11.11) is defined in Exercise 20 of Chapter 2].

52

THE ELEMENTS OF COMPLEX ANALYSIS

This is true for all P and all admissible choices of uk. It is clear that there is atmost one such A. In case such A exists, we call A the Riemann-Stieltjes integral of f with respect to g and denote it by

z

b a

z

or

f ( x ) dg( x )

b a

f dg .

We have already defined a path as a continuous function γ : [a, b] → C. The set {γ(t) : a ≤ t ≤ b} is called the trace of γ where γ : [a, b] → C is a path. Denote the trace of γ by {γ}. Note that the set {γ} is compact. Denote the length of γ by L(γ). A path γ is said to be rectifiable if L(γ) < + ∞. If γ is piecewise differentiable, then γ is rectifiable and L(γ) =

z

b a

| γ ′ | dt.

The following fact will be useful while defining the line-integral. Let γ : [a, b] → C be a rectifiable path with {γ} ⊂ Ω ⊂ C and let f : Ω → C be continuous function. Then f o γ (composite function) is also continuous on [a, b]. 4.3

LINE-INTEGRAL

Let γ : [a, b] → C be rectifiable. Let f be defined and continuous on the trace of γ. Then the line-integral of f along γ is defined by the expression

z

b a

f ( γ (t )) dγ (t ) =

Denote this integral by

z

γ

f ( z ) dz =

z z

b a

γ

( f o γ ) dγ .

f dz .

Example 1: Let γ : [0, 2π] → C be given by γ (θ) = eiθ 1 and define f (z) = for z ≠ 0. z We have to evaluate the integral of f over the circle {γ} i.e. 1 dz (0 ≤ θ ≤ 2π). γ z By definition, this integral is equal to 2π 1 2π ie iθ dθ = i dθ = 2πi. i θ 0 0 e Exercise: Let γ : [a, b] → C be a rectifiable curve. Define the reverse or opposite curve to be – γ : [a, b] → C such that – γ (t) = γ (a + b – t).

z z

z

53

LINE INTEGRAL AND CAUCHY’S THEOREM

Let f be continuous on {γ}. Show that (i) (ii)

z z z z –γ

γ

f =–

f ≤

γ

γ

f.

| f | | dz | ≤ V(γ) sup [| f (z) | : z ∈ {γ}].

We shall prove one important theorem for the line-integral which is analogous to the Fundamental Theorem of Calculus. We will need the concept of modulus of continuity. 4.4

MODULUS OF CONTINUITY

Let (X, d1) and (Y, d2) be metric spaces. Recall that f is uniformly continuous provided that for each positive ε there exists a positive δ such that d2 (f (x), f (y)) < ε when d1 (x, y) < δ. Consider the set (2) {d2 (f (x), f (y)) : d1 (x, y) ≤ δ} where δ ≥ 0. When this set is bounded, denote its supremum by wf (δ). The function wf assigning to each such δ the value wf (δ) is called the modulus of continuity of f. The expression (2) on the notion of uniform continuity is given by dela Vallée Poussin. Note that f is uniformly continuous iff the domain of wf does not reduce to 0 and wf is continuous at 0 (X ≠ φ). Theorem 1. Let f : Ω → C where Ω is a region in C and f is continuous on Ω. Let γ : [a, b] → Ω be rectifiable. Suppose that g is a primitive of f. Then (3)

z

γ

f dz = g[ γ(b)] – g[γ(a)]

(g is a primitive of f when g′ = f). Proof: Case I. Suppose γ′ is continuous on [a, b]. Then

z

γ

f dz =

z

b a

(g o γ )′ dt

and hence (3) holds. If the domain of γ is [0, 1], then replace γ by γ [a + (b – a)t], 0 ≤ t ≤ 1. Case II. Suppose Ω is an open disk and γ is taken arbitrary. Let Ω = Dr (c) = {z : | z – c | < r}. Since {γ} is compact we have | γ(t) – c | ≤ ρ < r, 0 ≤ t ≤ 1 for some ρ, 0 < ρ < 1.

54

THE ELEMENTS OF COMPLEX ANALYSIS

Let P be a partition of [0, 1] and {t0, t1, ..., tn} its associated sequence. Define Γ : [a, b] → C by tk + 1 – t t – tk + γ (t k + 1 ) Γ(t) = γ(tk) tk + 1 – tk tk + 1 – tk

z

Now

tk ≤ t ≤ tk + 1, k = 0, 1, ..., m – 1. m −1

Γ

f dz – Σ f [ γ (tk )] {γ(tk + 1) – γ(tk)}

z

m −1

= Σ 0

0

tk + 1 tk

[ f o Γ – ( f o Γ )(tk )] dΓ.

But Γ is a polygon so, from case I

z

Γ

f dz = g[γ(1)] – g[γ(0)].

Therefore m −1

(4) | {g[γ(1)] – g[γ(0)]} – Σ f [ γ (t k )] {γ(tk + 1) – γ(tk)} | 0

≤ wF [wγ(|| P ||)] L [γ] where F = f | Dρ (c) and wF and wγ are the modulus of continuity of the functions. Thus (3) follows from (4) and the theorem is proved for Γ restricted to each [tk, tk + 1]. Case III. The general case Consider Ω ≠ C. Since {γ} is compact, we have min {| γ(t) – w | : 0 ≤ t ≤ 1, w ∈ C – Ω} > 0. Denote the above expression by μ. Let n ∈ N (the set of natural numbers) be so large that wγ(2– n) < μ. Then | γ(t) – γ(tk) | < μ, tk ≤ t ≤ tk + 1 k k = 0, ..., 2n – 1, where tk = n . 2 Hence, the integral of f along γ[tk , tk + 1] is g[γ(tk + 1)] – g[γ(tk)] and summing we obtain (3). This completes the proof of the theorem. A curve Γ : [a, b] → C is said to be closed if γ(a) = γ(b). The following result is an immediate consequence of Theorem 1. Theorem 2. Let Ω be open in C and let γ be rectifiable in Ω. Let f : Ω → C be continuous with a primitive g : Ω → C. If γ is closed curve then

z

γ

f = 0.

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LINE INTEGRAL AND CAUCHY’S THEOREM

Example: Let f(z) = zn, where n is an integer ≠ – 1. Then for any closed path γ, we have

z

γ

z n dz = 0.

If n is negative, the result holds good for any closed path not passing through the origin. This is true because zn has the primitive

zn +1 . n +1

We know from the Fundamental Theorem of calculus that each continuous function has a primitive. But this is not true for functions of a complex variable. For example, let f (z) = | z |2 = x2 + y2. If g is a primitive of f then g is analytic. Write g = ξ + iη. Then x2 + y2 = g′ (x + iy) By C – R equations

∂ξ ∂η = = x2 + y2 ∂x ∂y ∂ξ ∂η =– = 0. ∂y ∂x

and But

∂ξ = 0 implies that ξ(x, y) = ψ(x) for some differentiable function ψ. ∂y

∂ξ = ψ′(x), which is a contradiction. This shows that f(z) = | z |2 does ∂x not have a primitive.

Thus x2 + y2 =

4.5

LOCAL PRIMITIVE

We would like to define a primitive for f on some open disc centred at z0. Before finding a primitive locally we prove Cauchy’s theorem for a rectangle. The proof is due to E. Goursat. We denote by T the rectangle and by δT the boundary of the rectangle (the path describing the boundary of the rectangle will be taken counterclockwise). Note that T is meant as the set of points inside and on the boundary of the rectangle, so T is not a region. In the following theorem we consider a function f which is analytic on a region which contains T. Theorem 3 (Cauchy-Goursat). Let T be a rectangle, and let f be analytic on T. Then

z

δT

f = 0.

Proof: Join the mid-points of the sides of the rectangle T to obtain four rectangles T1, T2, T3, T4 as shown in Fig. 4.I.

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THE ELEMENTS OF COMPLEX ANALYSIS

T1

T2

T3

T4

Fig. 4.I

Then Hence

z

z

4

f =∑

δT

i =1 4

f ≤∑

δT

i =1

z z

f.

δTi

f .

δTi

It follows that there is one rectangle, say T(1), among T1, T2, T3 and T4 such that

z

z

1 f . δT 4 δT Next, join the mid-points of the sides of the rectangle T(1) to obtain four rectangles as shown in Fig. 4.II. (1 )

f ≥

T(2)

Fig. 4.II

We denote one of the four rectangles thus obtained by T(2) and have the similar inequality.

z

z

1 f . δT 4 δT (1) By repeating this process, we obtain a sequence of rectangles T(1) ⊃ T(2) ⊃ T(3) ⊃ T(4) ... such that

z z

(2)

f ≥

1 4 1 f ≥ n Hence δT ( n ) 4 (n) Let Ln be the length of δT . Then δT ( n + 1 )

f ≥

Ln + 1 =

z z

δT ( n )

δT

1 L. 2 n

f .

f .

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LINE INTEGRAL AND CAUCHY’S THEOREM

It follows by induction that Ln =

1 L0, 2n

where L0 is the length of δT. Note that the diameter of T(n) tends to 0 as n → ∞. It follows from Cantor’s ∞

theorem that the intersection ∩ T ( n ) consists of a single point z0. Since f is complex n =1

differentiable at z0, there is a disk Dr(z0) such that for all z ∈ Dr (z0) we have f (z) = f (z0) + f ′(z0) (z – z0) + (z – z0) η (z, z0) lim η ( z, z0 ) = 0.

where

z → z0

It follows that T(n) is contained in Dr(z0) for sufficiently large n and we have

z

(5)

δT ( n )

f ( z ) dz =

z

δT ( n )

f ( z0 ) dz + f ′( z0 )

z

δT ( n )

+

( z – z0 ) dz

z

δT ( n )

( z – z0 ) η(z, z0) dz.

The first two integrals on the right side of (5) are zero.

z

Hence

δT ( n )

and we obtain the inequality 1 4n

z

δT

f ≤

f=

z

z

δT ( n )

δT ( n )

( z – z0 ) η(z, z0) dz

z

f =

( z – z0 ) η ( z, z0 ) dz

δT ( n )

≤ where the sup is taken for all z ∈ T(n). Thus (6)

z

δT

1 L diam (T(n)) sup | η (z, z0) | 4n 0

f ≤ L0 diam (T(n)) sup | η(z, z0) |.

The right side of (6) tends to 0 as n → ∞. Hence

z

δT

f =0

and this completes the proof of the theorem. We now find a primitive locally. Theorem 4. Let f be continuous on Dr (z0). Let

for each T ⊂ Dr (z0). Define

z

δT

f =0

F(z1) =

z

z1 z0

f

58

THE ELEMENTS OF COMPLEX ANALYSIS

where z1 is an arbitrary point in the disc Dr (z0) and the integral is taken along the sides of a rectangle T whose opposite vertices are z0 and z1. Then F is analytic on Dr(z0) and F ′(z) = f (z). Proof: We have F(z1 + h) – F(z1) =

z

z1 + h z1

f ( z ) dz.

The integral between z1 and z1 + h is taken as shown in Fig. 4.III. z1 + h z1

z0

Fig. 4.III

Since f is continuous at z1, there is a function φ(z) such that f(z) = f(z1) + φ(z) where lim φ ( z ) = 0. z → z1

F(z1 + h) – F(z1) =

Then

z

z1 + h z1

f ( z1 ) dz +

= hf (z1) +

z

z1 + h z1

z

z1 + h z1

φ( z ) dz.

φ( z ) dz.

Dividing both sides by h we have F( z1 + h) – F( z1 ) 1 z1 + h φ( z ) dz. = f ( z1 ) + (7) h h z1 The length of the path from z1 to z1 + h is bounded by | h1 | + | h2 |. Hence

z

z

1 z1 + h 1 f ( z ) dz ≤ (| h1 | + | h2 |) sup φ(z). h z1 |h| Note that the sup is taken for z on the path of integration. The expression on the right side of (8) tends to 0 as z → z1. Hence F( z1 + h) – F( z1 ) lim = f (z1). h→0 h This proves the theorem. It now follows from the above discussion that if f has a primitive on a disc Dr(z0), then the integral of f along any path between z0 and z in Dr(z0) is independent of the path. Thus according to Theorem 2 in the previous section we find. (8)

LINE INTEGRAL AND CAUCHY’S THEOREM

59

Theorem 5. If f is analytic on a disc D, then f has a primitive on D, and the integral of f along any closed path in D is 0. 4.6

CAUCHY’S THEOREM (HOMOTOPY FORM)

We first give the general definition of homotopy. Definition: Let Ω be an open set. Let γ0, γ1 be two closed paths in Ω. Assume that they are defined on the same interval [0, 1]. We say that γ0 is homotopic to γ1 in Ω if there exists a continuous function ψ : [0, 1] × [0, 1] → Ω such that (i) ψ(t, 0) = γ0 (t), ψ (t, 1) = γ1(t) (0 ≤ t ≤ 1); (ii) ψ(0, u) = ψ(1, u) (0 ≤ u ≤ 1). Thus, if we define γu : [0, 1] → Ω by γu (t) = ψ (t, u) then each γu is a closed curve and they form a continuous family of curves which start at γ0 and go to γ1. We now give another definition of homotopy which is less general than the definition given above. We say that the curve γ is piecewise smooth if γ(t) possesses a continuous derivative γ′(t) everywhere except at a finite number of points of [0, 1] and γ′(t) = x′(t) + iy′(t) ≠ 0 for all t, 0 ≤ t ≤ 1. Definition: Let γ0 = ψ(t, 0) and γ1 = ψ(t, 1) belong to the family of piecewise smooth closed curves in a region Ω. We say that γ0 is homotopic to γ1 in Ω if there exists a continuous function ψ = ψ(t, u) such that (i) ψ has continuous partial derivatives ∂ψ ∂ψ ≠ 0, ≠ 0, except at a finite number of points. ∂u ∂t (ii) ψ maps the square I 2 = {(t, u) : 0 ≤ t ≤ 1, 0 ≤ u ≤ 1} into Ω. This definition will be used. If γ0 is homotopic to γ1 with respect to Ω, and γ1 is a point, then γ0 is homotopic to a point with respect to Ω. Note that if γ0 and γ1 are homotopic in a region Ω then they are homotopic in any region Ω1 containing Ω. Remark: Suppose that two piecewise smooth, closed curves γ0 and γ1 and a region Ω are given. In order to exhibit the homotopy between γ0 and γ1 we have to find a family of piecewise smooth, closed curves ψ(t, u) such that

60

THE ELEMENTS OF COMPLEX ANALYSIS

(i) γ0 and γ1 belong to this family where γ0(t) = ψ(t, 0) and γ1(t) = ψ(t, 1). (ii) ψ : I 2 → C is continuous with ψ(I 2) ⊂ Ω such that ψ(t, u) piecewise smooth closed curve for every u and ψ(t, u) is piecewise smooth closed curve for every t. From an intuitive standpoint the notion of homotopy is simple. This means that we can transform γ0 into γ1 by continuous deformations given by ψ(t, u). Note that during the process of deformation ψ(t, u), 0 ≤ u ≤ 1, does not leave Ω. We illustrate this notion in Fig. 4.IV. Ω

Y0

Y1

Y2

Fig. 4.IV

In Fig. 4.IV, γ0 and γ1 are homotopic in Ω but γ0 and γ2 are not homotopic in Ω. γ2 is homotopic to a point in Ω. Examples: (i) Let γ0 (t) = 12 (cos 2πt + i sin 2πt), 0 ≤ t ≤ 1 and γ1(t) = cos 2πt + i sin 2πt, 0 ≤ t ≤ 1 be two circles. Then γ0 and γ1 are homotopic in any region Ω containing the closed disc D (0, 1). Take Then and Since Also

ψ(t, u) =

1+ u (cos 2πt + i sin 2πt). 2

ψ(t, 0) = γ0(t) ψ(t, 1) = γ1(t). | ψ(t, u) | ≤ 1 for 0 ≤ u ≤ 1, it follows that ψ(t, u) ⊂ D (0, 1) ⊂ Ω. ψ(0, u) = ψ(1, u).

So the curves γ0 and γ1 are homotopic in Ω containing D (0, 1). (ii) Let Ω be a convex set, and let γ0, γ1 be two piecewise smooth closed curves in Ω. Then γ0 and γ1 are homotopic in Ω. Define ψ(t, u) = uγ0(t) + (1 – u)γ1(t). It can be checked that ψ = ψ(t, u) is continuous and each curve ψ(t, u) is a closed curve. Also, ψ(t, 0) = γ1(t)

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LINE INTEGRAL AND CAUCHY’S THEOREM

ψ(t, 1) = γ0(t). In order to simplify the proof of the homotopy form of Cauchy’s theorem, we introduce the notion of “near-together”. Let Ω be an open set. Let γ0 and γ1 be defined on the same interval [0, 1]. We say that γ0 and γ1 are near-together if there is a partition 0 = u0 ≤ u1 ≤ u2 ≤ ... ≤ un = 1 and for each j = 0, 1, 2, ..., n – 1 there exists a disc Dj ⊂ Ω such that the images of each segment under the mappings of γ0 and γ1 are contained in Dj. In other words γ0([uj, uj + 1]) ⊂ Dj and γ1([ui, uj + 1]) ⊂ Dj. Lemma: Let γ0 and γ1 be two piecewise smooth closed curves in the open set Ω. Let γ0, γ1 be defined on the same interval [0, 1]. Assume that they are neartogether. Then and

z z γ0

f=

γ1

f.

Proof: Choose a partition 0 = u0 ≤ u1 ≤ ... ≤ un = 1 and for each j = 0, 1, 2, ..., n – 1, choose discs Dj ⊂ Ω. Let Fj be a primitive of f on Dj. Let zj = γ0(uj) and wj = γ1(uj). Then

z z γ0

f –

n –1

γ1

f = ∑ [ Fj ( z j + 1 ) – Fj(zj) – (Fj (wj + 1) – Fj(wj))] j=0

n –1

= ∑ [ Fj ( z j + 1 ) – Fj(wj + 1) – (Fj(zj) – Fj(wj)] j=0

i.e. (9)

z z γ0

f –

γ1

f = Fn – 1(zn) – Fn – 1(wn) – (F0 (z0) – F0(w0)).

Since γ0 and γ1 are closed, we have z0 = zn and w0 = wn. Let there be some disc D which contains z0 and w0. The primitives Fn–1 and F0 differ by a constant on D ⊂ Ω. Hence the right-side of (9) is equal to 0. This proves the lemma. Theorem 6 (Cauchy’s Theorem). Let Ω be an open set. Let γ0 and γ1 be two piecewise smooth closed curves in Ω, and assume that they are homotopic in Ω. Let f be analytic on Ω. Then

z z z γ0

f –

γ1

f.

In particular, if γ0 is homotopic to a point in Ω, then γ0

f = 0.

62

THE ELEMENTS OF COMPLEX ANALYSIS

Proof: Let ψ : [0, 1] × [0, 1] → Ω be the homotopy. Since ψ is continuous and I is compact, ψ is uniformly continuous and ψ(I 2) is a compact subset of Ω. By uniform continuity we can find partitions 0 = u0 ≤ u1 ≤ ... ≤ un = 1 0 = t0 ≤ t1 ≤ ... ≤ tn = 1 of these intervals, such that if Sik = small square [ui, ui + 1] × [tk, tk + 1], then the image ψ(Sik) is contained in a disc Dik which is itself contained in Ω. Let ψk be the continuous curve defined by ψk(t) = ψ(t, uk), k = 0, 1, ..., m. Then the continuous curves ψk, ψk + 1 are near-together. By Lemma we have 2

z z z z ψk

f =

ψk + 1

f.

Since ψ0 = γ0 and ψm = γ1, we conclude that γ0

f=

γ1

f.

This proves the theorem. Definition: We say that an open set Ω is star shaped if there is a point z0 in Ω such that for each z in Ω the line segment [z0, z] lies entirely in Ω. Note also that each star shaped set is connected. We now define simply connected set. Definition: We say that an open set Ω is simply connected if it is connected and every closed path in Ω is homotopic to a point. We are now in a position to construct primitives for a wider class of open sets. 4.7

GLOBAL PRIMITIVES

Theorem 7. Let Ω be a simply connected open set. Let f be analytic on Ω. Define F(z) =

z

z z0

f (ζ) dζ

where z is any point in Ω and z0 is a fixed point in Ω. Then (i) F(z) is independent of the path in Ω from z0 to z. (ii) F ′(z) = f (z). Proof: Since Ω is open and connected there always exist a path from z0 to any other point of Ω. Let γ0 and γ1 be two paths in Ω from z0 to a point z in Ω. Let – γ1 be the opposite (reverse) path of γ1 from z to z0. Then γ = {γ0, – γ1} is a closed path.

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LINE INTEGRAL AND CAUCHY’S THEOREM

Hence, by Theorem 6

z z z z z

0= i.e.

f=

γ

f –

γ0

γ1

γ0

f+

– γ1

f

f = 0.

This proves (i). In order to prove (ii) we need to prove the differentiability of F at a point z1. Let z1 ∈ Ω and ρ > 0 be such that D (z1, ρ) ⊂ Ω. If z is in D(z1, ρ), then we may choose a path z0 to z by passing through z1. Hence (10) F(z) = F(z1) +

z

z

f.

z1

It follows from Theorem 4 that the integral in the right side of (10) defines a local primitive for f in a neighbourhood of z1. Hence F ′(z) = f. Examples (i) Every star shaped region is simply connected. (ii) Let Ω = C – {θeiθ : 0 ≤ θ < ∞}. Then Ω is simply connected. (iii) The domain of the principal branch of the logarithm is simply connected.

EXERCISES 1. Let γ(t) = exp {(– 1 + i) t – 1} for 0 < t ≤ 1 and γ(0) = 0. Prove that γ is a rectifiable path and find V(γ). Sketch the trace of γ. 2. Define γ : [0, 1] → C by γ(t) =

R|t + it sin 1 , S| 0, t T

t≠0 t=0

Prove that γ is a path but is not rectifiable. Sketch this path. 3. Let γ1 and γ2 be the two polygons [1, i] and [1, 1 + i, i]. Express γ1 and γ2 as paths and evaluate

z

γ1

f ( z ) dz and

z

f ( z ) dz where f(z) = | z |2 .

γ2

4. Show that the function f (z) = z does not have a primitive on E = {z : Im z ≠ 0}. 5. Let γ be a path in C with terminal point z1 and initial point z0. Let a ≠ 0 and b be complex numbers. Find

z

γ

e az + b dz.

6. Let γ be any closed path in C. Show that

z

γ

exp (– z 2 ) dz = 0.

7. Let f be continuous on a region G. Show that if for any two points a ∈ G, b ∈ G, the integral

z

γ

f ( z ) dz is independent of the path γ joining a to b, then

every closed path γ0 in G.

z

γ0

f ( z ) dz = 0 for

64

THE ELEMENTS OF COMPLEX ANALYSIS

8. Find the integral y = x2.

z

γ

sin z dz from the origin to the point 1 + i, taken along the parabola

9. Let f be analytic in a region that contains γ. Show that

z

f ( z ) f ′ dz is purely

γ

imaginary. (The continuity of f ′(z) is taken for granted). 10. Let f be analytic and satisfies the inequality | f (z) – 1 | < 1 in a region Ω. Show that

z

f ′( z ) dz = 0 for every closed curve γ in Ω. (The continuity of f ′(z) is taken for f (z) granted). γ

11. Let P(z) be a polynomial and let C1 be the circle | z – a | = r. Evaluate 12. Let γ be the closed polygon [1 – i, 1 + i, – 1 – i, 1 – i]. Find 13. Let γ(t) = 2eit for – π ≤ t ≤ π. Evaluate 14. Evaluate

z

1+ i 0

z

γ

z

γ

z

P( z ) dz .

C1

dz . z

( z 2 – 1) – 1 dz.

( z 2 + z ) dz. By choosing two different paths of integration show that

the results are the same.

z

15. Show that

i 2

( z + 1) 2 dz ≤ 9 5 .

(Hint: Take as the path of integration the line z = i + (2 – i) t, 0 ≤ t ≤ 1 and determine max | (z + 1)2 | on this line. 16. Evaluate

z

2+i

1

( x 2 + iy) dz along

(i) the line y = x – 1 (ii) the line y = (x – 1)2. dz 17. Compute where γ (θ) = 2 | cos 2θ | eiθ, 0 ≤ θ ≤ 2π. 2 γ z +1 18. Compute

z z

19. Show that

γ

RS T

dz θe iθ , 0 ≤ θ ≤ 2π . 2 where γ(θ) = z +π 4 π – θ, 2 π ≤ θ ≤ 4 π 2

z

γ

( x 2 – iy 2 )dz ≤ 2.5

where (i) γ is the interval [– i, i] on the y-axis. (ii) γ is semi-circle z = cos φ + i sin φ, – 20. Show that

z

dz

π π ≤φ≤ . 2 2

≤ 4π/3 z +1 where γ is the circle γ(ϕ) = 2(cos ϕ + i sin ϕ), 0 ≤ ϕ ≤ 2π. γ

2

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LINE INTEGRAL AND CAUCHY’S THEOREM

21. Let γ be any closed path with graph not containing zero. Find sin z dz. γ z2 22. Let {γn} be the sequence of arcs,

z

FG H

γn(t) = 1 –

f (z) = z2 + z + 1. Find lim

n→∞

z

γn

IJ K

1 2t + + it, 0 ≤ t ≤ 1, n = 1, 2, ..., and n +1 n +1

f ( z ) dz.

5

APPLICATION OF CAUCHY’S THEOREM

There are several versions of Cauchy’s theorem. In this chapter, we first state the homological version of Cauchy’s theorem and postpone the proof to a later chapter (see Appendix II). By using Cauchy’s theorem we derive Cauchy’s formula. In fact, Cauchy’s formula will be quite sufficient for many applications. By paths we mean rectifiable curves. Circles are assumed oriented counterclockwise unless otherwise specified. 5.1

THE WINDING NUMBER

Let γ be a closed path in C and let α ∉ {γ}. Then the index (the winding number) of γ with respect to α denoted by n(γ, α), indicates how many times γ winds around α, and in which sense. It follows that n(γ, α) is a positive integer if γ winds counterclockwise around α and a negative integer if γ winds clockwise around α, n(γ, α) = 0 if α is outside γ. Example .α1

.α1

.α2 .α1

.α

.α .α

Example I

n(γ, α) = 1 n(γ, α1) = 0

n(γ, α) = – 1 n(γ, α1) = 0

66

n(γ, α) = 2 n(γ, α1) = 1 n(γ, α2) = 0

67

APPLICATION OF CAUCHY’S THEOREM

Lemma 1 If γ : [0, 1]→ C is a closed rectifiable path and α ∉ {γ}, then 1 dz is an integer. 2πi γ z – α Proof: Define F : [0, 1] → C by

z

F(t) = Hence,

z

t 0

γ ′(u ) du . γ (u ) – α

F(0) = 0, and F(1) =

z

γ

dz . z–α

γ ′( t ) (0 ≤ t ≤ 1). γ (t ) – α We now compute the derivative of the function d – F(t) (e (γ(t) – α)) dt = e– F(t) γ′(t) – F′(t) e– F(t) (γ(t) – α))

Also,

F′(t) =

LM N

= e – F ( t ) γ ′( t ) –

γ ′( t ) ( γ (t ) – α ) γ (t ) – α

OP Q

= 0. Hence, there is a constant function A such that e– F(t)(γ (t) – α) = A, so γ (t) – α = AeF(t). Since γ is a closed path, we have γ (0) = γ (1) and AeF(1) = γ (1) – α = γ (0) – α = AeF(0). Since γ (0) – α ≠ 0, we conclude that A ≠ 0, so that eF(0) = eF(1). Hence, there is an integer k such that F(1) = F(0) + 2πik. But F(0) = 0, so F(1) = 2πik and the lemma is proved. We now define the index (the winding number) of γ. Definition: Let γ be a closed path in C. Then for α ∉ {γ} 1 dz n(γ, α) = γ 2πi z – α is called the index (the winding number) of γ with respect to the point α. The following properties of the index can be easily checked.

z

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THE ELEMENTS OF COMPLEX ANALYSIS

Properties (a) Let γ1 and γ2 be closed paths having the same initial points, then

(i) n(γ1, α) = – n(– γ1, α) for every α ∉ {γ1}; (ii) n(γ1 + γ2, α) = n(γ1, α) + n(γ2, α) for every α ∉ {γ1}∪ {γ2}. (b) Let γ1 and γ2 be homotopic closed paths in C – {α}. Then

n(γ1, α) = n(γ2, α) for every α ∉ {γ1} ∪ {γ2}. Lemma 2 Let γ be a path. Then for α ∉ {γ1}, the function dz α→ γ z –α is a continuous function of α. Proof: Let z0 ∉ {γ}. We need to prove that

z

z FGH γ

1 1 – z – α z – z0

IJ dz K

tends to zero as α approaches z0. Consider the function t → | α – γ(t) |. This function is continuous and not zero. Hence, it has a minimum. Let δ be the minimum distance between the path and the point z0, i.e. δ = min | z0 – γ (t ) |. t

α – z0 1 1 – = . z – α z – z0 ( z – α ) ( z – z0 )

Now

Since | α – γ(t) | ≥ δ/2 if α is sufficiently close to z0, we find that α – z0 1 ≤ 2 | α – z0 | . ( z – α ) ( z – z0 ) δ /2

Hence (1)

z FGH γ

IJ K

1 1 1 – dz ≤ 2 | α – z0 | L( γ ) . z – α z – z0 δ /2

The right side of (1) tends to zero as α → z0 and thus the lemma is proved. Lemma 3 Let γ be a closed path. Let S be a connected set not intersecting γ. Then the function is constant for α in S. If S is unbounded, then this constant is zero. dz α→ γ z –α

z

69

APPLICATION OF CAUCHY’S THEOREM

Proof: We know from Lemma 1 that the integral is the winding number, and is therefore an integer. If a function takes its values in the integers and is continuous, it follows from Lemma 2 that it is constant on any curve. This implies that it is constant on a connected set. If S is not bounded, then for α arbitrarily large, the integrand has arbitrarily small absolute value, that is 1 is arbitrarily small. |z –α| An estimate of the integral shows that this constant must be zero. 5.2

STATEMENT OF CAUCHY’S THEOREM

Before stating Cauchy’s theorem in a broader set up we need some new definitions. Definition: Let Ω be an open set. A closed path γ in Ω is homologous to 0 in Ω if dz =0 γ z –α for every point α not in Ω. In other words, n(γ, α) = 0 for every α ∉ Ω. Let γ1, γ2 be closed paths in Ω. γ1 and γ2 are said to be homologous in Ω if n(γ1, α) = n(γ2, α) for every α ∉ Ω. The following results can be verified easily.

z

(i) Let γ1 and γ2 be closed paths in Ω. If γ1 and γ2 are homotopic, then they are homologous. (ii) Let γ1 and γ2 be closed paths in Ω. If γ1 and γ2 are near-together, then they are homologous.

Definition: Let γ1, γ2, ..., γn be curves and let m1, m2, m3, ..., mn be integers. A sum of the form n

γ = ∑ mi γ i i =1

is called a chain. We say that γ is a chain in Ω if each γi is in Ω. A chain is said to be closed if it is a finite sum of closed paths. We define

z

γ

n

f = ∑ mi i =1

z

γi

f

where γ is a chain. If γ is a closed chain where each γi is a closed path, then the index of γ with respect to a point α is defined as

70

THE ELEMENTS OF COMPLEX ANALYSIS

n(γ, α) =

1 2πi

z

γ

dz z–α

where α is not on the chain. We denote (i) γ1 ~ γ2 if n(γ1, α) = n(γ2, α) for every α ∉ Ω. (ii) γ1 ~ 0, if n(γ1, α) = 0 for every α ∉ Ω. Cauchy’s Theorem: Let Ω be an open set. Let γ be a closed chain in Ω and let γ ~ 0 in Ω. Then

z

f = 0 for every analytic function in Ω.

γ

Corollary: Let γ1 and γ2 be closed chains in Ω and let γ1 ~ γ2 in Ω. Then

z z γ1

5.3

f =

f.

γ2

SOME CONSEQUENCES OF CAUCHY’S THEOREM

We now derive two important consequences of Cauchy’s theorem. Theorem 1. Let γ be a closed chain in an open set Ω such that γ ~ 0 in Ω. Let f be analytic on Ω except at a finite number of points ξ1, ξ2, ..., ξn. Let γi (i = 1, 2, ..., n) be the boundary of the closed disc Di (ξi) contained in Ω. Assume that

Di ∩ Dj = φ if i ≠ j. Let and

mi = n(γ, ξi) Ω* = Ω – {ξ1, ξ2, ..., ξn}.

Then

γ ~ ∑ mi γ i in Ω*

and

n

i=1

z

z

n

γ

f = ∑ mi i =1

γi

f.

Proof: Let n

σ = γ – ∑ mi γ i i =1

and let α ∉ Ω. Then n

n(σ, α) = n(γ, α) – ∑ mi n (γi, α) i =1

= 0. If α = ξk for some k, then by Lemma 3

RS T

1 if i = k n(γi, ξk) = 0 if i ≠ k .

Hence

n(σ, ξk) = n(γ, ξk) – mk = 0.

71

APPLICATION OF CAUCHY’S THEOREM

This proves that σ ~ 0 in Ω*. By Cauchy’s theorem we conclude that

z

z

n

γ

f = ∑ mi

f.

γi

i =1

We illustrate the theorem in Fig. 5.I.

γ3 γ1

γ2

Fig. 5.I

and

z

γ

– f =–

z

γ1

γ ~ – γ1 – 2γ2 – γ3 (– f ) – 2

z

γ2

(– f ) –

z

γ3

(– f ) .

Theorem 1 will be applied in many cases when Ω is a disc and γ is a circle in Ω. For example, let ξ1, ξ2, ξ3, ..., ξn be points inside the circle, as shown in Fig. 5.II. Then

z

n

γ

f =∑

i =1

where γi is small circle around ξi.

z

γi

f

ξ2

ξ3

ξ1

ξn

Fig. 5.II

Theorem 2 (Cauchy’s Integral Formula). Let γ be a closed chain in an open set Ω such that γ ~ 0 in Ω. Let f be analytic on Ω. Then for every z0 in Ω and not on γ, 1 f(z) dz = n(γ, z0) f (z0 ). γ 2 πi z – z0

z

72

THE ELEMENTS OF COMPLEX ANALYSIS

Proof: Consider the auxiliary function g defined by f ( z ) – f ( z0 ) . g(z) = z – z0 Note that z0 is fixed and z is variable. Since f (z) – f (z0) and z – z0 are both differentiable in Ω, the quotient f ( z ) – f ( z0 ) g(z) = z – z0 is differentiable except at z = z0. By continuity we define f ( z ) – f ( z0 ) g(z0) = lim f ′( z 0 ) . z → z0 z – z0 Then g is continuous on Ω and is bounded in a neighbourhood of z0. Let Cr be the circle of radius r centred at z0. By Theorem 1 we have f ( z ) – f ( z0 ) dz = g( z ) dz = n( γ , z0 ) g( z ) dz. (2) γ γ Cr z – z0 Since g is bounded in the neighbourhood of z0 and since the length Cr tends to zero as r → 0, we find that the right side of (2) approaches 0 in absolute value as r → 0. The left side of (2) is independent of r and is therefore equal to 0. Hence f ( z0 ) f ( z) dz = dz = n(γ, z0) 2πif (z0). γ z – z γ z – z 0 0 Thus the theorem is proved.

z

z

z

5.4

z

z

APPLICATION OF CAUCHY’S INTEGRAL FORMULA

In the preceding section, we derived Cauchy’s integral formula in a broader set up. In this section, we shall apply the formula. In fact, for many applications the simpler form of Cauchy’s formula will be used. We restate the theorem. We denote the centre of the disc by α, its radius ρ > 0 and assume that f is analytic in a larger disc centred at α. Theorem 3. Let f be analytic on the disc Dr(α). Let C1 be the circle | z – α | = ρ < r. Then for any point z0 in Dρ(α) we have 1 f (z) (3) f (z0) = dz . 2 πi C1 z – z0 Applying Theorem 3, it will be shown that an analytic function possesses derivatives of all orders. This is a very remarkable result. Theorem 4. Let f be analytic in a region Ω. Then f ′(z), f ″ (z), ..., f n(z), ... all exist in Ω and are analytic. Proof: Let α be a point of Ω. Then it will be proved that f has all derivatives at α.

z

73

APPLICATION OF CAUCHY’S THEOREM

Let C1 be a sufficiently small circle with centre α. It follows from (3) that for any point z0 inside C1 we have 1 f (z) dz. (4) f (z0) = C 2 πi 1 z – z0 Differentiating (4) n times with respect to z0, we have n! f (z) dz . (5) f n(z0) = C 2 πi 1 ( z – z0 )n + 1 The validity of (5) will now be established. Choose h sufficiently small so that z0 + h is in the interior of the disc bounded by C1. Using (4) we obtain

z

z

f ( z0 + h) – f ( z0 ) 1 1 = h 2 πi h

z z z

1 = 2 πi 1 = 2 πi where

z

C1

f (z)

LM 1 – 1 OP dz Nz – z – h z – z Q 0

0

f (z) dz C1 ( z – z – h) ( z – z ) 0 0 f (z) dz + J C1 ( z – z ) 2 0

f (z) h dz . 2 2 πi C1 ( z – z0 ) ( z – z0 – h) Since z0 is inside C1, min | z – z0 | for z on C1 is positive. Denote this minimum by 2δ. It now follows that if | h | < δ then for z on C1, | z – z0 – h | ≥ | z – z0 | – | h | > δ. Since f is uniformly continuous on C1, there exists some constant M such that | f (z) | ≤ M for z on C1. Let ρ be the radius of C1. |h| M Then | J |≤ 2 πρ 2π (4 δ 2 ) δ | h | Mρ = . 4 δ3 Letting | h | → 0 we get | J | → 0. This proves the existence of f ′ and establishes (5) for n = 1. Note that the existence of f ′ was assured by our assumption that f is analytic. However, repeating the above argument, starting with (5) with n = 1, establishes the existence of f ″ and proves (5) for n = 2. Thus f ′ has a derivative f ″ and so f ′ is analytic. This proves that if f is analytic then f ′ is also analytic. Applying this to f ′ instead of to f we prove that f ″ is analytic. More generally, we find that the analyticity of f n implies that of f n + 1. Hence, by induction, it follows that an analytic function has derivatives of all orders which are analytic. We have J=

74

THE ELEMENTS OF COMPLEX ANALYSIS

z

f (n) ( z) 1 dz . 2 πi C1 z – z0 Integrating by parts n times we obtain (5). We now establish an important inequality which is called Cauchy’s inequality. Theorem 5 (Cauchy’ Estimate). In Theorem 4 assume that there exists a positive constant M such that | f (z)| ≤ M on Dr (α). Then Mn ! (6) | f n(α) | ≤ n . r Proof: In (5), let C1 be a circle of radius ρ with centre at z0 = α. Then

f n(z0) =

z

n! f ( z ) dz C 2 πi 1 ( z – α ) n + 1 n! M ≤ | dz | 2 π ρn + 1 C1 n !M = n . ρ Since this holds for all ρ < r and letting ρ → r we have Mn ! | f n (α) | ≤ n . r From Cauchy’s estimate we will prove an important theorem which is known as Liouville’s theorem. Before starting the theorem we need the definition of entire function. Definition: A function f is called entire if it is analytic on the whole of C. Theorem 6 (Liouville’s Theorem). A bounded entire function is constant. Proof: By hypothesis there is a constant M such that | f (z) | ≤ M. By (6), with n = 1, M for arbitrary r. | f ′(α) | ≤ r Hence, f ′(α) = 0. Since α is arbitrary, f ′(z) = 0. | f n (α ) | =

Also,

f (z) – f (0) =

z

z

0

z

f ′(ζ) dζ .

It now follows that f (z) = f (0) which proves the theorem. We are now in a position to appreciate Liouville’s theorem in the following application. Theorem 7 (Fundamental Theorem of Algebra). Let f(z) be a non-constant polynomial. Then there is a complex number z0 with f (z0 ) = 0.

75

APPLICATION OF CAUCHY’S THEOREM

Proof: Let f (z) = a0 + a1z + ... + anzn where an ≠ 0. Suppose that f (z) ≠ 0 for all z. Let g(z) = [f (z)]–1. This function is analytic on the whole of C. Write

F H

I K

b b1 b2 + 2 + ... + nn z z z where b1, b2, b3, ..., bn are appropriate constants. Observe that | f (z) | is large when | z | is large. Hence | g(z) | → 0 as | z | → ∞. It now follows that there is a number

f (z) = anzn 1 +

r > 0 such that | g(z) | < 1 if | z | > r. But g is continuous on D (0, r), so there is a constant M such that | g(z) | ≤ M for | z | ≤ M. Hence g is a bounded entire function and by Liouville’s theorem g must be constant. It follows that f must be constant which contradicts our assumption. Hence, the theorem is proved. Example: Let f be entire function. Suppose that there are constants A and m such that | f (z) | ≤ | z |m for | z | ≥ A. Then f is a polynomial of degree at most m. We verify the validity of this example. Write m ≤ k, φ(z) =

f ( z ) – ( a0 + a1 z + a2 z 2 + ... + ak – 1z k – 1 )

zk Observe that φ(z) is analytic everywhere. Now we have

ak – 1 a0 a + k 1– 1 + ... + ≤B k z z z for some constant B when | z | ≥ A. When | z | ≤ A, | φ(z) | is bounded. Applying Liouville’s theorem we find that φ(z) is a constant. This shows that f is a polynomial of degree at most m. In Theorem 4, we have proved that an analytic function has derivatives of all orders which are analytic and is represented by the formula (5). Using this result we prove a classical result which is known as Morera’s theorem. Theorem 8 (Morera’s Theorem). Let f be continuous in a region Ω and let | φ(z) | ≤ 1 +

z

γ

f = 0 for all closed curves γ in Ω. Then f is analytic in Ω. Proof: Let f be analytic in Ω. Define F(z) =

z

z z0

f (ζ) dζ

where z is any point in Ω and z0 is a fixed point in Ω. The hypothesis implies, as we have already seen in section 4.7, that f (z) is the derivative of an analytic function F(z). It follows from Theorem 4 (section 5.4)

76

THE ELEMENTS OF COMPLEX ANALYSIS

that the derivative of an analytic function is itself analytic, that is, f is analytic in Ω. This proves Morera’s Theorem.

EXERCISES

z

ez – e– z γ z4 where γ is one of the curves depicted below.

1. Compute

I

II

O

Exercise I

z FH

I K

1 dz z where γ is the unit circle γ(t) = cist, 0 ≤ t ≤ 2π.

2. Compute

γ

z+

3. Using Cauchy’s formula for f (n)(z), determine the value of P( z ) dz k γ ( z – c) where γ is the circle γ(ϕ) = c + r(cos ϕ + i sin ϕ), and P(z) is a polynomial and k a positive integer.

z

z

z 2 + 3z + 5 dz γ z +1 where γ is given by γ(t) = a + r(cos t + i sin t), 0 ≤ t ≤ 2π, r > 0 and a ∈ C.

4. Compute

z

3z 2 + 7 z + 1 dz in the following cases: c1 z +1 (i) C1 is the circle | z + 1 | = 1.

5. Determine the value of

(ii) C1 is the circle | z + i | = 1. (iii) C1 is the ellipse x2 + 2y2 = 8. z+4 dz in the following cases: 6. Determine the value of 2 c1 z + 2 z + 5 (i) C1 is the circle | z | = 1. (ii) C1 is the circle | z + 1 – i | = 2. (iii) C1 is the circle | z + 1 + i | = 2.

z

77

APPLICATION OF CAUCHY’S THEOREM

z

ez dz where C1 is the circle | z – 1 | = 3. 2 c1 ( z + 1) z +1 8. Determine the value of dz , where 3 2 c1 z – 2 z (i) C1 is the circle | z | = 1. (ii) C1 is the circle | z – 2 – i | = 2. 2z – i dz 9. Compute 2 γ z – 2z + 5 in the positive sense around the contour γ of a simply connected domain. Examine all possibilities that can arise in this case. 10. Suppose that f is analytic on D(0, 1) and suppose | f (z) | ≤ 1 for | z | < 1. Prove that | f ′(0) | ≤ 1. 7. Compute

z

z

11. Compute

z

γ

FG z IJ H z – 1K

n

dz

where γ(t) = 1 + eit, 0 ≤ t ≤ 2π and n is any positive integer. 12. Suppose that f is analytic on the closed unit disk D. Determine the value of the integral

zz D

f ( x + iy) dy dx.

13. Suppose that f is analytic on an open set G. Suppose that z0 ∈ G and f ′(z0) ≠ 0. 2 πi 1 dz where C1 is a small circle centred at z0. = Prove that C f ′( z 0 ) 1 f ( z ) – f ( z0 )

z

14. Suppose that f is analytic on D(0, 1) and | f (z) | ≤

1 for | z | < 1. Determine (1 – | z |)

the best estimate of | f n(0)|. 15. Prove that

z

R+z 1 dz = 1 2 πi γ r ( R – z ) z where γr has the equation z = reiθ, 0 ≤ θ ≤ 2π. Deduce that 1 2π

z

2π

0

16. Prove that 1 2 πi

R2 – r 2 dθ = 1, 0 < r < R. R + r 2 – 2 rR cos θ 2

z

f ( Re iθ )

2π

0

Re iθ

FR I –G Je HrK 2

(i Reiθ dθ) = 0 iφ

where C1 is the circle defined by z = Reiθ and z0 = retφ for r < R. 17. Suppose that f is analytic in the entire (finite) plane and suppose that f (z) is not a constant. Let R and M be any real numbers (no matter how large). Prove that | f (z) | > M for | z | > R.

78

THE ELEMENTS OF COMPLEX ANALYSIS

18. Suppose that f (z) is a polynomial of degree n > 0 and let M be any arbitrary positive real number (no matter how large). Prove that | f (z) | > M for | z | > R. 19. We recall the Weierstrass’ theorem for a real interval [a, b]. The theorem states that a continuous function can be uniformly approximated by polynomials. It is true that every continuous function on the closed unit disk be uniformly approximated by polynomials? Examine this. 20. Recall the definition of the sup norm, i.e. || f || = sup | f (z) | z ∈E

where E is any set. We say that {fn} is a Cauchy sequence (for the sup norm), if given ε, there exists N such that if m, n ≥ N, then || fn – fm || < ε. Let S be the closure of a bounded open set in C. Let f, g be continuous functions on S. Define their scalar product 〈 f, g〉 =

zz RSz z T

S

f ( z ) g( z ) dy dx

and define the associated L2-norm by || f ||2 =

S

| f ( z ) |2 dy dx

Prove that || f ||2 does define a norm. We define || f ||1 =

zz

S

UV W

1/ 2

.

| f ( z ) | dy dx .

Prove that f → || f ||1 is a norm on the space of continuous functions on S. This is called L1-norm. (i) Suppose that f is analytic on D(0, r) and let 0 < r1 < r. Prove that there exists constants c1, c2 > 0 such that || f || ≤ c1 || f ||1 ≤ c2 || f ||2 where || || is the sup-norm on the closed disk of radius r1 and L1, L2 norms also refer to the closed disk of radius r1. (ii) Suppose that {fn} is a sequence of analytic functions on an open set G, and let {fn} be L2-Cauchy. Prove that {fn} converges uniformly on compact subsets of G.

6

POWER SERIES

In this chapter, we give an outline of the basic properties of a power series. We first recall some elementary facts on infinite series in C. 6.1

INFINITE SERIES IN C

Consider the infinite series ∞

(1) Σ zn = z0 + z1 + z2 + ... n=0

where the zn are arbitrary complex numbers. Associated with this series is the sequence of its partial sums (2) sn = z0 + z1 + ... + zn – 1 + zn. ∞

We say that the series Σ zn converges to z if lim sn exists and is equal to z. n→∞

n=0

∞

In other words, the series Σ zn converges to z if for every ε > 0 there is an integer n=0

n0 such that

n

∑ zk − z < ε whenever n ≥ n0.

k=0

If this is the case, we say that z is the sum of the infinite series, that is, ∞

z = ∑ zk . k=0

∞

∞

n=0

n=0

If z = ∑ zn and w = ∑ wn are two convergent series, with partial sums n

n

k=0

k=0

sn = ∑ z k and tn = ∑ w k , then the sum converges and ∞

z + w = ∑ ( zn + wn ) . n=0

79

80

THE ELEMENTS OF COMPLEX ANALYSIS ∞

∞

n=0

n=0

We say that the series ∑ zn converges absolutely if ∑ | zn | converges. ∞

∞

n=0

n=0

Theorem 1. If ∑ zn converges absolutely, then ∑ zn converges. Proof: Let ε > 0 and put ∞

sn = ∑ z k . k=0

∞

By assumption, ∑ | zn | is convergent. Hence, there is an integer n0 such that n=0

∞

∑ | zn | < ε

n = n0

Thus if m > n ≥ n0, m

m

k = n +1

k = n +1

∑ zk ≤ ∑ | zk |

| sm – sn | =

∞

≤ ∑ | zk | < ε. k = n0

with

That is, {sn} is a Cauchy-sequence and so there is a complex number z lim sn = z.

n→∞

∞

∑ zn = z.

Hence

n=0

If a series does not converge, we say that it diverges or is divergent. ∞

∞

n=0

n=0

Theorem 2. Let ∑ zn and ∑ wn be two absolutely convergent series. Let n

cn = ∑ z k wn – k = z0 wn + z1 wn – 1+ ... zn w0. k=0

∞

∑ cn is absolutely convergent and

Then

n=0

∞

FG H

∞

IJ FG KH

∞

IJ K

(3) ∑ cn = ∑ zn . ∑ wn . n=0

n=0

n=0

Proof: We have already seen this theorem during a real analysis course. Put and

n

n

n

k=0

k=0

An = ∑ z k , Bn = ∑ w k , Cn = ∑ ck k=0

n

n

n

k=0

k=0

k=0

αn = ∑ | zk | , βn = ∑ | wk | , γn = ∑ | ck | .

81

POWER SERIES

It can be easily checked that γn is a monotonically increasing and bounded sequence. Also, (4) | A2n B2n – C2n | ≤ | α2n β2n – αn βn |. It follows from Cauchy-criterion that for sufficiently large n, the right-side of (4) is ≤ ε. lim A2n B2n = lim C2n. Hence n→∞

n→∞

FG ∑ z IJ . FG ∑ w IJ = ∑ c . H KH K ∞

That is,

∞

n=0

n

∞

n

n=0

n=0

n

We now recall the definitions of limit inferior and limit superior of a sequence of real numbers. Let {an} be a sequence of real numbers. We define lim inf an = lim an = lim [inf {an, an + 1, ...}]. n→∞

lim sup an = lim an = lim [sup {an, an + 1, ...}]. n→∞

Note that lim an and lim an always exist although they may be ± ∞. We assume that the reader is familiar with the root and the ratio tests for convergence of infinite series whose terms are real. Here we state the ratio test and the root test for series of complex terms. ∞

Theorem 3 (Ratio Test). Let ∑ z k be a series of non-zero complex terms. k=0

Let zn + 1

λ = lim inf

zn

n→∞

Λ = lim sup

and

zn + 1

n→∞

zn

.

Then ∞

(i) the series ∑ z k converges absolutely if Λ < 1; k=0 ∞

(ii) the series ∑ z k diverges if λ > 1; and k=0

(iii) the test gives no information if λ ≤ 1 ≤ Λ. ∞

Theorem 4 (Root Test). Let ∑ z k be a series of complex terms. Let k=0

r = lim sup n | zn | . n→∞

Then ∞

(i) the series ∑ z k converges absolutely if r < 1; k=0

82

THE ELEMENTS OF COMPLEX ANALYSIS ∞

(ii) the series ∑ z k diverges if r > 1; and k=0

(iii) the test gives no information if r = 1. 6.2

SERIES OF FUNCTIONS AND UNIFORM CONVERGENCE

Let E be a set, and f a bounded function on E. Define || f || = sup | f ( z ) | . z ∈E

Let {fn}, n = 0, 1, 2, 3, ..., be a sequence of functions on E. We say that {fn} converges uniformly on E to a function f if the following property is satisfied. For every ε > 0 there exists an integer n0 such that if n ≥ n0 , then || fn – f || < ε. We say that {fn } is a Cauchy-sequence for the sup norm, if for every ε > 0, there exists an integer n0 such that m, n ≥ n0, then || fn – fm || < ε. Note that for each z ∈ E, the inequality | fn (z) – f m (z) | ≤ || fn – fm || holds. Hence, for each z ∈ E, the sequence of complex numbers {fn (z)} converges. Theorem 5. Let {fn} be a sequence on E which is Cauchy-sequence. Then {fn} converges uniformly on E. Proof: Suppose that for each z ∈ E, lim f (z) = f(z).

n→∞ n

Let ε > 0 be given. Then there exists an integer n0 such that if m, n ≥ n0, then | fn (z) – fm (z) | < ε, for all z ∈ E. Let z ∈ E and let n ≥ n0. Choose m ≥ n0 sufficiently large (depending on z) such that | f (z) – fm (z) | < ε. Then | f (z) – fn (z) | ≤ | f (z) – f m (z) | + | fm (z) – fn (z) | < ε + || fm – fn || < 2ε. This is true for every n ≥ n0 and every z ∈ E, which completes the proof. ∞

We say that the series ∑ fk ( z ) converges uniformly on E if the sequence k=0

{sn} of partial sums defined by sn (z) = f0(z) + f1(z) + ... + fn (z) converges uniformly.

83

POWER SERIES ∞

We say that the series ∑ fk ( z ) converges absolutely if k=0 ∞

∑ | fk ( z ) | converges.

k=0

Theorem 6 (Weierstrass-M-test). Let {fn} be a sequence of functions defined on E. Let Mn be a sequence of non-negative real numbers such that || fn || ≤ Mn for all n. ∞

∞

n=0

n=0

∑ fn converges uniformly if ∑ Mn converges.

Then

Proof: Let m ≤ n. Let n

sn = ∑ fk . k=0

|| sn – sm || ≤

Then

n

n

k = m +1

k = m +1

∑ || fk || ≤ ∑ Mk .

∞

Since ∑ Mn converges, it follows from Theorem 5 that n=0

∞

∑ fn converges uniformly.

n=0

Theorem 7. Let E be a set of complex numbers, and let {fn} be a sequence of continuous functions on E. Suppose that {fn} converges uniformly to f on E. Then f is continuous on E. Proof: The proof is similar to as in the case of the set of real numbers. Let z0 ∈ E. Choose n sufficiently large such that (5) || fn – f || < ε. Since {fn} is continuous at z0, we can choose δ such that whenever | z – z0 | < δ we have (6) | fn (z ) – fn (z0) | < ε for sufficiently large n as chosen above. Thus (7) | f (z) – f(z0) | ≤ | f (z) – fn (z) | + | fn (z) – fn (z0) | + | fn (z0) – f (z0) |. The first and third terms on the right side of (7) are bounded by || f – fn || < ε. The second term on the right side on (7) is < ε by (6). Hence | f(z) – f(z0) | < 3ε, and the theorem is proved.

84

THE ELEMENTS OF COMPLEX ANALYSIS

6.3

POWER SERIES

A power series about a is an infinite series of the form ∞

n (8) ∑ an ( z − a) = a0 + a1 (z – a) + a2 (z – a)2 + ... n=0

where a and a0, a1, a2, ..., are complex numbers. If a = 0, we obtain as a particular case a power series ∞

n (9) ∑ an z = a0 + a1 z + a2 z2 + ... . n=0

Example (i) One of the simplest example of a power series is the geometric series ∞

∑ z n = 1 + z + z2 + z3 + ... .

n=0

Suppose that a power series (8) is given, we would like to consider all the points z in the complex plane for which the series converges. The following three possibilities arise: (i) The series converges only at z = a; (ii) The series converges in the entire complex plane; and (iii) There exists an open disc D (a ; r) such that the series converges inside the disc and diverges outside the disc. The circle | z – a | = r is called the circle of convergence, and its radius r is called the radius of convergence of the series (8). The radius of convergence of the power series (8) may be determined from the coefficients of the series. In fact, we have the following theorem: Theorem 8. With each power series (8) we can associate a real number (10) r =

RS 1 UV (0 ≤ r ≤ ∞), T lim sup | a | W n

1/ n

called the radius of convergence, which has the following properties: (i) The series converges absolutely in the open disc D (a, r); (ii) The series diverges outside the closed disc D (a, r); and (iii) The series converges uniformly in every closed disc D (a, ρ) where ρ > r. In other words, the series converges uniformly on every compact set E ⊂ D (a, r). Proof: Let

r=

RS 1 UV (0 ≤ r ≤ ∞). T lim sup | a | W n

1/ n

We want to show that r is the radius of convergence. If r = 0 then no matter how small we choose ρ = | z – a |, the expression | an |1/n ρ > kn for an infinite number of terms (k an arbitrary constant). Choosing k > 1, we find the sequence | an | | z – a |n

85

POWER SERIES

does not converge to 0 as n → ∞. Hence the series (8) diverges for | z – a | ≠ 0 and converges to a0 for z = a. If r > 0, i.e. if lim sup | an |1/n = 1/r, then for any ρ, 0 < ρ < r, choose ρ0, 0 < ρ < ρ0 < r such that (11) | an |1/n
ρ0 ρ0 r

IJ K

for sufficiently large n. If follows from (11) that

F I F I |ρ =|a |ρ G ρ J r. Then for an infinite number of terms of the series, we have 1 | an |1/n > ρ i.e., | an | | z – a |n = | an | ρn > 1. Hence, the series (8) diverges. (iii) Using Weierstrass M-Test, we find that the series (8) converges uniformly on every closed disc D (a, ρ) where ρ < r. Since E ⊂ D (a, ρ) with ρ < r, the series (8) converges on any compact set E ⊂ D (a, r). When r = ∞, the series (8) converges uniformly on any compact subset E of the complex-plane C. Let Ω be a region and let the open disc D (a, r) ⊂ Ω. We say that a function f defined in Ω is representable by power series in Ω if to every disc D (a, r) ⊂ Ω there corresponds a series (8) which converges to f(z) for all z ∈ D (a, r). The geometric series in Example (i) represents the function 1 f(z) = for | z | < 1. 1− z Examples: (ii) We define the series ∞ zn z z2 z3 + exp (z) = 1 + + + ... = ∑ . 1! 2 ! 3! n=0 n!

86

THE ELEMENTS OF COMPLEX ANALYSIS

It can be easily seen that the radius of convergence of this series is r = ∞ and defines a continuous function for all values of z. z3 z5 – ... – + 3! 5! ∞ z 2n + 1 n ( − 1 ) = ∑ (2n + 1) ! n=0 2 z z4 + and cos z = 1 – – ... 2! 4! ∞ z 2n = ∑ ( − 1) n (2 n) ! n=0 converges at every complex number and the convergence is uniform on each compact subset of C. These examples give extensions to the exp, sine and cosine functions for the complex numbers.

(iii) The series

(iv) The series

sin z = z –

∞

∑ n ! z n = 1 + z + 2z2 + 6z3 + ...

n=0

converges only at z = 0. z2 z3 zn + =z+ + ... 2 3 n =1 n converges for | z | < 1, the series diverges for | z | > 1. At z = 1, the series diverges and at z = – 1, the series converges. This example shows that a series may converge or diverge on the circle of convergence.

(v) The series

∞

∑

FG H

IJ K

FG H

IJ K

2

FG H

1 z −1 1 z −1 1 z −1 + 2 + 2 2 2 z +1 3 z +1 4 z +1 converges for Re (z) ≥ 0 and diverges for Re (z) < 0. Using the ratio test, we have

(vi) The series 1 +

fn + 1 ( z )

IJ K

3

+ ...

n2 z −1 . 2 fn ( z ) (n + 1) z + 1 It follows that the series converges absolutely for those values of z for which =

z −1 < 1. Thus the series converges absolutely in the region defined by z +1 | z – 1 | < | z + 1 |. In other words, the series converges absolutely for values of z which lie in the right half of the z-plane. The series diverges in the region defined by |z–1|>|z+1| i.e., for values of z which lie in the left half of the z-plane.

87

POWER SERIES

The test fails for values of z which lie on the perpendicular bisector of the line joining z = 1 and z = – 1.

z −1 1 1 = 1, the corresponding series 1 + 2 + 2 + ... z +1 2 3

Note that when

converges. Hence, the series converges for Re (z) ≥ 0 and diverges for Re (z) < 0. (vii) The series

FG z IJ + 1 FG z IJ H z + 1K 2 H z + 1 K 2

1 2

converges for Re (z) > –

2

FG H

z 23 + 3 3 z +1

IJ K

3

and diverges for Re (z) ≤ –

have

F H

FG H

z 34 + 4 4 z +1 1 2

IJ K

4

+ ...

. Using the root test, we

I K

1 z . n 1+ z It follows that the series converges absolutely for values of z which lie in the 1 region Re (z) > – . The series diverges for values of z which lie in the region 2 1 Re (z) < – . 2 n

| fn ( z ) | = 1 −

The test fails for values of z for which Note that when

z = 1. z +1

z = 1, the corresponding series z +1 1+

1 2 3 34 + + + ... 2 2 33 4 4

diverges. (viii) The series

1 + z + z2 + ... + zn + ... 1 converges uniformly to for 0 ≤ | z | ≤ ρ < 1, but not for | z | < 1. 1− z We have seen in Example (i) that this series converges absolutely to

1 1− z

for | z | < 1. Observe that the remainder 1 1 − zn +1 zn +1 = − 1− z 1− z 1− z becomes arbitrary large for real z = x < 1 and sufficiently close to 1. Thus for given ε > 0 we cannot find a η0 which is independent of z and such that | Rn (z) | < ε for all n > η0. Hence, the series is not uniformly convergent in the region | z | < 1. Rn (z) =

88

THE ELEMENTS OF COMPLEX ANALYSIS

Note that in the region 0 ≤ | z | ≤ ρ < 1, ∞

∞

∞

n=0

n=0

n=0

∑ | z n | = ∑ | z |n ≤ ∑ ρ n .

It follows from Weierstrass M-test that the series is uniformly convergent in the region 0 ≤ | z | ≤ ρ < 1.

EXERCISES 1. Find the radius of convergence for each of the following series: (i)

∞

∑ nn z n

(ii)

n =1

(iii) (v)

n =1

∞

∑ 2n zn

(iv)

n=0

∑ 2 −n z n

(vi)

n =1

(vii) (ix)

∑

n =1

(viii)

∑ z n!

(x)

n=0

n=0

∞

∑ n2 z n ∞

∑

n =1

∞

∞

∞

∑ (log n)2 z n

n =1

n! n z nn

∞

∑

n =1

F H

2n (xi) ∑ 2 z −

zn nn

n =1

∞

∞

∞

∑

1 2

I K

n

(xii)

(n !) 3 n z (3n) ! ( z − i) n n2

∞

∑ π n z 2n .

n=0

2. Prove that the radius of convergence of the power series ∞

∑

n =1

( − 1) n n( n + 1) z n

is 1, and examine the convergence for z = 1, z = – 1, and z = i. ∞

3. Suppose that f(z) =

∑ an z n have radius of convergence r > 0. Prove that the

n=0

following series have the same radius of convergence. (i)

∞

∑ nan z n

(ii)

n =1

(iii)

∞

∑ n 2 an z n

n =1

∞

∑ n α an z n for any positive integer α

n =1

(iv)

∞

∑ nan z n − 1.

n =1

4. Prove that if r is the radius of convergence of the series convergence of the series

∞

∑

n=0

an2

n

z is r2.

∞

∑ an z n , then the radius of

n=0

89

POWER SERIES

5. Suppose that

∞

∞

n=0

n=0

∑ an is a convergent series of complex numbers. Prove that ∑ an z n

is uniformly convergent on the domain of z such that π – λ < arg (z – 1) < π + λ and | z – 1 | ≤ δ where 0 < λ < π/2 and 0 < δ < 2 cos λ. 6. Suppose that an is a decreasing sequence of positive numbers approaching zero. Prove that

∞

∑ an z n is uniformly convergent on the domain of z such that | z | ≤ 1

n =1

and | z – 1 | ≥ δ where δ > 0. 7. Find the region of convergence and the sum of each of the following series: (i) 1 + (z – i) + (z – i)2 + (z – i)3 + ... (ii) z(1 – z) + z2 (1 – z) + z3 (1 – z) + ...

IJ K

FG H

2

IJ K

FG H

3

1 z +1 1 z +1 1 z +1 + ... + + 3 2 z − 1 22 z − 1 2 z −1 1 1 1 (iv) + ... + 2 + 3 2 2( z + i ) 2 ( z + i ) 2 ( z + i)3 8. Determine the region of convergence of each of the following: (iii)

(i) 1 +

FG Re (z) IJ H z +1 K Im ( z ) 1 F Im ( z ) I + G J z +1 3 H z +1 K

1 Re ( z ) 1 + 2 2 z + 1 32

1 (ii) 1 + 2 2

2

+ ... 2

2

+ ...

4 − z 2 (4 − z 2 )2 (4 − z 2 )3 + + + ... 12 22 33 9. Prove that the series

(iii)

z(1 – z) + z2 (1 – z) + z3 (1 – z) + ... converges uniformly for | z | ≤ ρ < 1 but not for | z | ≤ 1. 10. Find the region of convergence and the sum of the series z z z + + + ... . (0. z + 1) ( z + 1) ( z + 1) (2 z + 1) (2 z + 1) (3z + 1) Show that the series does not converge uniformly for | z | ≥ ρ > 1. 11. Find the region of convergence and the sum of the series

1 1 1 + ... . + + z ( z + 1) ( z + 1) ( z + 2) ( z + 2) ( z + 3) Find also the region of uniform convergence of the series. 12. Prove that the Riemann zeta function ζ defined by ζ (z) =

∞

∑ n−z

n =1

converges for Re (z) > 1 and converges uniformly for Re (z) ≥ 1 + ε where ε > 0 is arbitrary small.

7 7.1

LAURENT SERIES, SINGULARITIES

POWER SERIES REPRESENTATION OF ANALYTIC FUNCTIONS

Power series are basic in complex analysis. The following theorem illustrates this fact. Theorem 1. Let Ω be a region. If f is represented by the power series in Ω then f is analytic in Ω. Moreover, if ∞

(1) f (z) = ∑ an (z – a)n n=0

for z ∈ Dr (a), then for these z, ∞

n −1 (2) f ′ (z) = ∑ n an ( z − a ) . n =1

Proof: By the root-test we can show that the radius of convergence of both series in (1) and (2) are the same. Without loss of generality, we can take a = 0 and let g (z) be the sum of the series in (2). Fix w ∈ Dr (a) and choose ρ so that | w | < ρ < r. If z ≠ w, then ∞ ⎧⎪ z n − wn ⎫⎪ f ( z) − f (w) − nwn − 1 ⎬ . – g(w) = ∑ an ⎨ n =1 z−w ⎩⎪ z − w ⎭⎪ If n = 1, the expression in brackets in (3) is 0 and if n ≥ 2, we have the series

(3)

n –1

(4) (z – w) ∑ kwk −1 z n − k − 1 . k =1

The sum in (4) in absolute value is less than n(n − 1) n – 2 (5) ρ for | z | < ρ. 2 Thus (6)

f ( z ) − f ( w) − g( w ) ≤ | z – w z−w 90

∞

∑ n 2 an | ρn – 2 .

n=2

91

LAURENT SERIES, SINGULARITIES

Since ρ < r, the series in (6) converges. Thus the left side of (6) tends to 0 as z → w. This shows that f ′ (w) = g (w) and the proof is complete. Since f ′ satisfies the same hypothesis as f does, the theorem can be applied to f ′. It follows that f has derivatives of all orders, that each derivative is representable by power series in Ω. Thus we obtain ∞

(7) f (k) (z) = ∑ n (n – 1) ... (n – k + 1) an (z – a)n – k n=k

if (1) holds. Hence (1) implies that (8) n ! an = f (n) (a) (n = 0, 1, 2, ...). 7.2

TAYLOR SERIES

In Theorem 1, we have seen that power series with non-zero radius of convergence represent analytic functions. We shall now see that every analytic function can be developed in a convergent Taylor series. Theorem 2. Let f be analytic in Ω. Then f can be represented by a power series ∞

f (z) = ∑ an (z – a)n n=0

about each point a ∈ Ω. Proof: Let a ∈ Ω and let D (a ; r) be the greatest open disk contained in Ω. Since Ω is open, r is positive. Now choose ρ and ρ0 such that 0 < ρ < ρ0 < r and denote by γ the boundary of D (a ; ρ0), i.e. γ (t) = a + ρ0 cos t, 0 ≤ t ≤ 2π. By Cauchy’s formula, we have for each z ∈ D (a ; ρ) f (ξ) 1 dξ . (9) f(z) = 2πi γ ξ − z 1 1 1 1 Now . = = z ξ − z (ξ − w ) − ( z − w ) ξ − w 1 − − w ξ−w

z

| ξ – w | = ρ0, | z – w | ≤ ρ and z − w ≤ ρ/ρ0. ξ−w

where Then

and

FG H

IJ + ... K z − w F z − wI 1 1 L 1+ = +G M J ξ − z ξ − w MN ξ − w H ξ − wK

1 z−w z−w =1+ + z−w ξ−w ξ−w 1− ξ−w

2

2

OP PQ

+ ... .

92

THE ELEMENTS OF COMPLEX ANALYSIS

The geometric series in brackets converges uniformly with respect to ξ on γ and z ∈ D (a ; ρ). Since f (ξ) is continuous function on the compact set γ, f (ξ) is bounded on γ. Therefore

f (ξ) f ( ξ) f (ξ) = + ( z − w) + ... ξ−z ξ−w (ξ − w ) 2 converges uniformly and we can integrate the right side of (10) term by term and obtain (10)

z z z

f (ξ) 1 dξ γ 2πi ξ−z f (ξ ) f (ξ ) 1 1 dξ + ( z − w) = + ... γ γ ξ−w (ξ − w ) 2 2 πi 2 πi 1 f (ξ ) dξ (n = 0, 1, 2, ...) Setting an = 2 πi γ (ξ − w) n + 1 we find that f ( z) = a0 + a1 (z – a) + a2 (z – a)2 + ... . Using (8) and noting that f (z) =

1 2 πi

an =

z

γ

z

1 (n) f (ξ ) f (a) dξ = n +1 (ξ − w ) n!

we obtain ∞

(11) f (z) = ∑

n=0

1 (n) f (a) (z – a)n n!

( z − a) ( z − a) 2 (2) ( z − a) n (n) f ′ (a) + f (a) + ... f (a) + ... 1! n! 2! The right side of (11) is the Taylor series representation of f about a. This

= f (a) +

series converges uniformly in D (a; ρ), ρ < r. The reader can verify that this series representation of f (z) is unique. 7.3

ZEROS OF ANALYTIC FUNCTION

Let S be a set of points. Let a ∈ S. We say that a is isolated point of S if there exists a disk D (a ; r) (r > 0) such that D (a ; r) does not contain any point of S other than a. We say that S is discrete if every point of S is isolated. ∞

Let f be analytic near a and let ∑ an ( z − a) n be its Taylor series n=0

representation about a. We say that a is a zero of f if f (a) = 0. If f (a) = f ′ (a) = ... = f (n – 1) (a) = 0 and f (n) (a) ≠ 0, then we have a0 = a1 = a2 = ... = an – 1 = 0 and we say that a is a zero of order n of f. A zero of order one is called a simple zero.

93

LAURENT SERIES, SINGULARITIES

If a is zero of order m, then f may be represented as ∞

n f (z) = ∑ an ( z − a ) n=m

∞

= (z – a)m ∑ an ( z − a)n − m n=m ∞

= (z – a)m ∑ an + m ( z − a) n . n=0

∞

Denote h(z) by

n h(z) = ∑ an + m ( z − a) . n=0

f m ( a) ≠ 0. m! Now h is analytic and therefore continuous. Since h(a) ≠ 0, h(z) is different from zero near a. The factor (z – a)n = 0 only for z = a. Thus f(z) ≠ 0 near a. We thus obtain that the zeros of finite order of an analytic function are isolated. We now state a result and the proof is left to the reader. Theorem 3. Let f be analytic of Ω and let f (a) = f ′ (a) = ... f n (a) = ... = 0. Then f (z) = 0 in Ω. Example: Define the functions sin z and cos z by

h(a) = am =

Then

z3 z5 – ... + 3! 5! z2 z4 – ... and cos z = 1 – + 2! 4! The function sin z has simple zeros at z = 0, ± π, ± 2π, ... . The function 1 – cos z has second order zeros at z = 0, ± 2π, ± 4π, ... .

sin z = z –

7.4

LAURENT SERIES

Buy a Laurent series, we mean a series of the form ∞

(12) f (z) =

∑

n=−∞

an ( z − a )n .

Let E be the set of complex numbers. We say that the series (12) converges absolutely (uniformly) on E if the two series ∞

(13) f1 (z) = ∑ an ( z − a) n n=0

and ∞

(14) f2 (z) = ∑ a− n /( z − a)n n =1

94

THE ELEMENTS OF COMPLEX ANALYSIS

converges absolutely (uniformly) on E. In that case, we write f (z) = f1 (z) + f2 (z). Let 0 ≤ r1 < r2 ≤ ∞. We define the annulus, say annulus E = ann (a ; r1, r2) = {z : r1 < | z – a | < r2}. Notice that ann (a ; 0, r2) is a punctured disk. In the following theorem, we consider the Laurent series f (z) =

∞

∑

n=−∞

an z n and the annulus G defined by

ann (0 ; r1, r2) = {z : r1 ≤ | z | ≤ r2}. Theorem 4. Let f be analytic in the annulus G as above. Then f is representable by Laurent series ∞

f (z) =

∑ an z n

n=−∞

which converges absolutely and uniformly on s1 ≤ | z | ≤ s2 where r1 < s1 < s2 < r2. Let γ1 and γ2 be the circles of radius s1 and s2 respectively. Then the coefficients an are given by the formula

z z

f (ξ) dξ 1 , if n ≥ 0; 2 πi γ 2 ξ n + 1 1 f (ξ) dξ , if n < 0. (16) an = γ 2 πi 1 ξ n + 1 Moreover, the series representation is unique. Proof: By definition, f is analytic in the annulus r1 – ε < | z | < r2 + ε. It follows from Lemma 3 of Chapter 5 that the chain γ2 – γ1 is homologous to 0. Cauchy’s integral formula implies that (15) an =

z

z

f (ξ) dξ f (ξ) 1 1 − dξ γ γ 2 πi 2 ξ − z 2 πi 1 ξ − z = f2(z) + f1(z). Since f2 is analytic in the disk D (0, r2), it has a power series expansion about 0. By Theorem 2, we find that f (z) =

∞

n f2 (z) = ∑ an z n=0

where the coefficients an are given by (15). Write

FG H

ξ – z = – z 1−

Then | ξ/z | ≤

IJ K

ξ . z

r1 < 1, so the geometric series converges and we obtain s1

95

LAURENT SERIES, SINGULARITIES

F GH

FG IJ HK

1 1 1 ξ ξ . = 1+ + z (1 − ξ / z ) z z z

2

I JK

+ ... .

We can then integrate the series term by term and the desired result follows if we handle them as in the case of the derivation of Cauchy’s integral formula. If the function f is analytic in the annulus ann (a ; r1, r2), then f has a Laurent expansion. ∞

(17) f (z) =

an ( z − a)n

∑

n=−∞ ∞

∞

n=0

n =1

n n = ∑ an ( z − a ) + ∑ a− n /( z − a) .

The uniqueness of the Laurent series can be verified easily. The Laurent series is a generalization of the Taylor series. We call ∞

∞

n=0

n =1

∑ ( z − a) n the Taylor part and ∑ a− n /( z − a)n the principal part of the Laurent

series. If a–1 = a–2 = ... = a –n = ... = 0, then the principal part vanishes, and the Laurent series reduces to a Taylor series. Example: We want to find the Laurent series for the function 1 with centre z = 1. f (z) = 1 − z2 1 1 . =− ( z − 1) ( z + 1) 1 − z2

F H

1 1 1 ∞ z −1 = = ∑ − 2 z + 1 2 + ( z − 1) 2 n = 0

I K

n

( − 1) n (z – 1)n. n +1 n=0 2 This series converges for | z – 1 | < 2. ∞

= ∑

(18)

Similarly,

1 1 1 ∞ ⎛ 2 ⎞ = = ∑ ⎜− z + 1 ( z − 1) + 2 z − 1 n = 0 ⎝ z − 1⎟⎠

n

( − 2) n n +1 . n = 0 ( z − 1) This series converges for | z – 1 | > 2. Thus, from (18) we obtain

(19)

∞

= ∑

f (z) = –

∞ 1 (− 1) n + 1 = ∑ (z – 1)n – 1 ( z − 1) ( z + 1) n = 0 2n + 1

96

THE ELEMENTS OF COMPLEX ANALYSIS

1 − 12 1 1 + − ( z – 1) + (z – 1)2 ... 16 z −1 4 8 This series converges in 0 < | z – 1 | < 2. Similarly, from (19) we obtain =

∞ 1 (− 2)n =− ∑ n+2 ( z − 1) ( z + 1) n = 0 ( z − 1) 1 2 4 =– + ... + − ( z − 1) 2 ( z − 1) 3 ( z − 1) 4 This series converges for | z – 1 | > 2.

f (z) = –

7.5

ISOLATED SINGULARITIES

Let f be analytic in the annulus ann (a ; 0, r2), that is, in the punctured disk D′ (a ; r2). In this case, it may be possible to define f (a) in such a way that f is analytic in the entire disk D (a ; r2). In fact, we have the following theorem. Theorem 5. Let f be analytic in the punctured disk D′ (a ; r2 ). If f is bounded in a neighbourhood of a, then we can define f (a) in a unique way such that the function is also analytic at a. Proof: Suppose that f (z) is bounded in a certain neighbourhood of a. Let γ be the circle | z – a | = ρ and let | f (z) | ≤ M inside and on the boundary of γ.

z

f (ξ) 1 dξ , (n = 0, ± 1, ± 2, ...). 2 πi γ (ξ − a)n + 1 We see that the Cauchy’s inequalities M | an | ≤ n ρ is true for negative n, i.e. | a– n | ≤ M ρn. Taking ρ arbitrary small, we find that a– n is 0 for n = 1, 2, 3, ... . Hence, the Laurent series reduces to a Taylor series, and if we define f (a) = a0, f is analytic in the entire disk D (a ; r2) (The uniqueness follows from continuity). In this case, we say that a is a removable singularity of f. If f is unbounded near a, then at least one of the coefficients a–n of the principal part of the Laurent series must be different from zero. It may happen that the principal part of the Laurent series (17) has only a finite number of terms, Write

an =

∞

m

n f(z) = ∑ an ( z − a) + ∑

a− n

( z − a)n and a– m ≠ 0. Then f is said to have a pole of order m (or multiplicity m) at a. In this case, the function (z – a)m f(z) is analytic in a neighbourhood of a, and is given by the Taylor series n=0

n =1

97

LAURENT SERIES, SINGULARITIES

(z – a)m f (z) = a– m + a– m + 1 (z – a) + ... + ... + a– 1 (z – a)m – 1 + a0 (z – a)m + ... If the principal part of the Laurent series has an infinite number of terms, then a is called an essential singularity. Example 1: The function 1 3 f(z) = + 5 z( z − 2) ( z − 2) 2 has a simple pole at z = 0 and pole of fifth order at z = 2. Theorem 6. If f is analytic and has a pole at z = a, then | f (z) | → ∞ as z → a in any manner. Proof: Suppose f has a pole of order m at z = a. Write ⎧⎪ ∞ ⎫⎪ m+n + bm + bm − 1 ( z − a ) + ... + b1 ( z − a ) m − 1 ⎬ f (z) = (z – a)–m ⎨ ∑ an ( z − a ) ⎩⎪ n = 0 ⎭⎪

= (z – a)– m ψ (z) where ψ is analytic in | z – a | < r and ψ (a) = bm ≠ 0. Hence, 7.6

lim | f (z) | = lim

z→a

z→a

ψ(z) = ∞. ( z − a) m

LIMIT POINTS OF ZEROS AND POLES

Observe that the limit point of the sequence of zeros of a non-zero function, analytic in a region Ω, cannot be an interior point of Ω. If it is an interior point of Ω, then either f (z) = 0, or else it cannot be analytic at the limit point. It follows that the limit point of the sequence of zeros of a non-zero analytic function is a singularity of f (z). It is an isolated singularity. If an isolated singularity is neither a pole nor a removable singularity, it is called an essential singularity.

1 1 . It has zeros at z = (k = ± 1, ± 2, ...). z kπ The limit point of these zeros is the point z = 0. Hence, z = 0 is an isolated essential singularity of f(z). Let z1, z2, ..., zn, ... be a set of points having a limit point z0 in a region Ω. Let f be analytic in Ω except for poles at z1, z2, ... . Then z0 is also a singularity of f(z). The reason is that f is unbounded in the neighbourhood of z0. But z0 is not isolated because it is the limit point of poles. Hence, f (z) has a non-isolated essential singularity at z0. Consider the function f (z) = sin

Example 2: Consider the function f (z) =

1 . This function vanishes for cos 1/ z

1 = ± (2k + 1) π/2, (k = 0, 1, 2, ... ). In other words, this function vanishes for z

98

z=±

THE ELEMENTS OF COMPLEX ANALYSIS

2 2 1 (k = 0, 1, 2, ...). Hence, has poles at z = ± (k = 0, 1, (2 k + 1)π (2 k + 1)π cos 1z

2, ...). The point z = 0 is the limit point of these poles. Hence,

1 has a cos 1z

non-isolated essential singularity at z = 0. It can be easily checked that if f (z) has a pole of order m, then

1 has a f (z)

zero of order m at z = a, and conversely. The behaviour of a function f near an essential singularity is given by the following theorem. Theorem 7. Let f be analytic in the punctured disk D′ (a ; r2 ), and let a be an essential singularity of f. Then the image of an arbitrarily small disk D (a ; ε) is everywhere dense in the complex plane. In other words, the values of f on D (a ; ε) come arbitrarily close to any complex number. Proof: Suppose the theorem is false. There exists a complex number c and a positive number ρ such that | f (z) – c | > ρ for all z ε D′ (z ; ε). Consider the function 1 . g(z) = f ( z) − c g is analytic and bounded in D′ (a ; ε). Hence a is a removable singularity of g , and g possesses an analytic extension for the entire disk D (a ; ε). It, then, follows that 1 has a pole at a, which means that f(z) – c has a pole, contradicting the hypothesis g( z )

that f(z) has an essential singularity. This completes the proof. Picard has proved that, in an arbitrarily small neighbourhood of an essential singularity, f not only comes arbitrarily close to every complex number, but takes on every complex value except possibly one. For instance, the function e1/z omits the value 0, so it is necessary to allow for this one omission. The proof of Picard’s Theorem is beyond the scope of this book. 7.7

MEROMORPHIC FUNCTIONS

We say that f is meromorphic in the region Ω if the only singularities of f in Ω are poles. ez is meromorphic in the entire plane. z (ii) A rational function is meromorphic in the entire plane.

Examples: (i)

1 is meromorphic in the entire plane. The only sin z singularities are simple poles at z = kπ, (k = 0, ± 1, ± 2, ...).

(iii) The function

99

LAURENT SERIES, SINGULARITIES

Riemann Sphere: In complex analysis, we will be concerned with functions that become infinite as the variable approaches a given point. To discuss this situation we introduce the extended-plane C∞ ≡ C ∪ {∞}. Let f be a function defined on C∞. Let t =

1 , and define z

F 1I for t ≠ 0, ∞. H tK

ϕ(t) = f

We say that f has a removable singularity, a pole, or an essential singularity at infinity if ϕ has, respectively, a removable singularity, a pole, or an essential singularity at z = 0. If f has a pole at ∞ then the order of the pole is the order of the pole of ϕ at z = 0. We say that f is meromorphic on C∞ if f is meromorphic on C and is also meromorphic at infinity. We say that f is analytic on C∞ if f is analytic on C and is also analytic at infinity. Examples: (i) The function ez has an essential singularity at infinity. n

k (ii) The polynomial P (z) = ∑ ak z has a pole of order n at infinity. k=0

(iii) The rational function P (z)/Q(z) has a pole of order p – q at infinity if the degree p of P (z) is greater than the degree q of Q (z). The rational functions are meromorphic on the Riemann sphere S. Example: Find the Laurent series for the function e1/z in 0 < | z | ≤ ∞. Using this expansion show that for n = 0, 1, 2, 3, ... 1 x 1 exp (cos θ) cos (sin θ − nθ) dθ = . 0 n! π Note that 1/z is analytic for | z | > 0 and has a removable singularity at ∞. It follows that e1/z has removable singularity at ∞. Hence 1 1 1 + ... + + ... . e1/z = 1 + + 2 z 2!z k ! zk The coefficient of zk in this series is given by 1 ak = e1/ z z − k − 1 dz γ 2 πi where γ is any circle | z | = r > 0. Take r = 1 and z = eiθ. Then e1/z = exp (cos θ – i sin θ) = exp (cos θ) ⋅ exp (– i sin θ). The integral becomes 1 π exp (cos θ) exp ( − i (sin θ + kθ)) dθ . ak = 2π − π

z

z

z

100

THE ELEMENTS OF COMPLEX ANALYSIS

Since the integrand is an odd function, it follows that 1 π ak = exp (cos θ) cos (sin θ + kθ) dθ . 2π − π Note that the coefficient ak can be read off by inspection and setting k = – n we find that 1 π 1 exp (cos θ) cos (sin θ − nθ) dθ = . n! π 0

z

z

EXERCISES 1. Expand the following functions in a Taylor series about the point z = 1: 1 (i) f (z) = (ii) f (z) = log z. z+2 2. Find two distinct Laurent expansions for f (z) =

(3z + 1) around z = 1. Examine the z2 − 1

convergence of each series. 3. Give two different Laurent expansions for 1 f (z) = 2 z (z − i) around z = i. Examine the convergence of each series. 4. Give the Laurent expansion of f (z) = annuli (i) ann (0, 0, 1),

1 in each of the following z ( z − 1) ( z − 2)

(ii) ann (0, 1, 2),

(iii) ann (0, 2, ∞).

LM F 1 I OP in powers N H zKQ

5. Show that the coefficients in the Laurent expansion of f (z) = sin z + of z are given by the formula 1 2π cos nθ sin (2 cos θ) dθ . an = 2π 0 6. Determine the zeros of the following functions and their order

z

(iv) sin z . z 7. Prove that f (z) = tan z is analytic is C except for simple poles at z = π/2 + nπ, for each integer n. Determine the singular part of f at each of these poles. 8. Suppose that f : G → C is analytic except for poles. Show that the poles of f cannot have a limit point in G. 9. Expand in a Laurent series the following functions: (i) ez – 1,

(i) f (z) =

(ii) sin2 z,

1 − e2z about the origin. z4

(iii) sin z2,

101

LAURENT SERIES, SINGULARITIES

(ii) f (z) =

1 about z = 0 and z = 1. z ( z − 1)

(iii) f (z) =

e2z about z = 1. ( z − 1) 2

2

1 + ez about the origin. sin z + z cos z 10. Show that the singularities in each of the following functions is a pole. Find the location and order of each pole

(iv) f (z) =

(i) (ez – 1)–z, (ii) z–3 exp (z2)/2, (iii) z2/(z + 1)2 (z – 2). 11. Determine the singularities in each of the following functions. Find the singular part if it is a pole. Define f (0) if it is a removable singularity so that f is analytic at z = 0. 4 z2 + 1 (iii) 1 − ez z( z − 1) 1 (iv) zn sin 1 (v) z cos . z z 12. Show that an entire function has a removable singularity at infinity iff it is a constant.

(i)

log ( z + 1) z2

(ii)

13. Show that an entire function has a pole at infinity of order m iff it is a polynomial of degree m. 14. Show that the image of an entire function is dense in C. 15. Isomorphism: Let (X1, d1) and (X2, d2) be metric spaces. Let f : X1 → X2 be a function. A one-to-one correspondence f is called a homeomorphism if f and f –1 are both continuous. Let Ω be a region. Let f be complex differentiable on Ω. Let f be a homeomorphism Ω → f (Ω). It can be easily seen that f –1 is also complex differentiable on f (Ω). We call a complex differentiable homeomorphism an isomorphism. If f is an isomorphism Ω → f (Ω), then Ω is said to be isomorphic to f (Ω). If f (Ω) = Ω, then the isomorphism f is called an automorphism of Ω. Prove that the only analytic automorphism of C are the funcitons of the form f(z) = cz + d, where c, d are constants and c ≠ 0. 16. Let U and V be open sets. Suppose that f is meromorphic on U and suppose that ϕ : V → U is an analytic isomorphism. Let φ (z0) = w0, and let f has order n at w0. Prove that f o ϕ has order n at z0. In other words, the order is invariant under analytic isomorphisms. Note that n is a positive or negative integer.

8 8.1

RESIDUE THEOREM AND ITS APPLICATIONS

DEFINITION Let f has an isolated singularity at z = z0 and let ∞

f (z) = ∑ an (z – z0)n

(1)

n=−∞

be its Laurent expansion. We call a–1 the residue of f at z0, and write a–1 = Res (f , z0). 1/z Examples: (i) Res (e , 0) = 1.

FG F 1I , 0IJ = 0. H H zK K

(ii) Res cos

We give below some methods for calculating the residue of a function at a pole. If f (z) has a simple pole at a point z = z0, then ∞ a−1 + ∑ an (z – z0)n f (z) = z − z0 n = 0 where a–1 ≠ 0. Res (f , z0) = a–1 = lim (z – z0) f (z). z → z0

Example: Let e iz , (a real) z2 + a2 The function has simple poles at z = ai and z = – ai.

f (z) =

Res (f, ai) = lim ( z − ai ) z → ai

= e– a/2ai. Similarly, Res (f, – ai) = – ea/2ai. 102

e iz z 2 + a2

103

RESIDUE THEOREM AND ITS APPLICATIONS

In the case of a simple pole, another formula is obtained as follows: If f (z) has a simple pole at z = z0, we write f (z) = g (z) / h (z) where h has a simple zero at z0 and g (z0) ≠ 0. Res (f, z0) = Res

F g ,z I Hh K 0

g ( z) h ( z) g (z) = lim h ( z ) − h ( z ) (h ( z0 ) = 0) z → z0 0 z − z0 = g (z0) / h′ (z0).

= lim ( z − z0 ) z → z0

Example: Let 7 − 3z . z2 − z The function has simple poles at z = 0 and z = 1.

f (z) =

FG 7 − 3z IJ H 2z − 1K F 7 − 3z IJ Res (f, 1) = G H 2z − 1K

= – 7.

Res (f, 0) =

z=0

= 4. z =1

If f (z) has a pole of order m > 1 at a point z = z0, then ∞ a− m a−1 … + + + a (z – z0)n. (1) f (z) = ∑ ( z − z0 ) m z − z0 n = 0 n Multiplying both sides of (1) by (z – z0)m we get (z – z0)m f (z) = g (z) = a–m + … + a–1 (z – z0)m – 1 + a0 (z – z0)m + … + an (z – z0)m + n + … . This shows that the residue of f (z) at z = z0 is the coefficient of (z – z0)m – 1 in the Taylor expansion of g (z) = (z – z0)m f (z). It follows from Taylor’s Theorem that g ( m − 1) ( z0 ) . ( m − 1) ! Hence, we have the following result: If f has a pole of order m at z = z0 and put g(z) = (z – z0)m f (z), then

a–1 =

Res (f , z0) =

g ( m − 1) ( z0 ) . ( m − 1) !

104

THE ELEMENTS OF COMPLEX ANALYSIS

Example: Let 2z ( z + 4) ( z − 1) 2 The function has a pole of order two at z = 1.

f (z) =

8 d . ((z – 1)2 f (z)) = 25 z → 1 dz Theorem 1. Let f has an isolated singularity at z = z0. Let γ be a small circle (centred at z0) such that f is analytic on γ and its interior, except the centre z0. Then 1 Res (f, z0) = f (z) dz. 2πi Res (f, 1) = lim

z γ

Proof : Since f (z) has an isolated singularity at z = z0, then we can represent it by its Laurent series ∞ ∞ a− n f (z) = ∑ an (z – z0)n + ∑ n n=0 n = 1 ( z − z0 ) valid throughout the circle γ except z = z0 itself. This series converges uniformly and absolutely for z on the circle, we can integrate it term by term. The integral of (z – z0)n over the circle is equal to 0 for all values of n except n = – 1. In this case, the value of the integral is equal to 2πi (cf. Example 1, § 4.3). This completes the proof. Inserting Theorem 1 in Theorem 1 of Chapter 5 we obtain the generalized Residue Theorem. Theorem 2 (Residue Theorem). Let Ω be a region. Let γ be a closed chain in Ω such that γ is homologous to 0 in Ω. If f is analytic in Ω except at a finite number of points z1, z2, ... zn, then

z γ

n

f=

∑ i =1

mi Res (f, zi)

mi = η (γ, zi). Theorem 2 is very important and used quite often when Ω is simply connected region. In applications, Ω will be a disk, or inside of a rectangle, where the simply connectedness is obvious. where

8.2

APPLICATIONS OF THE RESIDUE THEOREM

This section is devoted to calculating certain integrals by means of Residue Theorem. (i) We consider integrals of the type (2)

I=

z

2π

0

R (cos θ, sin θ) dθ

where R (cos θ, sin θ) is a real rational function of cos θ and sin θ. Put z = eiθ,, (0 ≤ θ ≤ 2π). Then

105

RESIDUE THEOREM AND ITS APPLICATIONS

F H

I K

1 iθ –iθ 1 1 z+ ; (e + e ) = 2 2 z 1 1 iθ –iθ 1 z− . sin θ = (e – e ) = z 2i 2i We find that the integrand becomes a rational function of z. Since dθ = dz/iz, the given integral takes the form cos θ =

(3) I =

z

Example:

I K

dz . iz

f (z)

|z| = 1

F H

z z z

dθ (0 < p < 1). 1 − 2 p cos θ + p 2 Setting z = eiθ, we find dz/iz I= 1 1 |z| = 1 z+ 1− 2p + p2 2 z dz = . | z | = 1 i (1 − pz ) ( z − p )

I=

2π

0

The singularities of

F H

I K

1 1 are simple poles at z = > 1 and p (1 − pz ) ( z − p)

z = p < 1. The only singularity inside the unit circle is a simple pole at z = p < 1. Res (f, p) = lim (z – p) f (z) z→ p

1 z → p i (1 − pz ) 1 = . i (1 − p 2 ) Thus by Residue theorem

= lim

z

1 dθ 2 = 2πi i (1 − p 2 ) 0 1 − 2 p cos θ + p 2π = (0 < p < 1). 1 − p2 (ii) Consider the integrals of the type

(4) where f (z) =

I=

z

2π

∞

−∞

f (x) dx

g (z) is a rational function with no poles on the real axis. The degree h (z)

of h(z) is at least two units higher than the degree of g(z). We consider the corresponding complex integral

106

THE ELEMENTS OF COMPLEX ANALYSIS

z

γ

f (z) dz

around a path γ as shown in Fig. 8.I. Y g

–r

+r

O

X

Fig. 8.I

The boundary γ consists of the segment (– r, r) on the real axis and of the semi-circle s centred at the origin. If γ is large enough, all the poles of f (z) in the upper half plane will lie inside γ, and we have

z

f (z) dz =

γ

z

r

f (x) dx +

−r

= 2πi

∑

Im z j > 0

z

s

f (z) dz

Res (f, zj)

where zj, j = 1, 2, ... are the poles of f (z). Let r → ∞, then lim

r→∞

z z

r

−r

f (x) dx =

z

∞

f (x) dx.

−∞

(Since I is a convergent integral, lim

r→∞

r

−r

f (x) dx = alim →∞ b→∞

z

b

−a

f (x) dx).

Since the degree of h (z) is at least two units higher than the degree of g (z), there exists a constant M such that M | f (z) | < 2 r for sufficiently large r and for all z ∈ s. It follows that M πM | f (z) dz | ≤ πr 2 = . s r r Hence

z

lim

z

r→∞ s

z z

∞

and

−∞

Example:

I=

∞

−∞

f (z) dz = 0 f (x) dx = 2πi dx 1 + x4

∑

Im z j > 0

Res (f, zj).

107

RESIDUE THEOREM AND ITS APPLICATIONS

1 has four simple poles at the points 1 + z4 z1 = eπi / 4, z2 = e3πi / 4, z3 = e–3πi / 4, z4 = e–πi / 4. The first-two poles lie in the upper half plane

The function f (z) =

LM 1 OP N (1 + z )′ Q

Res (f, z1) =

4

z = z1

1 1 = e–3πi / 4 = – eπi / 4. 4 4 1 Res (f, z2) = e– πi / 4. 4

Similarly Thus

I=

z

∞

−∞

FG H

− e πi / 4 + e − πi / 4 dx = 2πi 4 1 + x4

IJ K

π . 2 (iii) Consider integrals of the type

=

(5)

I=

z

∞

f (x) eix dx

−∞

where f is real valued on the real axis. The real and imaginary parts of this integral are

z

∞

−∞

f (x) cos dx and

z

∞

f (x) sin x dx

−∞

respectively. We assume that f is analytic in the upper half plane except at z1, z2, ..., zn which do not lie on the real axis. We also assume that lim f (z) = 0 in the upper half plane. z→∞

Now take the rectangle γ with vertices B, B + iC, – A + iC, – A (Fig. 8.II). Choose A, B and C sufficiently large to ensure that all the singular points z1, z2, ..., zn lie inside the rectangle. Then

z

γ

n

f (z) eiz dz = 2πi ∑ Res (f (z) eiz, zj). j =1

–A + iC

–A

iC

B + iC

g

Fig. 8.II

B

108

THE ELEMENTS OF COMPLEX ANALYSIS

It follows that (6)

z

B

−A

n

f (x) eix dx = 2πi ∑ Res (f (z) eiz, zj ) –

z

j =1

B + iC

B

f (z) eiz dz +

z

B + iC

− A + iC

f (z) eiz dz +

z

− A + iC

−A

f (z) eiz dz.

The three integrals in the right side of (6) converges to 0 as A, B → ∞ independently and C → ∞. For, I1 = Put z = B + iy, then I1 =

z z

B + iC

f (z) eiz dz.

B

C

0

f (B + iy) ei(B + iy) i dy.

z

| I1 | ≤ M (B)

Hence

C

0

e–y dy ≤ M (B).

where M (B) = max | f (B + iy) |. 0≤ y≤C

Similarly

| I3 | = |

z

− A + iC

−A

f (z) eiz dz | ≤ M (A)

where M (A) = max | f (– A + iy) |. 0≤ y≤C

I2 =

Let

Put z = x + iC, then | I2 | = |

z z

B + iC

f (z) eiz dz.

− A + iC

B

−A

≤ M (C)

f (x + Ci) ei(x + iC) dx |

z

B

−A

e–C dx

= M (C) e–C (A + B). Since lim f (z) = 0 in the upper half plane, we can choose A and B sufficiently z→∞

large such that M (A) ≤ ε, M (B) ≤ ε and C sufficiently large such that M (C) e–C (A + B) < ε. It follows that | I1 | + | I2 | + | I3 | < 3ε. Thus we have

z

∞

−∞

n

f (x) eix dx = 2πi ∑ Res (f (z) eiz, zj). j =1

Example: I=

z

∞

−∞

cos x dx. 1 + x2

109

RESIDUE THEOREM AND ITS APPLICATIONS

In fact,

e iz has only one pole in the upper half plane i.e. a simple pole at 1 + z2

z = i.

FG e , iIJ = 1 . H 1 + z K 2ie iz

Res Therefore

z

2

π e ix 1 = 2 dx = 2πi 1+ x 2ie e

∞

−∞

and this gives the result. We have assumed that the function has no singularities on the real axis. If f has singular points on the real axis, say a simple pole at z = z1, then we modify the path γ by replacing the interval [z1 – ε, z1 + ε] on the real axis by the semi-circle Sε (z1) as in the Fig. 8. III.

g

g

g Z1

Z1 – e

Z1 + e

Fig. 8.III

z1. Let

Note that if f (z) has a simple pole at z1, then f (z) eiz has also a simple pole at f (z) eiz = g (z) +

where g (z) is the Taylor part. Then

z

Sε

Now

f (z) eiz dz =

z

Sε

z

Sε

a−1 z − z1

g (z) dz +

z

Sε

g ( z ) dz ≤ M (ε) πε

where M (ε) = max {| g (z) | : z ∈ Sε}. Since g (z) is continuous at z1, it follows that lim

z

ε → 0 Sε

and

z

Sε

g (z) dz = 0

a−1 dz = a–1 πi. z − z1

a−1 dz. z − zi

110

THE ELEMENTS OF COMPLEX ANALYSIS

Hence,

lim

z

f (z) eiz dz = a–1 πi.

ε → 0 Sε

lim

Thus,

ε→0

F H

= 2πi Example:

z

I= Since

z

z1 − ε

−∞

LM N

f ( x ) e ix dx +

Im z j > 0

0

z

lim

ε→0

Hence

z

FG H

z

f ( x ) e ix dx

∑ Res ( f ( z ) e iz , z j ) +

sin x is an even function, we have x 1 ∞ sin x 1 dx = Im I= 2 −∞ 2 x

half plane, therefore

z1 + ε

OP Q

I K

a−1 . 2

sin x dx. x

∞

Consider the function

z

∞

FG H

z

∞

−∞

IJ K

e ix dx . x

e iz . This function has no singular points in the upper z

z

IJ K

∞ sin x sin x dx dx + −∞ ε x x 1 = Im (πia–1). 2 e iz , 0 = 1. a–1 = Res z ∞ sin x dx = π / 2. lim ε x ε→0 ε

z

FG H

IJ K

∞

sin x dx is a continuous function of ε, hence x ∞ sin x dx = π / 2. 0 x (iv) Consider the integrals of the type

Note that

(7)

I=

ε

z

∞

0

z

xα f (x) dx (0 < α < 1)

g (z) is a rational function analytic on C, except for a finite number of h (z) poles, none of which lies on the positive real axis x > 0. We assume that the degree of h (z) is greater than the degree of g (z) by at least two units. Suppose α is real positive number. We define zα = eα log z where the log is defined on the simply connected set equal to the plane from which the axis x ≥ 0 has been deleted.

where f (z) =

111

RESIDUE THEOREM AND ITS APPLICATIONS

In order to evaluate the integral, we choose the closed path γ as in Fig. 8.IV. Then γ consists of the segment [ε, r] and the circle Sr (0) (taken counterclockwise), the segment [r, ε] and the circle Sε (0) (taken clockwise). Let z1, z2, ..., zn be the poles of f inside γ. Then

z

n

z α f (z) dz = 2πi ∑ Res [zα f (z), zj]. γ j =1

g

g

g e

x

r

g

g

Fig. 8.IV

We suppose that ε is sufficiently small and r sufficiently large so that all the singular points of f except the origin lie inside γ. Now

z

γ

zα f (z) dx =

z

r

xα f (x) dx +

ε

+

z

ε

s

z

Sr

zα f (z) dz

xα e2πiα f (x) dx –

z

Sε ( 0 )

zα f (z) dz.

n

= 2πi ∑ Res [zα f (z), zj] (1 – e2πiα)

or n

z

j =1 r

ε

xα f (x) dx

(8)

= 2πi ∑ Res (zα f (z), zj) –

As

ε → 0 and r → ∞ , we have

j =1

lim

ε→0 r→∞

z

r

ε

z

zα f (z) dz +

Sr

xα f (x) dx =

z

∞

0

z

Sε ( 0 )

zα f (z) dz.

xα f (x) dx.

We estimate the last two integrals in (8) as r → ∞ and ε → 0 respectively. Since |

z

Sr ( 0 )

A 2rπ r2 − α 2 Aπ = 1− α , r

zα f (z) dz | ≤

112

THE ELEMENTS OF COMPLEX ANALYSIS

therefore,

z

lim

r→∞

Also, since |

z z

zα f (z) dz | ≤

Sε ( 0 )

ε → 0 S (0) ε

Thus,

A 1− α

ε α z f (z) dz = 0.

lim

hence,

zα f (z) dz = 0.

Sr ( 0 )

2πε = 2Απεα,

n

(1 – e2πiα) I = 2πi ∑ Res (zα f (z), zj).

Example:

z

I=

j =1

∞

0

xα dx (0 < α < 1). x (1 + x )

We have

FG z − 1IJ H z(1 + z)′ K a

(1 – e2πiα) I = 2πi Res

= – 2πi (– 1)α 2πi 2 πie πiα = πiα − πiα 2 πiα e −1 e − e π = (eπi = – 1). sin πα (v) Consider the integrals of the type

Ι=

I=

z

∞

0

f (x) log x dx

where f (z) = g (z) / h (z) is a rational function, analytic on C except for finite number of poles, none of which lies on the positive real axis and the origin. We assume that the degree of h (z) is greater than the degree of g (z) by at least two units. Consider

z z

γ

f (z) log z dz, where γ is the same closed curve as in Fig. 8.I. If

z1, z2, ..., zr are singular points of f (z) log2 z, we have (as ε → 0 and r → ∞ ) ∞

0

i.e.

f (x) log2 x dx – n

4πi

z

z

∞

0

f (x) (log x + 2πi)2 dx

= 2πi ∑ Res (f (z) log2 z, zj) j =1

∞

0

f (x) log x dx – 4π2 n

z

∞

0

f (x) dx

= – 2πi ∑ Res (f (z) log2 z, zj). j =1

If f is real-valued function on the real axis, we obtain

z

∞

0

f (x) log x dx = –

1 Re 2

R| S|∑ T n

j =1

Res ( f ( z ) log 2 z , z j )

U| V| W

113

RESIDUE THEOREM AND ITS APPLICATIONS

z

and

∞

0

f (x) dx =

1 Im 2π

Example: I= The singularities of

z

RS ∑ T n

j =1

UV W

Res ( f ( z ) log 2 z, z j .

log x dx. 1 + x2

∞

0

1 are i and –i, we have 1 + z2

F log z , iIJ = π i Res G H1+ z K 8 F log z , – iIJ = −9π Res G H 1+ z K 8 2

2

2

2

and

2

Thus

z

and 8.3

I= ∞

0

z

∞

0

2

i.

log x dx = 0 1 + x2

dx = π / 2. 1 + x2

THE LOGARITHMIC RESIDUE

Definition: Let Ω be an open set. Let f be meromorphic in Ω and z0 ∈ Ω . The logarithmic residue of f at z0 is the residue of

f′ d = {log f (z)} f dz evaluated at z0. Suppose that f is analytic in Ω and has a zero of order m at z0. So f (z) = (z – z0)m g (z) where g (z0) ≠ 0. Hence

f′ m g′ + = z − z0 g f

(9)

g′ is analytic in the neighbourhood of z0. Now suppose that f has a pole of g order m at z0 ; i.e. f (z) = (z – z0)–m g (z) where g is analytic and g (z0) ≠ 0. It follows that and

(10) and again

f′ −m g′ + = f z − z0 g

g′ is analytic in the neighbourhood of z0. g

114

THE ELEMENTS OF COMPLEX ANALYSIS

Now from (9) But Res

Res

FG f ′ , z IJ = Res F m , z I + Res FG g′ , z IJ . GH z − z JK H g K Hf K 0

0

FG g′ , z IJ = 0 and therefore Hg K F f′ I Res G , z J = m. Hf K

0

0

0

0

Similarly, from (10) we have Res

FG f ′ , z IJ = – m. Hf K 0

We summarize the above discussion in the following theorem. Theorem 3. Let f be meromorphic on Ω with poles s1, s2, ..., sm and zeroes z1, z2, ..., zn. If γ is a closed chain in Ω homologous to zero in Ω and not passing through s1, s2, ..., sm; z1, ..., zn, then

z

n n 1 f′ = ∑ n( γ , z k ) − ∑ n( γ , s j ) . 2πi γ f k =1 j =1 In applications, γ is taken as a circle or a rectangle, and the points s1, s2, ..., sm, z1, z2, ..., zn lie inside γ. Earlier we have defined “interior”. In the following definition, we redefine “interior” which will be used in the subsequent discussion. Definition: Let γ be a closed path. We say that γ has an interior if n (γ, α) = 0 or 1 for every complex number α which does not lie on γ. The set of points α such that n(γ, α) = 1 is called the interior of γ. Theorem 4. (Rouche’s Theorem). Let γ be a closed path homologous to 0 in Ω. Assume that γ has an interior. Let f and g be analytic in Ω, and | f (z) – g (z) | < | f (z) | for z on γ. Then f and g have the same number of zeroes in the interior of γ. Proof : It follows from the assumption that f and g have no zero on γ. We have

g ( z) − 1 < 1 for z on γ. f ( z)

This gives that the values of

g are contained in the open disk centred at 1 f

and radius g . Then h o γ is a closed path contained in that disk. Thus, since f 0 lies outside the disk, we have n (h o γ, 0) = 0.

1.

Let h =

115

RESIDUE THEOREM AND ITS APPLICATIONS

If γ is defined on [a, b] then

z

z z

1 dz hoγ z b h ′ ( γ (t )) = γ′ (t) dt = 0 a h ( γ (t ))

n (h o γ, 0) =

h′ =0 γ h g′ f ′ − = 0. i.e. γ g f It now follows from Theorem 3 that f and g have the same number of zeros. This completes the proof of Rouche’s Theorem. Rouche’s theorem can be applied to give the proof of the Fundamental Theorem of Algebra. If p (z) = zn + a1 zn – 1 + ... + an, then p ( z) a a = 1 + 1 + … + nn n z z z p (z) lim and = 1. z→∞ zn So for sufficiently large number r we have

i.e.

z FGH

RS T

i.e.

IJ K

UV W

p (z) − 1 < 1 ( z = r) zn n p (z) − z n < z ( z = r) .

It follows from Rouches’ theorem that p (z) must have n zeros inside z = r. Example: Show that the equation ez = azn has n roots inside the unit circle, if a > e. Write f (z) = azn and g (z) = – ez. Then f (z) + g (z) = azn – ex. Suppose that γ is a positively oriented closed curve in a region Ω and we take γ to be a circle z = 1. Then on γ, f ( z) = a

and

z2 +… 2! 2 z +… ≤1+ z + 2! 1 1 + +… ≤1+1+ 2 ! 3! = e.

g (z) = 1 + z +

116

THE ELEMENTS OF COMPLEX ANALYSIS

Since a > e, it follows that g ( z ) < f ( z ) . Applying Rouche’s theorem we find that f (z) + g (z) and f (z) have the same number of zeros inside z = 1. Since f (z) = azn has n zeros, all at the origin, it follows from the Fundamental Theorem of Algebra that ez = ezn has n roots inside z = 1.

EXERCISES 1. Prove that

FG z , 0IJ = 0 H sin z K F sin z , 0IJ = 1 Res G Hz K

(ii) Res (tan z, π/2) = – 1

(i) Res (iii)

(iv) Res

2

FG 1 − cos z , 0IJ = 1 H z K 2 2

2. Compute the following integrals where C1 is the unit circle (counterclockwise) (i) (iii) (v)

z z z z z z z z

C1

C1

dz 2 z + 6iz

(ii)

ez dz cos πz ( z + 4) 3

C1

4

z + 5z 3 + 6z 2

(iv)

(iii)

∞

dx

−∞

x6 + 1

∞

x2

0

x6 + 1

(ii)

dx

(iv)

4. Compute the following integrals (i) (iii)

∞

cos x 4

−∞

x +1

∞

cos x

0

dx

( x 2 + 1) 2

(ii)

2π

0

( z 2 + 2) 3

C1

2z 2 − z

dz

z z z

∞

x2

−∞

x4 + 1

∞

x 2 dx

0

dx

( x 2 + 1) 2

∞

sin x

−∞

x3 + 1

dx

dx.

5. Prove that (i)

C1

dz

3. Evaluate the following integrals (i)

z z

ez dz sin z

dθ =π 2 1 + sin 2 θ

(ii)

z

2π

0

sin 2 θ dθ = 4 π (2 − 3 ) 1 1 + cos θ 2

117

RESIDUE THEOREM AND ITS APPLICATIONS

(iii)

(v)

z FH z

dθ

π

0

π

0

I K

2

=

8π

(iv)

3 3 1 cos θ 2 π dθ = 3 + 2 cos θ 5 1+

6. Show that

z

π/2

0

z

2π

0

z

2π

0

(iii)

z z

∞

log x 2 2

dx = − π / 4

(1 + x )

∞

(log x ) 2 dx = π 2 /8 1 + x2

10. Compute

a dθ

π

0

2

2

a + sin θ

π

=

1 + a2

2 πa dθ = , (0 < b < a) ( a + b cos θ) 2 ( a 2 − b 2 ) 3 / 2

0

0

0

π dθ = 2 1 + sin θ 2 2

dθ = 2 π/ 3 (2 − sin θ)

9. Prove that (i)

π/2

π (2 a + 1) dθ = , ( a > 0) 2 2 4 ( a 2 + a) 3/ 2 ( a + sin θ)

7. Prove that

8. Prove that

(vi)

z z

z

(ii) (iv)

z z

∞

log (1 + x )

0

1+ x

e ax

∞

−∞

2

1+ e

2

x

dx =

dx =

π log 2 2

π ,0 0.

14. Prove that for any real number a > 0.

118

THE ELEMENTS OF COMPLEX ANALYSIS

z

∞

cos x

−∞

x2 + a2

dx = πe − a / a

15. Prove that (i)

z z

∞

cos x

−∞

(x2 + a2 )

dx =

π (1 + a ) 2a 3 e a

, (a > 0)

cos x π dx = 2 (x 2 + a2 ) (x 2 + b2 ) a − b2 16. Prove that

(ii)

∞

0

z

∞

x2

0

17. Prove that

z z

18. Prove that

sin 2 x

∞

cos x

−∞

a2 − x2

F1 H be

b

−

I K

1 , (0 < b < a) ae a

dx = π/2

dx =

π sin a , (a > 0) a

π cos x dx = π / 2 −x −∞ e + e e − e−π/2 Use the indicated contour: ∞

x

– R + pi

R + pi

–R

R

19. Prove that (i) (ii)

z z

∞

0 ∞

0

π x e dx = , (0 < a < 1) sin π a 1+ x x xa

dx = 1+ x x

20. Show that

3

z

2π

0

π 3 sin

FG π a IJ , (0 < a < 3). H 3K

dθ 2π = , (0 < a < 1). 1 + a − 2 a cos θ 1 − a 2 2

9

CONFORMAL MAPPING

The main general theorem concerning conformal mapping is the Riemann Mapping Theorem: “If G is a simply connected region which is not the whole plane, then there exists an isomorphism of G with the open disk D1(0)”. The proof of this theorem is postponed to a later chapter (see Appendix 1). In this chapter, we shall give specific examples, where the mapping can be exhibited concretely. We begin with a simple example. Consider the function defined by w = f (z) = z2. Put z = reiθ and w = Reiφ Then R = r2 and φ = 2θ. We find that circles r = const. are mapped onto circles R = const. and rays θ = θ0 = const. onto rays ϕ = 2θ0 = const. In particular, the positive real axis (θ = 0) in the z-plane is mapped onto the positive real axis in the w-plane, and the positive imaginary axis (θ = π/2) in the z-plane is mapped onto the negative real axis in the w-plane. The angles at the origin are doubled under the mapping. The first quadrant 0 ≤ θ ≤ π/2 is mapped onto the entire upper half of the w-plane. Let z = x + iy and w = u + iv. Then u = x2 – y2, v = 2xy. Hence, the hyperbolas 2 2 x – y = c and 2xy = d are mapped by f into the straight lines u = c, v = d. One interesting fact is that for c and d not zero, these hyperbolas intersect at right angles, just as their images do. Now examine what happens to the lines x = c and y = d. Consider x = c where y is arbitrary. f maps x = c into u = c2 – y2 and v = 2cy. Eliminating y we get that x = c is mapped onto the parabola v2 = – 4c2 (u – c2). Similarly, f maps the line y = d onto the parabola v2 = 4d 2(u + d 2). The above discussion sheds some light on the nature of f (z) = z2 and likewise, it is useful to study the mapping properties of other analytic functions. 119

120

THE ELEMENTS OF COMPLEX ANALYSIS

Suppose γ : [a, b] → G is a differentiable path and that for some t0 in (a, b), γ′ (t0) ≠ 0. Then γ has a tangent at the point γ(t0). The slope of the line is tan (arg γ′ (t0)). If γ1 and γ2 are two paths with γ1(t1) = γ2(t2) = z0 (say) and γ1′ (t1) ≠ 0, γ2′ (t2) ≠ 0, then define the angle between the paths γ1 and γ2 at z0 to be arg γ2′ (t2) – arg γ1′ (t1). Suppose γ is a path in G and f : G → C is analytic. Then σ = f o γ is also a path and σ′ (t) = f ′ (γ (t)) γ′ (t). Let z0 = γ (t0) and suppose that γ′ (t0) ≠ 0 and f ′ (z0) ≠ 0, then σ′ (t0) ≠ 0 and arg σ′ (t0) = arg f ′ (z0) + arg γ′ (t0), i.e. (1) arg σ′ (t0) – arg γ′ (t0) = arg f ′ (z0). Now let γ1 and γ 2 be paths with γ 1 (t1) = γ 2 (t 2) = z 0 (say) and γ 1′ (t2) ≠ 0, γ 2′ (t2) ≠ 0. Let σ1 = f o γ1 and σ2 = f o γ2. Suppose also that the paths γ1 and γ2 are not tangent to each other at z0. In other words, suppose that γ1′ (t1) ≠ γ2′ (t2). Equation (1) yields (2) arg γ2′ (t2) – arg γ1′ (t1) = arg σ2′ (t2) – arg σ1′ (t1). Thus given any two paths through z0, f maps these paths onto two paths through w0 = f (z0), and when f ′ (z0) ≠ 0, the angles between the curves are preserved both in magnitude and direction. We summarize the above discussion in the following theorem. Theorem 1. Let f : G → C be analytic. Then f preserves angles at each point z0 of G where f ′ (z0) ≠ 0. 9.1

DEFINITION

Let f : G → C be such that it preserves angles and | f ( z ) – f ( z0 ) | lim exists, z → z0 | z – z0 | then f is called a conformal map. If f is analytic and f ′(z) ≠ 0 for any z, then f is conformal. The converse of this statement is also true. Example 1: The mapping w = f (z) = z2 is conformal except at z = 0. At z = 0, the angles are doubled under the mapping, because each ray arg z = c = const. maps into a ray arg w = 2c = const. as in Fig. 9.I. Y

X

–4

–2

–1

Fig. 9.I. Mapping w = z . 2

+1

+2

+4

121

CONFORMAL MAPPING

Example 2: The mapping w = f (z) = ez is conformal throughout C. In order to discuss the other properties of this mapping put z = c + iy where c is fixed, then f (z) = reiy for r = ec. That is, f maps the line x = c onto the circle with centre at the origin and radius ec. Also, f maps the line y = d onto the infinite ray {reid : 0 < r < d}. We now find the image of the vertical line segment x = c, – π < y ≤ π under the mapping given by w = f (z) = ez. Note that every point on the given line segment has the form z = c + iy, (– π < y ≤ π) hence, as y varies from – π to + π, eiy describes a complete circle, while | w | remains fixed at ec. In other words, as z varies on the given line segment, w describes a circle centred at w = 0 and having radius ec (see Fig. 9.II).

pi d

c

ec

Fig. 9.II

Observe that if y were allowed to vary over a larger domain (but, always, on the same vertical line), then w would repeat its trace on the same circle, and if we took the entire vertical line x = c, then the circle | w | = ec would be repeated infinitely many times. We summarize as follows: If we take all the horizontal lines between y = – π (not included) and y = π (inclusive), their images will be all the rays with angles of inclination ranging from – π to π. The totality of all such rays cover all the points in the w-plane except w = 0. On the other hand, if we take all the vertical line segments, as in the above example, contained between the lines y = – π and y = π, then their images will be all the circles, with positive radius centred at w = 0. The totality of all such circles will cover all the points in the w-plane except w = 0. From the preceding discussion we arrive at the following conclusion: Under the mapping w = f (z) = ez, the fundamental strip S : – π < y ≤ π, – ∞ < x < + ∞,

122

THE ELEMENTS OF COMPLEX ANALYSIS

is mapped onto the entire w-plane except its origin. Arguing similarly we can say that any horizontal strip Sα : α < y < α + 2π, – ∞ < x < + ∞ (α is any real number) is mapped onto the entire w-plane except its origin. Note that the function log z is the inverse of the function ez and this means that log z maps the z-plane (minus its origin) onto the fundamental strip – π < v ≤ π, – ∞ < u < + ∞ of the w-plane. We shall arrive at the same conclusion as above by considering the logarithmic transformation directly. Consider w = log | z | + i arg z, z ≠ 0 Then u = log | z |, v = arg z. Now, as z varies over all non-zero values, | z | varies between 0 and + ∞, hence log | z | varies from – ∞ to + ∞, and, therefore, – ∞ < u < + ∞. On the other hand, since, by definition, – π < arg z ≤ π, we have – π < v ≤ π. The last two relations involving u and v represent the fundamental strip of the w-plane. 9.2

LINEAR FRACTIONAL TRANSFORMATION

We shall now consider an important class of conformal mappings, starting with the simplest types of transformations in this class. The mapping (3) w = h (z) = z + b is called a translation. If w = h (z) = az with a ≠ 0, then the mapping is called a dilation. If w = h (z) = eiθ . z, then the mapping is called a rotation. If w = h(z) = 1/z, then it is called the inversion. A mapping of the form az + b (4) w = h(z) = cz + d is called a linear fractional transformation. If a, b, c and d also satisfy ad – bc ≠ 0 then the mapping is called a Möbius transformation. dz – b satisfies If h is a Möbius transformation then h–1 (z) = – cz + a h(h– 1 z) = h–1 (hz) = z. If h and g are both linear fractional transformations then it follows that h o g is also a linear fractional transformation.

123

CONFORMAL MAPPING

Consider the Möbius transformation az + b . h(z) = cz + d Then ad – bc . h′(z) = (cz + d ) 2 This shows that the condition ad – bc ≠ 0 implies that h1(z) ≠ 0. Hence, the mapping h is conformal. It follows from (4) that to each z for which cz + d ≠ 0 there corresponds precisely one complex number h(z). Suppose that c ≠ 0. Then to each z = – d/c where cz + d ≠ 0, there does not correspond a number h(z). Thus we attach a point ∞ to the w-plane. We denote C∞ = C ∪ {∞}. We consider h defined on C∞ where h(∞) = a/c and h(– d/c) = ∞. Note that when a = 0 = c or d = 0 = c then ad – bc = 0, hence we cannot take a = 0 = c or d = 0 = c. Since h has an inverse it maps C∞ onto C∞. Theorem 2. If h is a Möbius transformation then h is the composition of transformations of the types (i) translation, (ii) dilation and (iii) inversion. (Of course, some of these may be missing.) Proof : We have az + b , where ad – bc ≠ 0. h(z) = cz + d Suppose that c = 0. Hence

F aI z + F bI . H dK H dK F bI F aI h (z) = H K z, h (z) = z + H K , d d h(z) =

If

1

2

h2 o h1 = h. Suppose now that c ≠ 0 and put d 1 h1 (z) = z + , h2 (z) = , c z bc – ad a z, h4(z) = z + . h3 (z) = 2 c c Then h = h4 o h3 o h2 o h1. The fixed points of the mapping (4) are obtained from the equation az + b z= cz + d 2 i.e. cz + (d – a)z – b = 0. Hence, a Möbius transformation can have at most two fixed points unless h(z) = z for all z.

then

124

THE ELEMENTS OF COMPLEX ANALYSIS

Now let h be a Möbius transformation and let a, b, c be distinct points in C∞ with ξ = h(a), η = h(b), ζ = h(c). Suppose that g is another transformation with this property. Then g–1 o h has fixed points a, b, c. Hence g–1 o h = I = the identity transformation. Thus a Möbius transformation is uniquely determined by its action on three given points in C∞. Theorem 3. A Möbius transformation takes circles onto circles. Proof : It can be checked easily that the theorem holds for the transformation of the type w = az + b. We show that the theorem also holds for the transformation 1 z

of the type w = . The family of circles is represented by the equations (5)

α z z + β z + βz + γ = 0

where α and γ are real. If α = 0, then (5) reduces to a straight line. By using the transformation w = (6)

1 we find that (5) is mapped onto z

α + βw + βw + γww = 0.

When γ ≠ 0, (6) represents a circle and when γ = 0, (6) represents a straight line. Theorem 3 is then a consequence of Theorem 2. Example: Show that Möbius transformation w =

z −1 maps the half-plane z +1

Re z > 0 onto the unit disc | w | < 1. We discuss this mapping in three gradual steps, each of which is accompanied by a drawing in Fig. 9.IIA. It is easy to see that the given map is obtained by successive application of the maps: 1 ζ = z + 1, η = , w = 2η + 1. ζ Step I : Under the map ζ = z + 1, Re z > 0 is mapped onto the half-plane Re ζ > 1; see Fig. 9IIA (a). Step II : Under the map η = interior of the circle η −

1 , the half-plane Re ζ > 1 is mapped onto the ζ

1 1 = , but with the upper and lower halves of the 2 2

half-plane and of the circle interchanged, see Fig. 9.IIIA (b). Step III : Under the map w = – 2η + 1, the interior of the circle found in step II will be rotated through – π radians onto the interior of the circle η −

1 1 = , but 2 2

125

CONFORMAL MAPPING

the rotation will interchange the positions of the upper and lower halves of the disc. Then the rotation will be followed by a stretching (by a factor of 2) onto the interior of the circle | η + 1 | = 1, and, lastly, a shift through the vector 1 will yield the disc | w | < 1. W=

z–1 z+1

O

z=z+1

(c) W = –2x + 1

(a)

(b)

1 x= t

Fig. 9.III

9.3

DEFINITION

Let z2, z3, z4 be points in C∞. We define M : C∞ → C∞ by M(z) = If z2, z3, z4 are points in C

FG z – z IJ FG z Hz–z K Hz 3

2

4

2

z – z3 , if z2 = ∞; z – z4 z – z4 , if z3 = ∞; M(z) = 2 z – z4 z – z3 , if z4 = ∞. M(z) = z 2 – z3 M(z) =

IJ K

– z3 . – z4

126

THE ELEMENTS OF COMPLEX ANALYSIS

Observe that M is the only transformation where M(z2) = 1, M(z3) = 0, M(z4) = ∞. Definition The cross-ratio of four points is defined by ( z – z2 )( z3 – z4 ) . (z1, z2, z3, z4) = 1 ( z1 – z3 )( z2 – z4 ) Theorem 4. A Möbius transformation leaves a cross-ratio invariant. 1 This can be verified first for w = az + b, and then for w = . z Theorem 4 permits us to write down the Möbius transformation which maps the points z1, z2, z3 into the points w1, w2, w3 : w – w1 w2 – w3 z – z1 z2 – z3 . . . = w – w2 w1 – w3 z – z2 z1 – z3 Example 3: Find the Möbius transformation which maps the points z = – 1, 0, 1 into the points w = 0, i, 3i, respectively. Setting up the appropriate cross-ratios, we have ( w – 0) (i – 3i ) ( z + 1) ( 0 – 1) . = . ( w – i ) (0 – 3i ) ( z – 0 ) (– 1 – 1) After a little calculation, we get z +1 . w = – 3i z–3 9.4

SYMMETRY

We recall that the union C ∪ {∞} is the extended complex plane. Denote the extended complex plane by C∞. Let Γ be a circle through points z2, z3, z4. We say that the points z, z* in C∞ are symmetric with respect to Γ if (7)

(z*, z2, z3, z4) = ( z, z2 , z3 , z 4 ) .

We explain below what it means for z and z* to be symmetric. Let Γ be a straight line. Choose z4 = ∞. Then (7) reduces to z * – z3 z – z3 = . z2 – z3 z2 – z3 That is, | z* – z3 | = | z – z3 |. Hence, z and z* are equidistant from each point on the straight line Γ. z * – z3 z – z3 z – z3 Im = Im = – Im . Also, z2 – z3 z2 – z3 z2 – z3

127

CONFORMAL MAPPING

Hence, z and z* lie in different half planes determined by Γ. Thus, the line segment [z, z*] is perpendicular to Γ as in Fig. 9.IV. Suppose that Γ is a circle, i.e. Γ = {z : | z – α | = r} (0 < r < ∞). Let Γ passes through the points z2, z3, z4. It follows from equation (7) that (z*, z2, z3, z4) = ( z, z2 , z3 , z 4 ) . Since the Möbius transformations leave the cross-ratio invariant, we have z*

Γ

z

Fig. 9.IV

( z , z 2 , z 3 , z 4 ) = ( z – α , z 2 – α , z 3 – α, z 4 – α )

FG r r r I , , H z – α z – α z – α JK F r , z – α, z – α, z – αIJ =G K Hz –α F r + α, z , z , z IJ =G K Hz –α 2

= z – α,

2

2

3

2

3

2

4

2

4

2

2

Hence, i.e.

3

4

r2 +α z–α (z* – α) ( z – α ) = r2.

z* =

From this we have z* – α r2 = > 0. z – α | z – α |2 Thus z* lies on the ray {α + λ(z – α) : 0 < λ < α}. We illustrate this by a diagram (Fig. 9.V). A

α

z

z*

Fig. 9.V

128

THE ELEMENTS OF COMPLEX ANALYSIS

Let L be the ray from α through z. Let P be the perpendicular line to L at z and let P intersects Γ at A. Draw a tangent at A. Then the point of intersection of this tangent with L is the point z*. Theorem 5 (The Principle of Symmetry). The Möbius transformations preserve the symmetry ; that is, if a Möbius transformation M maps a circle C1 onto the circle C2, the z and z* symmetric with respect to C1 are mapped onto w and w* symmetric with respect to C2. Proof : It follows from Theorem 3 that the circles Γ1 passing though z and z* are mapped onto the circles Γ2 passing through w and w*. Since the Möbius transformations are conformal, the result follows. Other Examples Example 4: Find the Möbius transformation which maps the upper half of the z-plane into the interior of the unit circle in the w-plane (Fig. 9.VI). z-plane

w-plane

1

Fig. 9.VI

az + b cz + d be the required transformation. Since the unit circle in the w-plane is the image of the real axis in the z-plane, we have

Let

w=

b a = 1. d z+ c |a| | z | → ∞, we get = 1. |c|

|a| . |w|= |c| When

z+

a = eiθ and we have for all real values of z c b d z– – = z– – . a c This is possible iff the real axis in the z-plane is the perpendicular bisector of d b the line joining the points – and – . a c That is,

F I H K

F I H K

129

CONFORMAL MAPPING

Thus –

d b and – are conjugate complex numbers and we denote them by λ a c

and λ . Hence we have a z–λ . c z–λ z–λ (8) = eiθ . z–λ We can easily check that the upper half of the z-plane is not mapped onto the outside of | w | = 1. The point z = λ is mapped into the point w = 0, which is inside the unit circle | w | = 1. This completes the solution. As a special case, let eiθ = – 1, and let λ = i. Then z–i (9) w=– . z+i w–w Now Im (w) = 2i 1 z–i z +i – =– . 2i z + i z – i That is, we have z+z (10) Im (w) = . ( z + i )( z – i ) The denominator of the right side of (10) is a positive number. Thus I(w) is positive iff z + z is positive. But z + z = 2 Re (z). This shows that the transformation (9) maps the upper half of the z-plane onto the unit circle in the w-plane in such a way that the first quadrant of the z-plane (Re (z) > 0) is mapped onto the upper half of the circle (Im (w) > 0) and the second quadrant of the z-plane is mapped onto the lower half of the circle. The inverse transformation w –1 (11) z=–i w +1 maps the interior of the unit circle in the w-plane onto the upper half of the z-plane in such a way that the upper half of the circle is mapped onto the first quadrant of the z-plane. Example 5: Find a transformation which maps a sector of angle π/4 onto the interior of the unit circle. In fact, we cannot get the result only by taking the Möbius transformation. The result follows if we take the composition of transformations, namely t–i t = z4 and w = . t+i

w=

FG H

IJ K

130

THE ELEMENTS OF COMPLEX ANALYSIS

z4 – i z4 + i gives the desired result as illustrated in the Fig. 9.VII.

That is,

w=

O

O′ O′

z-plane

t-plane

w-plane

Fig. 9.VII

9.5

THE SCHWARZ-CHRISTOFFEL TRANSFORMATION

Consider a polygon [Fig. 9.VIII] in the w-plane having vertices at w1, w2, ..., wn and α1, α2, ..., αn the corresponding interior angles. Let the points w1, w2, ..., wn map respectively into the points x1, x2, ..., xn on the real axis of z-plane (Fig. 9.VIII). y

v

z

w1

α1

π α1

α4

α3 α2 w2

w4

w3 π α2

θ2

θ1 u

x1

x2

O

x

Fig. 9.VIII

The transformation (12)

w=K

z

( z – x1 ) α1 / π – 1 ( z – x 2 ) α 2 / π – 1 ... ( z – x n ) α n / π – 1

(where K is a complex constant) can be shown to map the interior of the polygon in the w-plane onto the upper half of the z-plane. The transformation can also be written as dw = K ( z – x1 )α1 / π – 1 ( z – x2 )α 2 / π – 1 ... ( z – x n )α n / π – 1 (13) dz Any three of the points x1, x2, ..., xn can be chosen at will. If xn is chosen at infinity the last factor in (12) and (13) is not present. Note that infinite open polygons can be considered as limiting cases of closed polygons. The points w1, w2, ... are mapped onto the points x1, x2, ... on the real axis. It follows from (13) that

131

CONFORMAL MAPPING

(14)

F α – 1I arg (z – x ) Hπ K α I Fα I +F H π – 1K arg (z – x ) + ... H π – 1K arg (z – x ).

arg (dw) = arg (dz) + arg (K) +

1

1

2

n

2

n

Suppose that as z in the z-plane moves along the real axis from the left toward x1, the point w in the w-plane moves along a side of the polygon toward w1. When z crosses from the left of x1 to the right of x1, θ1 = arg (z – x1) changes from π to 0 while all other terms in (14) remain unchanged. Hence arg (dw) decreases by (α1/π – 1) arg (z – x1) = α1 – π. In other words, the direction through the point w1 turns through the angle π – α1. Similarly, as z moves through x2, θ1 = arg (z – x1) and θ2 = arg (z –x2) change from π to 0 while other terms in (14) remain constant. Hence w2 turns through the angle π – α2 in the w-plane. By continuing the process, we see that as z moves along the real axis in the z-plane the point w moves along the sides of the polygon in the w-plane and vice-versa. It can be shown that the interior of the polygon is mapped onto the upper half plane by (13). For this purpose, it is sufficient to prove that the transformation maps the interior onto the unit circle. Suppose that the transformation w = f (z) maps the polygon P onto the unit circle | z | = 1 in the z-plane and that f (z) is analytic inside and on | z | = 1. Now we have to show that to each point a inside the polygon P there corresponds one and only one point z0 such that f (z0) = a. By Cauchy’s integral formula 1 dw = 1. 2πi P w – a Since w – a = f (z) – a, we have

z

z

1 f ′( z ) dz = 1. 2 πi | z | = 1 f ( z ) – a But f (z) – a is analytic inside | z | = 1. Hence, it follows from Theorem 3, § 7.3 that there is only one zero of f (z) – a inside | z | = 1. In other words, f (z0) = a. Observe that in order to get the required result we have used the known fact that the unit circle can be mapped onto the upper half plane. 9.6

THE TRANSFORMATIONS w = sin z AND w = cos z

Consider the interval of the real axis described by – π/2 ≤ x ≤ π/2, y = 0 Consider the mapping w = sin z. From the decomposition we have w = sin x cosh y + i sinh y cos x Then (15) u = sin x cosh y and v = cos x sinh y

132

THE ELEMENTS OF COMPLEX ANALYSIS

If y = 0, then cosh y = 1 and sinh y = 0. It follows that for any point on the given interval, the last equations become u = sin x and v = 0 Now, as x varies between – π/2 and π/2, sin x and therefore u varies between – 1 and + 1. Hence, under the mapping w = sin z, the given interval maps onto the interval – 1 ≤ u ≤ 1, v = 0, in the w-plane. We now show that the imaginary axis x = 0 is mapped onto the imaginary axis u = 0, under the mapping w = sin z. We have from (15) u = sin x cosh y, v = cos x sinh y When x = 0, u = 0 and v = sinh y. Then, as y varies from – ∞ to + ∞ on the imaginary axis, sinh y and hence v varies from – ∞ to + ∞, while u remains fixed at 0. This shows that the axis x = 0 is mapped onto the axis u = 0. We now consider the transformation w = cos z. Using the decomposition cos z = cos x cosh y – i sin x sinh y We can consider several special cases as considered for sin z. But cos z = sin (z + π/2) Hence, we can translate what we know about sin z into mapping properties of cos z. In other words, w = cos z can be expressed as the composite mappings: ζ = z + π/2, w = sin ζ We illustrate this process in the following example. Consider the interval I : – π ≤ x ≤ 0, y = 0. Under the mapping ζ = z + π/2, I is mapped onto the interval J : – π/2 ≤ R (ζ) ≤ π/2, Im (ζ) = 0 Then, using the result of the above example, we see that, under the mapping w = sin ζ, J is mapped onto K : – 1 ≤ u ≤ 1, v = 0 Hence, under the mapping w = cos z, the interval I is mapped onto the interval K. 9.7

RIEMANN SURFACES

In this section, we propose to introduce the idea of Riemann surface, with illustrative examples. Let w = z1/3. This multi-valued function may be split into the following three parts:

133

CONFORMAL MAPPING

R| w = r S| w = r Tw =r 0

(16)

1

2

e iθ / 3 , 0 ≤ θ < 2π 1/ 3 i ( θ + 2 π )/ 3 e , 0 ≤ θ < 2π 1/ 3 i ( θ + 4 π )/ 3 e , 0 ≤ θ < 2π 1/ 3

Note that each of these expressions defines a single-valued function that maps the entire z-plane onto “one third” of the w-plane. We can write this in the following form: w0 maps the z-plane onto the wedge 0 ≤ arg w ≤ 2 π/3 plus w = 0, w1 maps the z-plane onto the wedge 2 π/3 ≤ arg w < 4 π/3 plus w = 0, w2 maps the z-plane onto the wedge 4 π/3 ≤ arg w < 2 π plus w = 0 Observe that all three of the functions above are analytic everywhere except along the non-negative real axis. Now consider the functions defined by equations (16). We restrict their domain by deleting from it the negative real axis, and define ψ0 (z) = r1/3 eiθ/3, 0 < θ < 2π, r ≠ 0 ψ1 (z) = r1/3 ei (θ + 2π)/3, 0 < θ < 2π, r ≠ 0 ψ2 (z) = r1/3 ei (θ + 4π)/3, 0 < θ < 2π, r ≠ 0 Each of these functions is called a branch of the multi-valued function 1/3 w = z . The ray consisting of the non-negative real axis is called a branch cut; and the point z = 0 from which the branch cut emanates is a branch point of each branch. Write equations (16) in the following alternative form: w0 = r1/3 eiθ/3, 0 ≤ θ < 2π w1 = r1/3 eiθ/3, 2π ≤ θ < 4π w2 = r1/3 eiθ/3, 4π ≤ θ < 6π It can be easily seen that the last three equations can be grouped together and written into one: (17) w = r1/3 eiθ/3, 0 ≤ θ < 6π A careful examination of these three alternatives will reveal that one fact remains unaltered in all of them: The z-plane must be traversed three times before the w-plane can be covered entirely. However, equation (17) lends itself to a “geometrical construction” which leads to a Riemann surface for the function w = z1/3. Instead of travelling the z-plane three times, we take three copies of the z-plane, S1, S2, S3, forming a configuration that can be described as follows: [see Fig. 9.IX (a)]. As θ varies from 0 to 6π, trace the path of a point z = reiθ as it describes a continuous curve around the origin, say, a circle. The point z starts from

134

THE ELEMENTS OF COMPLEX ANALYSIS

the positive real axis and travels on S1 until its argument θ reaches 2π. At that instant, z ascends onto S2 and it continues to travel around the origin while it remains on this second level. When its argument reaches 4π, z ascends onto S3 and travels around the origin once again until θ reaches 6π. By this time, the function given by equation (17) has covered the w-plane completely, and the point z descends onto the first level S1. The actual configuration can be visualized in the form of the sheets of the surface joined along the non-negative real axis [see Fig. 9.IX]. +

4n

S3

+

2n

S2 –

4n

S1

+

O

S3 S2

O

S1

–

2n (a)

(b)

Fig. 9.IX

This scheme has enables us to have a one-to-one mapping from the threesheet Riemann surface onto the w-plane by means of the function given in equation (17). In a similar way one can visualize the Riemann surface of w = z1/2 (with two sheets), of w = z1/4 (with four sheets) and, in general, of w = z1/n (with n sheets). Example: Riemann Surface of the Natural Logarithm We recall that the multi-valued function (18) w = log z can be written as w = log | z | + i(arg z + 2 k π), (k = integer) Then, as k ranges over all integral values, we obtain the single-valued functions (19) ..., w–3, w–2, w–1, w0, w1, w2, w3, ... . Note that each of these maps the z-plane; except z = 0, onto a horizontal strip of the w-plane, having width 2π. Thus, in order to cover the entire w-plane we map the z-plane an infinite number of times, by using all the functions given in (19). We now employ an infinite number of copies of the z-plane ..., S–3, S–2, S–1, S0, S1, S2, S3, ... . each with its origin deleted and with a slit starting at the origin and running along the non-positive real axis. Consecutive sheets are then glued together along the slits so that they form a continuous sheet which resembles like a circular stair case of infinite height and depth. A point moving around the origin in a counterclockwise direction will be going up the staircase, and everytime it completes a circle (centred at the origin), its position will be on the next sheet up, directly above the starting point.

135

CONFORMAL MAPPING

Observe that the above geometrical scheme enables us to think of the multivalued function w = log z as a single valued, mapping its Riemann surface in a one-to-one fashion onto the entire w-plane.

EXERCISES 1. Show that the mapping w = z2 transforms every straight line into a parabola. 2. Consider the mapping w = z2. Find the area of the image of the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. Use J = | f ′ |2 where J is the Jacobian of the mapping. 3. Find the image of Re z > 0 and Im z > 0 under the mapping w = log z. 4. Find the image of the strip – π < y ≤ π under the mapping w = 1 + ez. 5. Let G = {z : 0 < | z | < 1}. Find the map which maps G conformally onto the open unit disk. 6. Let G be a region and suppose that f : G → C is analytic such that f (G) is a subset of a circle. Prove that f is constant. 7. Determine the fixed point, the rotation and dilation of the following mappings: (i) w = iz + 1 (ii) w = 2iz + 1 + i (iii) w = (2 + 4i) z + 2 8. Prove that if the complex numbers z1, z2, z3 and z4 lie on a circle, then their crossratio is real. 9. Determine the Möbius transformation which maps the following: (i) the half plane Re z ≥ 0 onto the half plane Im z ≥ 0. (ii) the upper half plane Im z ≥ 0 onto itself. (iii) the upper half plane Im z ≥ 0 onto the lower half plane Im z ≤ 0. (iv) the half plane Im z ≥ 0 onto the closed disk D (0, 1). 10. Suppose that T is a Möbius transformation with fixed points z1 and z2. Let S be a Möbius transformation. Prove that S–1 TS has fixed points S–1 z1 and S–1 z2. 11. Prove that if the two fixed points z1, z2 of a Möbius transformation T coincide, then T is of the form K w = T (z) = + z2 where K ∈ C. z – z1 12. (a) Show that a Möbius transformation T has 0 and ∞ as its only fixed points iff it is a dilation. (b) Show that a Möbius transformation T has ∞ as its only fixed point iff it is a translation. 13. Show that a Möbius transformation T satisfies T(0) = ∞ and T(∞) = 0 iff Tz = z1 z– 1 for some z1 ∈ C.

136

THE ELEMENTS OF COMPLEX ANALYSIS

14. Find the Möbius transformation which maps (i) the points a, b, c into 0, 1, ∞ (ii) the points 0, 1, ∞ into a, b, c (iii) the points – 1, 0, 2 into 0, 3, 6. 15. Find the Möbius transformation which maps (i) 1, i, – 1 onto i, – 1, 1 (ii) i, – 1, 1 onto – 1, – i, 1 (iii) – 1, – i, 1 onto – 1, 0, 1 (iv) – 1, 0, 1 onto – 1, i, 1 (v) – 1, i, 1 onto 1, i, – 1 16. Find the Möbius transformation which maps (i) 0, 1, ∞ onto 1, ∞, 0 (ii) 0, 1, ∞ onto 1, – 1, i (iii) 0, 1, ∞ onto – 1, 0, 1 (iv) 0, 1, ∞ onto – 1, – i, 1 17. Let z1, z2, z1 ≠ z2, be the two fixed points of a Möbius transformation T. Show that T may be written as: w – z1 z – z1 =K w – z2 z – z2 Show that the cross-ratio of z1, z2, z, w is constant. 18. Let z1, z2, z3, z4 be distinct complex numbers. Suppose that they lie on the same circle, in that order. Prove that | z1 – z 3 | | z2 – z 4 | = | z1 – z 2 | | z3 – z 4 | + | z2 – z 3 | | z4 – z 1 | 19. Recall: A function f with domain A and range B is given. We say that f is univalent if for x1, x2 ∈ A, the equation f (x1) = f (x2) implies that x1 = x2. We say that two regions B and B* in C are conformally equivalent if there exists a univalent function g : B → B* such that B* = g(B). The mapping g is called a conformal equivalence. Prove that a conformal equivalence mapping C onto C is necessarily a polynomial. 20. With reference to problem (19), show that a conformal equivalence mapping C onto C is necessarily a nonconstant polynomial function.

10 10.1

HARMONIC FUNCTIONS

DEFINITIONS OF HARMONIC FUNCTIONS

Let f be a (real or complex valued) differentiable function of the real variables x and y. Consider the differential

∂f ∂f dx + dy . ∂x ∂y The functions z = x + iy and z = x – iy have differentials

(1)

df =

(2) Thus

dz = dx + idy, dz = dx – idy.

1 1 ( dz + dz ), dy = ( dz − dz ) . 2 2i Substituting this in (1) we have

(3)

dx =

(4)

df =

FG H

IJ K

FG H

IJ K

FG H

IJ K

1 ∂f ∂f 1 ∂f ∂f −i dz dz + +i 2 ∂x ∂y 2 ∂x ∂y

Write

FG H

IJ K

∂ 1 ∂ ∂ ∂ 1 ∂ ∂ . = −i , = +i ∂z 2 ∂x ∂y ∂z 2 ∂x ∂y With this notation, we get the equation

(5)

(6)

df =

∂f ∂f dz + dz . ∂z ∂z

The condition

∂f ∂f +i =0 ∂x ∂y can now be written as (7)

∂f = 0. ∂z 137

138

THE ELEMENTS OF COMPLEX ANALYSIS

Definition: Let Ω be an open subset of C. A function f (x, y) of two real variables x and y defined in an open set Ω is said to be harmonic in Ω if it has continuous second partial derivatives and satisfies the Laplace’s equation (8)

∂2 f ∂2 f = 0. + ∂x 2 ∂y 2

∂ ∂ and with respect to the complex variables z = x + iy ∂z ∂z and z = x – iy, we find that

Differentiating

∂2 ∂2 ∂2 . + = 4 ∂x 2 ∂y 2 ∂z ∂z Thus the condition (8) is equivalent to

(9)

∂2 f = 0. ∂z ∂ z Note that the condition (10) expresses that f is a harmonic function. Observe that by (8), a necessary and sufficient condition for a complex valued function f = u + iv (u and v being real-valued) to be harmonic is that u and v are harmonic. Definition: If f : Ω → C is an analytic function, then v = Im f is the conjugate harmonic function of u = Re f, and u is the conjugate harmonic function of – v.

(10)

10.2

HARMONIC FUNCTIONS AND ANALYTIC FUNCTIONS Theorem 1. Let f : Ω → C be analytic. Then f is harmonic. Proof : Since f is analytic, it is infinitely differentiable, and by taking the

derivative

∂ of the relation ∂z ∂f = 0, we get ∂z ∂2 f = 0. ∂z ∂z

Hence f is harmonic. Corollary 1 : The real and imaginary parts of an analytic function are harmonic functions. Example: log | z | is harmonic in the whole plane C excluding the point z = 0. log z has a branch in some neighbourhood of each point z ≠ 0 and log | z | is the real part of such a branch. Theorem 2. Let Ω be an open subset of C. Let u (x, y) be a real harmonic function in Ω. Then in a neighbourhood of each point of Ω, u is the real part of an analytic function f which is determined upto addition of a constant.

139

HARMONIC FUNCTIONS

Proof : Since u is harmonic ∂2u = 0. ∂z ∂z ∂u ∂u is analytic in Ω. The differential form 2 dz has therefore ∂z ∂z primitive f locally. In other words, ∂u (11) df = 2 dz. ∂z The relation (11) shows that f is analytic. Taking the complex conjugate of the relation (11) we obtain

Hence

∂u dz . ∂z Adding (11) and (12) we get 1 d(f + f ) = du. 2 Thus u is equal to the real part of f with a real constant added if necessary. It remains to show that if two analytic functions f1 and f2 in a neighbourhood of the same point have the same real part, then their difference f = f1 – f2 is constant. We have (12)

df =2

d (f + f ) = 0 i.e.

∂f ∂f dz + dz = 0 ∂z ∂z

∂f ∂f = 0 and = 0. This completes the proof. ∂z ∂z Note the Theorem 2 says only that any real harmonic function is locally the real part of an analytic function. Given a real harmonic function u in an open set Ω, there does not necessarily exist an analytic function f in the whole of Ω, whose real part is equal to u. For example, when Ω = C ~ {0}, log | z | is not the real part of an analytic function in Ω. Theorem 3. Let u : Ω → C be harmonic. Then u is infinitely differentiable. Proof : Let z0 = x0 + iy0 be fixed in Ω. Choose δ such that D (z0; δ) ⊂ Ω. Then there is a harmonic function v in D (z0; δ) such that f = u + iv is analytic in D (z0; δ). Thus f is infinitely differentiable on D (z0; δ) and hence u is infinitely differentiable. The next result is analogous to Cauchy Integral Formula. Theorem 4 (Mean Value Theorem). Let u : Ω → R be a harmonic function

which implies that

and let D (a; r) ⊂ Ω. If γ is the circle | z – a | = r, then 1 2π (13) u (a) = u ( a + re iθ ) dθ . 2π 0

z

140

THE ELEMENTS OF COMPLEX ANALYSIS

Proof : Choose a disk D such that D (a ; r) ⊂ D ⊂ Ω. Let f be analytic on D where u = Re f. It follows from Cauchy’s Integral formula that 1 2π (14) f (a) = f (a + re iθ ) dθ . 0 2 πi By equating the real part of (13) we get the result. Theorem 5 (Maximum Principle). Let u be a continuous real valued function on a region Ω satisfying the Mean Value Theorem. Let a ∈ Ω such that u (a) ≥ u (z) for all z ∈ Ω, then u is a constant function. Proof : Define the set S by S = {z ∈ Ω: u (z) = u (a)}. Since u is continuous, the set S is closed in Ω. Fix z0 in S and choose r such

z

that D (z0 : r) ⊂ Ω. Let b be a point in D (z0; r) such that u (a) ≠ u (b), then u (a) > u (b). Since u is continuous, we have u (z) < u (a) = u (z0) for all z in a neighbourhood of b. In particular, let ρ = | z0 – b | and b = z0 + ρeiα where 0 ≤ α < 2π. Then there exists a proper interval say I of [0, 2π] such that α ∈ I and u (z0) > u (z0 + ρeiβ) for all β in I. Hence, by Theorem 2 1 2π u (z0) = u ( z0 + ρe iβ ) dβ < u ( z0 ) 0 2π which is a contradiction. Hence D (z0; r) ⊂ S and S is open. Since Ω is connected, S = Ω and the proof is complete. By considering the function – u and appealing to Theorem 5 we get the Minimum principle theorem. Theorem 6 (Minimum Principle). Let u be a continuous real-valued function on a region Ω satisfying the Mean Value theorem. Let a ∈ Ω such that u (a) ≤ u (z) for all z ∈ Ω, then u is a constant function.

z

10.3

HARMONIC FUNCTIONS ON A DISK

The maximum principle has the following consequences. If u (z) is harmonic on a region Ω which contains a closed and bounded set S, then it is uniquely determined by its values on the boundary of S. If u1 and u2 are two harmonic functions with the same boundary values, then u1 – u2 is harmonic with the boundary value 0. By the Maximum and Minimum principle we find that u1 – u2 must be identically zero on S. Now the problem arises of finding u when the boundary values are given. This type of problem is known as the Dirichlet problem for the Laplace equation in two variables. the Dirichlet problem for a circular disk can be solved by

141

HARMONIC FUNCTIONS

Poisson formula. We derive below the Poisson formula by using Cauchy’s integral formula. Recall that the Cauchy’s integral formula is given by

z

1 f (ζ) dζ 2πi γ ζ − z where γ is the circle represented by ζ = Reiφ (0 ≤ φ ≤ 2π) and the function f (z) = u (r, θ) + iv (r, θ) (z = reiθ) is analytic in a simply connected region Ω containing γ. Since dζ = iReiφ dφ = iζdφ we have from (15) (15)

f (z) =

(16)

f (z) =

1 2π

z

2π

f (ζ)

0

ζ dφ . ζ−z

Consider a point η outside γ where η = ζζ | z . Then the integral in (15) is analytic in the disk | z | ≤ R and the integral is zero. Hence

i.e. i.e.

z

f (ζ) dζ = 0 ζ−η 1 2π (ζ) f (ζ) dφ = 0 2π 0 ζ−η z 1 2π f (ζ) dφ = 0 (17) 2π 0 z −ζ Subtracting (17) from (16) and using the expression

z

(18)

z

1 2πi

γ

z ζ ζζ − zz − = ζ − z z − ζ (ζ − z ) ( ζ − z )

we obtain

z

ζζ−zz 1 2π f (ζ) dφ. 2π 0 (ζ − z ) ( ζ − z ) By the polar representations of z and ζ we find that the quotient in the integrand is equal to (19)

f (z) =

R2 − r 2 R2 − r 2 = ( Re iφ − re − iθ ) ( Re − iφ − re − iθ ) R 2 − 2 Rr cos (θ − φ) + r 2 Hence equating the real parts of (19) we obtain Poisson’s integral formula

z

R2 − r 2 1 2π dφ. u ( R, φ) 2 R − 2 Rr cos (θ − φ) + r 2 2π 0 This formula represents a harmonic function in terms of its values given on the boundary circle of the disk. On this circle this function is equal to u (R, φ), except at points where u (R, φ) is not continuous. The proof is given in the next section.

(20)

u (r, θ) =

142

THE ELEMENTS OF COMPLEX ANALYSIS

From (20) we derive an important series of the function u in terms of simple harmonic functions. We have seen that the quotient in the integrand of (20) was derived from (18), and the right side of (18) is the real part of ζ + z/ζ – z. Using geometric series expansion we obtain z 1+ n ∞ z ζ . (21) ζ + z/ζ – z = =1+ 2 ∑ z n =1 ζ 1− ζ iθ iφ Since z = re and ζ = Re , we have

FG IJ HK

(22)

=

F rI H RK

F zI Re G J H ζK

n

= Re

RS r TR

n n

e inθ e − inφ

UV W

n

(cos nθ cos nφ + sin nθ sin nφ).

From (21) and (22) we obtain

Re

FG IJ H K

∞ ζ+z r =1+ 2 ∑ ζ−z n =1 R

n

(cos nθ cos nφ + sin nθ sin nφ).

This expression is equal to the quotient in (20), and by inserting the series in (20) and integrating term by term we have (23)

∞

u (r, θ) = a0 + ∑

n =1

z

F rI H RK

n

(an cos nθ + bn sin nθ)

1 2π u ( R, φ) dφ; 2π 0 1 2π u ( R, φ) cos nφ dφ; (24) an = π 0 1 2π bn = u ( R, φ) sin nφ dφ (n = 1, 2, ...). π 0 Observe that for r = R, the series (23) becomes the Fourier series u (R, φ). Example: Find the potential u (r, θ) in the unit disk (r < 1) where the boundary values are where

a0 =

z z

u (1, φ) =

RS− φ / π (− π < φ < 0) T φ / π (0 < φ < π).

Since u (1, φ) is even, bn = 0, and from (24) we have 1 a0 = and 2 π φ 0 1 − cos nφ dφ φ / π cos nφ dφ + an= −π 0 π π 2 = 2 2 (cos nπ – 1). n π

LM N

z

z

OP Q

143

HARMONIC FUNCTIONS

Hence

RS T

2 2 an = − 4 / n π , when n is odd 0, when n is even

and the potential is

LM N

10.4

OP Q

r3 1 4 − 2 r cos θ + 2 cos 3θ + ... . 2 π 3

u (r, θ) =

CONSTRUCTION OF HARMONIC FUNCTIONS ON A DISK

In this section, we shall construct harmonic function on the unit disk with prescribed boundary value. We shall introduce first the method of Dirac sequences, which we use. We shall be concerned with periodic functions of period 2π, so we make that assumption from the very beginning. Definition: A Dirac sequence is a sequence of real valued functions {Qn} of real variable, periodic of period 2π, which satisfies the following conditions: (i) Qn (x) ≥ 0 for each n and x; (ii) {Qn} is continuous, and

z

2π 0

Qn (t ) dt = 1;

(iii) Let ε and δ be given. There exists n0 such that if n ≥ n0, then

z

–δ –π

Qn (t ) dt +

z

π δ

Qn (t ) dt < ε.

We illustrate Dirac Family {Qn} in Fig. 10.I. Note that the condition (iii) means that for sufficiently large n the area under the curve is concentrated near 0. Observe that Qn (– x) = Qn (x). Let f be a periodic function of period 2π. We define the convolution of f with Qn as Qn* f =

z

π −π

Qn ( x − t ) f (t ) dt =

z

π −π

f ( x − t ) Qn (t) dt.

Theorem 7. Let f be a continuous function, periodic of period 2π. Then the sequence of functions {Qn * f} converges uniformly to f.

–p

p

Fig. 10.I

144

THE ELEMENTS OF COMPLEX ANALYSIS

Proof : Write

z

wn (x) = Qn* f =

π

−π

f ( x − t ) Qn (t ) dt .

If follows from condition (ii) that f (x) = f (x) Hence (25)

z z

π

−π

π

wn (x) – f (x) =

−π

Qn (t ) dt =

z

π

−π

f ( x ) Qn (t ) dt.

( f ( x − t ) − f ( x )) Qn (t ) dt.

By the compactness of the circle, and the uniform continuity of f, it follows that | f (x – t) – f (x) | < ε whenever | t | < δ. Let | f (x) | ≤ M. Then choose n0 such that if n ≥ n0. −δ π ε . Qn (t ) dt + Qn (t ) dt < −π δ 2M Now

z

z z z z −∞

| wn (x) – f (x) | ≤

−π

+

δ

−δ

+

π δ

| f ( x − t ) − f ( x ) | Qn (t) dt

= I1 + I2 + I3. Since M is the bound for f, it follows that | f (x – t) – f (x) | ≤ 2M. Thus, I1 + I3 ≤ 2M Also

I2 = ≤

z z

δ

−δ δ

−δ

F H

z

−δ −π

Qn (t ) dt +

z

π δ

I K

Qn (t ) dt < ε.

| f ( x ) − f ( x − t ) |Qn (t) dt ε Qn (t ) dt ≤

z

π

−π

ε Qn (t ) dt ≤ ε.

This completes the proof of the theorem. Observe that if Qn has continuous derivatives, then

F H

I K

d d Qn * f. (Qn * f ) ( x ) = dx dx It was convenient to introduce the general Dirac family {Qn}, but we shall deal with families, indexed by r with 0 < r < 1 and r approaching to 1. In this context, we need to define the Poisson kernel which is of basic importance. The Poisson kernel is defined by 1 ∞ | k | ikθ (27) Pr (θ) = ∑r e . 2π k = − ∞

(26)

145

HARMONIC FUNCTIONS

ζ = reiθ, 0 ≤ r < 1, then

Let

1 + re iθ = (1 + ζ) (1 + ζ + ζ2 + ...) 1 − re iθ ∞

∞

k =1

k =1

k k ikθ = 1 + 2 ∑ζ = 1 + 2 ∑ r e .

FG 1 + re IJ = 1 + 2 ∑ r H 1 − re K iθ

Thus

Re

∞

iθ

k

cos kθ

k =1

∞

= 1 + 2 ∑ rk k =1

FG e H

ikθ

+ e − ikθ 2

IJ K

= 2π Pr (θ). Observe that 1 + re iθ 1 + r (e iθ − e − iθ ) − r 2 , = | 1 − re iθ |2 1 − re iθ

FG 1 + re IJ = 1 − r H 1 − re K 1 − 2r cos θ + r iθ

Re

2

iθ

2

.

Hence

1 (1 − r 2 ) . 2 π (1 − 2r cos θ + r 2 ) The Poisson kernel satisfies the following properties. Properties: (i) Pr (θ) ≥ 0 for all r, θ;

(28)

(ii)

Pr (θ) =

z

π

−π

Pr (θ) dθ = 1;

(iii) Let ε and δ be given. There exists ρ, 0 < ρ < 1, such that if ρ < r < 1, then

z

−δ

−π

Pr (θ) dθ +

z

π δ

Pr (θ) dθ < ε.

Proof : (i) By the equation (28), 1 − r2 ( | 1 – reiθ |)–2, and since r < 1, Pr (θ) ≥ 0. 2π (ii) For a fixed r, 0 ≤ r < 1, the series 1 ∞ | k | ikθ ∑r e 2π k = − ∞ converges uniformly in θ. Thus,

Pr (θ) =

z

π −π

Pr (θ) dθ =

∞

∑r

k=−∞

|k|

1 2π

z

π

e ikθ dθ = 1.

−π

By using (28) it can be easily verified that the property (iii) holds. The verification of (iii) is left to the reader.

146

THE ELEMENTS OF COMPLEX ANALYSIS

Note that with these properties we view Pr (θ) not as a function of r and θ but a family of functions of θ, indexed by r. In other words, we view {Pr} as a Dirac family, with r → 1. Write wr = Pr * f, where wr (θ) = w (r, θ). Then wr (θ) is a function on the open unit disk. By theorem 7, wr (θ) converges uniformly to f (θ) as r approaches 1. The next theorem shows that the Dirichlet problem can be solved for the open disk 0 ≤ r < 1 and 0 ≤ θ ≤ 2π. Theorem 8. The function (r, θ) → wr (θ) is harmonic on the open disk 0 ≤ r < 1 and 0 ≤ θ ≤ 2π. Proof : We recall that the Laplace operator in polar coordinates is given by ∂2 1 ∂ 1 ∂2 + + . ∂r 2 r ∂r r 2 ∂θ 2 By applying this operator to Poisson kernel Pr (θ), and differentiating the

Δ=

series

1 ∞ | k | ikθ ∑ r e term by term, we find that 2π k = − ∞

(29)

LMr F ∂ I + r FG ∂ IJ + ∂ OP P (θ) = 0. N GH ∂r JK H ∂r K ∂θ Q 2

2

2

2

2

r

That is, ΔP = 0, where P is a function of two variables, P (r, θ) = Pr (θ). Using (26), we obtain Δ ((Pr* f ) (θ)) = (ΔPr (θ)) * f = 0. This completes the proof of the theorem. Consider now the original periodic function f as a boundary value on the circle. Then the function u is defined by convolution for 0 ≤ r < 1, and by continuity for r = 1, and is given by 1 π u (reiθ) = Pr (θ − t ) f (e it ) dt. 2π – π Thus, the theorem shows that there exists a harmonic function u having the prescribed continuous boundary value f on the circle. Note that the results of this section can be extended to disks of arbitrary radius R by means of the Poisson kernel

z

Pr (θ) =

1 R2 − r 2 . 2 , 0 ≤ r < R. 2 π R − 2 R r cos θ + r 2

147

HARMONIC FUNCTIONS

In this case, one needs to prove: (a) the statements about a Dirac family; (b) the existence and uniqueness of a harmonic function with prescribed continuous boundary value on the closed disk of radius R. We leave it as an exercise. In many applications, it may happen that the boundary value function f has a finite number of discontinuities. In that case, assuming that f is bounded, a similar analysis of proof will show that the theorem is true. We state the theorem and the verification is left to the reader. Theorem 9. Let f be a bounded real valued function, periodic of period 2π. Let f be continuous on a compact set E. Then the sequence of functions {Qn * f } converges uniformly on E to f. Further, the function u defined by 1 π u (reiθ) = Pr (θ − t ) f (e it ) dt 2π – π is harmonic on the open disk 0 ≤ r < 1 and 0 ≤ θ ≤ 2π.

z

10.5

SOME PHYSICAL APPLICATIONS OF CONFORMAL MAPPING

We have seen in this chapter that there are physical problems in which we need to solve Dirichlet problems with piecewise constant boundary values. In particular, these Dirichlet problems can be solved by using Fourier series. In general situations, conformal mapping is useful because a conformal map carries harmonic functions to harmonic functions. Consequently, it may be possible to solve a complicated problem by solving a simpler problem and then transforming the solution by conformal mapping. This process is used in solving some problems regarding fluid flow, electrostatics, and heat. In order to explain how it is applied, we need some physical terminology, which we present first for fluid flow. Terminology of Fluid Flow Consider a fluid that is flowing in a channel or lake at uniform depth, and in horizontal layers, so that a horizontal cross-section at any level represents the whole flow. We may then idealize the flow as two-dimensional and take it as the flow of a plane sheet of two-dimensional fluid in some region. We represent a velocity vector v by v = pi + qj that specifies the velocity at each point. We assume that the fluid is incompressible and non-viscous. The term non-viscous means that there is no internal friction, and no tendency for the fluid to stick to the boundary of the region in which it is flowing. We also assume that the flow is irrotational (i.e. curl v = 0), which in physical language says that there are no local whirlpools in the fluid. Under these ∂φ assumptions, the velocity vector v is the gradient of a potential φ, so that p = and ∂x ∂φ q= , where φ is harmonic. Then φ is called the velocity potential, and the curves ∂y

148

THE ELEMENTS OF COMPLEX ANALYSIS

φ = constant are equipotentials. The conjugate harmonic function ψ is called the steam function, and its level curves ψ = constant are orthogonal to the equipotentials. Since, the derivative of φ along an equipotential is zero, the flow is directed along

z FGH

IJ K

∂φ ds along ∂n A an arc A of an equipotential is proportional to the amount of fluid crossing the arc A per unit time. The analytic function f = φ + iψ is the complex potential, f ′ is the complex velocity, and | f ′ (z) | is the speed of z. Since the function f (z) is also analytic, it follows that – ψ and φ are the velocity potential and stream function of another flow, whose streamlines are the equipotentials of the original flow. If f ′ (z0) = 0, the speed is zero at z0, then z0 is called a stagnation point. On the other hand, if f ′ (z) → ∞ as z → z0. We may conclude that the speed must be very large near z0. If the flow is taking place in a region whose boundary contains a curve C across which fluid cannot flow, then the velocity vector at points of C must be tangent to C. Therefore a rigid boundary is always a streamline. It follows that we can have an interesting flow inside a simple closed curve C only if there are singular points on C, points at which fluid is being supplied or removed. Note that for an unbounded region (for example, the strip between two parallel lines) different parts of the boundary can be different streamlines, with a singular point at ∞. Example: (a) Consider the function f (z) = αz (α is real and positive) as representing a flow. Then its velocity potential is φ = αx and its stream function is ψ = αy. The streamlines, which are the curves of the family. ψ = c are the horizontal lines y = c/α. Observe that the flow described by f is parallel to the real axis. Its velocity is V = α [see Fig. 10.II(a)].

the curves of constant ψ, which are the streamlines. The integral

x

Fig. 10.II (a)

(b) We now examine a flow around 90° turn and for that we consider the function w = z1/2.

149

HARMONIC FUNCTIONS

We know that, under this function, the upper half of the z-plane is mapped onto the first quadrant of the w-plane. Its inverse function z = x + i y = (u2 – v2) + i (2uv) transforms the first quadrant of the w-plane [Fig. 10.II (b)] into the upper half of the z-plane [Fig. 10.II (a)]. Then, it follows from the first part of this example that the velocity potential in this case is given by φ = αx = α (u2 – v2) and the stream function becomes ψ = αy = 2α uv. Hence, the streamlines ψ = c are the rectangular hyperbola c uv = 2α [see Fig. 10.II(b)], and the complex potential is f = w2. Finally, the velocity of the flow at any point (u, v) of the first quadrant is V = f ′ = 2α (u – iv) and hence its speed is | V | = 2α (u2 + v2)1/2. V

Fig. 10.II (b)

Some Special Flows We shall first construct the flows generated by some simple analytic functions. We can then use conformal mapping to construct more complicated flows, using the principle that a conformal map transforms the solution of a Dirichlet problem to the solution of another Dirichlet problem.

150

THE ELEMENTS OF COMPLEX ANALYSIS

Uniform Flows Consider the simplest flow which has complex potential z. We could have taken the complex potential as cz, c > 0 but we can suppose that the units have been

dw = 1. Thus the velocity is constant, parallel dz to the real axis, and from left to right. If we take the real axis as rigid boundary, we have a uniform flow in the upper half plane. Suppose we take a line y = k as a rigid boundary, then we have a flow in the channel between two parallel lines. In any of these cases the flow is referred to as a uniform flow. The streamlines are the lines y = constant. The following question arises: Where does the fluid come from, and where does it go ? We have to think of a source of fluid at ∞ and a sink, where fluid is being removed, also at ∞. It can be arranged in such a way that the fluid leaves the source, flows to the right, and then disappears at ∞. In order to give a mathematical description of sources and sinks, it will be easier to consider a source and a sink separately at finite points.

chosen so that c = 1. It follows that

Sources and Sinks We begin by considering the complex potential w = m log z. Since, this is not a single valued function, we have to consider it only in the plane cut from 0 to ∞ (i.e. along the positive real axis). The complex velocity is m = (m/r) (cos θ – i sin θ) z Thus the speed is m/r. Since, the velocity vector is (m/r) (i cos θ + j sin θ), it follows that the flow is outward along rays from the origin (θ = constant). Hence, the cut along the positive real axis is along a streamline, and its presence is neglected. Now suppose that the fluid have unit density and let the fluid crosses a circle | z | = r > 0 in unit time. This is given by

z

∂φ ds = ∂ n |z|=r

=

z z

2π 0

∂ ( m log r ) r dθ ∂r

2π 0

m d θ = 2 πm ,

so that the same amount of fluid crosses (in unit time) each circle centred at the origin. Hence, we say that m log z is the complex potential of a source at 0, of strength m. A sink is defined as a negative source, its complex potential is – m log (z – z0) if it is located at a finite point z0.

151

HARMONIC FUNCTIONS

We illustrate the above facts in the form of a solved example. Example: Let fluid emanates at a constant rate from an infinite line source perpendicular to the z-plane at z = 0. (a) Show that the speed of the fluid at a distance r from the source is m/r where m is a constant. (b) Show that the complex potential is m log z. (c) Consider portion of the line source of unit length as shown in Fig. 10.III. Suppose Vr is the radial velocity of the fluid at a distance r from the source. Let σ be the density of the fluid which is incompressible.

Fig. 10.III

We then have: Mass of the fluid per unit time emanating from line source of unit length = mass of fluid crossing surface of cylinder of radius r and height 1 = (Surface Area) ⋅ (Radial Velocity) ⋅ (Fluid Density) = (2πr ⋅ 1) (Vr) ⋅ σ = 2πr Vr σ. If this is taken as constant k, then k m = Vr = 2π r σ r k where m = is called the strength of the source. 2πσ ∂φ m = , it follows by integrating that (b) Since Vr = ∂r r φ = m log r where the constant of integration has been omitted. But m log r is the real part of m log z which is therefore the required complex potential. 10.6

SOME OTHER PHYSICAL INTERPRETATIONS

One can interpret every flow problem as a problem of the flow of heat or electrostatic potentials. The physical terminology is, of course, different. The heat problem can

152

THE ELEMENTS OF COMPLEX ANALYSIS

be considered as the problem of steady-state temperatures in a two dimensional lamina with its flat faces insulated. This problem of steady-state temperature can also be considered in a solid, homogeneous cylinder all of whose cross sections parallel to a given plane have the same shape (not necessarily circular), and with boundary conditions imposed on the edges of the lamina or the surface of the cylinder. A simple example of this model is the door of a refrigerator, with given distributions of temperature on the inside and outside surfaces. The temperature T satisfies Laplace’s equation ∇2 T = 0 The curves T = constant are called isothermals. Suppose that U is the harmonic function conjugate to T, then the curves U = constant are the lines along which heat flows. The quantity of heat crossing an arc A per unit time is proportional to

z FGH A

IJ K

∂T ds. ∂n

∂T = 0 on A, no heat is flowing through A and in this case we say that A is ∂n insulated. We now solve one problem related to an insulated boundary. Exercise: Prove that a conformal map transforms an insulated boundary to an insulated boundary. Solution: Let w = f (z) map a region Ω, bounded by a curve C, to a region Ω1, bounded by a curve γ, in the w-plane, so that w = u + iv. Let φ (u, v) be harmonic in

When

Ω1. We shall prove that if

∂φ = 0 on an arc γ1 of γ, then ∂n

FG ∂ IJ φ [u (x, y), v (x, y)] = 0 H ∂n K

∂φ = 0 implies that no heat flows ∂n across γ1, and therefore γ1 is an isothermal. We can identify this problem with a fluid flow problem considering γ1 as a streamline and φ as velocity potential. Note that the conformal map transforms φ to a harmonic function and its harmonic conjugate ψ to a streamline. Thus, the conformal map transforms the orthogonal trajectory

on the corresponding arc C1 of C. The condition

φ = constant to an equipotential where Observe that the value of the condition

∂φ = 0. ∂n

∂φ is not preserved under a conformal map, but ∂n

∂φ = 0 is preserved. ∂n

153

HARMONIC FUNCTIONS

Electrostatic Problems If a solid is perfect conductor, all charge is located on its surface. It follows that if the surface is represented by the simple closed curve C in the z-plane, the charges are in equilibrium on C and hence C is equipotential line. In other words, a conductor is the cross section of an actual conductor, and a point charge is the cross section of a uniform line charge perpendicular to the plane. Then the potential φ is harmonic in regions that contain no charges. Note that if φ + iψ is an analytic function f, then the force on a unit charge is proportional to | f ′ (z) |. Curves ψ = constant are lines of force, and conductors are equipotentials, φ = constant. Capacitor: Two conductors having charges of equal magnitude but of opposite sign, have a difference of potential which we denote v. The quantity ζ defined by η = ζv depends only on the geometry of the conductors. The conductors themselves form what is called a capacitor. Illustration: The harmonic function in the unit disc with boundary values α on the upper semicircular boundary and β on the lower can be interpreted in various ways (a) Describe a flow from a source at –1 to a sink at +1 ; (b) Give the temperatures in the disc when the upper boundary is held at temperature α and the lower boundary at temperature β; and (c) Describe the electric field in the disc when the upper boundary is at potential α, the lower boundary is at potential β, and there are bits of insulation at ± 1. Observe that in case (c) we have a capacitor.

EXERCISES 1. Denote by Δ Δ=

FG ∂ IJ + FG ∂ IJ H ∂x K H ∂y K 2

2

Prove that Δ= 4δδ where

δ=

∂ ∂ ,δ= . ∂z ∂z

2. Suppose that f is analytic and let f = u – iv be the complex conjugate of f. Show that δf = 0.

154

THE ELEMENTS OF COMPLEX ANALYSIS

3. Suppose that f : U → V is an analytic isomorphism and let ϕ be a harmonic function on V, which is the real part of an analytic function. Prove that ϕ o f is harmonic. 4. Let f and g be two harmonic conjugate functions. Prove that (i) F = f 2 – g2 and G = 2fg (ii) φ = ef cos g and ψ = ef sin g are harmonic functions. 5. Suppose that U and V are conjugate harmonic functions and suppose that ψ (x) = U 2 (x, y) + V 2 (x, y). Prove that ψ (x) = e ax + b where a and b are real numbers. 6. Suppose that U and V are conjugate harmonic functions and let φ (x) =

V ( x, y) . U ( x, y)

Prove that φ (x) = tan (αx + β) where α and β are real numbers. 7. Let f (z) = log z. Put z = reiθ. Then f (z) = log r + iθ, where u = log r and v = θ. Draw the level curves u = constant and v = constant. Show that they intersect orthogonally. 8. Suppose that u is harmonic and bounded in 0 < | z | < r. Prove that the origin is a removable singularity provided that u (0) is defined properly and u is harmonic in | z | < r. 9. Suppose that u (z) is harmonic in 0 < | z | < r and suppose lim zu (z) = 0. Prove that z→0

u is of the form u (z) = α log | z | + v (z) where α is a constant and v is harmonic in | z | < r. 10. Suppose that U (ζ) is piecewise continuous and bounded for all real ζ. Prove that HU (z) =

1 π

z

∞

−∞

y U (ζ) dζ ( x − ζ) 2 + y 2

is harmonic in the upper half plane where U (ζ) represents the boundary values at points of continuity. This is called Poisson’s integral for the half plane. 11. Let f (z) be harmonic and bounded in the upper half plane and let f (z) be continuous on the real axis. Prove that f (z) can be represented as a Poisson integral. 12. Let Py (t) =

y , y > 0. Prove that {Py} is a Dirac family. π (t 2 + y 2 )

13. Define φ (x, y) = Py* f (x) =

z

∞

Py ( x − t ) f (t ) dt

−∞

155

HARMONIC FUNCTIONS

where x and y are real and > 0. Prove that ϕ is harmonic. More precisely, prove that Δ

FG y IJ = 0, Δ is the Laplace operator. H (t − x ) + y K 2

2

14. Harnack’s theorem: Suppose that {wn} is an increasing sequence of harmonic functions on the open disk D. Prove that either the sequence converges uniformly on compact subsets of the disk, or the sequence converges pointwise to ∞. 15. Suppose that {wn} is a sequence of harmonic functions on the open disk D. Suppose also that {wn} converges uniformly on compact subsets of the disk D. Prove that the limit of the sequence is harmonic. 16. Prove that a harmonic function satisfies the mean value property. 17. Harnack inequality: Let

k (θ, z) = Re

1 2π

z

iθ iθ

I JK

1 − | z |2 +z = − iθ . −z | e − z| 2

1−|z| 1 + | z | , | z | < 1. ≤ k (θ, z ) ≤ 1+|z| 1−| z|

Show that 18. Let v (z) =

Fe GH e

2π 0

F H

v ( a + re iθ ) k θ,

I K

z−a dθ, | z – a | < r. r

Show that if v (z) ≥ 0, | z – a | ≤ r, then

r −|z − a| r +|z − a| ≤ v (z) ≤ v (a) , | z – a | < r. r +|z −a| r −|z − a| (Hint: Use Harnack inequality.) 19. Let G be a region in C and let a ∈ G. Let F denotes the set of functions ϕ harmonic on G such that ϕ (a) = 1, ϕ (z) > 0 for z ∈ G. Define λ (z) = inf {ϕ(z) : ϕ ∈F} and μ (z) = sup {ϕ (z) : ϕ ∈ F} for each z ∈ G. Prove that 0 < λ (z) ≤ 1 ≤ μ (z) < ∞, z ∈ G. Prove also that λ and μ are continuous on G. (Hint: Consider the sets {λ(z) > 0} and {λ(z) = 0}. By the aid of exercise 18 show that each set is open.) 20. Subharmonic Functions: Let Ω and Ω1 be regions such that Ω ⊆ Ω1. Suppose that v (z) is a continuous real valued function defined on Ω. We say that the v (z) is subharmonic in Ω if for any harmonic function u (z) in Ω1, the difference v – u satisfies the maximum principle in Ω1. We say that a function is subharmonic at a point z0 if it is subharmonic in a neighbourhood of z0. Observe that if v is subharmonic in a neighbourhood of each point z ∈ Ω, then it is subharmonic in Ω. Observe also that every harmonic function is subharmonic but the converse is not true. Let Ω be a region and let Dr (z0) be an open disk centred at z0 and radius r. Suppose that v (z) is continuous. Prove that v (z) is subharmonic in Ω iff 1 2π v (z0) ≤ v ( z 0 + re iθ ) dθ 2π 0 holds for every Dr (z0) ⊆ Ω. v (a)

z

156

THE ELEMENTS OF COMPLEX ANALYSIS

21. Using the results in the above exercise establish the following properties: (i) if v is subharmonic, then Kv is also subharmonic, K is a constant and ≥ 0; (ii) if v1 and v2 are subharmonic, then v1 + v2 is also subharmonic; (iii) if v1 and v2 are subharmonic in Ω, then v = max (v1, v2) is also subharmonic in Ω. 22. Prove that the following functions are subharmonic: (i) | x | (ii) | z | α, (α ≥ 0) (iii) log (1 + | z |2). 23. Let v be continuous and let v has continuous partial derivatives of second order. Prove that v is subharmonic iff Δv ≥ 0.

11

WEIERSTRASS FACTORIZATION THEOREM PART I

11.1

METRIC ON C (G, Ω)

Let G be an open set in C and let (Ω, d) be a complete metric space. We denote by C (G, Ω) the set of all continuous functions from G to Ω. In order to introduce a metric on C (G, Ω), we first prove a theorem about open subsets of C. Theorem 1. Let G be open in C. Then there exists a sequence {En} of compact subsets of G such that ∞

G = ∪ En. n =1

Further, the sets En can be chosen to satisfy the following properties: (i) En ⊂ int. En + 1. (ii) E ⊂ G and E compact implies E ⊂ En for some n.

Proof : Define

RS T

UV W

1 . n Since, En is the intersection of two closed subsets of C and En is bounded, it follows that En is compact. The set

En = {z : | z | ≤ n} ∩ z : d ( z, C − G ) ≥

RS T

UV W

1 n +1 is open and contains En. Also, S is contained in En + 1. Thus the property (i) holds. It can be easily verified that S = {z : | z | < n + 1} ∩ z : d ( z, C − G) >

∞

∞

n =1

n =1

G = ∪ En and G = ∪ int. En. If E is a compact subset of G, the sets {int. En} form an open cover of E. Hence (ii) follows. 157

158

THE ELEMENTS OF COMPLEX ANALYSIS ∞

We now define the metric on C (G, Ω). Suppose that G = ∪ En where each En is compact and En ⊂ int. En + 1. Define (1) ρn (f, g) = sup {d (f (z), g (z)): z ∈ En} where f and g belong to C (G, Ω). Also, define (2)

∞

ρ (f, g) = ∑

n =1

F 1I H 2K

n

n =1

ρ n ( f , g) . 1 + ρ n ( f , g) ∞

Observe that the series in (2) is dominated by ∑

n =1

F 1I H 2K

n

and hence it converges.

Lemma 1: Let (X, d) be a metric space and define d ( x, y) μ (x, y) = . 1 + d ( x, y) Then μ is a metric on X. A set is open in (X, d) iff it is open in (X, μ) ; a sequence is a Cauchy sequence in (X, d) iff it is a Cauchy sequence in (X, μ). The proof is left to the reader. Theorem 2. (C (G, Ω), ρ) is a metric space. Proof : It can be easily seen that ρ (f, g) = ρ (g, f). Since each ρn satisfies the triangle inequality, it follows from the preceding ∞

lemma that ρ also satisfies the triangle inequality. Since G = ∪ En, it follows that n =1

f = g whenever ρ (f, g) = 0. Theorem 3. Define the metric ρ as in (2). Let ε > 0 be given. Then there is a δ > 0 and a compact set E ⊂ G such that for f and g in C (G, Ω) (3) sup {d (f (z), g (z)): z ∈ E} < δ ⇒ ρ(f, g) < ε. Conversely, let δ > 0 and a compact set E be given. Then there is an ε > 0 such that for f and g in C (G, Ω), (4) ρ (f, g) < ε ⇒ sup {d (f (z), g (z)): z ∈ E} < δ. Proof : Let ε > 0 be fixed and let m be a positive integer such that

F 1I H 2K

n

1 ε 2 n = m +1 Put E = Em. For suitably chosen δ > 0 we have t 1 < , where 0 ≤ t < δ. 1+ t 2 ∞

∑


0 such that u < δ. 0 ≤ u < 2m ε ⇒ 1− u t < 2m ε ⇒ t < δ. Hence 1+ t ρ m ( f , g) < 2m ε and this gives Thus if ρ (f, g) < ε then 1 + ρ m ( f , g) ρm (f, g) < δ. Hence, the theorem is proved. Lemma 2: Let {fn} be a sequence in (C (G, Ω), ρ). The sequence {fn} converges to f iff {fn} converges to f uniformly on all compact subsets of G. The proof is left to the reader. Theorem 4. C (G, Ω) is a complete metric space. Proof : Suppose {fn} is a Cauchy-sequence in C (G, Ω). Then the restrictions of the functions fn to E give a Cauchy-sequence in C (E, Ω) where E is compact and E ⊂ G. Hence {fn (z)} is a Cauchy-sequence in Ω. Thus there exists a point f (z) in Ω such that lim fz (z) = f (z). n→∞

We now got a function f : G → Ω and have to prove that f is continuous and ρ (fn, f) → 0 Let E be compact and fix δ > 0. Choose N so that sup {d (fn (z), fm (z)): z ∈ E} < δ

160

THE ELEMENTS OF COMPLEX ANALYSIS

holds for n, m ≥ N. If z is arbitrary in E but fixed then there is an integer m ≥ N so that d (f (z), fm (z)) < δ. By triangle inequality, we get d (f (z), fn (z)) < 2δ for all n ≥ N. Since N does not depend on z we have sup {d (f (z), fn (z)): z ∈ E} → 0 as n → ∞. Hence {fn} converges uniformly on every compact set in G. Thus f is continuous. Finally, it follows from Lemma 2 that ρ (fn, f) → 0. 11.2

SPACES OF ANALYTIC FUNCTIONS

The class of all analytic functions in G will be denoted by H (G) where G is an open subset of C. In the following theorem we will prove that H (G) is closed in C (G, C). Theorem 5. Let { fn } be a sequence in H (G). Suppose that f belongs to C (G, C) such that fn → f. Then f is analytic and fnk → f k for each integer k ≥ 1. Proof : Consider a disk D ⊂ G and take a triangle T inside D. Since T is compact, {fn} converges uniformly over T. Hence

lim

z z z z T

fn =

T

f.

But since each fn is analytic, we have lim

T

fn =

T

f = 0.

Thus f is analytic in every disk D ⊂ G and this gives that f is analytic in G. (k ) To prove that fn → f (k), consider D = D (a: r) ⊂ G. Then there is a number

r1 > r such that D (a; r1) ⊂ G. Let γ: | z – a | = r1 be a circle. Then by Cauchy’s integral formula we have for z ∈ D, fn( k ) (z) – f (k) (z) =

k! 2 πi

z

γ

fn (ζ) − f (ζ) dζ. (ζ − z ) k + 1

Hence

k ! 2 π Mn r1 2 π (r1 − r ) k + 1 where Mn = sup {| fn (ζ) – f (ζ) | : | ζ – a | = r1} and | z – a | ≤ r. Since fn → f, lim Mn = 0. Hence, it follows from (5) that (5)

| fn( K ) (z) – f (K) (z) | ≤

fn( K ) – f (K) uniformly on D (a; r).

161

WEIERSTRASS FACTORIZATION THEOREM

Now let E be an arbitrary compact subset of G and let 0 < r < d (E, ∂G). Then n

there are points a1, a2, ..., an in E such that E ⊂ ∪ D (aj ; r). Since fn( K ) → f (K) j =1

uniformly on each D (aj ; r), the convergence is uniform on E. Hence, the theorem is proved. Observe that we have considered H (G) as a subset of C (G, C) and the metric on H (G) is the metric which it inherits from C (G, C). Since C (G, C) is complete metric space we have the following results. Corollary 1: H (G) is a complete metric space. ∞

Corollary 2: Let fn : G → C be analytic. If ∑ fn ( z ) converges uniformly on n =1

compact sets to f (z) then ∞

(k ) f (K) (z) = ∑ fn ( z ) . n =1

Note that the above result has no analogue in the theory of functions of a real variable. 11.3

WEIERSTRASS FACTORIZATION THEOREM

The notion of convergence in H (G) can be used to prove Weierstrass Factorization theorem. Before discussing the theorem we first define an infinite product of complex numbers and its convergence. Infinite Products: An infinite product of complex numbers (6)

∞

z1 z2 ... zn... = Π zn n =1

is obtained by taking the limit of the partial products Zn = z1 z2 ... zn. ∞

Π zn is said to converge to the number z if lim Zn = z and is different from

n =1

n→∞

zero. Observe that if one of the numbers zn is zero, then the limit is zero, and the convergence would not depend on the whole sequence of factors. Thus we define. Definition: The infinite product (6) is said to converge iff at most a finite number of the factors are zero, and if the partial products formed by the non-vanishing factors tend to a finite limit which is different from zero. We write zn =

Zn and omit the zero factors. Thus in a convergent product Zn − 1

the general factor zn tends to 1. Hence we prefer to write all infinite products in the form

162

THE ELEMENTS OF COMPLEX ANALYSIS ∞

Π (1 + zn )

(7)

n =1

so that zn → 0 is a necessary condition for convergence. Consider the power series expansion of log (1 + z) about z = 0, i.e. z2 + ... . 2 It has radius of convergence 1. If | z | < 1 then

log (1 + z) = z –

1−

If | z |
0. It now follows from inequality (8) that 3 | log (1 + un (x)) | ≤ | un (x) | 2 for all n > n0 and x ∈ X. Proof : The hypothesis implies that there is an integer n0 such that | un (x) |
0 there is an integer N such that | zn | ≥ r for all n ≥ N. It now follows from (9) that the series ∞

Σ | 1 − E pn

n =1

FG z IJ | Hz K n

converges uniformly on compact sets in C. ∞

By Theorem 7, the infinite product ∏ E pn n =1

FG z IJ converges in H (C). Observe Hz K n

that {pn} can be found so that (9) holds for all r. For any r, | zn | > 2r for all n ≥ N. Hence

FG r IJ < 1 for all n ≥ N. Thus if p = n – 1 for all n, the tail end of the series H | z |K 2 n

n

(9) is dominated by Σ (1/2)n. Hence (9) converges. There is an advantage in taking the constant pn as small as possible. The resulting function f (z) in (10) is then called the canonical product corresponding ∞

1 < ∞, and the canonical product is n = 1 | zn |

to {zn}. For instance, choose pn = 0, if ∑ ∞

FG H

Π 1−

n =1

IJ K

z . zn

We now state the Weierstrass Factorization theorem. Theorem 9. Let F be an entire function and suppose that F (0) ≠ 0. Let z1, ..., zn ... be the zeros of f, listed according to their multiplicities. Then there is an entire function g and a sequence of non-negative integers {pn} such that ∞

(11)

E pn F (z) = eg (z) nΠ =1

FG z IJ . Hz K n

Proof : Let f be the product in Theorem 8, formed with the zeros of F. Then F/f has removable singularities in C and hence F/f is an entire function. Further, F/f has no zero and since C is simply connected, there is an entire function g such that F (z)/f (z) = eg(z) The result now follows. Note that if F has a zero of order m at z = 0, the theorem holds for F (z)/zm.

166

THE ELEMENTS OF COMPLEX ANALYSIS

We now give an application of the Weierstrass Factorization Theorem to sin π z. The zeros of sin π z =

∞ 1 1 iπz (e – e–iπz) are the integers z = ± n. Since Σ 2i n =1 n

∞ diverges and Σ 12 converges, we choose pn = 1 and obtain a representation of the n =1 n form

F H

I K

z z/n e . n Taking the logarithmic derivatives of both sides of (12) we have

(12) sin π z = [exp g (z)] Π 1 − n≠0

(13) π cot πz =

FG H

IJ K

1 1 1 + g′ ( z ) + Σ + . n≠0 z − n z n

The convergence is uniform over compact subsets of C which does not contain the points z = n. Also (14) π cot πz =

FG H

IJ K

1 1 1 + Σ + . z n≠0 z − n n

Comparing, we conclude that g′ (z) = 0. Hence g is constant. Since lim

z→0

= π, we have exp (g(z)) = π. Thus

F H

sin πz z

I K

z z/n e . n The terms of the infinite product corresponding to n and – n can be arranged and we obtain

sin π z = πz Π 1 − n≠0

∞

FG H

sin πz = πz Π 1 − n =1

IJ K

z2 . n2

The convergence is uniform over compact subsets of the plane. We now discuss an infinite series expansion, due to Mittag-Leffler, which is related to the Weierstrass product. The main idea is to represent a meromorphic function f (z) by a series, each term of which contains the principal part of f at one of the singularities. Observe that the familiar partial fraction expansion for a rational function is an example of a Mittag-Leffler expansion, just as the factorization theorem for polynomials is an example of the Weierstrass expansion. Suppose it is required to construct a function with simple poles of residue 1 at z = 1, 2, 3, ... . One would like to represent such a function by ∞

Σ

k =1

1 . z−k

167

WEIERSTRASS FACTORIZATION THEOREM

This representation is not satisfactory, because the series diverges for every value of z. We represent such function by ∞

Σ

k =1

FG 1 + 1 IJ . H z − k kK

Note that the series converges except at points where the denominator vanishes. We define the Mittag-Leffler primary terms for n = 1, 2, ... by (15) L (w, n) = and

1 + 1 + w + w2 + ... + wn – 1 w −1

L (w, 0) =

1 w −1

L (w, n) =

wn w −1

It follows that

and (16) | L (w, n) | ≤ 2 | w |n, | w | ≤

Write, for n = 1, 2, 3, ...,

FG H

1 2

(17) E (w, n) = (1 – w) exp w +

w2 wn + ... + n 2

IJ K

and E (w, 0) = 1 – w. Observe that there is a close relation between L (w, n) and E (w, n). Since, the first term of (15) is

1 , we find that w −1

z

w 0

L (ζ, n) dζ = log E ( w, n), | w | < 1.

Note that the path of integration is taken along the radius from 0 to w. Using (16) we get the estimate of the integral 2 | w |n + 1 1 ,|w|≤ . n +1 2 This is the basic inequality of Weierstrass expansion theory. Theorem 10 (Mittag-Leffler). Let {αk} be sequence of distinct complex numbers such that | αk | → ∞. Let {βk} be another sequence of complex numbers such that (18) | log E (w, n) | ≤

168

THE ELEMENTS OF COMPLEX ANALYSIS

| βk | < ∞. k = 1 | α |n + 1 k Then there exists a meromorphic function f whose only finite singularities are simple poles at αk with residue βk. The function is represented by

(19)

∞

Σ

FG H

IJ K

βk z ,n . L k =1 α αk k ∞

(20) f (z) = Σ

Proof : Let r > 0. Choose N so that | αk | > 2r for k > N. Then for | z | < r and k > N,

z 1 < . Hence, it follows from (16) that αk 2 |

FG H

IJ K

2 | βk | r n βk z L ,n |≤ . αk αk | α k |n + 1

It now follows from (19) that the series ∞

Σ

k=N

FG H

βk z L ,n αk αk

IJ K

converges uniformly in | z | < r. Observe that the remaining terms of the series for f (z) differ only by a polynomial form N −1

Σ

k =1

N −1 βk βk 1 = Σ = 1 k αk z / αk − 1 z − αk

and hence have simple poles at z = αk with residue βk. This completes the proof of the theorem. The above theorem is not the most general form of the Mittag-Leffler theorem. In its general form, the theorem states that a meromorphic function f (z) can be constructed whose only poles are at {αk} with the prescribed principal part. The proof of the general theorem is similar, in principle, to the proof of Theorem 10. The proof is given here for the sake of completeness. Theorem 11. Let {αk} be a sequence of distinct complex numbers such that | αk | → ∞. Let {Pk} be polynomials without constant term. Then there exists a meromorphic function f whose only poles are at {αk} with the prescribed principal part

Pk

FG 1 IJ . Hz−α K k

169

WEIERSTRASS FACTORIZATION THEOREM

The function f is represented by

LM F 1 I − Q (z)OP + ψ (z) MN GH z − α JK PQ

f (z) = Σ Pk k

k

k

where Qk is some polynomial, and ψ is an entire function. The series converges absolutely and uniformly on any compact set not containing the poles at {αk}. Proof : We assume without loss of generality that αk ≠ 0 for all k. We represent

FG 1 IJ Hz−α K

Pk

k

in a power series at the origin. Choose ρk such that Qk (z) is the sum of the terms of −1 the degree ρk in this series. It follows that Qk is the polynomial of degree ρk. For

|z|≤

| αk | we find that 2

F 1 IJ − Q (z) PG Hz−α K k

k

k

z ≤ Ak αk

where Ak is some fixed number depending on Pk

ρk

FG 1 IJ . Choose ρ sufficiently Hz−α K k

k

large so that

( Ak )1/ ρk → 0 as | αk | → ∞ | αk | and also such that the ρk form an increasing sequence ρ1 < ρ2 < ... . Write (21)

LM F 1 I − Q (z)OP MN GH z − α JK PQ L F 1 I − Q (z)OP + Σ LMP F 1 I − Q (z)OP . = Σ MP G MN H z − α JK PQ MN GH z − α JK PQ

Σ Pk k

k

k

∞

N

k =1

k

k

k

n =1

k

k

k

Given a radius r, let | αN | ≥ r. Note that the first sum on the right of (21) is a finite sum. The second sum on the right of (21) is dominated by

(22)

Σ k

FG A IJ z H|α | K k

k

ρk

ρk

r . 2 By taking the ρk-th root of the coefficients, we find that the radius of convergence is equal to ∞. Thus the series provided

|z|≤

170

THE ELEMENTS OF COMPLEX ANALYSIS

LM F 1 I − Q (z)OP MN GH z − α JK PQ

Σ Pk k

k

k

converges absolutely and uniformly for z in any compact set not containing the poles at {αk}. Observe that the finite sum on the right of (21) has the desired poles, and the r infinite sum on the right of (21) has no poles for | z | < . Since, this is true for every 2 r, this completes the proof of the theorem. Example: Show that

FG H

IJ K

1 1 1 +Σ + z k z − kπ kπ where the summation extends over k = ± 1, ± 2, ± 3, ... . Consider the function 1 z cos z − sin z cot z – = z z sin z 1 Note that the function cot z – has simple poles at z = kπ, k = ± 1, ± 2, z ± 3, ... and the residue at these poles is cot z =

FG z cos z − sin z IJ = 1. H z sin z K 1 lim F cot z − I = 0, H zK lim ( z − kπ)

z → kπ

Since

z→0

1 it follows that cot z – has removable singularity at z = 0. z 1 It can be easily checked that cot z – is bounded on circles γN having centre z 1 at the origin and radius rN = N + π. Hence, it follows from Mittag-Leffler 2 theorem that

F H

I K F 1 + 1 IJ . 1 cot z – = Σ G z H z − kπ kπ K k

EXERCISES 1. Prove that π (1 + zn) converges absolutely iff π(1 + | zn | ) converges. 2. Let 0 < | α | < 1 and let | z | ≤ r < 1; prove that

α +|α|z 1+ r ≤ (1 + α z ) α 1 − r

171

WEIERSTRASS FACTORIZATION THEOREM

3. (Blaschke products). Let {αn} be a sequence of complex numbers with 0 < | αn | < 1 and let Σ (1 – | αn | ) < ∞. Prove that

F GH

∞

| αn | αn − z n =1 αn 1 − αn z

B (z) = Π

I JK

converges in H (D (0, 1)) and that | B (z) | ≤ 1. 4. Let αn = 1 – n–2 and let B be the Blaschke product with zeros at these points α. Prove that lim B (r) = 0 if 0 < r < 1. In fact, prove the estimate. r →1

N −1

| B (r) |
0) , n = 1 nρ

Π

(ii)

∞

F H

Π 1+

n =1

i n

I K

(iii)

∞

n=2

7. Suppose {fn} is a sequence in H (G) where G is an open set. Let in H (G) to f (z). Prove that ∞

LM N

Σ f k ( z ) Π fn ( z )

k =1

converges in H (G) to f ′ (z). ∞

F H

n≠k

FG H

Π 1−

i n

2

IJ . K

∞

Π fn ( z ) converges

n =1

OP Q

I K

z –z/n e n =1 n converges absolutely and uniformly on every compact set.

Π 1+

8. Prove that

9. Prove that

10. 11. 12. 13. 14.

f (z) =

∞

Π (1 + q2n – 1 ez) (1 + q2n – 1 e–z)

n =1

where | q | < 1 is analytic in the whole complex plane and satisfies f (z + 2 log q) = q–1 e–z f(z). Find all entire functions f such that | f (z) | = 1 whenever | z | = 1. Suppose that f is an entire function, and n is a positive integer. Prove that there is an entire function g such that gn = f iff the orders of the zeros of f are divisible by n. Prove that there is a bounded analytic function f on D (0, 1) such that each point of the unit circle is a singularity. Let zn → ∞ and let βn be arbitrary complex numbers. Prove that there exists an entire function f (z) such that f (zn) = βn. (The Gamma function). Γ (z) is the meromorphic function on C with simple poles at z = 0, – 1, ... defined by ∞

F H

I K

−1

z ez/n, n =1 n where γ is constant chosen so that Γ (1) = 1. Show that 0 < γ < 1. Γ (z) = e–γz/n

Π 1+

172

THE ELEMENTS OF COMPLEX ANALYSIS

15. Show that Γ (z) Γ (1 – z) = π cosec πz where z ≠ an integer. 16. Show that

F H

π Γ (2 z ) = 2 2 z − 1 Γ ( z ) Γ z +

17. Prove that ∞

FG H

sin π(z + p) = eπz cot πp Π 1 + −∞

1 2

I K

IJ K

z e−z/n + p . z+p

where p ≠ an integer. 18. Prove that the residue of Γ (z) at – p is (– 1)p/p! 19. Prove that Γ′/Γ (z) =

z LMN ∞

0

LMUse 1 = N z+m

OP Q

e− 1 e − zt − dt t 1 − e− 1

z

∞ 0

exp [− t ( z + m)] dt

20. (Hankel’s Integral) Let

ze z e− z amd F (z) = ez − 1 1 − e− z 1 dz Let H (s) = F (z) z 2 C z z where the contour C is given by G (z) =

z

ke

–e

Exercise 11.I

(i) Prove that H is an entire function (ii) Prove by putting z = – w and letting ε → 0 that ∞ dt H (s) = – (eπ is – e– π is) G (t ) t 2 0 t 1 21. Let ζ (s) = Σ s for Re (s) > 1. Prove the following: n ∞ dt (i) Γ (s) ζ (s) = G (t ) t s . 0 t (ii) ζ (s) is analytic except for a simple pole at s = 1. (iii) The residue of ζ (s) at s = 1 is 1. (iv) ζ (s) has zeros of order 1 at the negative integers.

z

z

OP Q

11

EXTENSION OF THE MAXIMUM MODULUS PRINCIPLE PART II

We first recall Weierstrass Factorization Theorem. The theorem states that every entire function f can be represented in the form

FG H

zm eg (z) Π 1 −

z zn

IJ exp F z + 1 z K GH z 2 z n

2 2 n

+ ... +

1 z kn kn znkn

IJ K

where kn → ∞ sufficiently fast and g is an entire function and the zn are taken in order of increasing modulus. This representation is not very useful from the application point of view. We give below another representation which is more useful than Weierstrass factorisation theorem. The following Hadamard’s representation is for entire functions that are of finite order in the following sense. | f ( z ) | ; then f is said to be of order μ if μ is the greatest Let M (r) = max |z|=r

lower bound of numbers s for which lim sup r–s log M (r) < ∞. Roughly speaking, this means that M (r) is not much larger than exp (rμ), and sometimes about that large. 11.4

HADAMARD’S PRODUCT REPRESENTATION

If f is of order μ with zeros at 0 and zn, then ∞

FG H

f (z) = zm eP(z) Π 1 − n =1

z zn

IJ exp LM z + 1 F z I K MN z 2 GH z JK n

2

+…+

n

FG IJ OP H K PQ

1 z p zn

p

where p ≤ μ and P (z) is a polynomial of degree at most p. Examples (i): sin π z is of order 1; Hadamard’s product with p = 1 is

FG H

IJ K

2 ∞ sin π z = π z Π 1 − z . n =1 n2

This is called Euler’s product for the sine function. 173

174

THE ELEMENTS OF COMPLEX ANALYSIS

(ii) The reciprocal of the gamma function is also of order 1, its Hadamard product is

FG H

IJ K

∞ 1 z = zeγz Π 1 + e−z/n , n = 1 Γ ( z) zn

where γ is Euler’s constant,

F H

I K

1 1 1 + + … + − log n 2 3 n (for the definition and other properties of the gamma function, see 11.8). 1+ γ = nlim →∞

11.5

THE EFFECT OF ZEROS, JENSEN’S FORMULA

An entire function whose maximum modulus M (r) increases rapidly as r → ∞, may have very few zeros or even none at all. One can think of exp (zn) or exp (ez). On the other hand, if the function has too many zeros zn in a disk of radius r, the product Π (1 – z / zn) may not converge. One can think of zn = n. Hadamard’s product representation quoted above suggests that if there are more zeros, the order of M (r) will be larger. This may seem paradoxial because of our expectation that a function with many zeros may be small in modulus. The actual situation is quite different. There is a principle that says that when a function has many zeros, its modulus must frequently be large in order to compensate; if the modulus is not large, an analytic function with many zeros may vanish identically. We are familiar with one case (see Example page 81) when the principle can be quantified. In other words, if M (r) = O (rn) as r → ∞, then the function is a polynomial of degree at most n and therefore at most n zeros. Based on the same principle we present below some more results. Jensen’s Theorem Assume that f is analytic in a disk z ≤ R, but not identically zero. Assume also that f (0) ≠ 0. Let f have zeros {zk}, k = 1, 2, 3, ..., n in z ≤ R, and let f (z) ≠ 0 on | z | = R. Then 1 2π (1) log | f (Reiθ) | dθ – log f ( 0) 2π 0 n R = ∑ log . j =1 zj

z

Proof : We have seen in Chapter 8 (see 8.3) that, if z0 is a zero of f, then the function f ′ / f has a simple pole at z0 with residue 1. It follows that 1 2πi

z

z =R

g (z)

f ′ ( z) dz = ∑ g (zn), f (z)

175

EXTENSION OF THE MAXIMUM MODULUS PRINCIPLE

where g is analytic in z ≤ R and zn are the zeros of f inside z = R. This suggests that the right side of (1) is connected with

z

f ′ ( z) 1 (log z) dz, f ( z) 2πi C but g (z) = log z is not analytic. In order to remove this difficulty we consider the

I=

same integral when C is the circle z = R with a cut along the positive real axis from 0 to R. If f has zeros on the cut, we replace f (z) by f (zeiλ) with λ chosen so that f (zeiλ) has no zeros on the cut. Note that there exists such a λ, since there are only finitely many zeros to avoid; and the change does not affect the moduli of the zeros. The function log z is determined in the cut disk by taking log (–1) = π i. It follows now that the integrand in I is analytic except for simple poles at the zeros of f inside C, and we obtain N

I = ∑ log zj. j =1

On the other hand, we can write I as the sum of three integrals, one along the circle, one along the lower side of the cut, and one along the upper side. The integral along the upper side is

z

f ′ ( x) 1 R (log x) dx; 0 2 πi f ( x) the integral along the lower side is (2)

(3)

−

and their sum is

1 2 πi

z

z

R

0

(log x + 2πi)

f ′ ( x) dx; f ( x)

f ′ ( x) dx = log f (0) – log f (R). f ( x) The integral around the circle can be written in the following form:

(4)

−

R

0

z

1 [(log z)] [log f (z)]' dz. 2πi z = R Integrating (5) by parts, we have

(5)

z

1 1 dz (log z) log f (z) z = R − log f (z) . z = R 2πi 2πi z Putting the integrals (4) and (6) together and taking real parts, we have (6)

(7)

N

∑ log z j = Re I = log f ( 0 ) – log f ( R)

j =1

+ Re

RS 1 (log z ) log f (z) T 2 πi

z =R

UV − 1 W 2π

z

2π

0

log f (Reiθ) dθ.

We now need to compute change in (log z) log f (z) around z = R.

176

THE ELEMENTS OF COMPLEX ANALYSIS

When z goes around the circle z = R, log z starts as log R and ends as log R + 2πi, whereas log f (z) starts as log f (R) and ends as log f (R) + 2πi N, if there are N zeros. Thus the change in [log f (z)] (log z) around the circle z = R is (8)

(log R + 2πi) {log f (R) + 2πiN} – (log R) {log f (R)}

m

r

= 2πi log f ( R) + N (log R + 2πi ) Dividing (8) by 2 π i and taking real parts, we have

1 (log z) log f (z) z 2πi It now follows from (7) and (9) that

= log f ( R) + N log R

(9)

Re

(10)

∑ log z j = log f (0 ) + N log R –

=R

N

j =1

that is,

1 2π

(11)

z

2π

0

1 2π

z

2π

0

log | f (Reiθ) | dθ,

N

R

j =1

zj

log | f (Reiθ) | dθ – log f (0) = ∑ log

.

This proves Jensen’s theorem. 11.6

SOME CONSEQUENCES OF JENSEN’S THEOREM

Observe that if f has no zeros, then log f is harmonic and Jensen’s theorem reduces to the mean value theorem for harmonic functions (see Theorem 4, Chapter 10). Jensen’s formula also shows that if | f (Reiθ) | is not large, there cannot be too many zeros in z < R. Let n (t) denotes the counting function of the zeros (that is, the number of zeros of modulus less than t). The right side of (1) (Jensen’s formula) can be written as

z

R

0

t–1n (t) dt.

This is due to the fact that n (t) is a step function that jumps by k when t crosses a value for which the circumference z = t contains k zeros. We consider the case when no circle z = t contains more than one (simple) zero and the general case can be carried out in the same way. We have n (0) = 0, since f (0) ≠ 0, and n (t) = 0 for 0 < t < z j . Then n (t) = 1 from z1 to z2 , n (t) = 2 from z2 to z3 , and so on. Thus

z

R

0

af

n t dt = t

z

z2 z1

dt +2 t

z

z3 z2

dt +…+ N t

z

R zN

dt t

= log z2 – log z1 + 2 (log z3 – log z2 ) +... + N (log R– log z N )

177

EXTENSION OF THE MAXIMUM MODULUS PRINCIPLE

= – log z1 – log z2 – ... – log z N + N log R = log

R R R + log + … + log z1 z2 zN

Note that if f (0 ) = 1 and f ( z ) ≤ M for | z | ≤ R, then

z

n (t ) dt ≤ max log f ( z ) . z =R t This inequality puts a restriction on the number of zeros that such a function

(12)

R

0

can have in a disk z < r < R. This can be easily seen by considering the following simple case: n (R / 2) log 2 = n (R / 2)

z

z

R

R /2

dt t

n (t ) dt ≤ log M t In order to understand and give more precise statements regarding inequality (12) we solve below two exercises. ≤

R

R/ 2

Exercise 1: Let f be an entire function and f ( z ) ≤ Ae

B Z

where A and B

are constants, B < 1. Let f has zeros zn with z n = n (n = 1, 2, 3,...). Then f (z) ≡ 0. Suppose that f (z) ≠ 0. It follows from Jensen’s formula that f (0) ≠ 0,

z

z

n (t ) 1 2 π log | f (Reiθ) | dθ – log f ( 0) , dt = 0 t 2π 0 where n (t) is the number of zeros zk with z k ≤ t. If f (0) = 0, then we can consider f ( z) the function , where p is the order of the zero at 0. Then Jensen’s formula zp gives R n (t ) 1 2 π log | f (Reiθ) | dθ – p log R – log f ( 0) . dt = 0 2π 0 t Since f has at least one zero of modulus k, k = 1, 2, ..., [R], we may take roughly that n (t) is at least t, so R

z

and

z

1 2π

z

z

R

0

2π

0

t–1 n (t) dt ≥ R,

log | f (Reiθ) | dθ ≤ log A + BR,

therefore we have a contradiction for large R if B > 1. Using the same ideas as above we proceed as follows: We have n (t) = 0 for 0 ≤ t < 1, n (t) ≥ 1 for 1 ≤ t < 2, n (t) ≥ 2 for 2 ≤ t < 3, and so on. Hence n (t) ≥ [t] (the integral part of t), and

178

THE ELEMENTS OF COMPLEX ANALYSIS

[R] – 1 – log [R] =

z

[ R]

1

z

t −1 dt ≤ t

z

n (t ) dt t

R

0

1 2π log | f (Reiθ) | dθ – p log R – log f ( 0) . 2π 0 Since log | f (Reiθ) | ≤ log A + BR, it follows that [R] – 1 – log [R] ≤ log A + BR – p log R – log f (0). But this is impossible for large R if B < 1. ≤

2

Exercise 2: Let f be an entire function with f ( z ) ≤ A exp ( B z ) . Let f has at least n zeros on each circle z = n (n = 1, 2, 3, ...). Then f (z) ≡ 0 provided B < 1/4. Assume (as in the above Exercise 1) that f (0) ≠ 0. Then n (t) ≥ 1 + 2 + 3 + ... + [t] = Hence

z

R

0

[t ] ([t ] + 1) t2 or about . 2 2

t –1 n (t) dt is approximately

However,

1 2

z

R

0

t dt =

R2 . 4

z

1 2π log | f (Reiθ) | d θ ≤ log A + BR2, 2π 0 therefore we will get a contradiction if B < 1/4. Using the same ideas as above we proceed as follows: (t − 1) t for R ≥ t ≥ 1, We have n (t) ≥ 2 R n (t ) 1 [ R] (t – 1) dt dt ≥ 0 2 1 t 1 ≥ ([R] – 1)2, 4 1 2π 1 ([R] – 1)2 ≤ log | f (Reiθ) | dθ whereas 2π 0 4 ≤ log A + BR2.

z

z

z

We now consider the case when f has infinitely many zeros in z < 1. Suppose now that f is analytic and bounded in the disk z < 1. Let f ( z ) ≤ M, and number the zeros in order of increasing modulus. Then by Jensen’s formula, when R < 1, R 1 2π ∑ log (13) ≤ | log | f (0) | | + log | f (Reiθ) | dθ zj 2π 0 zj ≤ R

z

≤ | log | f (0) | | + log M.

179

EXTENSION OF THE MAXIMUM MODULUS PRINCIPLE

Since the right side of (8) is independent of R, we can let R → 1 and get the inequality ∞ 1 ∑ log ≤ | log | f (0) | | + log M. (14) j =1 zj Since z j → 1 as j → ∞ (otherwise there would be a limit point of zeros inside the unit disk,) inequality (14) shows that the moduli of the zeros cannot approach 1 so slowly that the series would diverge. In other words, the series ∞

∑ (1 − z j ) must converge. This can be seen easily since

j =1

log

1 zj

= – log [1 – ( z j ) ]

1 (1 − z j )2 + 2

= (1 − z j ) + ≥ 1 – zj . 11.7

PHRAGMEN-LINDELÖF THEOREM

In this section, we present some result which extend the Maximum Principle by easing the requirement of boundedness on the boundary. We write a complex number in the form S = α + it with real α, t. Phragmen-Lindelöf Theorem Let f be analytic in a strip α1 ≤ α ≤ α2 and bounded in absolute value by 1 on the sides of the strip. Let there be a number n ≥ 1 such that

e j in the strip.

f (S) = 0 e s

η

Then f is bounded by 1 in the whole strip. Proof : By assumption, for sufficiently large t , f (α + it ) ≤ e

t

λ

where λ > η.

Choose an integer k ≡ 2 (mod 4) such that k > λ. If S = reiθ, then Sk = rk (cos kθ + i sin kθ) and kθ is near to π. Consider the function K

hε (S) = h (S) = f (S) e ε S , with ε > 0.

180

THE ELEMENTS OF COMPLEX ANALYSIS

Then for S in the strip we have h (S ) ≤ e

t

λ

k

e εr cos kθ

T

O

a1

a2

–T

Fig. 11.I

It follows that for large T the function h (S) is bounded by 1 on the horizontal segment t = T, α1 ≤ α ≤ α2. It is also clear that | h (S) | is bounded by 1 on the boundary of the rectangle, as shown in Fig. 11.I. Hence k

f ( S ) ≤ e −εr cos kθ

inside the rectangle. This is true for ε > 0, and so f ( S ) ≤ 1 inside the rectangle.

This proves the theorem. Our next theorem gives the batter result regarding the behaviour of the function, and so we assume that the function is analytic in a whole strip. First Convexity Theorem α. Let

Let S = α + it. Let f be analytic and bounded on the strip a ≤ α ≤ b for each Mf (α) = M (α) = sup f (α + it ) . Then log M (α) is a convex function of α. Proof : We need to prove that M (α)b – a ≤ M (a)b – α M (b)α – a. Consider first the case when M (a) = M (b) = 1. We shall prove that M (α) ≤ 1. Suppose that f ( S ) ≤ K in the strip. For ε > 0, let

1 . 1 + ε ( S − a) Then the real part of 1 + ε (S – a) is ≥ 1. It follows that | hε (S) | ≤ 1. Also, for t ≠ 0

hε (S) =

181

EXTENSION OF THE MAXIMUM MODULUS PRINCIPLE

and therefore

hε ( S ) ≤

1 , ε t

f ( S ) hε ( S ) ≤

K . ε t

K . On the boundary of the rectangle with ε K sides at α = a, α = b with top and bottom ± , we find that fhε ≤ 1. Hence fhε ε ≤ 1 on the whole of the rectangle. Letting ε → 0 we get

Let ε be small, and choose t = ±

f ≤ 1 on the strip.

In order to consider the general case, let b−s b−a

s−a b−a

g (S) = M ( a) M (b) . Then g is entire, has no zeros, and 1/g is bounded on the strip. It follows that g ( a + it ) = M (a) and g ( b + it ) = M (b)

for all t. Thus Mf | g (a) = Mf | g (b) = 1. Making use of the first part of the proof, it follows that | f | g | ≤ 1.

Hence f ≤ g and this completes the proof.

Corollary (Hadamard Three Circle Theorem) Let f be analytic on an annulus ξ ≤ z ≤ η, centred at the origin. Let M (ρ) = sup f ( z ) . z =ρ

Then log M (ρ) is a convex function of log ρ. In other words, log

FG ηIJ log M (ρ) ≤ log FG ηIJ log (ξ) + log FG ρ IJ log M (η). H ζK H ρK H ξK

Proof : Let φ (s) = f (es). Then φ is analytic and bounded on the strip a ≤ α ≤ b, where ea = ξ and eb = η. Applying the first convexity theorem we get the result. In order to state the next theorem we need to define the growth exponent of f. Let f be analytic in the neighbourhood of a vertical line α + it, with fixed α, and assume that

182

THE ELEMENTS OF COMPLEX ANALYSIS v

f (α + it) = 0 ( t ) for some positive number v. The infimum of all such v is called the growth exponent of f. Denote this by ψ (α). It follows that f (α + it) = 0 ( t

ψ (α ) + ε

)

for every ε > 0, and ψ (α) is the least exponent which makes the inequality true. Second Convexity Theorem Let f be analytic in the strip a ≤ α ≤ b. Assume that for each α, f (α + it) grows at most like a power of t , and let ψ (α) be the least number ≥ 0 for which f (α + it) = 0 ( t )

ψ (α ) + ε

for every ε > 0. Assume also that t

ζ

f (α + it) = 0 (e ) in the strip, with ζ, 1 ≤ ζ. Then ψ (α) is a convex function of α . In particular, ψ (α) is continuous on [a, b]. Proof : It follows from Phragmen-Lindelöf Theorem that there is a uniform K such that f (α + it) = 0 ( t

K

) in the strip.

Write b−s s−a [ ψ (a) + ε] + [ψ (b) + ε]. b−a b−a Note that Mε (s) is the formula for the straight line segment between ψ (a) + ε and ψ (b) + ε . It can be easily seen that the function

Mε (s) =

− M (s)

f (s) ( −is) ε is bounded on the strip. Since we get now that ψ (α) ≤ Mε (α) for each α in the strip, and every ε > 0, the theorem follows. We conclude this section by observing that Phragmen-Lindelöf theorem gives bounds on the middle. 11.8

THE GAMMA FUNCTION

The gamma function is defined initially by

z

dt , Re z > 0. t In order to see that Γ is analytic for Re z > 0, we need to know that the integral converges uniformly on each compact subset of the right-hand half plane. The proof is not very interesting and so we omit them. We shall now establish the following useful result: (15)

Γ (z) =

∞

0

t z e −t

183

EXTENSION OF THE MAXIMUM MODULUS PRINCIPLE

(16) Γ (z + 1) = zΓ (z) for Re z > 0. In order to show that (16) holds, we consider

z

R

t z – 1 e − t dt.

0

Integrating by parts (differentiating e–t and integrating tz–1) we get

z

1 R z –t t e dt. z 0 When R → ∞, the integrated terms drop out because lim e–R Rz = 0, and the integral becomes Γ (z + 1). R 0

z–1 e–t tz

+

R→∞

Thus we obtained Γ (z + 1) = zΓ (z), when Re z > 0. This is called the function equation for the gamma function. It is easy to check that when n is 0 or a positive integer, then Γ (n + 1) = n!

dt because this t expression is invariant under “multiplicative translations”. This phrase means: Let f be any function which is absolutely integrable on 0 < t < ∞. Let c be a positive number. Then Integrals like (15) are called Mellin transforms. We write

z

z

∞ dt dt = f (t ) . 0 0 t t This can be verified by the change of variables formula. Note that, by replacing t by nt where n is a positive integer,

(17)

1 = ns

z

∞

∞

f (ct )

e–nt ts

0

dt , for Re s > 0. t

∞

∑ n − s is called the Riemann zeta function of the complex variable s.

n =1

We shall now obtain a very useful formula for the gamma function. Let (18)

Γ (p) =

z

∞

0

t p – 1 e–t dt, p > 0.

Setting t = y2 in (18), we have

z z zz

Γ (p) = 2 Similarly, we have

Γ (p) = 2 Thus (19)

Γ (p) Γ (q) = 4

∞

2

y2p – 1 e − y dy.

0

∞

0

2

x2q – 1 e − x dx

∞

∞

0

0

x2q – 1 y2p – 1 e − ( x

2

+ y2 )

dx dy

184

THE ELEMENTS OF COMPLEX ANALYSIS

Now, writing the double integral in polar co-ordinates in (19), we get Γ (p) Γ (q) = 4 =4 (20)

z z

∞

0 ∞

0

2

e − r r dr

π/ 2

0 2

(r cos θ)2q –1 (r sin θ)2p – 1 dθ

r2p + 2q – 1 e − r dr

= 2Γ (p + q)

If we take p = q =

z

z

π/ 2

z

π/ 2

0

(cos θ)2q–1 (sin θ)2p – 1 dθ

(cos θ)2q–1 (sin θ)2p – 1 dθ.

0

1 , we get 2

z

1 π. 2 Next, let x = sin2 θ, 1 – x = cos2 θ, dx = 2 sin θ cos θ dθ; ∞

0

(21)

2

z

π/ 2

0

2

e − x dx =

(cos θ)2q – 1 (sin θ)2p – 1 dθ =

z

1

0

xp – 1 (1 – x)q – 1 dx.

The right-hand integral in (21) is called the beta function B (p, q). Now let x =

y , then 1+ y

B (p, q) =

z

∞

0

y p −1 dy. (1 + y) p + q

It now follows from (20) that (22) B (p, q) =

Γ ( p) Γ ( q ) Γ ( p + q)

If 0 < p < 1 and q = 1 – p, then (23)

z

∞

0

y p −1 dy = Γ (p) Γ (1 – p). 1+ y

The left-hand integral of (23) is equal to

π when 0 < p < 1 (see worked sin πp

out example 27). Hence we obtain π (24) Γ (z) Γ (1 – z) = , 0 < z < 1. sin π z In particular, we take z = 1/2 to get Γ

F 1I = H 2K

π . Since Γ (1 – z) = ∞ when

z = 0, 1, 2, ..., it follows that Γ (z) is never 0. Using (24) it can be seen that Γ is analytic on the line Re z = 0 except at z = 0. This is due to the fact that (1 + z) is analytic and not zero for Re z = 0, and so is Γ (– z) = – π / [Γ (1 + z) sin πz] except at z = 0.

185

EXTENSION OF THE MAXIMUM MODULUS PRINCIPLE

(Stirling’s Formula) Stirling’s formula is the asymptotic development of the gamma function. The following is the statement giving an exact error term:

F H

(25) log Γ (z) = z −

I K

1 1 log z – z + log (2π) – 2 2

z

∞

0

P1 (t ) dt, z+t

1 , a periodic function of period 1 with average 0 over a 2 period ([t] denotes the largest integer ≤ t). We do not prove Stirling formula in this form. Stirling formula with expressions that involve factorials are proved at the end of the book (see Exercise 53). This formula is useful in probability and statistical mechanics. We now prove another simple formula which is called Euler’s summation formula.

where P1 (t) = [t] – t +

Euler’s Formula Let f be any continuously differentiable function of a real variable. Then n

(26)

∑

f (k) =

k=0

z

n

0

f (t) dt +

1 (f (n) + f (0)) + 2

z

n

0

P1 (t) f ′(t) dt.

Proof : Note that P1′ (t) = 1 for t ≠ an integer. Integrating by parts (with u = P1 (t) and dv = f ′(t)), we have

z k

P1 (t) f ′(t) dt = P1 (t) f (t) |kk − 1 −

k −1

z

k

k −1

f (t) dt

z

k 1 (f (k) + f (k – 1)) – f (t) dt. 2 k −1 We take the sum from k = 1 to k = n. Adding the integral

=

z

n

f (t) dt and

0

1 (f (n) + f (0)) 2

n

we get the sum

∑

f (k) on the left side of (26).

k=0

We now apply Euler’s formula for the function f (t) = log t. We have log n! =

z

n

1

log t dt +

1 log n + 2

Hence n! ~ nn + (1/2) e–n τ (n),

z

n

1

t–1 P (t) dt

186

THE ELEMENTS OF COMPLEX ANALYSIS

where lim τ (n) = A exists, by using the reasoning used in proving the convergence n→∞

of the integral of the type

z

sin x dx. x Some more problems are solved on these topics and interested readers can see the solved exercises at the end of this book. ∞

0

EXERCISES 1. Suppose that f is an entire function of order less than 2, that is,

f ( z ) ≤ A exp ( B z

2 −ε

) , ε > 0,

where A and B are constants. Assume that f (m + n) = 0 for all integral m and n. Then prove that f (z) = 0. 2. Suppose a1, a2, ..., an are the zeros and b1, b2, ..., bn are the poles of f in the disk

z ≤ R. Prove that log Rm – n

b1 ... bn 1 f (0) = a1 ... am 2π

z

2π 0

log f ( re iθ ) dθ ε z

3. Suppose that f is the entire function such that f ( z ) ≤ c (ε) e for every ε > 0. Suppose also that f is bounded on the real axis. Then show that f is a constant. 4. Let V be the right-half plane. Let f be continuous on the closure of V and analytic on V. Suppose that there are constants A > 0 and α < 1 such that

f ( z) ≤ A e z

α

for all z in V. Suppose also that f is bounded by 1 on the imaginary axis. Then prove that f is bounded by 1 on V. Prove also that the result does not hold if α = 1. 5. Find the radius of convergence of ∑ zn Γ (n + 1/2) / n ! 6. Prove that for Re z ≥ 0, Γ ′ ( z) = Γ ( z)

z LMN et ∞ 0

−t

−

e − zt 1 − e −t

OP dt. Q

12

ELLIPTIC FUNCTIONS

We assume that the reader is familiar with the preliminary notions of abstract algebra, which are very essential for the general theory of elliptic functions. We begin with some basic definitions. 12.1

GROUPS

A group is a set G together with operation defined between a, b ∈ G ((a o b) or ab) satisfying (i) for all a, b ∈ G, ab ∈ G; (ii) a (bc) = (ab) c; (iii) there exists an element e ∈ G (called the identity element) such that a ⋅ e = e ⋅ a = a for all a ∈ G; (iv) to each element a ∈ G, there exists an element a–1 ∈ G (called the inverse to a ) such that aa–1 = a–1 a = e. A set G satisfying just the first two axioms is called a semi-group. If G is a group and if for all a, b ∈ G, ab = ba; then G is called abelian. A subset S of a group G is called a subgroup if S is itself a group with respect to the operation in G. A lattice is the complex plane C is a subgroup which is free of dimension 2 over the set of integers, and which generates C over the reals. Let w1, w2 be a basis for a lattice L over the set of integers. We write L = [w1, w2]. Such a lattice is illustrated in Fig. 12.I Let L = [w1, w2] and let β ∈ C. The set consisting of all points β + t1w1 + t2 w2, 0 ≤ ti ≤ 1 is called a fundamental parallelogram with respect to the given basis for the lattice. 187

188

THE ELEMENTS OF COMPLEX ANALYSIS

Note that we can also take the values 0 ≤ ti < 1 to define a fundamental parallelogram.

W1 W2

Fig. 12.I

If z1 and z2 are two complex numbers such that z2 – z1 = m 2w, where 2w is fixed complex number and m is an integer. Then z2 is called congruent to z1 (modulus 2w) and is denoted by z2 ≡ z1 (mod ⋅ 2w). If z1 and z2 are two complex numbers such that z2 – z1 = m 2w1 + n 2w2, where w1 and w2 are complex numbers and m, n are integers, then z2 is called congruent to z1 (modulus 2w1, 2w2) and is denoted by z2 ≡ z1 (mod 2w1, 2w2). Let a function f be defined on a region Ω. If f (z1) = f(z2) for all z1, z2 ∈ Ω where z2 – z1 = m 2w for at least one complex number 2w (m an integer), then the function f is called periodic with period 2w. Note that exp z, exp mz, sin z and sin

2π z are periodic functions with period l

2πi, 2πi , 2π and l respectively. m If the periods of a periodic function f are expressible in the form m2w1, then f is called simply periodic. If the periods of a periodic function f are expressible in the form m2w1 + n2w2, where m and n are integers, then f is called doubly periodic. 12.2

ELLIPTIC FUNCTIONS

An elliptic function is a meromorphic function on the complex plane C which is doubly periodic. In other words, an elliptic function f with respect to the lattice L is a meromorphic function on the complex plane C where f (z + w) = f (z) for all z ∈ C and w ∈ L. Observe that f is periodic iff f (z + w1) = f (z ) = f (z + w2).

189

ELLIPTIC FUNCTIONS

We state some useful properties of elliptic function and the proof is left to the reader. We assume that 0 < arg (2w2/2w1) < π. Properties: (i) The sum, difference, product and quotient of any two co-periodic elliptic functions are also elliptic with the same period. (ii) The set of all co-periodic elliptic functions forms a field. (iii) A rational function of co-periodic elliptic functions is an elliptic function with the same period. (iv) The derivative of an elliptic function is an elliptic function with the same period. (v) An elliptic function which is entire is a constant. Theorem 1. Let P be a fundamental parallelogram for the lattice L. Let ∂P denotes the boundary of P. Suppose that the elliptic function f has no poles on ∂P. Then the sum of the residues of f in P is 0. Proof : Let P (Fig. 12.II) be a fundamental parallelogram. z0 + 2w2

z0 + 2w1 + 2w2

z0

z0 + 2w1 Fig. 12.II

By Cauchy’s theorem, the sum of the residues

Σ Res f = Now

z

∂P

f ( z ) dz =

z

z 0 + 2 w1

z0

f ( z ) dz +

+ =

=

z

z

z 0 + 2 w1

z0

z0

z

z

z z

∂P

f ( z ) dz.

z 0 + 2 w1 + 2 w2

z 0 + 2 w1 + 2 w2

z0

z 0 + 2 w1

f ( z ) dz

z 0 + 2 w1

z 0 + 2 w2

f ( z ) dz +

+ z 0 + 2 w1

1 2πi

z

z 0 + 2 w2

z0

f ( z ) dz +

z

z0

z 0 + 2 w2

f ( z ) dz

f ( z + 2 w1 ) d ( z + 2 w1 )

f ( z + 2 w2 ) d ( z + 2 w2 ) +

{ f ( z ) − f ( z + 2 w2 )} dz +

z

z 0 + 2 w2

z0

z

z0

z 0 + 2 w2

f ( z )dz

{ f ( z + 2 w1 ) − f ( z )} dz .

190

THE ELEMENTS OF COMPLEX ANALYSIS

Since f (z + 2w1) = f (z), f (z + 2w2) = f (z), it follows that

z

f ( z ) dz = 0.

∂P

Thus the theorem is proved. Theorem 2. Let P (Fig. 12.II) be a fundamental parallelogram. Suppose that the elliptic function f has no zero or pole on ∂P. Let {zi} be the singular points of f inside P. Suppose also that the function f has order mi at zi. Then

Σmi = 0. Proof : Since f is an elliptic function, then f ′/ f is also an elliptic function. It follows from Theorem 1 that 0=

1 2πi

z

f ′( z ) dz f (z)

∂P

f ′( z ) at singularities inside P. f ( z)

= sum of the residues of = Σmi.

Theorem 3. Let the hypotheses of Theorem 2 hold. Then Σ mi zi ≡ 0 (mod ⋅ 2w1, 2w2). Proof : Since Res zi

f ′( z ) = mi zi, then f ( z)

z

1 z ⋅ f ′( z ) dz = Σ mi zi. 2πi ∂P f ( z ) We now compute the integral over the boundary of the parallelogram.

1 2 πi

z

∂P

z

f ′ (z) 1 dz = f ( z) 2 πi

LM N

z

z0 + zw1

z0

+

z

LM N 1 L = 2w 2 πi MN = 1 2 πi

z0 + 2 w1

z0

1

z

{z − ( z + 2 w2 )}

z 0 + 2 w2

z0

z

f ′( z ) dz + f (z)

z

z 0 + 2 w2

z

z0 + 2 w1 + 2 w2

f ′( z ) dz + f (z)

f ′( z ) dz − 2 w2 f ( z)

z

z

z0 + 2 w1

z0

z

z0 + 2 w1 + 2 w2

z0 + 2 w1

f ′( z ) dz + f (z)

z 0 + 2 w2

z0

f ′( z ) dz f (z)

z0

z 0 + 2 w2

{z + 2 w1 − z}

f ′( z ) dz f (z)

1 [2 w1[log f ( z )]zz00 + 2 w2 − 2 w2 [log f ( z )]zz00 + 2 w1 2 πi 1 = (4πimw1 + 4πi nw2) 2πi = m2w1 + n2w2. =

z

z

OP Q

z

f ′( z ) dz f ( z)

f ′( z ) dz f (z)

OP Q

OP Q

191

ELLIPTIC FUNCTIONS

Hence, we conclude that

Σ mi zi ≡ 0 (mod ⋅ 2w1 ⋅ 2w2). 12.3

WEIERSTRASS’ ELLIPTIC FUNCTIONS

In order to define Weierstrass’ Elliptic function we need an important lemma. Lemma: If μ > 2, then

Σ Σ′ m n

1 (2 mw1 + 2nw2 )μ

converges absolutely. Note that the summation Σ Σ′ extends over all positive and negative integers m, n not simultaneously equal to zero. Proof : Let Sk be the sum of the terms

1 . Note that we consider (2 mw1 + 2nw2 )μ

two alternatives i.e. either m = ± k, – k ≤ n ≤ k or n = ± k, – k ≤ m ≤ k. 1 Sk ≤ Σ k | ( m 2 w + n 2 w ) |μ 1 2 1 ≤ 8k (h = min ( | 2w1 |, | 2w2 |) ( kh) μ 8 1 = μ μ −1 . h k Hence, Σ Sk is absolutely convergent for μ > 2. Let 1 . s ≤ m | s 2 w + t 2 w |μ 1 2 t≤n

Smn = Σ Then

N

Smn ≤ Σ S′k 1

where S′k is the sum obtained by adding absolute values of the terms of Sk and k = max (m, n). Thus the lemma is proved. The Weierstrass function is defined by P (z) =

RS T

1 1 1 + ΣΣ ′ − 2 2 2 z ( z − Ω mn ) Ω mn

UV W

where Ωmn ≠ 0, Ωmn = m2w1 + n2w2 ≡ 0 (mod, 2w1, 2w2). Note that the summation ΣΣ ′ extends over all positive and negative integers m, n except m = 0, n = 0 simultaneously. Observe that the series expression for P shows that it is meromorphic with pole of order two at Ωmn. We now show that the series for P (z) converges absolutely and uniformly for every z except z = Ωmn.

192

THE ELEMENTS OF COMPLEX ANALYSIS

2Ω mn z − z 2 1 1 − = ( z − Ω mn ) 2 Ω 2mn Ω 2mn ( z − Ω mn ) 2

≤
1 would be 2

and the Taylor series of f about ∞

f (x) = Σ

n=0 ∞

= nΣ= 0

F H 1 F 1 x− I H 2K n!

f ( n ) (1/2) 1 x− n! 2 n

I K

n

FI HK

∞

1 k! Σ ak k = n ( k − n )! 2

k−n

Since everything is positive in the above expression, it follows that ∞

F H

I K

n

FI HK

k−n

1 1 1 k! x− . k=0 n = 0 n! 2 ( k − n)! 2 Note that the inner series in above equation (*) is the binomial series for

(*)

k

f (x) = Σ ak Σ

LMF x − 1 I + 1 OP NH 2 K 2 Q

k

= xk,

∞

k therefore f (x) = Σ ak x , k > 1. k=0

But this is the original power series for f, now with x > 1, although we assumed to begin with that f was not analytic in any disc of radius greater than 1, centred at 0. Thus we got a contradiction and this shows that f cannot be continued directly across x = 1. Observe that the argument is unaffected if we assume that all the coefficients, except for a finite number, are positive. ∞

The proof of the Vivanti-Pringsheim theorem is easy if we assume that Σ ak k=0

diverges. The following example shows this.

208

THE ELEMENTS OF COMPLEX ANALYSIS ∞

k Example: Suppose that Σ ak z has radius of convergence 1. If ak ≥ 0 and if k=0

∞

Σ ak diverges, then f (x) → ∞ as x → 1 along the radius (0, 1).

k=0

∞

Proof : Since ak ≥ 0 and Σ ak diverges, it follows that k=0

N

Σ ak > M

k=0

where M is a given large number and N is large enough. Now, if 0 < x′ < 1, ∞

N

N

k=0

k=0

k=0

f (x) = Σ ak x k ≥ Σ ak x k ≥ x N Σ ak > MxN By taking x sufficiently close to 1, we have M . f (x) > 2 But M is arbitrary, so f (x) → ∞. We now give an illustration which can be expressed by saying that f must have at least one singular point on its circle of convergence. ∞

k Example: (i) If f (z) = Σ ak z is analytic in | z | < R and also at z = r. Then it k=0

is always possible to make a direct analytic extension of f to some points outside | z | ≤ r. (ii) Continuing as above in (i), prove that if f is analytic in | z | ≤ r, this cannot be the disc of convergence of f. Proof : (i) If f is analytic at z = r, then f is analytic in some neighbourhood of the point r. If we expand f about any point x0 with 0 < x0 < r, then a disc | z – x0 | < r – x0 + δ will be in the region of analyticity of f where δ > 0, is sufficiently small (see Fig. 13.V).

O

x0

Fig. 13.V

r

209

ANALYTIC CONTINUATION, DIFFERENTIAL EQUATIONS

(ii) If f is analytic in | z | ≤ r, it is analytic in an open disc about each point of the circumference given by | z | = r. It follows from the Heine-Borel theorem that a finite number of the discs cover the circumference | z | = r; their union is an open set containing the trace of | z | = r and therefore contains a circle, centred at 0, of radius greater than r. 13.4

NOWHERE-CONTINUABLE POWER SERIES ∞

We have just seen that the sum of power series Σ ak z k with ak ≥ 0 cannot be k=0

continued directly past the real positive point on the circle of convergence. Now we shall be concerned with conditions that make it impossible to continue the sum of a power series beyond the disc of convergence in any direction at all. There are functions that are analytic for | z | < r but cannot be continued analytically outside this disc. Such a function is said to have a natural boundary on the circle | z | = r. We shall first verify that such functions do exist. ∞

Example: Prove that f (z) = Σ z n! cannot be continued outside the unit disc. n =1

Note that the Vivanti-Pringsheim theorem shows that f cannot be continued past the point 1. Now consider f (zein! π p/q), where p and q are integers. When n is sufficiently large, n! π p/q is an even integer and ein! π p/q = 1 Hence, the power series represented by f (rein! π p/q), has positive coefficients when n is sufficiently large. Then f (zein! π p/q), cannot be continued past 1 on the radius z = x. In other words, f cannot be continued along the ray arg z = π p/q. Since these rays intersect the unit circumference in a dense set, f cannot be analytic at any point of the circumference. We shall now introduce some new concepts which will be useful in understanding Hadamard’s gap theorem. If a function is defined by a power series with a finite radius of convergence, then we can extend the function beyond the disc of convergence by grouping the terms of the series. This can be done by considering only a subsequence of the partial sums. A power series for which this can be done is said to be overconvergent. We shall now construct an example of an overconvergent power series and the key to the construction is the following lemma: Lemma: Let {nk} be a sequence of positive integers such that nk + 1/nk ≥ λ 1 > 1 and let p be a positive integer such that p > . Then the polynomial λ −1 n pk (z) = [ z p (1 + z )] k , when expanded in powers of z, contains no power of z that appear in any other Pj ( j ≠ k).

210

THE ELEMENTS OF COMPLEX ANALYSIS

Proof : Note that the highest power of z in Pk (z) has exponent pnk + nk ; the p +1 , lowest power in Pk + 1(z) has exponent pnk + 1. Since nk + 1/nk > λ > (1/p) + 1 = p it follows that the difference of these exponents is pnk + 1 – (p + 1)nk > 0. In order to construct an example of an overconvergent power series we define a function f by ∞

f (z) = Σ ak Pk ( z ) , k =1

where

1 is the coefficient with largest modulus in the binomial expansion of ak

(1 − z ) nk . It follows from Lemma (with – z instead of z) that Pk (z) contains no powers that appear in any other Pj (z). Thus, if we replace each Pk (z) by its expansion in powers of z, we get a power series Σck zk whose partial sums of order (p + 1) nk are just the partial sums of the series ΣPk (z). Observe that Σ ck zk has an infinite number of terms with coefficient 1, and none with coefficient of modulus greater than 1. It follows that the radius of convergence of Σ ck zk is 1. We now show that ΣPk(z) converges in a set that extends outside the circle | z | = 1; this will mean that a sequence of partial sums of Σ ck zk (namely, those of index nk) converges in a set that contains points z with | z | > 1 ; in other words, this will mean that Σ ck zk is over convergent. This will be possible if there are points ξ outside the unit disc for which | ξp(1 – ξ) < α < 1, since then Σ ak Pk (ξ) will be dominated by Σ | ak | α nk with | ak | ≤ 1. The existence of such points is obvious, since if | 1 – ξ | < ε (0 < ε < 1), then p P | ξ (1 – ξ) | < | ξ | ε < a provided | ξ | < (α/ε)1/p. This allows | ξ | = 1 + δ with δ > 0 if 1 > α > ε. Hence, we conclude that the series Σ c k z k is overconvergent in a neighbourhood of the point 1 provided that

nk + 1 nk

> λ > 1 and p >

1 . The series λ −1

Σ ck z has long gaps. The ratio of the lowest exponent following a gap to the highest exponent preceding that gap is k

FG n IJ FG p IJ > 1. H n K H p + 1K k +1 k

Note that the sequence of partial sums that converges outside the unit disc consists of the partial sums that end at a gap. Note also that the existence of long gaps is necessary for overconvergence, in the sense that an overconvergent series (whose radius of convergence is 1) is always the sum of a series that converges in a larger disc and a series with long gaps.

211

ANALYTIC CONTINUATION, DIFFERENTIAL EQUATIONS

We illustrate the above mentioned fact by an example. Example: If p = 1 and λ > 2, then the overconvergent series just constructed has a sequence of partial sums that converges at least in S=|z|

1 . We need to show that λ −1

Σ ak Pk (z) = Σ ak [ z (1 – z )]nk converges when | 1 – z | < 1. We know that the series converges for | z | < 1 ; the series also converges for | 1 – z | < 1, since the series is unchanged if z is replaced by 1 – z. We now state and prove Hadamard’s gap theorem. Theorem 4 (Hadamard’s Gap Theorem). If an = 0 except for n = nk, where nk + 1 nk

∞

≥ λ > 1, then f (z) = Σ an z n , with a finite radius of convergence, cannot be n=0

continued beyond the circle of convergence. ∞

Proof : Suppose that the radius of convergence of Σ an z n is 1. Since f (zeiα) n=0

has the same gaps in its power series as f does, it is sufficient to prove that f cannot be analytic at z = 1. Assume that f is analytic at z = 1, we need to arrive at a contradiction. Consider now the function w (z) = f with p > λ/λ + 1.

RSL 1 z (1 + z)O UV PQ W TMN 2 p

nk

n

k 1 p expands into a sum of z (1 + z ) 2 powers of z that do not occur in the expansions of the corresponding sum for any

It follows from Lemma that each

212

THE ELEMENTS OF COMPLEX ANALYSIS

other value of nk. Hence, the series by which we defined w is just the Maclaurin series of w with its terms grouped. It follows that w is analytic whenever Maclaurin series of w converges. Now, since Σak zk has radius of convergence 1, the set E where w is analytic contains at least the set where 1 p z (1 + z ) < 1 2 This set contains at least the intersection of the sets where | z | < 1 and | z + 1 | < 2. Note that the second of these sets contains all the points of the disc | z | ≤ 1, except for z = 1. But by our assumption f is analytic at z = 1, the point 1 also belongs

1 p z (z + 1) = 1 at z = 1. Since E contains a disc of positive radius ε 2 centred at 1, and hence a disc, centred at 0, of radius 1 + δ > 1 (see Fig. 13.VII). It follows that the Maclaurin series of w has radius of convergence greater than 1 and hence that to E, since

Σak

LM 1 r (r + 1)OP N2 Q p

nk

converges for some r > 1. Since 12 r p (r + 1) > 1 when r > 1, it follows that Σak ζk converges for some ζ > 1. This is a contradiction since this series had the radius of convergence 1. Hence the theorem is proved. |z+1|=2

|z|=1+d

|z|=1 –1

O

Fig. 13.VII

1

|z–1|=e

ANALYTIC CONTINUATION, DIFFERENTIAL EQUATIONS

13.5

213

DIFFERENTIAL EQUATIONS

In the preceding discussion we found means to extend the domain of definition of an analytic function. In fact, one can define analytic function in a broader set up and we refer (for detailed discussion) the last chapter “Global Analytic Functions” of Ahlfors [1]. The theory of global analytic functions makes it possible to study, with a great degree of generality, the complex solutions of ordinary linear differential equations. Keeping in view the application of the topic we confine ourselves to ordinary linear homogeneous differential equations of the second order. These are of the form (2) w″(z) + a(z) w′(z) + b(z) w(z) = 0. If (z – z0) a(z) = p(z) and (z – z0)2 b(z) = q(z) are analytic at z = z0, this point is said to be a regular singularity of (2) if (3) | p(z0) | + | q(z0) | + | q′(z0) | > 0. The point z = z0 is said to be an ordinary point of (2) if (4) | p(z0) | + | q(z0) | + | q′(z0) | = 0. If p(z) or q(z) have a singularity at z = z0, the point z0 is said to be an irregular singularity of (2). By using the methods of series substitutions for w(z) we will solve (2). There is no loss of generality if we take z0 = 0. Write ∞

(5)

α k w(z) = z Σ wk z

(6)

k p(z) = za(z) = Σ pk z ;

(7)

k q(z) = z2b(z) = Σ qk z .

k=0

where ∞

k=0 ∞

k=0

Multiplying (2) by z and substituting ζ = ln z, (2) becomes (8) w″(eζ) – w′(eζ) + p(eζ) w′(eζ) + q(eζ) w(eζ) = 0 and (5) – (7) become 2

∞

(9)

( k + α )ζ w(eζ) = Σ wk e

(10)

p(eζ) = Σ pk e kζ

(11)

q(eζ) = Σ qk e kζ

k=0 ∞

k=0 ∞

k=0

214

THE ELEMENTS OF COMPLEX ANALYSIS

and ∞

(12)

w′(eζ) = Σ (k + α) wk e(k + α) ζ

(13)

w″(eζ) = Σ (k + α)2 wk e(k + α)ζ.

k=0 ∞

k=0

Taking the Cauchy product of (10) with (12) and (9) with (11) we get ∞

RS T

k

UV W

Σ ( k + α ) 2 wk − ( k + α ) wk + Σ [( m + α ) pk − m wm + qk − m wm ] e(k + α)ζ = 0.

k=0

m=0

Hence, for k ≥ 1 (14) [(k + α)2 + (p0 – 1) (k + α) + q0] wk k −1

= – Σ [( m + α ) pk − m + qk − m ] wm m=0

and for k = 0 [α2 + (p0 – 1) α + q0] w0 = 0. Since (2) is second order differential equation, there should be two arbitrary constants in the general solution of equation (2). Suppose that w0 is one of them. Then (15) α2 + (p0 – 1) α + q0 = 0. This is called the indicial equation. The two values of α say, α1 and α2 determined by (15) are called the exponents of the regular singularity. Write ϕ (α) = α2 + (p0 – 1)α + q0. Then (14) becomes (16)

k −1

ϕ (k + α) wk = – Σ [( m + α ) pk − m + qk − m ] wm . m=0

It follows that any wk can be determined in terms of {pk}, {qk} and the preceding wk provided (17) ϕ (k + α) ≠ 0 for k > 0. If the indicial equation does not have a double root (α1 ≠ α2) and (17) holds, (16) will give a distinct solution for each root of (15). If w0 = 0 where w0 is not arbitrary, (14) gives [(1 + α)2 + (p0 – 1) (1 + α) + q0]w1 = 0 and for w1 arbitrary, the same indicial equation (15) is satisfied by (1 + α). Continuing in this way, it can be easily seen that the indicial equation must hold if there are any arbitrary constants in the series solution (5). It follows that there will be one arbitrary constant in each distinct solution. Thus there will be two arbitrary constants in the general solution of (2).

215

ANALYTIC CONTINUATION, DIFFERENTIAL EQUATIONS

When 4 q0 = (p0 – 1)2, the roots of the indicial equation (15) become

1 − p0 . 2 It now follows that for this α, (16) can be used to successively determine the coefficients in (5) for one solution with one arbitrary constant. Again if the roots of the indicial equation (15) differ by an integer, (16) can be used to determine successively the coefficients in (5) for one solution. Note that α is equal to the root with the larger real part. In each of these cases (7) cannot be used to find a second solution of (2) where α is given by (18). Observe that in the case where the roots differ by an integer, the successive determinations is not possible when ϕ (k + α) = 0. In both these cases a first solution can be found. Denote this first solution by w f1 ( z ) . By the (18)

α=

help of the first solution, the second solution can also be determined. We indicate below how this is done. Take

w (z) = f (z) w f1 ( z ) in (2).

A little calculation will show that r(z) = A + B

z

{

exp −

z

} w dz(z)

adz

2 f1

where A and B are constants of integration. A = 1, B = 0 gives the original solution, while A = 0, B = 1 gives a new second solution, say w f2 ( z ) where (19)

w f2 ( z ) = w f1 ( z )

z

e

exp −

z

adz

j w dz(z) . 2 f2

Since a(z) – is free of singularity at z = 0,

p0 p( z ) − p0 = z z

z

adz − p0 ln z

is also free of singularity at z = 0. Suppose that w0 ≠ 0 in (5). Then g( z ) dz w f2 ( z ) = w f1 ( z ) p + 2 α (20) z 0 where g (z) is analytic and free of singularity at z = 0. Hence

z

(21)

∞

k g (z) = Σ gk z . k=0

If α and α – n (where n is a non-negative integer) are the roots of the indicial equation

216

THE ELEMENTS OF COMPLEX ANALYSIS

p0 – 1 = – 2α + n i.e. p0 + 2α = n + 1, then by inserting (21) into (20) we have

z

∞

w f2 ( z ) = w f1 ( z ) Σ gk k=0

i.e. (22)

w f2 ( z ) = w f1 ( z )

RS Σ g T ∞

k≠0

k

z k − n − 1 dz

UV W

zk − n + gn log z . k−n

When the indicial equation has double root, (22) becomes by putting n = 0, (23)

RS T

∞

w f2 ( z ) = w f1 ( z ) Σ gk k =1

UV W

zk + g0 log z . k

It now follows that g(z) in (21) becomes

e

z

z p0 + 2 α exp − a( z ) dz (24)

g(z) =

j

w 2f1 ( z )

Suppose that z = 0 is an ordinary point of (2). In this case w(z) is analytic and free of singularity at z = 0. Taking α = 0 = p0= q0 = q1, (15) is automatically satisfied and (14) reduces to (25)

k −1

k (k – 1)wk = – Σ (mpk – m + qk – m)wm, (k ≥ 1). m=0

The case k = 1 is trivial. We have k=2 – 2w2 = q2w0 + p1w1, – 3.2w3 = q3w0 + (p2 + q2)w1 + 2p1w2 , k=3 k = 4. – 4.3w4 = q4w0 + (p3 + q3) w1 + (2p2 + q2) w2 + 3p1w3, and so on. It can be easily checked that by successive substitutions all the coefficients wk can be expressed in terms of w0 and w1. It now follows that w0 and w1 are the two arbitrary constants in the general solution of (2). 13.6

SOLUTIONS AT INFINITY

In order to determine the nature of solutions of (2) for large values of | z |, we make 1 the variable transformation z = . Since u

dw dw = − u2 ; du dz 2 d 2w 3 dw 4 d w u u = 2 + dz 2 du du 2

217

ANALYTIC CONTINUATION, DIFFERENTIAL EQUATIONS

then (2) takes the form

LM N

F I OP H KQ

FI HK

d 2w 2 1 1 dw 1 1 w = 0. + 4b + − 2a 2 du u u u du u u We say that the point at infinity (z = ∞) is an ordinary point of (26). In other words, we say that the point at infinity (z = ∞) in the z-plane is an ordinary point if 2z – z2 a(z) and z4 b(z) are analytic and free of singularities at z = ∞. This is expressed as

(26)

(27)

∞

−k a(z) = Σ a− k z k=0

∞

−k b(z) = Σ b− k z . k=0

We say that the point at infiniy (z = ∞) is a regular singularity if u = 0 is a regular singularity of (26). In other words, we say that the point z = ∞ is a regular singularity of (2) if za (z) and z2 a(z) are analytic and free of singularities at z = ∞. This is expressed by putting a0 = b0 = b–1 = 0 in (27). In order to get a solution of (2) valid for large | z | we make a series substitution of the form 1 ∞ (28) w = α Σ w− k z − k in (2). z k=0 Then an indicial equation for α is obtained and a second solution is correspondingly found from the second root of this equation provided the roots do not differ by a non-negative integer. In other cases, the second solution is obtained as before by using the first solution. 13.7

THE HYPERGEOMETRIC DIFFERENTIAL EQUATION

In practice, most of the second-order linear homogeneous differential equations possess three regular singularities. We can place the three singularities at prescribed points and it is not difficult to choose them at 0, 1 and ∞. We have in mind the hypergeometric differential equation which has regular singularities with exponents α1 and α2 of values 0, 1 – c at z = 0; 0, c – a – b at z = 1, a, b at z = ∞. This equation is (29) z(1 – z) w″ + {c – (a + b + 1)z]w′ – abw = 0. Here with

∞ c − ( a + b + 1) z = Σ pk z k k=0 1− z p0 = c and pk = c – (a + b + 1) for k > 0,

p(z) = za (z) =

∞ ab z = Σ qk z k k=0 1− z with q0 = 0 and qk = ab for k > 0. Thus the indicial equation in (15) reduces to ϕ (α) = α2 + (c – 1) α = α (α + c – 1)

q (z) = z2 b (z) = –

218

THE ELEMENTS OF COMPLEX ANALYSIS

and (16) takes the form (k + α) (k + α + c – 1) wk k −1

= – Σ {( m − α ) [c − ( a + b + 1)] − ab} wv v=0

where

(k + 1 + α) (k + α + c) wk + 1 – (k + α) (k + α + c – 1) wk = {ab – (k + α) [c – (a + b + 1)]} wk

and (30) (k + 1 + α) (k + c + a) wk + 1 = (k + a + α) (k + b + α) wk. In particular, we have (α + 1) (α + c) w1 = (α + a) (α + b) w0 (α + 2) (α + c + 1) w2 = (α + a + 1) (α + b + 1) w1 (α + 3) (α + c + 2) w2 = (α + a + 2) (α + b + 2) w2 and so on. By successive substitutions we have for k ≥ 1, (α + a + k + 1) ... (α + a) (α + b + k + 1) ... (α + b ) (31) wk = w0. (α + c + k − 1) ... (α + c ) (α + k ) ... (α + 1) i.e. v (α + a + k ) v (α + b + k ) w. (32) wk = v (α + c + k ) v (α + k + 1) 0 where v (x + k) = x (x + 1) ... (x + k + 1). Furthermore, v (x + k + 1) = (x + k) v (x + k) i.e. v (y + 1) = yv (y) where y = x + k. It follows that v (y) is proportional of Γ (y). Hence (32) becomes Γ (α + a + k ) Γ (α + b + k ) wk = w. Γ (α + c + k ) Γ (α + k + 1) 0 The two roots of the indicial equation are 0 and 1 – c. The solution of (29) corresponding to α = 0 and w0 = Γ(c)/Γ(a) Γ(b) is called hypergeometric function. We denote by (33)

F (a, b, c, z) =

∞ Γ (a + k ) Γ(b + k ) Γ (c ) zk. Σ Γ ( a) Γ (b) k = 0 k ! Γ (c + k )

This series converges for | z | < 1. If c is not an integer the other solution corresponding to α = 1 – c will be of the form z1 – c F (1 + a – c, 1 + b – c, 2 – c, z). Hence, the general solution of (29) is of the form AF (a, b, c, z) + Bz1 – c F (1 + a – c, 1 + b – c, 2 – c, z). 13.8

SOME SIMPLE CONSEQUENCES OF THE FUNCTION F (a, b, c, z) IN (33)

We will obtain different types of functions for special choice of the parameters a, b, c.

219

ANALYTIC CONTINUATION, DIFFERENTIAL EQUATIONS

(i) Setting b = c in (33), we get ∞

F (a, b, c, z) = Σ

k=0

Γ (a + k ) k z. Γ ( a) k !

Γ (a + k ) = a (a + 1) ... (a + k – 1) Γ ( a)

Since

= (– 1)k (– a) (– a – 1) ... (– a – k + 1). Hence

F(a, b, c, z) =

1 . a (1 − z )

(ii) Setting a = b = 1, c = 2 in (33), we get zk 1 = − log (1 – z). k=0 1+ k z ∞

F(1, 1, 2, z) = Σ (iii) Setting c = b and z =

F H

F a, b, b, lim

Since

a→∞

F H

I K

z in (33), we get a

∞ Γ (a + k ) z k z . = Σ k k = 0 a Γ ( a) k ! a

Γ (a + k ) =1 a k Γ ( a)

I K

z = ez. a We can express the derivative of F(a, b, c, z) as a hypergeometric function and leave to the reader to verify the following relation

hence

lim F a, b, b,

a→∞

ab F(a + 1, b + 1, c + 1, z). c It can be easily seen that the hypergeometric function F (a, b, c, z) = F (b, a, c, z) is symmetric in the first two parameters. Also, if b is a negative integer, say – m, then Γ (b + k ) will be zero for k > m ; otherwise it is evaluated from the residues of Γ (b) Γ (z) and thus (33) becomes a polynomial. (34)

F′(a, b, c, z) =

(35)

F(a, – m, c, z) =

Γ (c) m Γ ( a + k ) m ! ( − 1) k z k Σ Γ ( a) k = 0 Γ (c + k ) k ! ( m − k ) !

Setting a = m + q in (35), we get

F I H K

Γ (c) m m Γ ( a + k ) Σ (– z)k. Γ ( a) k = 0 k Γ (c + k ) This is called a Jacobi polynomial and is denoted by Jm (q, c, z).

(36)

F(m + q, – m, c, z) =

220

THE ELEMENTS OF COMPLEX ANALYSIS

13.9

BESSEL’S DIFFERENTIAL EQUATION

Bessel’s differential equation of order n is given by (37) z2 w″ + z w′ + (z2 – n2) w = 0. If n ≥ 0, a solution of the equation (37) is (38) Jn (z) =

RS T

UV W

zn z2 z4 1 − + − ... 2 n Γ(n + 1) 2(2n + 2) 2.4 (2n + 2) (2n + 4)

Jn (z) is called Bessel’s function of the first kind of order n. If n ≠ an integer, the general solution of (37) is given by (39) w = AJn (z) + BJ– n (z) where A and B are arbitrary constants. We list some useful properties of Bessel functions and leave the verification to the reader (i)

exp

RS 1 z F t − 1I UV = T2 H t K W

∞

Σ Jn (z) t n .

n=−∞

This is called the generating function for the Bessel functions of the first kind for integer values of n. Note that if n is an integer then J–n (z) = (– 1)n Jn(z) and ∞

k (38) fails to give the general solution. By taking the series of the form (ln z) Σ ak z , k=0

the general solution in this case can be obtained. (ii) Recursion Formula: z Jn – 1 (z) – 2n Jn (z) + z Jn + 1 (z) = 0 Jn (z) =

(iii) where n is an integer. (iv)

Jn (z) =

1 2 πi

z

γ

1 π

z

π

0

cos (nθ − z sin θ) dθ ,

t − n − 1 exp

RS 1 z F t − 1I UV dt , n = 0, ± 1, ± 2, ± 3, ..., T2 H t K W

where γ is any simple closed curve enclosing t = 0. (v)

Jn (z) =

zn 1.3.5 ... (2 n − 1)π

z

1

−1

e izt (1 − t 2 )

n−

1 2

dt .

If n is a positive integer, a second solution to Bessel’s differential equation (37) is given by (40) Sn (z) = Jn (z) ln z – –

FI HK

1 n − 1 ( n − k − 1) ! z Σ 2 k=0 k! 2

FI HK

1 ∞ ( − 1) k z Σ 2 k = 0 ( k !) ( n + k ) ! 2

2k + n

2k − n

{ρ (k) + ρ (n + k)}

221

ANALYTIC CONTINUATION, DIFFERENTIAL EQUATIONS

1 1 1 + + ... + and ρ (0) = 0. 2 3 k This is called Bessel’s function of the second kind of order n or Neumann’s function. For a full discussion of the function of the second kind one may refer “A treatise on the theory of Bessel functions” by G.N. Watson [14].

where ρ (k) = 1 +

13.10

LEGENDRE’S DIFFERENTIAL EQUATION

Legendre’s differential equation of order n is given by (41) (1 – z2) w″ – 2zw′ + n (n + 1) w = 0. The general solution of the equation is

RS n (n + 1) z + n (n − 2) (n + 1) (n + 3) z − ...UV 4! T 2! W R (n − 1) (n + 2) z + (n − 1) (n − 3) (n + 2) (n + 4) z + B Sz − 3! 5! T 2

w = A 1−

(42)

4

3

5

UV W

− ...

If n is zero or a positive integer, polynomial solutions of degree n are obtained and these polynomial solutions are called Legendre polynomials. We also find that they can be expressed by Rodrigues’ formula 1 dn 2 (43) Pn (z) = n (z – 1)n. 2 n ! dz n We list some useful properties of Legendre polynomials and leave the verification to the reader. ∞ 1 (i) = Σ Pn ( z ) t n 1 − 2 zt + t 2 n = 0 This is called the generating function for Legendre polynomials. (ii) Pn(z) =

UV W

RS T

n ( n − 1) n − 2 n(n − 1) (n − 2) (n − 3) n − 4 (2 n) ! z − ... z + zn − n 2 2 (n !) 2(2n − 1) 2.4(2 n − 1) (2 n − 3)

Recursion formula: (iii) (n + 1) Pn + 1 (z) – (2n + 1) z Pn (z) + n Pn – 1 (z) = 0

z

1 (t 2 − 1) n dt 2 πi γ 2 n ( t − z ) n + 1 where γ is any simple closed curve enclosing the pole t = z. (iv) Pn (z) =

(v)

z

1

−1

R| 02 P ( z ) P ( z ) dz = S |T 2n + 1 m

n

if m ≠ n if m = n

If n is not a positive integer, there are two infinite series solutions obtained from (42). These solutions to Legendre’s equation are called Legendre functions.

222

THE ELEMENTS OF COMPLEX ANALYSIS

EXERCISES 1. Show that the series (i)

∞

Σ

n=0

zn 2n + 1

and

(ii)

∞

Σ

n=0

(z − i)n (2 − i ) n + 1

are analytic continuations of each other.

FG H

IJ K

n

1 ∞ z+i is an analytic continuation of f2 (z) = Σ 1+ i n=0 1+ i showing the regions of convergence of the series.

2. Prove that f1 (z) =

zn +1 n = 0 3n Find an analytic continuation of f (z) which converges for z = 3 – 4i.

3. Let

f(z) =

∞

Σ

4. Prove that the series z1 ! + z2 ! + z3 ! + ... has the natural boundary | z | = 1. 5. By use of analytic continuation, show that

F 1I = − 2 H 2K

Γ −

π.

Legendre Functions 6. Prove that (i)

z z

1

Pm ( z ) Pn ( z ) dz = 0

−1

if m ≠ n

2 if m = n. −1 2n + 1 7. Establish the following recurrence relations for the Legendre polynomials (i) nPn – (2n – 1) zPn + 1 + (n – 1) Pn – 2 = 0 (ii) zP′n – P′n – 1 = nPn (iii) P′n – zP′n – 1 = nPn – 1 (iv) P′n + 1 – P′n – 1 = (2n + 1) Pn (v) (z2 – 1) P′n = nzPn – nPn – 1 8. Prove that (ii)

1

Pm ( z ) Pn ( z ) dz =

P2n (0) = (– 1)n

FG H2

2n ! (n !) 2

2n

P2n + 1(0) = 0 Pn ( x ) dx = π [Pn (0)]2 9. Prove that −1 1 − x2 Bessel Functions

z

1

IJ K

∞

Σ zn, n=0

223

ANALYTIC CONTINUATION, DIFFERENTIAL EQUATIONS

10. Show that (i) J′0 (z) = – J1 (z) (ii) (iii)

z z

z 3 J 2 ( z ) dz = z3 J3 (z) + C, C a constant z 3 J 0 ( z ) dz = z3 J1 (z) – 2z2 J2 (z) + C

(iv) J1/2 (z) = (v) J–1/2 (z) = 11. Prove that

2 /π z sin z 2 /π z cos z

(i) J3/2 (z) sin z – J– 3/2 (z) cos z = 2 /π z3 1 (ii) J′n (z) = {Jn – 1 (z) – Jn + 1(z)} 2 (iii) If Re (z) > 0, then

z

∞

1

0

2

e − zt J 0 (t ) dt =

z +1

12. Establish the following recurrence relations for the functions. (i) z (Jv – 1 + Jv + 1) = 2vJv (ii) vJv = z(Jv – 1 – J′v ) (iii) 2J′v = Jv – 1 – Jv + 1 (iv) vJv = z(Jv + 1 + J′v ) 13. Prove that z Jn – 1 (z) – 2n Jn (z) + z Jn + 1 (z) = 0 Hypergeometric Functions 14. Show that (i) F

F 1 , 1 , 3 , z I = sin H2 2 2 K z 2

−1

z

(ii) zF (1, 1, 2, – z) = ln (1 + z) (iii) F

F 1 , 1, 3 , − z I = tan H2 2 K z 2

−1

z

d F (a, b, c, z) = ab F (a + 1, b + 1, c + 1, z). c dz 16. By using power series solve each of the following differential equations. Find also the region of convergence.

15. Prove that

(i) Y″ + 2Y′ + Y = 0 (ii) Y″ + 2Y = 0 (iii) zY″ + 2Y′ + zY = 0 (iv) (1 – z2) Y″ + 2Y = 0 17. Solve by power series method Y″ + z2 Y = 0 subject to the conditions Y(0) = 1, Y ′ (0) = 1. Find also the region of convergence.

224

THE ELEMENTS OF COMPLEX ANALYSIS

18. Solve by power series method zY ″ + (1 + z)Y ′ – Y = 0 and determine the region of convergence. 19. Show that the differential equation Y ″ – zY ′ + nY = 0 has two independent series solutions representing entire functions given by Y=1–n+ Y=z–

z 2 n ( n − 2 ) 4 n( n − 2 ) ( n − 4 ) 6 z − z + ... + 2! 4! 6!

(n − 1) 3 (n − 1) (n − 3) 5 (n − 1) (n − 3) (n − 5) 7 z + z − z + ... 3! 5! 7!

14

APPROXIMATION BY RATIONAL FUNCTIONS AND POLYNOMIALS

14.1 UNIFORM APPROXIMATION We begin this chapter by recalling some useful and fundamental theorems on approximation. We assume that the reader is familiar with the Weierstrass theorem on approximation and it is given in almost all standard books on Real Analysis. For completeness, we first state and prove the Weierstrass theorem and then discuss the approximation by rational functions and polynomials in the complex setting. We know that for many applications it is convenient to “approximate” continuous functions by functions of an elementary nature. The most reasonable definition that one can use to make the word “approximate” is that every point of the given domain the approximating function shall not differ from the given function by more than the preassigned error. The above definition is also referred to as uniform approximation and is connected with uniform convergence. Theorem 1. Let A = [a, b] ⊆ R be compact. Suppose that f is continuous on A. Then f can be uniformly approximated on A by continuous piecewise linear functions. Proof: Since A is a compact set hence f is uniformly continuous on A. Therefore, given ε > 0, we divide A = [a, b] into subintervals by the points ck, k = 0, 1, 2, ..., n, with a = c0 < c1 < c2 < ... < cn = b so that ck – ck–1 < δ (ε). Connecting the points (ck, f (ck)) by line segments, and defining the resulting continuous piecewise linear function by ϕε, we see that ϕε approximates f uniformly on A within ε. For illustration see Fig. 14.I. f+e f ye f–e Fig. 14.I

225

226

THE ELEMENTS OF COMPLEX ANALYSIS

We now discuss in detail the Bernstein theorem. This theorem gives us a constructive method of finding a sequence of polynomials which converges uniformly on [0, 1] to the given continuous function. By using Bernstein theorem, we will prove Weierstrass theorem on approximation. Definition: Let f : [0, 1] → R. We define the nth Bernstein polynomial for f as (1) Bn(x) = Bn(x; f )

F nI

∑ f FH n IK GH kJK x n

=

k

k

(1 − x ) n − k .

k =0

Observe that the value Bn(x; f ) of the polynomial at the point x is calculated

F 1 I , f F 2 I , ..., f (1) with certain non-negative weight factors H nK H nK F nI g (x) = G J x (1 − x ) . H kK

from the values f (0), f

n−k

k

k

It can be easily checked that gk takes its maximum value at the point Write

F kI

n

(2)

(u + t)n =

∑ GH nJK u t

k n−k

,

k =0

where

FG nIJ denotes the binomial coefficient FG nIJ = n ! . H kK k ! (n − k ) ! H kK Note that (3) (4)

FG k − 1IJ = (n − 1) ! = k FG nIJ , H n − 1K (k − 1) ! (n − k) ! n H kK FG k − 2IJ = (n − 2) ! = k (k − 1) FG nIJ . H n − 2K (k − 2) ! (n − k) ! n (n − 1) H kK

Setting u = x and t = 1 – x in (2), we get n

(5)

1=

F nI

∑ GH kJK x

k

(1 − x )n − k .

k =0

By replacing n by n – 1 and k by j in (5), we have n −1

(6)

1=

F n − 1IJ x (1 − x ) j K

∑ GH j =0

j

n −1− j

.

Multiplying both sides of (6) by x and applying the identity (3), we get n −1

(7)

x=

∑ j =0

FG IJ H K

j +1 n x j +1 (1 − x ) n −( j +1) . n j +1

k . n

227

APPROXIMATION BY RATIONAL FUNCTIONS AND POLYNOMIALS

Setting k = j + 1 in the right side of (7), we have n

(8)

x=

∑ k =1

FG IJ HK

k n k x (1 − x )n − k . n k

Note that (8) can be written n

(9)

x=

∑ k =0

FG IJ HK

k n k x (1 − x ) n − k . n k

Replacing n by n – 2 in (5) and using the identity (4) and by a similar calculation as above, we get n

(n2 – n) x2 =

∑ (k

2

− k)

k =0

Hence, we have (10)

F1 − 1 I x H nK

2

+

1 x= n

F nI x (1 − x) H kK k

k F kI ∑ FH n IK GH nJK x (1 − x ) 2

n

k

n−k

n−k

.

.

k =0

Multiplying (5) by x2, (6) by – 2x, and adding them to (7), we get

F 1 I x (1 – x) = ∑ FH x − k IK FG nIJ x (1 − k ) H nK n H kK n

n

(11)

k

n−k

.

k =0

It follows from (5) that Bn(x; f0) = f0(x) = 1, from (8) that Bn(x; f1) = f1(x) = x. We also find from (10) that the nth Bernstein polynomial for the function f2(x) = x2 is

F H

I K

1 2 1 x + x n n which converges uniformly on [0, 1] to f2. We now prove Bernstein theorem on approximation. Theorem 2. Let f : [0, 1]→ R be continuous. Let the sequence of Bernstein polynomials for f be defined by

Bn(x, f2) = 1 −

Bn(x) = Bn(x ; f ) =

F nI

∑ f FH n IK GH kJK x n

k

k

(1 − x ) n − k .

k =0

Then Bn(x) converges uniformly on [0, 1] to f. Proof : Multiplying (5) by f (x), we get

F nI

n

f (x) =

∑ f ( x ) GH kJK x k =0 n

Now

f (x) – Bn(x) =

k

R

U F nI

∑ ST f ( x) − f FH n IK VW GH kJK x k

k

(1 − x ) n − k .

k =0

Hence n

(12)

(1 − x ) n − k .

| f (x) – Bn(x) | ≤

∑ k =0

f ( x) − f

F k I FG nIJ x (1 − x) H n K H kK k

n−k

.

228

THE ELEMENTS OF COMPLEX ANALYSIS

k is near x, then the corresponding n k term in the sum (12) is small, since f is continuous at x. If is not near x, the factor n involving f is less than 2M where M is the bound for f. In order to estimate the k expression in (12) we consider those values of k for which x – is small and those n k for which x – is large. n Let ε > 0. Choose n so large that (13) n ≤ sup {(δ(ε))–4, M 2/ε2}, and break (12) into two sums. We first consider the sum taken over those k for which Note the following facts. If k is such that

x−

Hence

k < n −1/ 4 ≤ δ( ε ). n

F nI HK

n

F nI HK

∑ ε k x k (1 − x ) n − k ≤ ε ∑ k x k (1 − x ) n − k = ε. k k =1

We now consider the sum taken over those k for which x−

i.e.

k ≥ n −1/ 4 n

Fx − kI H nK

2

≥ n −1/ 2 .

By using (11) we find that

F nI HK

k n k ∑ 2 M k x (1 − x ) − k

Fx − kI H n K F nI x (1 − x) = 2M∑ . F x − k I H kK H nK F k I FG nIJ x (1 − x) ≤ 2M n Σ x − H n K H kK R1 U ≤ 2 M n S x (1 − x )V Tn W 2

n−k

k

2

k

n

2

k =1

≤

1 M , since x (1 – x) ≤ 4 2 n

k

n−k

229

APPROXIMATION BY RATIONAL FUNCTIONS AND POLYNOMIALS

on [0, 1]. Hence, for n chosen in (13) we have | f (x) – Bn(x) | < 2ε. Thus the theorem is proved. We will see below that Weierstrass approximation theorem is a direct consequence of Bernstein theorem. Theorem 3. Let f : [a, b] → R be continuous. Then f can be uniformly approximated by polynomials. Proof : Let f be defined on [a, b]. Define another function g on [0, 1] by g(t) = f ((b – a) t + a) where t ∈ [0, 1]. The function g is continuous and hence can be approximated by Bernstein polynomials. By changing the variable we get a polynomial approximation to the function f. n

k Suppose that f (z) is analytic in | z | ≤ ρ and let p(z) = Σ ak z be a polynomial. k =0

By using the Taylor’s theorem for analytic function it can be seen that given an ε > 0, | f (z) – pn(z) | ≤ ε for | z | ≤ ρ. If f (z) is not analytic, we cannot expand f (z) in power series. But Weierstrass theorem asserts that we can approximate functions which are merely continuous. Approximation of analytic functions by analytic functions is totally different. Let Ω be a region bounded by a simple closed curve γ. Suppose that f (z) and p(z) are two functions analytic in Ω and on γ. Suppose also that | f (z) – p(z) | ≤ ε on γ. By maximum principle, this inequality and hence approximation persists throughout Ω. It follows from Cauchy’s inequality that n ! L (γ ) ε (14) | f (n)(z) – p(n)(z) | ≤ 2 π δ n +1 where z belongs to a point set, say E ⊂ Ω and the distance of those points from γ is no less than δ. In (14) we fix E and n and allow ε → 0. It then follows from (14) that the nth derivative of the approximant is also an approximation to the nth derivative of the approximate. Thus we see that the approximation over regions carries with it the simultaneous approximation of all the derivatives. Suppose that f is continuous on [a, b]. We know that we can approximate f by a polynomial. The following question naturally arises. Let x1, x2, ..., xn be n points of [a, b] and let ε > 0 be given. Can we find a polynomial p(x) such that | f (x) – p(x) | ≤ ε, x ∈ [a, b] and p(xi) = f (xi), i = 1, 2, ..., n ? In the complex setting the following theorem of Walsh answers this question. Theorem 4. Let E be a point set which is closed and bounded in the complex plane C. Let z1, z2, ..., zn be n distinct points of E. Suppose that f (z) is defined on E and is approximable by polynomials there. Then f(z) is approximable by polynomials p and satisfy the property p(zi) = f (zi), i = 1, 2, ..., n.

230

THE ELEMENTS OF COMPLEX ANALYSIS

Proof : Let ε > 0 be given. Choose a polynomial p(z) such that | f (z) – p(z) | ≤ ε, z ∈ E. Define n

q(z) =

∑ (f (z ) – p(z )) u (z) k =1

where

uk(z) =

k

k

k

v( z ) ( z − zk ) v′ ( zk ) n

v(z) = Π ( z − zk ).

and

k =1

It follows that q(zk) = f (zk) – p(zk), k = 1, 2, ..., n. n

M = Σ max | uk ( z ) |.

Write Now

k =1 z ∈E

n

max | q( z ) | ≤ Σ | f ( z k ) − p( zk ) | max | uk ( z ) | ≤ ε M . z ∈E

k =1

z ∈E

Observe that M depends on z1, z2, ..., zn and E only. Set p1(z) = p(z) + q(z) Then p1(zk) = p(zk) + q(zk) = f (zk), k = 1, 2, ..., n. Further | f (z) – p1(z) | ≤ | f (z) – p(z) | + | q(z) | ≤ ε + M ε, z ∈ E. Example: Suppose that f (z) is analytic in | z | ≤ ρ. Since f (z) is expressed in terms of power series which is uniformly convergent for | z | ≤ ρ. Hence f (z) is approximable there by polynomials. Let z1, z2, ..., zn be distinct points in | z | ≤ ρ. Then we can find a polynomial p(z) such that | f (z) – p(z) | ≤ ε, | z | ≤ ρ and p(zk) = f (zk), k = 1, 2, ..., n. We say that the power series representation of analytic functions provides us with a very specific procedure of approximation by polynomials, but this process can be applied in a circular disk contained in the set in which the given function is analytic. We will now discuss Runge’s theorem which provides a global approximation by rational functions. We will present here the simple version of Runge’s theorem and refer to [8] for generalized version of the theorem. Initially, we need some new definitions. 14.2

LOCALLY ANALYTIC FUNCTIONS

Let E be a set which is dense in itself. Let f (z) be a single-valued function defined on E. We say that f is locally analytic on E if given any z0 ∈ E, there is a neighbourhood N of z0 and a power series

231

APPROXIMATION BY RATIONAL FUNCTIONS AND POLYNOMIALS ∞

Σ ak ( z − z 0 ) k

k =0

such that

∞

f (z) = Σ ak ( z − z 0 ) k k =0

for all z ∈ N ∩ E. When E is a region, the concept of a locally analytic function on E reduces to that of an analytic function on a region. Example: Suppose we divide the z-plane into infinitely many closed squares A1 , A2 , ..., An , ... with sides of unit length parallel to the coordinate axes. Let E = A1 ∪ A2 ∪ ... ∪ An ∪ ... be the set of interior points of these squares. Then the function defined by f (z) = zn, z ∈ An, is locally analytic on E. Let E = Ω be an arbitrary open set. The set Ω is a region if it is connected, but otherwise Ω is the union of countably many disjoint regions. Suppose we divide

1 parallel to 3n the coordinate axes where n is any positive integer. Here the origin is a vertex of one of the squares. Let A be any of these squares such that A and the eight squares bordering A are all contained in Ω as shown in the Fig. 14.II. Then the union of all such squares minus its boundary is an open set Ω′n contained in Ω. the z-plane into infinitely many closed squares with sides of length

Ω

Ω′

Ω

Ω′

Fig. 14.II

If Ωn is the intersection of Ω′n with the open square – 3n < x < 3n, – 3n < y < 3n then Ωn is a bounded open set such that Ωn ⊂ Ω . Recall that a Jordan curve is a homeomorphic image of the unit circle. We will use some facts of Jordan curves without discussing the properties. In the above example, the boundary of Ωn consists of a finite number of closed rectifiable Jordan

232

THE ELEMENTS OF COMPLEX ANALYSIS

curves, each of which is made up of a finite number of line segments. Moreover, Ωn ⊂ Ω n +1 and given any compact set F ⊂ Ω, we can find an integer N(F) > 0 such that F ⊂ Ωn for all n > N(F). We summarize these properties by saying that {Ωn} is an increasing sequence of open sets approximating Ω. We now prove a theorem on approximation by rational functions. Theorem 5. Let Ω be any open set. Suppose that f(z) is locally analytic on Ω. Then there exists a sequence of rational functions {rn(z)} converging uniformly to f (z) inside Ω. Proof : Let {Ωn} be the increasing sequence of open sets approximating Ω as mentioned above. Let γn denotes the boundary of Ωn. Then

(15)

1 2π i

f (z) =

z

γ n+1

f (ζ) dζ ζ−z

for all z ∈ Ωn , where the integral along γn+1 is the sum of integrals along the separate closed rectifiable Jordan curves forming γn+1. Since f (z) is continuous on γn+1, it follows that | f (ζ′) – f (ζ″) | < ε where | ζ′ – ζ″ | < δ(ε′), ζ′, ζ″ ∈ γn+1. Divide all the curves making up γn+1 into arcs Γk, k = 1, 2, ..., N(ε′). Let ζk′(n+1) be the initial and ζk″(n+1) the final point of Γk. Let lk be the length of Γk where lk < min {δ(ε′), ε} for all k = 1, 2, ..., N(ε′). Observe that the final point of the arc Γk coincides with the initial point of the arc Γn+1, except for those k for which Γk and Γk+1 lie on different curves making up γn+1. We now approximate the integral (15) by the sum

(16)

S(n+1)(z) =

1 N ( ε ′ ) f (ζ′ (kn +1) ) Σ (ζ ″ (kn +1) −ζ′ (kn +1) ). ( n +1) k = 1 2π i ζ′ k − z

We have | f (z) – S(n+1)(z) |

z

=

1 2π i

=

1 N (ε ′) Σ 2 π i k =1

γ n +1

1 N ( ε ′ ) f (ζ′ (kn +1) ) ( n +1) f (ζ) (ζ ′ k −ζ ″ (kn +1) ) Σ dζ− 2 π i k =1 ζ′ (kn +1) − z ζ−z

z

Γk

LM f (ζ) − f (ζ′ N ζ − z ζ′

( n +1) ) k ( n +1) −z k

OP d ζ Q

233

APPROXIMATION BY RATIONAL FUNCTIONS AND POLYNOMIALS

=

1 N (ε ′) Σ 2 π i k =1

z

Γk

LM f (ζ) − f (ζ′ N ζ−z

( n +1) k

)

+ f (ζ ′ (kn +1) )

OP Q

ζ′ (kn +1) − ζ dζ . (ζ − z ) (ζ′ (kn +1) − z )

| f ( z) | Mn+1 = max z∈

Set

γ n +1

and let dn+1 be the distance between Ω n and γn+1. Then | f (z) – S(n+1) (z) |

(17)


0 by choosing ε sufficient small. Put ε = εn where εn → 0 as n → ∞. We thus obtain a sequence of rational functions {S (n+1)(z)} which converges uniformly to f (z) on every closed set Ω1 , Ω2 , ... and hence uniformly to f (z) inside Ω. The proof is complete by choosing {rn(z)} = {S(n+1)(z)}. In order to prove Runge’s theorem we need a lemma. Lemma: Let G be the complement of a compact set F and let ζ ∈ G. Suppose that r (z) =

P (z) , P (ζ) ≠ 0, ( z − ζ) k

where P (z) is a polynomial whose degree does not exceed k. Let η belongs to the same component of G as ζ. Then, given ε > 0, there exists a rational function r1(z) =

P1 ( z ) , P1(η) ≠ 0, ( z − η) k1

such that | r1(z) – r (z) | < ε, z ∈ F. Here the degree of the polynomial P1(z) does not exceed k1. Proof : Let g be the component of G containing the points ζ and η. Let Γ ⊂ g be a Jordan curve joining ζ and η and let ρ be the distance between Γ and F. Divide 1 Γ into arcs τ1, τ2, ..., τm of diameter less than ρ. Let ζ = ζ0, ζ1, ..., ζm = η be the 2 points of division. We construct the rational function

234

THE ELEMENTS OF COMPLEX ANALYSIS

LM F MN GH

ζ − ζ1 P( z ) 1− r1(z) = k ( z − ζ) z − ζ1

IJ K

OP PQ

n1 k

=

P2 ( z ) , ( z − ζ1 ) n1k

where the degree of P2(z) does not exceed n1k. Let M = max | f ( z ) |. Then z ∈F

L F ζ − ζ I OP | r (z) – r (z) | = | r (z) | M1 − G MN H z − ζ JK PQ LF kI 1 O ≤ M MG J F ρ / ρI + ...P H K NH1 K 2 Q

n1 k

1

1

−1

1

n1

2k for all z ∈ F. 2 n1 By choosing n1 sufficiently large, we find that ε | r1(z) – r (z) | < , z ∈ F. m By repeating the argument, we obtain the required rational function at the mth step. We now turn to the subject of approximation by polynomials and prove Runge’s theorem. Theorem 6. Suppose that Ω is a union of countably disjoint simply connected regions which do not contain the point at infinity. Suppose also that f(z) is locally analytic on Ω. Then there exists a sequence of polynomials {Pn(z)} which converges uniformly to f(z) inside Ω. Proof : Let {εn} be a sequence of positive numbers which converge to zero. Suppose that {Ωn} is an increasing sequence of open sets approximating the open set Ω. Suppose also that

0 and δ > 0 such that for | ξ – β | < δ, f (z) = ξ has exactly m simple roots in D(z0, ε). Proof : Since the zeros of an analytic function are isolated, f (z) = β has no

1 ρ. Also, f ′(z) ≠ 0 for 0 < | z – z0 | 2 < 2ε. Let γ(t) = z0 + ε exp (2πit), 0 ≤ t ≤ 1, and let σ = f o γ. Now β ∉ {σ}, thus there is a δ > 0 such that D(β, δ) ∩ {σ} = φ. Hence, |β–ξ| 0 and ε
0 such that D(β, δ) ⊂ f (U), where β = f (z0). It follows from the Lemma 1 that we need to find an ε > 0 and a δ > 0 such that D(z0, ε) ⊂ U and D(β, δ) ⊂ f (D (z0, ε)). Let X1 and X2 be metric spaces. Let f : X1 → X2. Suppose that f (U) is open in X2 whenever U is open in X1, then f is called an open map. Let f be one-one and onto. We define the inverse map f –1 : X2 → X1 by f –1(w) = z where f (z) = w. It follows that f –1 is continuous when f is open. In other words, for U ⊂ X1, (f –1)–1 (U) = f (U). Corollary: Let Ω be a region. Let f : Ω → C be one-one and analytic. Suppose that f (Ω) = X2. Then f – 1: X2 → C is analytic and ( f –1)′ (w) = [f ′(z)]– 1 where f(z) = w. Schwarz Lemma: Let D(0,1) be an open disk centred at 0 and radius 1. Let f be analytic in D(0, 1). Suppose that (i) | f (z) | ≤ 1 for z ∈ D(0, 1) (ii) f (0) = 0. Then (iii) | f (z) | ≤ | z | for z ∈ D(0, 1) (iv) | f ′(0)| ≤ 1; if equality holds in (iii) for one z ∈ D(0, 1) ~ {0}, or if equality holds in (iv), then f (z) = λ z, where λ is a constant, | λ | = 1. Proof : Define g : D(0, 1) → C by g(z) = f (z)/z for z ≠ 0 and g(0) = f ′(0) ; then g is analytic in D(0, 1). By Maximum Modulus Theorem | g(z) | ≤ r–1 for | z | ≤ r and 0 < r < 1. As r → 1, we have | g(z) | ≤ 1 for all z ∈ D(0, 1). In other words, | f (z) | ≤ | z | and | f ′(0) | = | g(0) | ≤ 1. If | f (z) | = | z | for some z ∈ D(0,1), z ≠ 0, or | f ′(0) | = 1. Then g assumes its maximum value inside D. Thus, again by Maximum Modulus theorem, g(z) = λ for some constant λ with | λ | = 1. Hence, f (z) = λz and the proof is complete. We need Schwarz’s Lemma to characterize the conformal maps of the open unit disk D(0, 1) onto itself. Let us introduce a class of such maps. If | α | < 1, we define the Möbius transformation z–α . (1) qα(z) = 1 – αz

239

APPENDIX 1—RIEMANN MAPPING THEOREM

Note that qα maps D(0,1) onto itself and is one-to-one. Let θ be a real number. Then | qα(eiθ) | =

e iθ – α e iθ – α = = 1. e – iθ – α 1 – α e iθ

We summarize the above discussion in Lemma 2. Lemma 2: If | α | < 1, define qα as in (1). Then qα is one-to-one map of D(0, 1) onto itself. The inverse of qα is q– α. Furthermore, we have q′α(0) = 1 – | α |2, and q′α(α) = (1 – | α |2)– 1. Since qα is analytic for | z |
δn for all ξ ∈ γ(t), (3) gives (4)

| f (z′) – f (z″)|
0 there corresponds a δ > 0 such that | f (z′) – f (z″)| < ε for all f ∈ F and all z′ and z″ ∈ En for which | z′ – z″ | < δ. Now let {fm} be a sequence in F. Let us choose a countable dense subset {z1} of Ω. By hypothesis {fm(z)} is bounded at each z ∈ Ω. Hence {fm} has a subsequence which converges. Denote this subsequence as {fm, 1} which converges at z1. From {fm, 1} we can extract a subsequence denoted by {fm, 2} which also converges at z2. Proceeding in this manner we obtain sequences {fm, i} which converges at zi where {fm, i} is a subsequence of {fm, i – 1}. Thus the sequence {fm, n} which is called the diagonal sequence converges at every one of the points zi. Now we have to prove that {fm, n} converges uniformly on each En. We fix En and also fix ε > 0. Choose δ as in (5). Then there are points z1, z2, ..., zp of the set {zi} such that p

En ⊂ ∪ {D(zi, δ)} i =1

and there is an integer N such that (6) | fμ, μ (zi) – fv, v (zi) | < ε where μ > N, ν > N, and 1 ≤ i ≤ p. We have for every z ∈ En there corresponds a zi such that 1 ≤ i ≤ p and | z – zi | < δ. Then (7) | fμ, μ(z) – fv, v (z) | ≤ | fμ, μ (z) – fv, v(zi) | + | fμ, μ(zi) – fv, v(zi) | + | fμ, μ (zi) – fv, v(z) |. If μ > N and ν > N, the second term in the right of (7) is less than ε. The first and third term in the right of (7) is less than ε by our choice of δ. Hence (8) | fμ, μ (z) – fv, v (z) | < 3ε for all z ∈ En, if μ > N and ν > N. Thus {fm, n} converges uniformly on each En and hence on each compact subset E of Ω. The proof of the theorem is now complete.

241

APPENDIX 1—RIEMANN MAPPING THEOREM

Definition (Conformal Equivalence) : Let Ω1 and Ω2 be two regions. We say that Ω1 is conformally equivalent to Ω2 if there is a function f ∈ H(Ω1) such that f is one-one and f (Ω1) = Ω2. Note that this is an equivalence relation. It follows from Liouville’s theorem that C is not equivalent to any bounded region. It also follows from the definitions that if Ω1 is simply connected and Ω1 is equivalent to Ω2 then Ω2 is simply connected. Theorem 2 (Riemann Mapping). Every simply connected region Ω in C (Ω ≠ C ) is conformally equivalent to the open unit disk D(0, 1). Before carrying out the proof of this important theorem we would like to point out that the only property of a simply connected region which will be used is that every analytic function which has no zero in such a region has an analytic square root there. Proof : Suppose Ω is a simply connected region in C. Let w0 be a complex number and w0 ∉Ω. Let Φ be the class of all ψ ∈ H(Ω) such that ψ is one-one and map Ω into D(0, 1). We need to prove that some ψ ∈ Φ maps Ω onto D(0, 1). We first prove that Φ is non-empty. Since Ω is simply connected there is a function f ∈ H (Ω) such that f 2(z) = z – w0 in Ω. If f (z1) = f (z2), then f 2(z1) = f 2(z2). Hence z1 = z2 and thus f is one-one. By similar argument we find that f (z1) = – f (z2), where z1, z2 ∉ Ω Since f is an open mapping, f (Ω) ⊃ D(a, r) where 0 < r < | a |. The disk D(– a, r) therefore does not intersect f (Ω). By putting ψ = r/(f + a) we see that ψ ∈ Φ. We now prove that if ψ ∈ Φ, if ψ(Ω) does not cover all of D(0, 1), and if z0 ∈ Ω, then there is a ψ1 ∈ Φ with | ψ′1(z0) | > | ψ′(z0) |. Define qα as defined in (1) i.e. qα(z) =

z–α . 1 – αz

It follows from Lemma 2 that for α ∈ D(0, 1), qα is one-one mapping of D(0,1) onto D(0, 1), its inverse mapping is q–α. Suppose ψ ∈ Φ, α ∈ D(0, 1) and α ∉ ψ(Ω). Then qα o ψ ∈ Φ, and qα o ψ has no zero in Ω. Thus there is a function g ∈ H(Ω) such that g2 = qα o ψ. We find that g is one-one and since Φ is non-empty we have g ∈ Φ. Let ψ1 = g o qβ where β = g(z0). It follows that ψ1 ∈ Φ. Denote w2 = s(w). We now have

242

THE ELEMENTS OF COMPLEX ANALYSIS

ψ = qα o s o g = q– α o s o q– β o ψ1. Since ψ1(z0) = 0, by Chain rule we get ψ′(z0) = F′(0) ψ′1(z0), where F = q– α o s o q– β. Hence F{D(0, 1)} ⊂ D(0, 1) and F is not one-one in D(0, 1). Therefore, by using Schwarz Lemma, we get | F′(0) | < 1. Hence | ψ′(z0) | < | ψ′1(z0) |. We now fix z0 ∈ Ω, and put η = sup {| ψ′(z0) | : ψ ∈ Φ}. We saw in the preceding discussion that any function h ∈ Φ for which | h′(z0) | = η will map Ω onto D(0, 1). The proof of the theorem will be complete if we prove the existence of such a function h. Since | ψ(z) | < 1 for all ψ ∈ Φ and z ∈ Ω. It follows from Theorem 1 that Φ is a normal family. By the definition of η it follows that there is a sequence {ψn} in Φ such that | ψ′n(z0) | → η, and since Φ is normal. We can extract a subsequence {Pn} from {ψn} which converges uniformly on compact subsets of Ω to a limit function h ∈ H(Ω). By theorem 5 of Chap. 10, | h′(z0) | = η. Since Φ is non-empty, η > 0, so h is not constant. Since Pn(Ω) ⊂ D(0,1), we have h(Ω) ⊂ D (0, 1). Hence by the open mapping Theorem we get h(Ω) ⊂ D(0, 1). Now it remains to prove that h is one-one. Let z1 and z2 be fixed points belonging to Ω. Put α = h(z1) and αn = Pn(z1). Let D (z2, r) be a closed circular disk contained in Ω such that z1 ∉ D (z2, r). Suppose also that h – α has no zero on the boundary of D (z2, r). Since the zeros of h – α have no limit point in Ω, the above supposition makes sense. The function ψn – αn

converges to h – α uniformly in D (z2, r), they have no zero in D(0, 1). Since they are one-one and have a zero at z1, it follows from Rouche’s Theorem that h – α has no zero in D(0, 1). Thus, in particular, we have h(z1) ≠ h(z2). Hence h ∈ Φ, and the proof of the Riemann mapping theorem is complete.

APPENDIX

2

HOMOLOGICAL VERSION OF CAUCHY’S THEOREM

We recall some definitions. Let Ω be an open set. Let γ, τ be closed paths in Ω, γ and τ are said to be homologous in Ω if n(γ, α) = n(τ, α) for every α ∉ Ω. A closed path γ in Ω is homologous to 0 in Ω if n(γ, α) = 0 for every α ∉ Ω. Let γ1, γ2, ..., γn be curves, and let m1, m2, ..., mn be integers. A sum of the form n

γ = ∑ mi γ i i =1

is called a chain. We say that γ is a chain in Ω if each γi is in Ω. A chain is said to be closed if it is a finite sum of closed paths. We define

z

γ

n

f = ∑ mi i =1

z

γ

f

where γ is a chain. If γ is a closed chain where each γi is a closed path, then the index of γ with respect to a point α is defined as dz 1 n(γ, α) = γ 2π i z – α where α is not on the chain. We write γ~τ if n(γ, α) = n(τ, α) for every α ∉ Ω and γ~0 if n(γ, α) = 0 for every α ∉ Ω.

z

243

244

THE ELEMENTS OF COMPLEX ANALYSIS

Let Ω be an open set. Let γ, τ be two paths in Ω. Suppose that γ, τ are defined on the same interval [0, 1]. We say that γ, τ are “near-together” if there exists a partition 0 = u0 ≤ u1 ≤ u2 ≤ ... ≤ un = 1 and for each j = 0, 1, 2, ..., n – 1, there exists a disc Dj contained in Ω such that γ({uj, uj + 1]) ⊂ Dj and τ([uj, uj + 1)] ⊂ Dj. Let Ω be an open set. Let γ be a curve in Ω, defined on [0, 1]. Let 0 = u0 ≤ u1 ≤ u2 ≤ ... ≤ un = 1 be a partition of [0, 1]. Let γj : [uj, uj + 1] → Ω be the restriction of γ to the smaller interval [uj, uj +1]. Then the chain γ1 + γ2 + ... + γn is called a subdivision of γ. By another parameterization, if τj is obtained from γj, then the chain τ1 + τ2 + ... + τn is also a subdivision of γ. Note that the chain γ and τ1 + τ2 + ... + τn do not differ from each other. We illustrate this in Fig. II.I τ1

τ2

τ3

τ4

r

Fig. Appendix II.I

Suppose that {τji} is a subdivision of γj where γ = ∑ mj γj. Then

∑ ∑ m j τ ji j

i

is called a subdivision of γ. We say that a path is rectangular if every curve of the path is either a horizontal line segment or a vertical line segment. In the following lemma we shall show that every path is homologous with a rectangular path. Lemma: Let Ω be an open set. Let γ be a closed path in Ω. Then there exists a rectangular closed path τ with the same terminal points, and such that γ, τ are “near together” in Ω.

245

APPENDIX 2—HOMOLOGICAL VERSION OF CAUCHY’S THEOREM

In particular, γ ~ τ in Ω, and if f is analytic on Ω, then

z z γ

f =

τ

f.

Proof : Let γ be defined on an interval [0, 1]. Choose a partition 0 = u0 ≤ u1 ≤ ... ≤ un = 1 of [0, 1] such that the image of each small interval ([uj, uj + 1]) is contained in a disc Dj on which f has a primitive. Replace the curve γ on the interval [uj, uj +1] by the rectangular curve as shown in Fig. II.II. We now set zj = γ(uj) in Fig. II.II. zn zj + 1 zj z2 z1 Dj

z0 Fig. Appendix II.II

Note that if γ is a closed path, then the rectangular path constructed in the lemma is also a closed path. The closed rectangular path is illustrated in Fig. II.III.

Fig. Appendix II. III

We now prove a very important theorem which has nothing to do with analytic functions. The proof of this remarkable theorem is due to Artin. We shall see that Homological version of Cauchy’s theorem follows immediately from this theorem. Theorem 1. Let Ω be an open set. Let γ be a rectangular closed chain in Ω. Assume that γ ~ 0 in Ω. Then there exists rectangles T1, T2 , ..., Tn contained in Ω N

such that a subdivision of γ is equal to ∑ m j ∂ Tj where ∂Tj is the boundary of Tj j=1

(oriented counterclockwise) and mj are some integers. Proof : Let the rectangular chain γ be given. We draw lines parallel to both axes passing through the sides of the chain (Fig. II.IV). Then these vertical and

246

THE ELEMENTS OF COMPLEX ANALYSIS

horizontal lines decompose the plane into finite rectangles and some unbounded regions which may be considered as rectangular regions extending to infinity in the vertical and horizontal direction. Let Tj be one of the finite rectangles. Choose a point αj from the interior of Tj and let mj = n(γ, αj).

Fig. Appendix II.IV

It follows that for some rectangles mj = 0, and for some rectangles mj ≠ 0. Let mj ≠ 0. j = 1, 2, 3, ..., N for rectangles Tj, j = 1, 2, ..., N and let ∂Tj be the corresponding boundary (oriented counterclockwise). We shall first show that every rectangle Tj such that mj ≠ 0 is contained in Ω. By our hypothesis αj must lie in Ω, because n(γ, α) = 0 for α ∉ Ω. Since Ω is connected, the index (winding number) is constant on Ω, and hence it is constant on the interior of Tj. This shows that the index (winding number) is not zero and the interior of Tj is contained in Ω. It is obvious that if a boundary point of Tj is on γ, then it is in Ω. If a boundary point of Tj is not on γ, then it follows from Lemma 2 of Chapter 5 that mj ≠ 0. Thus the rectangle Tj with its boundary is contained in Ω. We now prove that some subdivision of γ is equal to N

∑ m j ∂ Tj .

j =1

In order to prove this we need to find first an appropriate subdivision of γ. Since the vertical and horizontal line cut γ at several points, we can find a subdivision τ of γ such that every curve in τ is some side of a finite rectangle, or the finite side of the infinite rectangles. The subdivision τ is the sum of such sides as described above. It is now sufficient to prove that τ = ∑ mj ∂Tj. Let γ0 be the closed chain defined by N

γ0 = τ – ∑ m j ∂ Tj . j =1

247

APPENDIX 2—HOMOLOGICAL VERSION OF CAUCHY’S THEOREM

We claim that n(γ0, α) = 0 for every α ∉ γ0. We consider various possible cases to prove our claim. If α lies in one of the infinite rectangles, then it follows from Lemma 3 of Chapter 5 that n(γ0, α) = 0. Suppose α lies in some infinite rectangle T. If T is one of the rectangles Tj, say T = Tk, then n(∂Tk, α) = 1 and n(∂Tk, α) = 0 if j ≠ k. Applying Lemma 3 of Chapter 5 again and using the definition of mj we get n(γ0, α) = n(γ, α) – mk n(∂Tk, α) = 0. Suppose that T is one of the rectangles Tj, it follows from the definition that n(γ, α) = 0, and by Lemma 3 of Chapter 5 that n(Tj, α) = 0 for all j = 1, 2, ..., N. Thus in this case also we have n(γ, α) = 0. Suppose α lies on the boundary of a finite rectangle or an infinite rectangle, but not on γ0. By the continuity of the index (winding number) we find that n(γ0, α) = 0. We thus proved that if γ0 is a closed chain defined by N

γ0 = τ – ∑ m j ∂ T j j =1

then n(γ0, α) = 0 for α ∉ γ0. It remains to prove that γ0 = 0. Suppose that γ0 ≠ 0. Write γ0 = mσ + γ0 where σ is a horizontal or vertical segment, γ0 is a chain of vertical and horizontal segments other than σ and m is a non-zero integer. We have to consider two cases. In the first case σ is the side of some finite rectangle and in the second case σ is the side of two infinite rectangular regions. We illustrate these two cases in the following Fig. II.V. α α′

σ

T

α′

σ

α

Fig. Appendix II.V

In the first case, the chain γ0 – m ∂ T does not contain σ. Suppose that α is a point inside T and α′ is a point on the opposite side from α but near to σ. Then there is a line segment which joins α and α′ but does not intersect γ0 – m ∂ T. It now follows from the connectedness and the continuity of the line segment that

248

THE ELEMENTS OF COMPLEX ANALYSIS

n(γ0 – m ⋅ ∂T, α) = n(γ0 – m ⋅ ∂T, α′). Note that by definition, n(m ⋅ ∂T, α) = m, and n(m ⋅ ∂T, α′) = 0, because α′ is outside T. Since n(γ0, α) = n(γ0, α′) = 0, it follows that m = θ. In the second case, the two regions adjacent to σ are both infinite. We shall see that this is not possible. By Lemma 3 of Chapter 5 n(γ0, α) = n(γ0, α′) = 0. Hence,

m i.e. (1)

m

z

σ

z FGH σ

dz + z–α

z

γo

dz =m z–α

IJ K

1 1 dz = – z – α z – α′

z

σ

z FGH γo

dz + z – α′

z

γo

dz z – α′

IJ K

1 1 dz . – z – α′ z – α

It can be easily seen that the left side of (1) is equal to a fixed non-zero multiple of 2πi. The right side of (1) is a continuous function of α, α′ where α, α′ approach each other on a line segment passing through σ. This can happen only when m = 0. We thus proved that γ0 = 0 and therefore a subdivision of γ is equal to

∑ mj ∂Tj.. This proves the theorem. Cauchy’s Theorem (Homological Version) Let Ω be an open set. Let γ be a closed chain in Ω and let γ ~ 0 in Ω. Assume that f is analytic on Ω. Then

z

γ

f = 0.

Proof : By the Lemma, it is sufficient to prove the theorem to the case when γ is a rectangular closed chain. Applying Theorem 1 we see that Cauchy’s theorem follows easily. We know that if f is analytic on Ω, then

Thus

z

z

∂Tj

N

∑ m j ∂ Tj

j =1

This prove that

z

γ

f =0

f = 0. f = 0.

SOLUTIONS TO SOME SELECTED EXERCISES

1. Let (X, d) be a metric space. Prove that (X,

d ) is also a metric space. 1+ d

Solution: (i)

d ( x, y ) = 0 iff d(x, y) = 0 iff x = y 1 + d ( x, y )

(ii)

d ( y, x ) d ( x, y) = since d(x, y) = d(y, x) 1 + d ( x, y) 1 + d ( y, x )

The triangle inequality (iii)

d ( x, y ) d ( x, z ) d ( y, z ) ≤ + 1 + d ( x, z ) 1 + d ( x, y) 1 + d ( y, z )

can be verified as follows: Multiplying both sides of (iii) by (1 + d(x, y)) (1 + d(y, z)) (1 + d(z, x)) we obtain d(x, z) ≤ d(x, y) + d(y, z) + 2d(x, y) d(y, z) + d(x, y) d(y, z) d(z, x) Since d is a distance and is ≥ 0 (iii) follows. 2. Let E and F be two compact subsets of the metric space (X, d). Define the distance dist. (E, F) by dist. (E, F) = inf {d (x, y) : x ∈ E, y ∈ F}. Show that if E ∩ F = φ, then dist. (E, F) > 0. Solution: Clearly, dist. (E, F) ≥ 0. We will show that if E ∩ F = φ, then dist. (E, F) ≠ 0. The proof is by contradiction. Suppose that dist. (E, F) = 0. Then for a given n there 1 . Taking n = 1, 2, ... we get a sequence n {xn} and a sequence {yn}. Since E is compact and xn ∈ E, we can find a subsequence of {xn} converging x0 ∈ E. Similarly, there exists a subsequence of {yn} converging to y0 ∈ F.

exists xn ∈ E and yn ∈ F such that d(xn, yn) ≤

Now

d(x0, y0) = lim ( x n , yn ) = 0, so x0 = y0 and E ∩ F ≠ φ. n→∞

249

250

THE ELEMENTS OF COMPLEX ANALYSIS

3. Prove that the mapping x →

1 , x ∈ (0, 1) is continuous but not uniformly continuous. x

1 is continuous for x ≠ 0 in (0, 1). x Let ε > 0 be given. Then, however small we might choose δ(ε), there exists x1, x2 ∈ (0, 1) with | x1 – x2 | < δ(ε) such that

Solution:

1 1 – x1 x 2

Taking x2 = x1 +

>ε

1 δ(ε), we have 2 | x2 – x1 | < δ(ε) and

F H

I K

1 1 1 1 – = δ(ε)/x1 x1 + δ(ε ) . x1 x 2 2 2

But

lim x1

x1 → 0

1 2

cx

δ( ε )

h = ∞.

+ 12 δ(ε )

1

Hence, it is possible to choose x1 ∈ (0, 1) such that

RS 1 δ(ε) / x F x H T2 1

1

+

I UV > ε KW

1 δ( ε ) 2

4. Show that f (z) = | z | is nowhere differentiable. Solution: Let w = f (z) = u + io = ( x 2 + y 2 ) = r. ∂w ∂r = = 1. ∂r ∂r This shows that the rate of change in the radial direction is 1. But the derivative in the direction of the tangent

Now

∂f ∂r ∂w ∂r ∂r = =0 = 0. ∂r ∂θ ∂r ∂θ ∂θ Since the derivative in the radial direction is different from the derivative in the direction of the tangent. Hence f (z) = | z | is nowhere differentiable.

5. Show that the function f (x, y) = x2 + iy2 is uniformly continuous in the open disc D (0, 1) but not uniformly continuous in C. Solution: Consider | f (x, y) – f (x1, y1) | = | x2 + iy2 – x12 – iy12 | = | (x + x1) (x – x1) + i(y + y1)(y – y1) | ≤ | (x + x1) (x – x1) | + | (y + y1)(y – y1) | ≤ 2(| x – x1| + | y – y1 |), since | x + x1 | ≤ 2, (y + y1) ≤ 2 in D(0, 1).

251

SOLUTIONS TO SOME SELECTED EXERCISES

Hence | f (x, y) – f (x1, y1) | ≤ ε ( x – x1 ) 2 + ( y – y1 ) 2 ≤ ε/4.

for

This shows that f (x, y) is uniformly continuous in D (0, 1). But f (x, y) is not uniformly continuous in C. For, choose x + x1 = y + y1 = M > 0 Then | f (x, y) – f (x1, y1) | = M | (x – x1) + i(y – y1) | = M ( x – x1 ) 2 + ( y – y1 ) 2 Thus, however small we choose ( x – x1 ) 2 + ( y – y1 ) 2 , | f (x, y) – f (x1, y1) | > ε for sufficiently large M. 6. Show that

FG H

IJ K

∂ ∂ ∂ ∂ ∂ ∂ + = ,i – = ∂z ∂z ∂x ∂z ∂z ∂y

∂2 ∂2 ∂2 + = 4 ∂z ∂z ∂x 2 ∂y 2

and Solution: (1) (2)

FG IJ H K ∂ 1F ∂ ∂I = G +i J ∂z 2 H ∂x ∂y K

∂ 1 ∂ ∂ = –i ∂z 2 ∂x ∂y

By adding (1) and (2) we obtain ∂ ∂ ∂ = + ∂x ∂z ∂z and by subtracting (2) from (1) we have

∂ ∂ ∂ – =–i ∂z ∂z ∂y

i.e.,

FG ∂ – ∂ IJ = ∂ H ∂z ∂z K ∂y ∂ ∂ ∂ F ∂I ∂ F ∂I + = G J+ G J ∂x H ∂x K ∂y H ∂y K ∂x ∂y F ∂ ∂ IF ∂ ∂ I F ∂ ∂ IF ∂ ∂ I = G + JG + J +i G – JG – J H ∂z ∂z K H ∂z ∂z K H ∂z ∂z K H ∂z ∂z K i

2

Now

2

2

2

2

=

∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 + + 2 – – + 2 = 4 ∂z ∂ z ∂ z 2 ∂ z 2 ∂z ∂z ∂z ∂z ∂ z 2 ∂z 2

252

THE ELEMENTS OF COMPLEX ANALYSIS

7. Prove that the function f (z) = u + iv where f (z) =

x 3 (1 + i ) – y 3 (1 – i ) x 2 + y2

z ≠ 0, f (0) = 0, satisfies C – R equations at the origin, but f ′ (0) does not exist. Solution:

and

u(x, y) =

x 3 + y3 x 2 + y2

v(x, y) =

x 3 + y3 x 2 + y2

u(0, 0) = 0, v(0, 0) = 0 ux = lim

u( x, 0) – u( 0, 0) =1 x

uy = lim

u(0, y) – u(0, 0) =–1 y

x→0

y→0

Similarly vx = 1, vy = 1 Hence, C – R equations are satisfied at the origin. f ′ (0) = lim

But

z→0

f ( z ) – f ( 0) z–0

[ x 4 + y 4 + ixy( y 2 – x 2 )](1 + i ) z→0 ( x 2 + y 2 )2

= lim

Letting z → 0 along y = x, we have f ′ (0) = f ′ (0) =

But

1 (1 + i) 2

FG ∂u + i ∂v IJ H ∂x ∂x K

=1+i ( 0, 0 )

Hence f ′ (0) is not unique. This shows that f (z) is not analytic at the origin but C – R equations are satisfied there. 8. Prove that f (z) = z z is nowhere analytic Solution: f (z) = z z = x2 + y2 = u + i0 The first four partial derivatives

∂u ∂v ∂u ∂v = 2x, = 2y, = 0, =0 ∂x ∂y ∂y ∂x are continuous everywhere. But C – R equations are satisfied only at the origin. Hence z = 0 is the only point at which f ′ (z) exists. Thus f (z) = z z is nowhere analytic. 9. Evaluate

z

1+ i 0

( z 2 + z ) dz. By choosing two different paths of integration show that

the results are the same.

253

SOLUTIONS TO SOME SELECTED EXERCISES

Solution: Choose two paths γ1 and γ2 as shown in the figure. i

1+i r2

Now

O

z

( z 2 + z ) dz =

γ1

=

z

r1

r1 1 0

( x 2 + x ) dx +

F H

z

1 1 0

[(1 + iy) 2 + (1 + iy)]2 idy

I K

1 1 1 3 + + 2– i– 3 2 3 2

5 2 i– 3 3 Let z = (1 + i) t, 0 ≤ t ≤ 1. Then dz = (1 + i)dt and

=

z

( z 2 + z ) dz = (1 + i )

γ2

z

1 0

[(1 + i ) 2 t 2 + (1 + i )t ] dt

(1 + i )3 (1 + i ) 2 + 3 2 5i 2 – = 2 3 =

10. Evaluate

z

z 2 dz along

(i) the line segment joining the points (1, 1) and (2, 4) (ii) the curve x = t, y = t2 joining the points (1, 1) and (2, 4). Solution: (i) The equation of the line joining (1, 1) and (2, 4) is y = 3x – 2. Any point on this line is given by z = x + i (3x – 2), 1 ≤ x ≤ 2. Hence dz = dx + idy = (1 + 3i) dx

B(2, 4)

A (1, 1) O

254

THE ELEMENTS OF COMPLEX ANALYSIS

and

z

Thus

z2 = x2 – y2 + 2ixy = (– 8 + 6i) x2 + (12 – 4i)x – 4 AB

f ( z ) dz =

z

2

1

[(– 8 + 6i ) x 2 + (12 – 4i ) x − 4] (3 + i ) dx

86 – 6i 3 (ii) Any point on the arc of the given curve is

=–

z = t + it2, 1 ≤ t ≤ 2 dz = 1 + 2it dt Now integrating f (z) = z2 along the curve, we have

z

γ

f ( z ) dz =

z

2

1

(t + it 2 ) 2 (1 + 2it ) dt

86 . 3 This shows that the value of the integral along the two paths joining (1, 1) and (2, 4) is the same.

= – 6i –

11. Show that

z

γ

( x 2 – iy 2 ) dz ≤ 2.5

where (i) γ is the interval [– i, i] on the y-axis (ii) γ is the semi-circle z = cos ϕ + i sin ϕ, – π/2 ≤ ϕ ≤ π/2 Solution: (i) We will use the inequality

z

γ

f ( z ) dz ≤ max | f (z) | L (γ) z ∈γ

where L (γ) is the length of γ. On [– i, i], x2 = 0 and | iy2 | ≤ 1. Hence | x2 + iy2 | ≤ 1. Since L (γ) = 2, it follows that

z

γ

( x 2 – iy 2 ) dz ≤ 1.2 < 2.5

(ii) In this case x = cos ϕ, y = sin ϕ, dz = (– sin ϕ + i cos ϕ) dϕ Now Hence

z

z

γ

γ

( x 2 – iy 2 ) dz =

( x 2 – iy 2 ) dz =

2 4 + i 3 3

2 5 < 2.5 3

255

SOLUTIONS TO SOME SELECTED EXERCISES

12. Prove that homotopy with respect to a domain Ω is a transitive property. Solution: Let γ1, γ2, γ3 be three closed curves in a domain Ω. Assume that γ1 and γ2 are homotopic with respect to Ω, with the family of curves γ12(t, u), 0 ≤ t ≤ 1, and 0 ≤ u ≤ 1 generating the homotopy, so that γ12 (t, 0) = γ1 and γ12 (t, 1) = γ2. Similarly, we assume that γ2 and γ3 are homotopic with respect to Ω where γ23 (t, 0) = γ2 and γ23 (t, 1) = γ3. Now, by choosing properly the parameter t it can be arranged that γ12 (t, 1) = γ23 (t, 0). Define γ13 (t, u) by γ13

R|γ (t, u) = S |Tγ

1 2 1 ≤ u ≤1 23 (t , 2u – 1) for 2

12 (t ,

2u )

for 0 ≤ u ≤

γ13 (t, u) is continuous in 0 ≤ t ≤ 1, 0 ≤ u ≤ 1 and γ13 (t, 0) = γ12 (t, 0) = γ1,

F 1I = γ H 2K

γ13 t,

12

(t, 1) = γ23 (t, 0) = γ2,

γ13 (t, 1) = γ23 (t, 1) = γ3. Note that they all are within the domain Ω. Hence γ1 is homotopic to γ3 with respect to Ω. 13. Evaluate

z

ez dz 2 γ z +1 if γ is the circle of radius 1 at (i) z = i and (ii) z = – i. Solution:

z

(i)

γ

ez dz = z2 + 1

z

γ

e z dz z+i z–i

z

Identify z0 as i and f (z) as

z

γ

e . Applying Cauchy’s integral formula we obtain z+i

ei e z dz = 2π i f (z0) = 2πi z+i z–i 2i = π(cos 1 + i sin 1)

(ii) Identify z0 as – i and f (z) as

ez . z –1

Applying Cauchy’s integral formula we obtain

z

γ

e z dz e– i = 2 π if ( z 0 ) = 2 πi z–i z+i – 2i

= – π(cos 1 – i sin 1)

256

THE ELEMENTS OF COMPLEX ANALYSIS

14. Prove that Liouville’s theorem remains true if we replace the boundedness of f by the boundedness of

z

2π

| f (re iθ ) | dθ.

0

Solution: The proof is the same as for Liouville’s theorem except that we estimate | f (z) – f (0) | by

|z| 2 π( R – | z |)

z

2π

| f ( Re iθ ) | dθ.

0

15. Prove that if f is analytic in | z | < 1 and | f (z) | ≤ 1 – | z |, then f (z) = 0. Solution: If ρ > 1, we have | f (ρeiθ) | ≤ 1 – ρ. For any given z with | z | < ρ, Cauchy’s formula gives | f (z) | ≤

1 2π

z

2π 0

(1 – ρ)ρ dθ ρ–|z|

Fixing z and letting ρ → 1 we find that | f (z) | = 0; hence f (z) = 0. 16. Is it true that an analytic function f satisfies | f (z) | ≥

1 for | z | < 1 ? 1–|z|

Solution: Suppose that there is such an f. If | f (z) | >

1 , then f (z) ≠ 0. Thus 1–|z|

1/f (z) satisfies the hypothesis of Exercise and hence is identically zero. This is a contradiction. 17. Assume that | f (z) | ≤ M on the circumference of a square whose side is L. Let z0 be the centre of the square. If f is analytic in the square, prove that 8M . πL Solution: There is no loss of generality if we take z0 = 0. Then

| f ′ (z0) | ≤

f ′ (0) =

1 2πi

z

C

f (ζ) dζ ζ2

where C is the boundary of the square. The length of C is 4L, and | ζ | ≥ so 1 4 8M 2 LM 2 = . πL 2π L 18. Expand in a Laurent series the function

| f ′ (0) | ≤

f (z) = Solution: and

1 z 2 ( z – 1)

about z = 0 and z = 1.

1 1 =– = – 1 – z – z2 – ... z –1 1– z 1 2

z ( z – 1)

=–

1 z

2

–

1 – 1 – z ... z

L on C, 2

257

SOLUTIONS TO SOME SELECTED EXERCISES

1 1 = 2 z [1 – (1 – z )]2 = [1 + (1 – z) + (1 – z)2 + ...]2

= 1 + 2(1 – z) + 3(1 – z)2 + ... = 1 – 2(z – 1) + 3(z – 1)2 – ... 1

and

2

z ( z – 1)

=

1 – 2 + 3(z – 1) – 4(z – 1)2 + ... ( z – 1)

19. Prove that the Riemann zeta function ζ defined by ∞

ζ(z) = ∑ n − z n =1

converges for Re z > 1 and converges uniformly for Re z ≥ 1 + ε where ε > 0 is arbitrary small. Prove also that ζ is analytic for Re z ≥ 1 + ε. Solution: 1 1 1 = z log n = x log n z n e e

= ∞

Since ∑

n =1

1 n

1+ ε

1 e

iy log n

1 1 1 = ≤ e x log n n x n1 + ε ∞

converges, we find by the Weierstrass M test that ∑

n =1 ∞

uniformly for Re z ≥ 1 + ε. We also see that each term of the series ∑

n =1

1 converges nz 1 is analytic nz

∞

1 is uniformly convergent for Re z ≥ 1 + ε. Hence by nz Theorem 1 of Chapter 7,

function and since ∑

n =1

∞

ζ(z) = ∑

n =1

1 nz

is analytic for Re z ≥ 1 + ε. 20. Expand the function f (z) = log (z + 2) in a power series and determine its radius of convergence. Solution: We have f ′ (z) = =

1 1 1 = z + 2 2 1 + 2z

F GH

I JK

z z2 z3 1 – + ... 1– + 2 2 4 8

Integrating it term by term we obtain

1 1 z2 1 z3 z– + – ... 2 4 2 8 3 Using f (0) = log 2, we find C = log 2. f (z) = C +

258

THE ELEMENTS OF COMPLEX ANALYSIS

The radius of convergence is given by 1 1 1 1 1 = lim n n . = lim n R n 2 n 2

i.e.

R = 2 lim

n

n = 2.

21. Prove that (i) Res (tan z, π/2) = – 1 (ii) Res

FG 1 – cos z , 0IJ = 1 H z K 2 3

Solution: (i)

tan z =

sin z cos ( π / 2 – z ) = cos z sin ( π / 2 – z )

At z = π/2, cos (π/2 – z) = 1 and sin (π/2 – z) has a simple zero. Hence z = π/2 is a simple pole of tan z. Thus Res (tan z, π/2) = lim ( z – π / 2) z → π/2

(ii)

1 – cos z z 2 / 2 ! – z 2 / 4 ! + ... = z3 z3

= Hence

cos ( π / 2 – z ) =–1 sin ( π / 2 – z )

1 z – + ... 2!z 4!

F 1 – cos z , 0I = 1 . H z K 2

Res

3

22. Prove that cot z =

∞ 1 1 + 2z ∑ 2 2 2 z m =1 z – m π

Solution: Let

f (z) = cot z –

1 z cos z – sin z = z sin z z

f (z) has simple poles where sin z = 0, i.e. z = m π, m = ± 1, ± 2, ± 3, ... Also,

lim

z→0

z cos z – sin z =0 z sin z

Thus we define f (0) = 0 Res (f (z), m π) = lim ( z – mπ) z → mπ

z cos z – sin z =1 z sin z

Using earlier results we obtain cot z =

∞ 1 1 + 2z ∑ 2 2 2 z m =1 z – m π

259

SOLUTIONS TO SOME SELECTED EXERCISES

It remains to show that f (z) is bounded on the contour Γ as shown in the figure. y D

C

x (m + 1)π

O mπ

A

B

ABCD is a square with center as origin and having sides of length 2m π + π. The

F H

poles ± π, ± 2π, ..., ± mπ lie within the contour. Along BC x = m +

| cot z | = i

F H

I K

e 2 iz + 1 – e–2y + 1 = < 1. e 2 iz – 1 – e–2y – 1

Along AD, x = – m +

1 π, the same result holds. 2

Along CD,

y= m+

F H

| cot z | ≤

and

I K

1 π . Now 2

I K

1 π 2

1 + exp [– (2 mπ + π )] 1 – exp [– (2 mπ + π)]

→ 1 as m → ∞ Along AB, similarly | cot z | → 1 as m → ∞ 1 Also → 0 on Γ z 1 Hence f (z) = cot z – z is bounded on Γ. 23. If β is not an integer, prove that

F H

∞

∑

n=–∞

Solution: Consider

I K

1 π cot πβ = β β2 – n2

z

1 π cot πz dz 2 πi γ β 2 – z 2 where γ is the boundary of the rectangular domain 1 1 – M+ < x < M + , – M < y < M and M ≥ 2 is an integer (see the figure below). 2 2

J=

F H

I K

260

THE ELEMENTS OF COMPLEX ANALYSIS

iM –M

M

– iM

The residue of

π cot πz 1 at z = n, an integer is 2 . If | β | < M , it follows from β2 – z 2 β – n2

the residue theorem that J=

M

∑

n=–M

π cot πβ 1 – β β2 – n2

Note that the last term arises from the residue at – β and β. We have proved in the above exercise 22 that | cot π z | ≤ cot h π, z on γ The length of γ is 8M + 2, and since | z | ≥ M on γ, we find that |J|≤

8 M + 2 . π cot h π , M ≥ 2, | β | < M. 2π M 2 – | β |2

This tends to 0 as M → ∞ and hence the results follows.

z

24. Evaluate

∞ 0

x 2 dx . ( x 2 + 1) 2

Solution: We have x2

∞

∫0

(1 + x 2 ) 2

dz =

1 2

∞

∫–∞

x2 (1 + x 2 ) 2

dx

We will use the formula

z

∞ –∞

R( x ) dx = 2 πi

∑ Res ( R, s ) j

Im s j > 0

where sj are the poles of R(z). The function R(z) =

z2 has the only pole of order 2 at z = i in the upper half ( z 2 + 1) 2

plane. Hence R(z) can be expressed in Laurent series about z = i, i.e. R(z) =

a– 2 a– 1 z2 = + + a0 + a1 (z – i) + a2(z – i)2 + ... 2 2 2 ( z – i) (1 + z ) ( z – i)

so that R(z) (z – i)2 = a– 2 + a– 1 (z – i) + a0 (z – i)2 + a1 (z – i)3 + ... is analytic at z = i. d a– 1 = [R(z) (z – i)2]z = i dz

261

SOLUTIONS TO SOME SELECTED EXERCISES

LM N

d z2 dz ( z + 1) 2

=

z

z

Thus

π 0

∞

R( x ) dx = – 2 πi

–∞

z=i

1 i =– 4i 4

= Hence

OP Q

F i I = π/2 H 4K

x2 dx = π/4 (1 + x 2 ) 2

25. Prove that (i) (ii)

z z

2π

dθ =π 2 1 + sin 2 θ

0 π 0

dθ = 8π / 3 3 (1 + 1/2 cos θ) 2

Solution: (i) Let z = eiθ, dz = iz dθ sin θ =

z

dθ = 1 + sin 2 θ

2π 0

z

| z| =1

=i

I K dz

=–i

The simple poles of

F H

1 1 z– 2i z

iz 1 –

z

1 4

cz – h

1 2 z

dz

| z| =1

z

z 1–

1 4

ez

2

–2+

1 z2

j

4 z dz z – 6z 2 + 1 4

| z|=1

4z are z 4 – 6z 2 + 1

z1 = 2 – 1, z2 = 1 – 2 , z3 = 2 + 1 and z4 = – 1 – But z1 and z2 are inside | z | = 1. Hence

z

Now

Res

FG Hz

Similarly Hence

4z dz = 2πi z – 6z 2 + 1 4

|z|=1

4

z

| z |=1

z j ∈| z | = 1

Res

IJ K

FG Hz

4

4z , zj – 6z 2 + 1

IJ K

4z 4z 1 =– , z1 = lim ( z – z1 ) 4 2 2 z → z z – 6z + 1 – 6z + 1 2 2 1

Res i

∑

2,

FG Hz

4z 4

2

− 6z + 1

, z2

FG H

IJ = – 1 K 22 IJ K

1 1 4z − =π 2 = i 2 πi − z 4 − 6z 2 + 1 2 2 2 2

262

THE ELEMENTS OF COMPLEX ANALYSIS

(ii) cos θ is an even function. Hence I=

z FH π

0

dθ 1+

I K

1 cos θ 2

Setting z = eiθ, we find 1 2

z

π −π

F H

dθ

I K

1 1 + cos θ 2

= − 8i = − 8i

z z

2

=

dθ

π

−π

z

| z |=1

1+

LM N

I K

1 cos θ 2

iz 1 +

2

dz

F H

1 1 z+ 4 z

| z| =1

z dz z + 8z + 18z 2 + 8z + 1

| z| =1

z dz ( z + 4 z + 1) 2

4

I OP KQ

2

3

2

2

inside | z | = 1 but z2 is outside | z | = 1. Res Hence

FG H (z

2

IJ K

z , z1 = 3 / 18 + 4 z + 1)

I = – 8i 2πi

z

1 2

z FH

z are z1 = 3 − 2 and z2 = – 3 − 2 , both of order 2. z1 is ( z + 4 z + 1) 2

The poles of

26. Prove that

1 2

=

2

Σ Res

z j ∈| z | = 1

FG H (z

2

z , zj + 4 z + 1) 2

IJ K

= 16π 3 / 18 = 8π / 3 3 ∞ 0

(log x ) 2 dx = π3/8 1 + x2

Solution: Let C be the curve as shown in the figure here and let C1 and C2 be the semi-circles of radii ε and R respectively with centre at the origin. y C2 C1 C –R

Consider

2

e

–e

z

C

R

(log z ) 2 dz z2 +1

(log z ) has a simple pole z = i which lies inside C z2 + 1

x

263

SOLUTIONS TO SOME SELECTED EXERCISES

F (log z) , iI = lim (z − i) (log z) = − π / 8i GH z + 1 JK ( z + i )( z − i ) F π I=− π (log z ) dz = 2 πi G − z +1 H 8i JK 4 2

Res

z z

Hence

2

2

C

2

2

3

2

(log z ) 2 dz = 2 C z +1

Now

2

z→i

z

z)2 dz + z2 +1

– ∈ (log

−R

+

z

R

ε

z

(log z ) 2 dz 2 C1 z + 1

(log z ) 2 dz + z2 +1

z

C2

(log z ) 2 dz z2 +1

Setting z = – x in the first integral and z = x in the third integral on the right, we find

z

R

ε

(log x + πi ) 2 dx + x2 +1

z

(log z ) 2 dz + 2 C1 z + 1

z z

R

ε

(log x ) 2 dx x2 +1

(log z ) 2 dz = − π 3 /4 2 C2 z + 1 Let ε→ 0 and R→ ∞. It can be easily checked that the integrals around C1 and C2 approach zero. Hence +

z

i.e. i.e. i.e. Hence 27. Prove that

(log x + πi ) 2 dx + 0 x2 +1 ∞ (log x ) 2 2 dx + 2 πi 0 x2 +1

z z z z

(log x ) 2 π3 dx = − 2 0 4 x +1 ∞ log x ∞ dx π3 2 dx π − − = 2 2 0 x +1 0 x +1 4 2 3 ∞ (log x ) ∞ log x π3 π 2 dx + 2 πi dx − =− 2 4 0 0 x +1 2 x2 +1 3 ∞ (log x ) 2 ∞ log x π 2 dx + 2 πi dx = 2 0 0 x +1 4 x2 +1 3 ∞ (log x ) 2 π 2 dx = 2 0 8 x +1 ∞

z z z z

z

∞

0

∞

z

π xa , (0 < a < 1) dx = sin πa x (1 + x ) y

R G E

F

e

A

B

C

D

x

264

THE ELEMENTS OF COMPLEX ANALYSIS

Solution: Consider

z

z a −1 dz where Γ is the contour indicated in the figure shown here. Note that Γ 1+ z

z = 0 is a branch point and hence the contour Γ is chosen in this fashion. the simple pole z = – 1 which is inside Γ.

F z , e I = lim (z + 1) z = e GH 1 + z JK 1+ z a −1

Res

z

z a−1 has 1+ z

a −1

πi

(a – 1)πi

z →−1

z a −1 dz = 2πi e(a – 1)πi Γ 1+ z

Hence

z z z z z z z z

i.e.

AB R

i.e.

0

+

BED

+

x a −1 dx + 1+ x

+

DC

= 2πi e(a – 1)πi

( Re iθ ) a −1 iRe iθ dθ 1 + Re iθ

2π

0

( xe 2 πi ) a −1 dx + 2 πi R 1 + xe ε

+

CFA

(εe iθ ) a −1 iεe iθ dθ = 2πi e(a–1)πi 2π 1 + εe iθ 0

Let ε→0 and R→∞. It can be easily checked that the second and fourth integrals tend to zero. Thus

z

∞

0

i.e.

{1 – e2πi (a – 1)}

Hence

z

∞

0

z z

x a −1 dx + 1+ x

e 2 πi ( a −1) . x a −1 dx = 2πi e(a–1)πi ∞ 1+ x 0

∞

0

x a −1 dx = 2πi e(a–1)πi 1+ x

2πi π x a −1 = dx = aπi − aπi sin aπ e −e 1+ x

28. Prove that (i)

z z

∞

0

e−x

2

cos 2 α

cos ( x 2 sin 2α ) dx =

π cos α 2

π sin α 2 0 Solution: Consider the contour Γ as shown in the figure here.

(ii)

∞

e−x

2

cos 2 α

sin ( x 2 sin 2α ) dx =

y B C1 α O

A

x

265

SOLUTIONS TO SOME SELECTED EXERCISES

The contour Γ consists of the line OA, the arc C1 of the circle | z | = r from A to B, (0 ≤ α ≤ π/4) and the line BO.

z

(1)

Γ

z

2

e − z dz =

r

0

2

e − x dx +

z

2

e − z dz +

C1

z

0

r

exp {−(ρe iα ) 2 }e iα dρ

(on BO, z = ρeiα, 0 ≤ ρ ≤ r) 2

Since e − z is analytic within and on contour Γ, we have

z z

2

Γ

e − z dz = 0

It can be checked that 2

e − z dz → 0 as r → ∞

C1

Hence, letting r → ∞, we obtain from (1)

i.e. (2)

z

z

∞ 0

e −ρ

2

∞ 0

exp ( − ρ 2 e 2 iα ) e iα dρ =

cos 2 α

z

∞ 0

2

e − x dx = π / 2

[cos (ρ2 sin 2α) – sin (ρ2 sin 2α)] 2ρ =

π /2e − iα

Equating real and imaginary parts on the two sides of (2) and replacing the variable of integration ρ by x, we get (i) and (ii). Note that, putting α = π/4 in (i) and (ii) we obtain

29. Prove that

z

z

1 0

∞ 0

cos ( x 2 ) dx =

z

∞ 0

sin ( x 2 ) dx =

π 2 2

dx 2π = 2 3 1/ 3 (x − x ) 3

Γ Γ1

Solution: Let Γ and Γ1 be the contours as shown in the figure here. 1 . z = 0 and z = 1 are the branch points of 2 ( z − z 3 )1/ 3 We have

z

|z|=r

f ( z ) dz =

z

0 2π

re iθ i dθ → 0 as r → 0 (r 2 e 2 iθ − r 3 e 3iθ )1/ 3

266

THE ELEMENTS OF COMPLEX ANALYSIS

and

z

| z −1 | = ρ

z

f ( z ) dz =

−π

ρe iθ i dθ → 0 as ρ → 0 (1 + ρe ) ( −ρe iθ )1/ 3 iθ 2 / 3

π

Here f (z) is analytic for every finite z except on the cut from 0 to 1. Hence

z

Γ

f ( z ) dz −

z

Γ1

f ( z ) dz = 0

The Residue* of f (z) at z = ∞ is

lim {− z f ( z )} = lim

z →∞

Hence But

z z

z →∞

Γ

Γ

−z

z

c − 1h 1 z

1/ 3

= – (– 1)–1/3

f ( z ) dz = – 2πi (– 1)–1/3 f ( z ) dz =

z

Γ

f ( z ) dz = (1 − e 2 πi / 3 )

z

1 0

dx ( x 2 − x 3 )1/ 3

= – 2πi (– 1)–1/3 Therefore

z

1 0

dx 2 πi 2π = 3 1/ 3 = πi / 3 − π i / 3 (x − x ) e −e 3 2

30. Find the number of zeros of the function F(z) = sin z + 2iz2 in the rectangle | x | ≤ π/2, | y | ≤ 1. Solution: Apply Rouche’s Theorem with 2 iz2 as f and sin z as g. For z = x ± i, | sin z | = (sin2 x cosh2 1 + cos2 x sinh2 1)1/2 < cosh 1 < 1.5431 For z = π/2 + iy, | y | ≤ 1, | sin z | = cosh y ≤ cosh 1. But 2 | z2 | ≥ 2. Hence | sin z | < | 2 iz2 | on the boundary of the rectangle, and there are two zeros inside. Note that one zero is at 0. 31. A Möbius transformation that transforms the real axis to the real axis can be represented with real coefficients a, b, c, d. We assume that none of the coefficients is zero. Let w =

az + b . Note that the cz + d

inverse image of 0 is real, so b/a is real; the image of ∞ is real, so d/c is real. Now we represent the transformation as w=

(a / b) z + 1 lz + 1 = (c / d ) z + 1 nz + 1

where l and n are real. *Let w = f (z)dz where f is a rational function. Write t = 1/z. The residue of w at ∞ is defined to be the residue of – t2 f (1/t) at t = 0.

267

SOLUTIONS TO SOME SELECTED EXERCISES

32. Find the Möbius transformation which maps the unit disc of the z-plane into the unit disc of the w-plane.

az + b . cz + d Observe that w = 0 and w = ∞ correspond to symmetric points α and 1 / α where | α | < 1. Then b = – aα, c = – αd. It follows that Let

w=

F a I (z − α) . H d K (αz − 1)

w= − But when | z | = 1,

Hence

(z − α) = 1 and therefore (αz − 1) a − = eiθ (θ real) d

w = e iθ

z−α , | α | < 1, θ real. αz − 1

33. Find the Möbius transformation which maps the upper half of the z-plane into the upper half of the w-plane. We have seen above that Möbius transformation that maps the real axis to the real axis can be represented with real coefficients. Conversely, a transformation with real coefficients maps the real axis to the real axis. Hence, the Möbius transformation in this case must be az + b w= where a, b, c, d are real. cz + d In order to make the upper half-plane correspond to the upper half-plane, we have to make z = i correspond to a point w with Im w > 0. Then

ai + b ac + bd − bci + adi = ci + d c2 + d 2 must have positive imaginary part, so ad – bc > 0. 34. Find the Möbius transformation which maps the right-hand half of the z-plane into the right-hand half of the w-plane. In order to transform the right-hand half plane to the right-hand half plane, we first transform the right-hand half plane to the upper-half ζ-plane by ζ = iz aζ + b Then w= cζ + d transforms the upper half plane to the upper half plane, and multiplication by – i carries this back to the right-hand half plane. Thus, aiz + b az − bi = , ad – bc > 0. w= −i ciz + d ciz + d Hence, the required Mobius transformation is az − bi . w=i cz − di

268

THE ELEMENTS OF COMPLEX ANALYSIS

35. Prove that if f is analytic and Re f < 0 in the disk | z | ≤ 1, then | f ′(0) | ≤ – 2 Re f (0). Solution: Let f (0) = A. Then w = f (z) maps the z disk | z | < 1 into the left-hand w-plane, and 0 to A. Now we map the half plane to the unit disk in a t-plane by A−w . t = g(w) = A+w Then g | f (z) | maps the unit disk into the unit disk and 0 to 0. By Schwarz’s lemma (see Appendix I),

i.e. But

d g f ( z) ≤ 1 dz | {g′ [f (0)] f ′(0)}z=0 | ≤ 1

− ( A + w) − ( A − w ) − ( A + A) = , ( A + w)2 ( A + w) 2 1 , A = f (0). g′(A) = − A+A g′(w) =

It follows that f ′( 0 ) ≤1 A+A

| f ′(0) | ≤ | A + A | = | 2 Re f (0) | = – 2 Re f (0). 36. Let f be analytic in | z | ≤ 1. Suppose that 0 < | f (z) | < 1 in | z | ≤ 1 . Then prove that | f ′(0) | ≤ 2/e Solution: Since f (z) ≠ 0 in | z | ≤ 1, there is an analytic branch of ψ(z) = log f (z) in | z | ≤ 1. Then Re ψ(z) < 0 because | f (z) | < 1. It follows from Exercise that | ψ′(0) | ≤ – 2 log | f (0) |, | f ′( 0 ) | ≤ – 2 log | f (0) |, | f (0) | | f ′(0) | ≤ – 2 | f (0) | log | f (0) |. The maximum of – t log t for 0 < t < 1 is 1/e at t = 1/e. Hence 2 | f ′(0) | ≤ e 37. Find the image of – π/2 + kπ ≤ Re z ≤ π/2 + kπ (k = Constant) under the mapping ω = sin z Solution: Write the mapping ω = sin z into three mappings:

e iz − e – iz f ogoh 2 1 1 g− . where h = iz, g = eh and f = 2i 8 Since h = iz is a rotation by π/2, the image of the strip ST1 = {z : – π/2 + kπ ≤ Re z ≤ π/2 + kπ} ω = sin z =

FG H

IJ K

269

SOLUTIONS TO SOME SELECTED EXERCISES

under the map h = iz is again a strip ST2 = {z : – π/2 + kπ ≤ Im z ≤ π/2 + kπ}. The image of the strip ST2 under the map g = eh is the half plane H ~ {0}, where H = {z : – π/2 + kπ ≤ arg z ≤ π/2 + kπ} Now it remains to find the image of H ~ {0} by f. Note that if k is an even integer, H = {z : – π/2 ≤ arg z ≤ π/2} and if k is an odd integer, H = {z : – π/2 ≤ arg z ≤ 3π/2} Setting f = u + iv and z = reiθ, we have

F I H K F 1 I sin θ , u= r+ H rK 2 f (z) =

Hence

F I H K F 1I cos θ v= r– H rK 2

1 1 i 1 z− re iθ − r − iθ =– 2i z r 2

It follows that u2 v2 − = 1. sin 2 θ cos 2 θ i.e. the image of the ray {reiθ; θ is constant} is the hyperbola whose axes are sin θ and cos θ.

F 2u I + F GH r + JK GH 2

Also,

1 r

2v 1 r −r

I JK

2

= 1,

i.e. the image of a circle | z | = r is the ellipse with axes

F H

I F K H

I K

1 1 1 1 r+ , −r . 2 r 2 r

When k is an even integer, v is positive for 0 < r < 1 but v is negative for r > 1 ; u is positive for 0 ≤ θ ≤ π/2 and negative for – π/2 ≤ θ < 0. Thus, when k is an even integer, the image of H ∩ D′ (0, 1) under the map f is the upper half plane Im z > 0 and the image of H – D′ (0, 1) is the lower half plane Im z ≤ 0. Hence, the image of the strip ST1 with k even under the map w = sin z is the entire complex plane C. 1+ z is an isomorphism of disc D(0, 1) with the strip 1− z – π/2 < Im z < π/2.

38. Prove that log

Solution: z1 = 1 – z is an isomorphism D(0, 1) → D(1, 1);

1 is an isomorphism z1 1 D(1, 1) → Re (z2) > ; 2 z2 =

270

THE ELEMENTS OF COMPLEX ANALYSIS

z3 = – 1 + 2z2 is an isomorphism 1 Re (z2) > → Re (z3) > 0; 2 Write z4 = log z3 = log | z3 | + i arg z3 is an isomorphism Re (z3) > 0 → – π/2 < Im (z4) < π/2 Hence

FG H F 2 IJ = log FG 1 + z IJ = log G1 + H 1 − zK H1 − zK

z4 = log z3 = log (1 + 2z2) = log 1 +

2 z1

IJ K

is an isomorphism D(0, 1) → {z : – π/2 < Im z < π/2} 39. Find the image of the given domain D under the following conformal mappings 1 (i) w = z + ; D : | z | < 1, Im z > 0 z 1 − cos z ; D : 0 < Re z < π /2 (ii) w = 1 + cos z

Solution: (i) Setting w = u + iv and z = reiθ, we find

F H

u= r+

I K

F H

I K

1 1 cos θ and v = r − sin θ r r

1 < 0 implies r v < 0 with v taking all negative values. u ≥ 0 for 0 < θ ≤ π/2 and u ≤ 0 for π/2 ≤ θ < π, u taking all real values.

Now Im z > 0 iff 0 < θ < π and | z | < 1 iff r < 1. Thus sin θ > 0 and r –

Hence, w = z +

1 maps the region (| z | < 1) ∩ (Im z > 0) onto the lower half-plane z

Im z < 0.

F GH

1 − cos z e iz − 1 (ii) Write the mapping w = = – iz 1 + cos z e +1

I JK

2

as the composition of mappings :

z1 − 1 and z3 = – z22 = w z1 + 1 arg z1 = arg eiz = arg eix–y = x = Re z.

z1 = eiz, z2 =

Thus 0 < Re z < π/2 implies 0 < arg z1 < π/2. It can be easily checked that Re z > 0 is mapped onto | z2 | < 1 by z1 − 1 z2 = z1 + 1

271

SOLUTIONS TO SOME SELECTED EXERCISES

and it is an isomorphism of 0 < arg z1 < π/2 onto the region (| z2 | < 1) ∩ (Im z2 > 0) or (| z2 |) < 1) ∩ (0 < arg z2 < π) Since | z3 | = | z2 |2 and arg z3 = π + 2 arg z2 it follows that 0 < arg z2 < π iff π < arg z3 < 3π and | z2 | < 1 iff | z3 | < 1 Finally, we find that the mapping 1 − cos z maps 0 < Re z < π/2 onto w= 1 + cos z (| w | < 1) ∩ (– π < arg w < π) 40. Let z1, z2, z1 ≠ z2, be the two fixed points of a Möbius transformation T. Show that T may be written as: w − z1 z − z1 =k w − z2 z − z2 where k is a complex constant. Show that the cross ratio of z1, z2, z, w is constant. Solution: Let z1 and z2 be the fixed points. Then z1 = T(z1) and z2 = T(z2). Let T1 = Hence

z − z1 . T maps z1 → 0 and z2 → ∞ and z − z2 1 T1 o T o T1–1 maps 0 → 0 and ∞ → ∞. T1 o T o T1–1 is given by

T1 o T o T1–1(u) = ku It follows that T = T1–1 o ku o T1 w − z1 z − z1 i.e. =k defines T. w − z2 z − z2 We will consider the particular case of the second part. We will determine z1, z2 and k for the transformation 5z − 2 w= z+3 Its fixed points are given by 5z − 2 . z= z+3 i.e. z=1±i Hence

w − (1 + i ) = w − (1 − i )

5z −2 z +3 5z − 2 z +3

− (1 + i ) − (1 − i )

4−i z− . = 4+i z− Thus

k=

15 − 8i 17

5 + 3i 4−i 5 − 3i 4+i

272

THE ELEMENTS OF COMPLEX ANALYSIS

41. Find a transformation which maps the semi-infinite strip onto the upper half plane.

B

A

y a2 = p/2 –1

a1 = p/2

1

x

Solution: Here α1 = π/2 = α2 as shown in the figure here. We choose x1 = 1 and x2 = – 1 and apply the Schwarz-Christoffel transformation. Thus we have dw = K (z + 1)(π/2)/π–1 (z – 1)(π/2)/π–1 dz = K (z + 1)–1/2 (z – 1)–1/2 Hence

w= K

z

dz ( z 2 − 1)

= K cosh–1 z + D

where D is a constant. To determine K and D, we have z = 1 for w = 0, this gives D = 0 and z = – 1 for w = π i, this gives K = 1. Thus w = cosh–1 z. 42. Let f be analytic in | z | ≤ 1. Suppose that | f (z) | is maximized for | z | ≤ 1 at z0 with | z0 | = 1. Prove that f ′(z0) ≠ 0 unless f is constant. Solution: Suppose that f ′(z0) = 0. Then in a neighbourhood U : | z – z0 | < ε of z0, (ε is small) f is analytic in U. Now f can be represented by Taylor series with centre at z0. We have, for f ≠ 0,

LM N

∞

OP Q

n−m f (z) – f (z0) = (z – z0)m am + Σ an ( z − z 0 ) (am ≠ 0, m ≥ 2). n = m +1

When ε is small, the part of U inside | z | = 1 is nearly half a disk, and arg z varies by nearly π on the part E of ∂U which is inside | z | = 1. It follows that arg [f (z) – f (z0)] varies by nearly kπ on E(m ≥ 2), and this makes w = f (z) go outside the disk | w | ≤ f (z0) |. Hence, | f | takes values greater than | f (z0) | at some points z with | z | < 1. This contradicts the hypothesis that | f | has its maximum at z0 for | z | ≤ 1. 43. Suppose that U and V are conjugate harmonic functions and let V ( x, y) φ (x) = U ( x, y) Prove that φ (x) = tan (αx + β) where α and β are real numbers. Also, determine U(x, y) and V(x, y).

273

SOLUTIONS TO SOME SELECTED EXERCISES

Solution: Since U and V are conjugate harmonic functions, hence f (z) = U (x, y) + iV (x, y) is analytic and

V ( x, y) = tan (arg f (z)) U ( x, y) V i.e. arg f (z) = arc tan U = arc tan φ (x) Now arg f (z) = Im log f (z) ∂ [Im log f ( z )] = 0 Hence ∂y Since log f (z) is differentiable when f (z) ≠ 0, it follows by C – R equations that ∂ ∂ Re log f ( z ) = Im log f ( z ) = 0 ∂x ∂y Therefore, Re log f (z) = g (y) ∂ ∂ g( y) = − Im log f ( z ) and g′(y) = ∂y ∂x ∂ arc tan φ ( x ) =− ∂x ∂ Since g′(y) depends only on y, and – arc tan φ(x) only on x, we find that ∂x ∂ g′(y) = – arc tan φ (x) = – α (a real constant). ∂x Thus arc tan φ (x) = αx + β i.e. tan (αx + β) = φ (x) It remains to determine U and V. Since g(y) = – αy + δ and g(y) = Re log f (z) = log | f (z) | we have | f (z) | = eg(y) = e– αy+δ We know that f (z) = | f (z) | cis arg f (z) Hence f (z) = e–αy + δ cis (αx + β) = eδ cis βe–αy cis αx Thus U(x, y) = Ce–αy cos αx and V (x, y) = Ce–αy sin αx where C = eδ cis β 44. Find the conjugate function of u = x2 – y2 Solution: ∂u ∂u = 2 x, = − 2 y. Hence a conjugate The function u = x2 – y2 is harmonic. Also, ∂x ∂y of u will satisfy

F I H K

274

THE ELEMENTS OF COMPLEX ANALYSIS

∂v ∂v = 2 x, = 2y ∂y ∂x Integrating the first equation of (3) with respect to y we have (4) v = 2xy + k (x) where k (x) depends only on x. Substituting (4) in the last equation of (3) we obtain k′(x) = 0 and so k (x) = constant = C. Hence, the conjugate function of x2 – y2 is 2xy + C and the analytic function f (z) is given by f (z) = (x2 – y2) + i (2xy + C) = z2 + iC. −y x . 45. Prove that the harmonic conjugate of 2 is 2 2 x + y2 x +y (3)

FI HK

1 x 2 = Re z x +y 1 We have f (z) = z is analytic in C except at the origin i.e. z = 0.

Solution:

2

1 is harmonic in C except at z = 0. But the harmonic z x 1 is Im . That is, the harmonic conjugate of 2 is x + y2 z

Threfore the real part of conjugate of Re

F 1I H zK

FI HK

−y . x + y2 2

46. Prove that under the transformation w = f (z) is harmonic function φ (x, y) remains harmonic where f (z) is analytic and f ′(z) ≠ 0. Solution: Let φ (x, y) be transformed into a function φ [x (u, v), y(u, v)] by the transformation. We will first prove that

F GH

2 2 ∂2φ ∂2 φ + 2 = | f ′(z) |2 ∂ φ + ∂ φ 2 2 ∂x ∂y ∂u ∂v 2

We have

I JK

∂φ ∂φ ∂u ∂φ ∂v + . = ∂u ∂x ∂v ∂x ∂x

∂φ ∂φ ∂u ∂φ ∂v + . = ∂y ∂u ∂y ∂v ∂y

FG IJ H K

FG IJ H K

∂ 2 φ ∂φ ∂ 2 u ∂u ∂ ∂φ ∂φ ∂ 2 v ∂v ∂ ∂φ + + + = ∂u ∂x 2 ∂x ∂x ∂u ∂ v ∂x 2 ∂x ∂x ∂v ∂x 2

=

LM N

OP Q ∂φ ∂ v ∂v L ∂ φ ∂u ∂ φ ∂v O + + + P M ∂v ∂x ∂x N ∂u ∂v ∂x ∂v ∂x Q

∂φ ∂ 2 u ∂u ∂ 2 φ ∂u ∂ 2 φ ∂v + + 2 2 ∂u ∂x ∂x ∂u ∂x ∂v ∂u ∂x 2

2

2

2

2

275

SOLUTIONS TO SOME SELECTED EXERCISES

LM N

∂ 2 φ ∂φ ∂ 2 u ∂u ∂ 2 φ ∂u ∂ 2 φ ∂v + + = ∂u ∂y 2 ∂y ∂u 2 ∂y ∂v ∂u ∂y ∂y 2

Similarly

+ Adding, (5)

F GH

I JK

LM N

OP Q

∂φ ∂ 2 v ∂v ∂ 2 φ ∂u ∂ 2 φ ∂v + + ∂v ∂y 2 ∂y ∂u ∂v ∂y ∂v 2 ∂y

OP Q

F I GH JK F ∂u I O ∂ φ LM ∂u ∂v + ∂u ∂v OP +G J P + 2 H ∂y K PQ ∂u ∂v N ∂x ∂x ∂y ∂y Q F ∂v I O +G J P H ∂y K PQ

∂φ ∂ 2 v ∂ 2 v ∂ 2 φ ∂ 2 φ ∂φ ∂ 2 u ∂ 2 u + 2 + + + 2 = 2 2 ∂u ∂x ∂v ∂x 2 ∂y 2 ∂y ∂x ∂y

LMF ∂u I MNGH ∂x JK ∂ φ LF ∂v I MG J + ∂v MNH ∂x K +

∂2φ ∂u 2

2

2

2

2

2

2

Now, by C – R equations

2

∂u ∂v ∂v ∂u , =− . Also, since u and v are harmonic, = ∂y ∂x ∂x ∂y

∂2u ∂2u ∂2v ∂2v + 2 = 0, 2 + 2 = 0 2 ∂x ∂y ∂x ∂y

FG ∂u IJ + FG ∂u IJ = FG ∂v IJ + FG ∂v IJ = FG ∂u IJ + FG ∂v IJ H ∂x K H ∂y K H ∂x K H ∂y K H ∂x K H ∂x K 2

Then

2

=

∂u ∂v +i ∂x ∂x

∂u ∂v ∂u ∂v = 0. + ∂x ∂x ∂y ∂y Hence (5) reduces to (6)

2

2

2

2

= | f ′(z) |2

F GH

2 2 ∂2φ ∂2 φ + 2 = | f ′(z) |2 ∂ φ + ∂ φ 2 ∂x ∂y ∂u 2 ∂v 2

It follows from (6) that if

2

I JK

∂2φ ∂2 φ = 0 and f ′(z) ≠ 0 then + ∂x 2 ∂v 2

∂2φ ∂2φ + =0 ∂u 2 ∂v 2 47. Determine a function which is harmonic in the upper half plane, Im (z) > 0 and which takes the prescribed values

F(x) =

RS1 x > 0 T0 x < 0

Solution: This is a Dirichlet problem for the upper half plane as shown in the figure

276

THE ELEMENTS OF COMPLEX ANALYSIS

y (x, y) r θ ϕ=0

ϕ=1

x

In other words, we have to solve for φ (x, y) the boundary-value problem

∂2φ ∂2 φ + = 0, y > 0 ; ∂x 2 ∂y 2

RS T

1 x>0 lim φ ( x, y) = F(x) = 0 x>0 Since Aθ + B is the imaginary part of A log z + B where A and B are real constants, hence Aθ + B is harmonic. y→ 0 +

We have to determine A and B under the conditions ϕ = 1 for x > 0 ϕ = 0 for x < 0 Thus (i) 1 = A(0) + B, (ii) 0 = A(π) + B

F I H K

y 1 tan–1 x π One can also solve this type of problem by using Poisson’s formula for the half plane.

Hence

A = – 1/π, B = 1 and φ = 1 –

48. Let ψ(θ) = 0 for 0 < θ < π and 1 for π < θ < 2π. Find a series for the function u(r, θ) which is harmonic in the unit disk with these boundary values. Using the series compute u (1/2, π/2) numerically. Solution: 1 0 − in θ 1 e dθ, a0 = an = − π 2π 2 For n ≠ 0,

z

1 [1 − ( − 1) n ] 2 π ni 1 1 1 in θ e Σ ψ(θ) = − 2 πi odd n =1 n 1 1 1 in θ Σ (e − e − in θ ) = − odd n = 1 n 2 πi 1 2 sin n θ Σ = − 2 π odd n n an = −

277

SOLUTIONS TO SOME SELECTED EXERCISES

u (r, θ) =

Hence, u

1 2 ∞ sin (2 m + 1) θ 2 m +1 r − Σ 2 π m = 0 (2 m + 1)

F 1 , πI = 1 − 2 H 2 2K 2 π

∞

Σ

m=0

( −1) m 2 2 m +1 (2 m + 1)

1 2 − tan −1 (1 / 2) 2 π = 0.2048... .

=

49. Prove that if | z |
max (1, | z |), then gm (z) – gm – 1 (z) = log m + z {log m – log (m – 1)} – log (z + m) =

z

1 0

z ( z + 1) t dt ( m − t ) ( m + tz )

Hence, if m > max (1, | z |) | gm (z) – gm – 1 (z) | ≤ | z (z + 1) | (m – 1)– 1 (m – | z |–1) It now follows that ∞

Σ { gm ( z ) − gm − 1 ( z )}

m=2

converges and hence lim gn (z) exists. Thus Γ (z) exists for all z ∈ Ω0. n→∞

Define the beta function by (13)

B (z, w) =

z

1 0

t z − 1 (1 − t ) w − 1 dt

55. Prove that B (z, w + 1) = B (z, w) ⋅ w/(z + w) for Re z > 0 and Re w > 0. Solution: (z + w) B (z, w + 1) – w B (z, w) = (z + w) =

z

1 0

z

1 0

t z − 1 (1 − t ) w dt − w

z

1 0

t z − t (1 − t ) w − 1 dt

{zt z − 1 (1 − t ) w − t z w(1 − t ) w − 1} dt

= [t z (1 − t ) w ]10 = 0

281

SOLUTIONS TO SOME SELECTED EXERCISES

56. Prove that Γn (z) = nz B (z, n + 1) for Re z > 0 and n is a natural number. Solution: B (z, n + 1) = = But

B (z, 1) =

n B (z, n) z+n n n −1 1 ... B (z, 1) z + n z + n −1 z +1

z

1 0

t z − 1 dt =

1 z

Hence, the result follows from (10). Observe that lim {n z B ( z, w + n + 1) = Γ ( z ).

(14)

n→∞

57. Prove that lim {nz B (z, w + n + 1)} = Γ (z) for Re z > 0

n→∞

and n is a natural number. Solution: For Re w ≥ 1, | n2 B (z, w + n + 1) – nz B (z, n + 1) |

z z

= nRe z | ≤ nRe z

1 0

1

0

t z − 1 {(1 − t ) w − 1} (1 − t ) n dt |

t Re z − 1 |(1 − t ) w − 1 | (1 − t ) n dt ,

| (1 – t)w – 1 | = |

and

z

t 0

w (1 − x ) w − 1 dx |

≤ t | w | (0 ≤ t ≤ 1) Hence, by (13) we have (15) | nz B (z, w + n + 1) – nz B (z, n + 1) | ≤ | w | nRe z B (Re z + 1, n + 1) It now follows from (14) that lim {nRe z + 1 B (Re z + 1, n + 1)} = Γ (Re z + 1)

n→∞

Hence, the right side of (15) tends to zero as n → ∞. This proves the result under the restriction Re w ≥ 1. But this restriction is not necessary as can be seen from the following argument. Consider z fixed for a moment Then lim {nn B (z, w + n + 1)} = g (w) (say) n→∞

Now whenever g (w) exists, g (w) = g (w – k) for any positive integer k. This shows that Re w ≥ 1 is not necessary. 58. Prove that B (z, w) = Γ (z) Γ (w)/Γ (z + w) for Re z > 0 and Re w > 0

282

THE ELEMENTS OF COMPLEX ANALYSIS

Solution: Let F (w) = B (z, w) Γ (z + w)/Γ(w) Considering z fixed, we have F (w) = F (w + 1) and F (w) = F (w + n + 1). It can be checked from the previous results that F (w) = lim F (w + n + 1) n→∞

= lim

n→∞

RSn T

z

z

u z − 1e − u du for Re z > 0

B ( z, w + n + 1)

Γ ( z + w + n + 1) n ! nw Γ ( w + n + 1) n ! nz + w

UV W

= Γ (z) 59. Prove that Γ (z) =

∞ 0

Solution: Since ex ≥ 1 + x and e–x ≥ 1 – x for every real number x, it follows that (16) 1 ≥ ex (1 – x) ≥ (1 + x) (1 – x) = 1 – x2 (x < 1). Now let n be a natural number. Then, by (16) if u ≤ n, then

F H

1 ≥ eu/n 1 − and hence (17)

1≥

F1 − u I ≥ FG1 − u IJ H nK H n K n

eu

2

I FG K H

u u2 ≥ 1− 2 n n

I JK

n

(0 ≤ u ≤ n)

2

Now (1 – x)n ≥ 1 – nx (x ≤ 1). Hence,

I K

F H

u n

F H

u n

1 ≥ eu 1 −

n

u2 n

≥1−

(0 ≤ u ≤ n),

so that 0 ≤ e–u – 1 −

(18)

I K

n

≤ e− u

Using (13), we have nz B (z, n + 1) = =

z z

1 0

u2 n

(nt ) z − 1 (1 − t ) n ndt

F H

n

0

(0 ≤ u ≤ n)

uz −1 1 −

u n

I K

n

du

It can be checked easily that

z

∞ 0

u z − 1e − u du exists if and only if Re z > 0

Thus it follows from (18) that |

z

∞ 0

u z − 1e − u du – nz B (z, n + 1) |

≤

z

n 0

|RS |T

F H

u Re z − 1 e − u − 1 −

u n

I K

n

|UV du + |W

z

∞ n

u Re z − 1e − u du

283

SOLUTIONS TO SOME SELECTED EXERCISES

z

≤ n–1

∞

u Re z + 1 e − u du

0

Hence, the result follows. 60. Let G = {z : Re z ≥ a} where a > 1. Prove that if ε > 0 then there is a number δ, 0 < δ < 1, such that for all z in G. (19)

|

z

β α

(e t − 1) − 1 tz – 1 dt | < ε

whenever δ > β > α. Solution: Since et – 1 ≥ t for all t ≥ 0 then for 0 < t ≤ 1 and z in G | (et – 1)– 1 tz – 1 | ≤ ta – 2 Since a > 1 the integral

z

t

t a − 2 dt is finite. Hence δ can be found to satisfy (19).

0

61. Let F = {z : Re z ≤ A} where – ∞ < A < ∞. Prove that if ∈ > 0 then there is a number k > 1 such that for all z in F

z

|

β α

(e t − 1) − 1 t z − 1dt | < ε

whenever β > α > k. Solution: If t ≥ 1 and z is any point in F then there is a constant C such that | (et – 1)–1 tz – 1 | ≤ (et – 1)– 1 tA – 1 ≤ Ce1/2t (et – 1)–1 1/2t t Since e (e – 1)– 1 is integrable on [1, ∞] the required number k can be found. Proceeding similarly we can obtain the following results. 62. Let H = {z : a ≤ Re z ≤ A} where 1 < a < A < ∞ then the integral

z

∞ 0

(e t − 1) − 1 t z − 1 dt

converges uniformly on H. 63. If E = {z : Re z ≤ A} where – ∞ < A < ∞ then the integral

z

∞

1

(e t − 1) − 1 t z − 1 dt

converges uniformly on E. 64. Prove that ζ (z) Γ (z) =

z

∞ 0

(e t − 1) − 1 t z − 1 dt for Re z > 1.

Solution: It follows from the above results that this integral is an analytic function in the region {z : Re z > 1}. Thus it is sufficient to prove that ζ (z) Γ (z) equals this integral for z = x > 1. It follows from exercises 60 and 61 above that there are numbers α and β, 0 < α < β < ∞, such that

z z

α 0

(e t − 1) − 1 t x − 1 dt < ε/4,

∞ β

(e t − 1) − 1 t x − 1 dt < ε/4

284

THE ELEMENTS OF COMPLEX ANALYSIS n

∞

k =1

k =1

Σ e − kt ≤ Σ e − kt = (et – 1)– 1

Since for all n ≥ 1,

z z

∞

Σ

n =1 ∞

Σ

n =1

α 0

e − nt t x − 1 dt < ε/4,

∞

β

e − nt t x − 1 dt < ε/4 ∞

ζ (z) Γ (z) = Σ n– z Γ (z)

Using the relation

n =1

z

∞

= Σ | ζ (x) Γ (x) –

we have

∞

≤ε+| Σ

n =1

z

z

n =1

∞ 0

0

e − nt t z − 1 dt

(e t − 1) −1 t x − 1 dt |

z

β α

∞

e − nt t x − 1 dt −

β α

(e t − 1) − 1 t x − 1 dt |

But Σ e – nt converges to (et – 1)–1 uniformly on [α, β], so the result is proved. By using the above result we would like to extend the domain of definition of ζ to {z : Re z > – 1}. Let us consider the Laurent expansion of the function ∞ 1 1 1 (20) = − + Σ an zn z e − 1 z 2 n =1 where a1, a2, ... are some constants. The function neighbourhood of t = 0. This implies that

z FGH 1

0

LM 1 − 1 OP is bounded in a Ne − 1 t Q t

IJ K

1 1 z–1 − t dt e −1 t t

converges uniformly on compact subsets of {z : Re z > 0} and hence represents an analytic function on the right half plane. Thus (21)

ζ (z) Γ (z) =

z

1 0

FG 1 − 1IJ t He −1 tK

z −1

t

dt +

1 + z −1

z

∞

1

tz −1 dt. et − 1

By using exercises 62 and 63 we see that each of these summands except

1 , is z −1

analytic in {z : Re z > 0}. Thus we may define ζ (z) for Re z > 0 in the following manner. ζ (z) =

1 Γ (z)

z

LM F 1 − 1I t MN GH e − 1 t JK t

0

t

z−t

dt +

1 + z −1

z

∞

1

tz −1 dt et − 1

OP PQ

285

SOLUTIONS TO SOME SELECTED EXERCISES

Hence ζ is meromorphic in the right half plane with a simple pole at z = 1 whose residue is 1. Let 0 < Re z < 1, then

1 =− z −1 Inserting this in (21) we obtain

z

F 1 He

∞

z

∞

1

t z − 2 dt

I K

1 z −1 t dt, 0 < Re z < 1. 0 t In fact, our aim is to establish Riemann’s Functional Equation.

(22)

ζ (z) Γ (z) =

t −1

−

65. ζ (z) = 2 (2π) z–1 Γ (1 – z) ζ (1 – z) sin

F 1 πzI for – 1 < Re z < 0. H2 K

Considering again the Laurent expansion of

LM 1 − 1 + 1 OP ≤ Ct N (e − 1) t 2 Q

1 ez − 1

we find that

t

for some constant C and all t in [0, 1]. Thus

z FH

1

1

t −1

−

I K

1 1 z −1 + t dt t 2

0 e is uniformly convergent on compact subsets of {z : Re z > – 1}. Also, since

F 1 − 1I = 1 H e −1 t K

lim t

t

t →∞

there is a constant C1 such that

F 1 − 1I ≤ C , t ≥ 1. H e −1 t K t 1

t

This implies that

z FGH ∞

1

IJ K

1 1 z −1 t dt − e –1 t t

converges uniformly on compact subsets of {z : Re z < 1}. Using these results we find that (23)

ζ(z) Γ(z) =

z FH 1

0

I K

1 1 1 z −1 1 − + t dt − 2z e −1 t 2 t

z FH ∞

I K

1 1 z −1 − t dt for 0 < Re z < 1. 1 e −1 t Since both integrals on the right side of (23) converge in the strip – 1 < Re z < 1, we can define ζ(z) in {z : – 1 < Re z < 1}.

+

t

The question naturally arises: What happens at z = 0 ? Since 1/2 z appears on the right side of (23), is it true that ζ have a pole at z = 0 ? It can be easily checked that this is not true. Dividing both sides of (23) by Γ(z) we will get one term 1 1 = 2z Γ( z ) 2 Γ( z + 1) which is analytic at z = 0.

286

THE ELEMENTS OF COMPLEX ANALYSIS

Thus if ζ is so defined in the strip {z : – 1 < Re z < 1} it is analytic there. Combining this with (21) we find that ζ(z) is defined for Re z > – 1 with a simple pole at z = 1. Now if – 1 < Re z < 0 then

z

1 = z Inserting this in (23) we obtain

–

(24)

z FH

∞

t z −1 dt

1

I K 1 1 1 F e + 1I i F 1 itI + = G = cot J H2 K 2 2 2 H e − 1K e −1 F 1 itI = 2 − 4 it Σ 1 for t ≠ 0. cot H 2 K it t + 4n π F 1 − 1 + 1I 1 = 2 Σ 1 H e −1 t 2K t t + 4n π

ζ(z) Γ(z) =

∞

0

1 1 1 z −1 t dt, − 1 < Re z < 0 − + e −1 t 2 t

t

We have

t

t

∞

n =1 2

2

2

∞

Thus

n =1 2

t

Inserting this in (24) we get ζ (z) Γ (z) = 2

(25)

∞

=2Σ

n =1

z

∞

z

∞

FG Σ H t

2

IJ K

∞

1 t z dt + 4n 2 π 2

n =1 2

0

2

tz dt t + 4n 2 π 2 2

0

∞

= 2 Σ (2 πn) z −1 n =1

z

∞ 0

= 2 (2π)z–1 ζ(1 – z)

tz dt , t +1 2

z

∞ 0

tz dt, t2 + 1

valid for – 1 < Re z < 0. Substituting s = t2 for the real x with – 1 < x < 0 we have (26)

z

∞ 0

1 tz dt = 2 2 t +1

z

∞ s1/ 2 ( x −1) 0

s +1

ds

LM N

=

1 1 π cosec π (1 − x ) 2 2

=

1 1 π sec πx 2 2

F I H K

OP Q

But (27)

1 Γ (1 − x ) sin πx = Γ( x ) π =

F I F I H K H K

Γ (1 − x ) 1 1 2 sin πx cos πx π 2 2

287

SOLUTIONS TO SOME SELECTED EXERCISES

It follows from (24), (26) and (27) that (28)

ζ(z) = 2(2π) z – 1 Γ (1 – z) ζ (1 – z) sin

F 1 πzI H2 K

valid for – 1 < Re z < 0. In fact, we established Riemann’s functional equation for real x with – 1 < x < 0. But since both sides of (28) are analytic on the strip – 1 < Re z < 0, (28) follows. Following similarly we can show that (28) holds for – 1 < Re z < 1. 66. If Re z > 1 then ∞

ζ (z) = Π

n =1

F 1 I GH 1 − p JK −z n

where {pn} is the sequence of prime numbers. Solution: Using the geometric series we find that (29)

∞ 1 = Σ pn− mz −z m=0 1 − pn

for all n ≥ 1. Taking the product of the terms using (29) we obtain (30)

n

Π

F 1 I for 1 ≤ k ≤ n where n ≥ 1 and GH 1 − p JK

F 1 I= Σ n GH 1 − p JK

k =1

∞

−z n

j =1

−z k

−z j

Note that the integers n1, n2, n3, ... are all the integers which can be factored as a product of powers of the prime numbers p1, p2, ..., pn alone. Allowing n → ∞ we get the required result. Weierstrass Elliptic Function P (z) We recall the following definitions. An analytic function f is called periodic if there exists a (complex) constant w ≠ 0 such that f (z + w) = f (z) for all z ∈ C. The constant w is called a period of the function. If there does not exist an integer n such that w/n is also a period, then we call w a fundamental period. If Im ( w1 w2 ) = 0, where w1 ≠ w2, then f is called simply periodic. Note that w1 and w2 are distinct periods. If Im ( w1 w2 ) ≠ 0, where w1 ≠ w2, then f is called doubly periodic. The doubly periodic analytic functions are called elliptic functions. Observe that if h is any analytic function for which the series ∞

Σ h (z – nw) = f (z)

−∞

converges uniformly, then f will be periodic with period w. If w is a sum of any integer multiples of w1 and w2 with Im ( w1 w2 ) ≠ 0, then

Σ h (z – w) = p (z) w

Note that the summation over all possible w yields p (z), a doubly periodic function

288

THE ELEMENTS OF COMPLEX ANALYSIS

Note also that h (z) + Σ [h (z – w) – h (w)] = p (z) w≠0

is doubly periodic. Setting h (z) = 1/z2 in the above expression we obtain the Weierstrass elliptic function

RS T

1 1 1 + Σ − 2 2 2 w ≠ 0 ( z − w) z w

UV = P (z). W

67. Show that P (z) satisfies the differential equations (31) 6 P2 – P″ = 30 k4 and (32) 4 P ″ – P ′2 – 60 k4 = 140 k6 where P ′ and P ″ are the first and second derivatives of P and 1 1 , k6 = Σ 4 w ≠ 0 w w6 Solution: By definition

(33)

k4 = Σ

w≠0

RS T

1 1 1 + Σ − z 2 w ≠ 0 ( z − w)2 w 2

P (z) =

1 ( z − w)3

– P ′ (z) = 2 Σ w

RS 1 + Tz

=2

3

Σ

w≠0

1 ( z − w)3

UV W

UV W

and (34)

P ′ (z) = 6 Σ w

1 (z − w) 4

RS 1 Tz

=6

4

+ Σ

w≠0

1 (z − w)4

UV W

where the summation Σ includes the origin (w = 0). w

Note that P has a double pole at the origin and P ″ has a fourth order pole at the origin. Thus the fourth-order pole of P2 at the origin can be eliminated by subtracting P ″/6 from P 2. We find 1 2 (35) P 2 = 4 + 2 Σ 2 + Σ 22 z z and for all small z (36) and

R|S F I F T| H K H LM 1 − 1 OP N ( z − w) w Q

1 1 z z +3 = 2 1+ 2 2 w w ( z − w) w Σ2 = Σ

w≠0

= 2z Σ

w≠0

2

I K

2

1 1 + 3z 2 Σ 4 w≠0 w w2

2

U|V W|

289

SOLUTIONS TO SOME SELECTED EXERCISES

Since (37) (38)

Σ

w≠0

1 = 0 for m = odd integer wm

lim Σ 2 = 0

z→0

lim

z→0

1 Σ 2 = 3k4 z2

Σ 2 = 3k4 z2 Hence, it follows from (34) and (35) that 12 (39) 6 P 2 – P ″ = 2 Σ 2 + 6 Σ 22 − 6 Σ 4 z where 1 (40) Σ4 = Σ w ≠ 0 ( z − w) 4 Since 6 P 2 – P ″ is an elliptic function it is in fact a constant if we could show the existence of lim [6 P 2 – P ″]. z→0

It follows from (37), (38) and (40) that the right side of (39) becomes 2 6 P0 − p″ 0 = 36 k4 – 6 k4 = 30 k4

as z → 0. Also since z = 0 is the only singularity of 6 P 2 – P ″ we find that 6 P 2 – P ″ = 36 k4 for all z In order to prove (32) we need to multiply (31) by 2P ′ and after integrating we get (41) 4P 3 – P ′2 – 60 k4 P = constant This constant can be determined by taking the limit of the left side of (41) as z → 0 Writing – P ′ (z) = 2 we have (42)

P ′2 = 4

RS 1 + Tz 3

Σ

w≠0

RS 1 + 2 Σ Tz z 6

3

UV RS W T

1 1 = 2 3 + Σ3 3 ( z − w) z

3

+ Σ 32

UV W

1 + Σ 2 , we have z2 1 3 3 (43) 4 P 3 = 4 6 + 4 Σ 2 + 2 Σ 22 + Σ 32 z z z Thus (41) becomes

Since

UV W

P (z) =

RS T

UV W

60 k 12 8 12 Σ 2 − 3 Σ 3 − 2 4 + 2 Σ 22 + 4 Σ 32 − 4 Σ 32 − 60 k 4 Σ 2 4 z z z z The last four terms of the above expression approach zero as z → 0. The first two terms in the above expression can be evaluated as follows:

290

THE ELEMENTS OF COMPLEX ANALYSIS

For small z

|RS T|

F I F I H K H K

1 1 z z +3 ≅ 2 1+ 2 2 w w (z − w) w

Hence

Σ2 = Σ

(44)

w≠0

Also, since and

Σ

w≠0

LM 1 N ( z − w)

2

−

2

+4

F zI H wK

3

+5

F z I |UV H w K W| 4

OP Q

1 ≅ 3 k4 z2 + 5 k6 z4 w2

1 1 = Σ =0 w3 w ≠ 0 w5

|RS |T

F I F I H K H K

1 1 z z +6 ≅ − 3 1+ 3 3 w w (z − w ) w

2

+ 10

F zI H wK

3

|UV |W

Hence 1 ≅ 3k4 z + 10k6 z3 ( z − w)3

– Σ3 = − Σ

(45)

w≠0

Thus lim

z→0

RS 36k Tz 2

4

+ 60 k6 +

24 k 4 60 k + 80 k6 − 2 4 2 z z

UV = 140k W

6

It now follows that 4 P 3 – P ′2 – 60k4 P = 140 k6 Elliptic Integrals: The integration of the above equation gives (46)

z

dP

P ∞

3

{4 P − 60 k 4 P − 140 k6 }

=z

We have already seen that P is a function of z. Hence, this equation gives z as a function of P. In other words this is the inverse function to the elliptic function P (z). This inverse function is called an elliptic integral. We denote this by w dw (47) z = P– 1 (z) = 3 ∞ {4 w − 60 k w − 140 k } 4 6

z

Note that such an integral cannot be expressed in closed form in terms of algebraic, trigonometric, inverse trigonometric, logarithmic, and exponential functions unless the cubic has repeated roots. We give below three types of elliptic integral. The integral w du (48) z= , |k|