A Complete Solution Guide to Complex Analysis 9789887415503, 9789887415510, 9887415510

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A Complete Solution Guide to Complex Analysis

by Kit-Wing Yu, PhD [email protected]

c 2020 by Kit-Wing Yu. All rights reserved. No part of this publication may be Copyright reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the author. ISBN: 978-988-74155-0-3 (eBook) ISBN: 978-988-74155-1-0 (Paperback)

ii

About the author

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

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iv

Preface

There are many books entitled Complex Analysis. For examples, Ahlfors [1], Bak and Newman [4], Freitag and Rolf [9], Gamelin [10], Lang [14], and Stein & Shakarchi [24]. (I believe that you can find more if you search your library website.) As an introductory textbook for complex analysis, I would like to recommend the book by Bak and Newman [4]. The reasons are that this book gives readers some important, insightful and interesting background to build up the theory of complex analysis. Furthermore, their presentation is quite clear and concise, so I believe that you can grasp the main concepts and skills in an easier way. Finally, the book contains a lot of helpful examples and problems. There are a total of 300 exercises in the third edition [4], but only 225 of them are provided by solutions. (The exercises without solutions are marked an asterisk.) In my opinion, some solutions are a bit “brief” and I guess that some students still have difficulties when reading them. To provide assistance, I decide to write a solution manual for this book and I hope that students / instructors can benefit from this solution book. Before you read this book, I have a gentle reminder for you. As a mathematics instructor at a college, I understand that the growth of a mathematics student depends largely on how hard he/she does exercises. When your instructor asks you to do some exercises of Bak and Newman’s book, you are not suggested to read my solutions unless you have tried your best to prove them yourselves. The features of this book are as follows: • It covers all the 300 exercises with detailed and complete solutions. As a matter of fact, my solutions show every detail, every step and every theorem that I applied. • There are 34 illustrations for explaining the mathematical concepts or ideas used behind the questions or theorems. • Different colors are used in order to highlight or explain problems, lemmas, remarks, main points/formulas involved, or show the steps of manipulation in some complicated proofs. (ebook only) • Necessary lemmas with proofs are provided. • Useful or relevant references are provided to some questions for interested readers. Finally, if you find such any typos or mistakes, please send your valuable comments or opinions to [email protected] v

vi so that I will post the updated errata on my website https://sites.google.com/view/yukitwing/ from time to time.

Kit-Wing Yu January 2020

List of Figures

1.1

The graph of z1 z2 and z1 z2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

1.2

The locations of the roots. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8

1.3

The product of the 4 diagonals of a regular 5-gon. . . . . . . . . . . . . . . . . .

9

1.4

The graph of lemniscate of Bernoulli. . . . . . . . . . . . . . . . . . . . . . . . . .

10

1.5

The graph of S in the plane C. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17

1.6

S is a circle when T is a circle. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

19

1.7

S is a circle minus (0, 0, 1) when T is a line. . . . . . . . . . . . . . . . . . . . . . P The effect of f (z) = 1z on . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

19

1.8

21

The function |f (z(t))| has a local maximum at z = 1 on the line 1 + t exp( π2 ). . .

76

.

77

The function |f (z(t))| has

.

77

7.1

The graph of the set S. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

84

8.1

The geometry of the circle C and the annulus B. . . . . . . . . . . . . . . . . . . 102

8.2

The “inside” of a simple closed polygonal path. . . . . . . . . . . . . . . . . . . . 105

8.3

A decomposition of a closed polygonal path. . . . . . . . . . . . . . . . . . . . . . 106

8.4

The circular arc C connecting −|z| and z clockwise. . . . . . . . . . . . . . . . . 107

6.1 6.2 6.3

The function |f (z(t))| has

a local minimum at z = 1 on the line 1 + t exp( 33π 18 ). an inflection point at z = 1 on the line 1 + t exp( 3π 4 ).

10.1 The regular closed curve C surrounding ±1. . . . . . . . . . . . . . . . . . . . . . 135 π 4. . . . . . . . square CN with vertices ±(N + 21 ) ± i(N + 21 ). rectangle CN with vertices (N + 21 ) ± i(N + 12 )

11.1 The contour CR with central angle

. . . . . . . . . . . . . . . . . 148

11.2 The

. . . . . . . . . . . . . . . . . 152

11.3 The

and −(N − 21 ) ± i(N + 12 ). . 153

11.4 The rectangle ΓR with vertices at ±R and ±R + 2πi. . . . . . . . . . . . . . . . . 154

12.1 The formations of the contours C(0; R) and C(0; 2). . . . . . . . . . . . . . . . . 164 12.2 The closed contour formed by γ1 , γ2 , γ3 and γ4 . . . . . . . . . . . . . . . . . . . . 165 13.1 The sector S and the disc D(z0 ; δ). . . . . . . . . . . . . . . . . . . . . . . . . . . 173 13.2 The rectangles R and R′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 13.3 The angles of ∠γ1 , γ2 and ∠f (γ1 ), f (γ2 ). . . . . . . . . . . . . . . . . . . . . . . . 177 13.4 The conformal mapping of the region between the two circles and the annulus. . 183 13.5 The conformal mapping of H onto a rectangle. . . . . . . . . . . . . . . . . . . . 184 vii

List of Figures

viii

14.1 The shape of a Norman window N . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 16.1 The level curves of u(x, y) = k for k ∈ (0, 21 ) ∪ ( 12 , 1). . . . . . . . . . . . . . . . . 205 16.2 The temperature problem with the prescribed boundary values. . . . . . . . . . . 206 16.3 The angle Arg (ω 2 − 1). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 19.1 The intersections of y = x and y = tan x. 19.2 The intersections of y =

sin2 x x2

. . . . . . . . . . . . . . . . . . . . . . 227

and y = tan x − x. . . . . . . . . . . . . . . . . . . 229

Contents

Preface

v

List of Figures

viii

1 The Complex Numbers

1

2 Functions of the Complex Variable z

23

3 Analytic Functions

37

4 Line Integrals and Entire Functions

47

5 Properties of Entire Functions

55

6 Properties of Analytic Functions

67

7 Further Properties of Analytic Functions

83

8 Simply Connected Domains

101

9 Isolated Singularities of an Analytic Function

109

10 The Residue Theorem

121

11 Applications of the Residue Theorem to the Evaluation ...

139

12 Further Contour Integral Techniques

163

13 Introduction to Conformal Mapping

173

14 The Riemann Mapping Theorem

187

15 Maximum-Modulus Theorems for Unbounded Domains

197

16 Harmonic Functions

201

17 Different Forms of Analytic Functions

211

18 Analytic Continuation; The Gamma and Zeta Functions

221

ix

Contents 19 Applications to Other Areas of Mathematics

227

Index

237

Bibliography

239

CHAPTER

1

The Complex Numbers

Problem 1.1 Bak and Newman Chapter 1 Exercise 1.

Proof. (a) It is clear that 1 6 − 2i 6 − 2i 3 1 1 = × = = − i. 6 + 2i 6 + 2i 6 − 2i 36 + 4 20 20 (b) It is easy to see that (2 + i)(3 + 2i) 4 + 7i 4 + 7i 1 + i −3 + 11i 3 11 = = × = = − + i. 1−i 1−i 1−i 1+i 2 2 2 √

(c) Since − 21 + 23 i = cis 2π 3 , we have √    1 √3  4 6π + 2π 2π 2π  2π  1 3 = cis = cis i = cis 4 × = cis 2π + =− + i. − + 2 2 3 3 3 3 2 2 (d) We have

 i,    −1, in = −i,    1,

if n = 4m + 1 for some m ∈ Z; if n = 4m + 2 for some m ∈ Z; n = 4m + 3 for some m ∈ Z; if n = 4m for some m ∈ Z.

We have completed the proof of the problem.



Problem 1.2 Bak and Newman Chapter 1 Exercise 2.

Proof. Suppose that (x + iy)2 = −8 + 6i. By the discussion on [4, p. 3], we see that s s √ √ −8 + 64 + 36 8 + 64 + 36 = ±1 and y = ± · sgn (6) = ±3. x=± 2 2 √ Hence the two values of −8 + 6i are ±(1 + 3i), completing the proof of the problem. 1



Chapter 1. The Complex Numbers

2

Problem 1.3 Bak and Newman Chapter 1 Exercise 3.

Proof. We have q√ √ √ √ − 32i ± ( 32i)2 − 4 × (1) × (−6i) √ √ − 32i ± −32 + 24i = = −2 2i ± −8 + 6i. z= 2 2 By Problem 1.2, we see immediately that √ √ √ z = −2 2i ± (1 + 3i) = 1 + (3 − 2 2)i or − 1 − (3 + 2 2)i. This completes the proof of the problem.



Problem 1.4 Bak and Newman Chapter 1 Exercise 4.

Proof. Suppose that z1 = a + bi and z2 = c + di, where a, b, c and d are real. Furthermore, we suppose that P (z) = a0 z n + a1 z n−1 + · · · + an−1 z + an , where a0 , a1 , . . . , an ∈ R and n ∈ N ∪ {0}. (a) Then z1 + z2 = (a + c) + (b + d)i = (a + c) − (b + d)i = a − bi + c − di = z1 + z2 . (b) We have z1 z2 = (a + c)(b + d)i = ac − bd + (ad + bc)i = ac − bd − (ad + bc)i = z1 · z2 . (c) Since a0 , a1 , . . . , an ∈ R, we have ak = ak for all 0 ≤ k ≤ n. By repeated applications of parts (a) and (b), we have P (z) =

n X k=0

ak z n−k =

n X

ak z n−k =

k=0

n X k=0

ak · z n−k =

n X k=0

ak · z n−k = P (z).

(d) If z = A + Bi, then we have z = A − Bi = A + Bi = z. We complete the proof of the problem.



Problem 1.5 Bak and Newman Chapter 1 Exercise 5.

Proof. Since P (z) = 0 if and only if P (z) = 0, it follows from Problem 1.4(c) that P (z) = 0 if and only if P (z) = 0. This completes the proof of the problem.  Problem 1.6 Bak and Newman Chapter 1 Exercise 6.

Proof. Using rectangular coordinates, let z = a + bi. Then z 2 = (a2 − b2 ) + 2abi so that p |z 2 | = (a2 − b2 )2 + 4a2 b2 = a2 + b2 = |z|2 .

Using polar coordinates, if z = rcis θ, then we have z 2 = r 2 cis 2θ so that |z 2 | = r 2 = |z|2 . We end the analysis of the problem. 

3 Problem 1.7 Bak and Newman Chapter 1 Exercise 7.

Proof. Suppose that z = rcis θ. (a) For every n ∈ Z, we have z n = r n cis nθ so that |z n | = r n = |z|n . (b) Since z = rcis (−θ), we have z · z = r 2 cis (θ − θ) = r 2 = |z|2 .

√ (c) Let z = a + bi. Then we have Re z = a, Im z = b and |z| = a2 + b2 . Since a2 ≤ a2 + b2 and b2 ≤ a2 + b2 , we get |Re z| ≤ |z| and |Im z| ≤ |z|. The fact a2 + 2|a| · |b| + b2 ≥ a2 + b2 implies that (|a| + |b|)2 ≥ a2 + b2 which reduces to exactly |Re z| + |Im z| ≥ |z|. Now we study when an equality occurs. For example, |Re z| = |z| if and only if |a|2 = a2 +b2 if and only if b = 0. The case for |Im z| = |z| is similar. Next, |z| = |Re z| + |Im z| if and only if a2 + b2 = |a|2 + 2|a| · |b| + |b|2 if and only if |a| · |b| = 0 if and only if |a| = 0

or

|b| = 0.

In conclusion, an equality occurs if and only if Re z = 0 or Im z = 0. This completes the proof of the problem.



Problem 1.8 Bak and Newman Chapter 1 Exercise 8.

Proof. (a) By Problem 1.7(b) and then Problem 1.4(a), we have |z1 + z2 |2 = (z1 + z2 )(z1 + z2 ) = (z1 + z2 )(z1 + z2 ).

(1.1)

By expanding the right-hand side of the equation (1.1) and then use Problems 1.7(a) and (c), we get |z1 + z2 |2 = |z1 |2 + |z2 |2 + 2Re (z1 z2 ) 2

2

≤ |z1 | + |z2 | + 2|z1 z2 |

(1.2) (1.3)

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

which gives the triangle inequality. (b) Note that we apply Problem 1.7(c) to get the inequality (1.3), so the equality (1.3) occurs if and only if Im (z1 z2 ) = 0 if and only if z1 z2 ∈ R. (c) Replace z1 by z1 − z2 in the triangle inequality, we know that |z1 − z2 + z2 | ≤ |z1 − z2 | + |z2 | which is equivalent to |z1 | − |z2 | ≤ |z1 − z2 |. This completes the proof of the problem.



Chapter 1. The Complex Numbers

4

Problem 1.9 Bak and Newman Chapter 1 Exercise 9.∗

Proof. (a) Let z1 = a + bi and z2 = c + di, where a, b, c, d ∈ Z. Thus z1 z2 = ac − bd + (ad + bc)i, where ac − bd, ad + bc ∈ Z. Since |z1 z2 | = |z1 | · |z2 |, we have |z1 z2 |2 = |z1 |2 · |z2 |2 so that (a2 + b2 )(c2 + d2 ) = (ac − bd)2 + (ad + bc)2 .

(1.4)

If we define u = ac − bd and v = ad + bc, then we get the desired result. (b) Since all a, b, c and d are nonzero, without loss of generality, we may assume they are all positive. Besides the formula (1.4), we also note thata (a2 + b2 )(c2 + d2 ) = (ac + bd)2 + (ad − bc)2 .

(1.5)

Now ad + bc > 0 and ac + bd > 0, we follow from the two formulas (1.4) and (1.5) that it suffices to check that at least one of ad − bc and ac − bd is nonzero. Assume that ad = bc

and ac = bd.

(1.6)

Then the product of the same sides of the equations (1.6) gives a2 cd = b2 cd which implies a2 = b2 . Similarly, the product of the opposite sides of the equations (1.6) gives abd2 = abc2 so that c2 = d2 . This is a contradiction to the hypotheses and hence we have ad − bc 6= 0 or ac − bd 6= 0. (c) We label u = ac − bd,

v = ad + bc,

s = ac + bd and

t = ad − bc.

Direct computation gives u2 − s2 = (u − s)(u + s) = (−2bd)(2ca) = −4abcd. Since a, b, c and d are nonzero, we have u2 6= s2 . Similarly, we can show that u2 − t2 = (u − t)(u + t) = (a2 − b2 )(c2 − d2 ). By the hypotheses, we know that u2 6= t2 . Furthermore, we also have v 2 6= s2 and v 2 6= t2 . Otherwise, u2 + v 2 = s2 + t2 will imply either u2 = s2 or u2 = t2 , a contradiction. Consequently, the sets {(ac − bd)2 , (ad + bc)2 } and {(ac + bd)2 , (ad − bc)2 } are distinct. (d) The assumption a, b, c, d 6= 0 means that the vectors from 0 to the complex numbers z1 and z2 are not parallel to the real axis or the imaginary axes. If we write z1 = r1 cis θ1

and z2 = r2 cis θ2 ,

(1.7)

then both θ1 and θ2 are not a multiple of π2 . Next, notice that a2 6= b2 if and only if a 6= ±b. Therefore, the vector from 0 to z1 is not parallel to the lines y = ±x. In other words, θ1 is not a multiple of π4 . Similarly, c2 6= d2 if and only if θ2 is not a multiple of π4 . a

This comes from the fact that |z1 z2 | = |z1 | · |z2 |.

5 – Geometric interpretation of part (b). We obtain directly from the representations (1.7) that z1 z2 = r1 r2 cis (θ1 + θ2 )

and z1 z2 = r1 r2 cis (θ1 − θ2 ).

Assume that both the vectors from 0 to z1 z2 and z1 z2 were parallel to the real axis or the imaginary axis. Then the previous analysis shows that both θ1 ± θ2 are multiples of π2 , i.e., nπ mπ and θ1 − θ2 = θ1 + θ2 = 2 2 but they imply that (m + n)π θ1 = 4 which is a contradiction. Thus at least one of the vectors from 0 to z1 z2 and z1 z2 is not parallel to the axes. Consequently, we have cis (θ1 + θ2 ) 6= 0 or

cis (θ1 − θ2 ) 6= 0.

(1.8)

Finally, we recall from the equations (1.4) and (1.5) that ac − bd and ad − bc are exactly the real and imaginary parts of z1 z2 or z1 z2 respectively, so the results (1.8) ensure that both u and v are nonzero. – Geometric interpretation of part (c). To show that {u2 , v 2 } and {s2 , t2 } are distinct, it is equivalent to showing that (|u|, |v|) 6= (|s|, |t|)

and (|u|, |v|) 6= (|t|, |s|).

(1.9)

In Figure 1.1, we see easily that (|u|, |v|) = (|s|, |t|) if and only if the difference between the arguments of z1 z2 and z1 z2 is a multiple of π.

Figure 1.1: The graph of z1 z2 and z1 z2 .

Chapter 1. The Complex Numbers

6

Assume that Arg (z1 z2 ) = Arg (z1 z2 ) + mπ for some m ∈ N. Then we have 2θ2 = (θ1 + θ2 ) − (θ1 − θ2 ) = Arg (z1 z2 ) − Arg (z1 z2 ) = mπ

so that θ2 = mπ 2 which is a contradiction by the description at the beginning of part (d). Next, it is also true that (|u|, |v|) = (|t|, |s|) if and only if the difference between the arguments of z1 z2 and z1 z2 is a multiple of π2 . In this case, we have Arg (z1 z2 ) = Arg (z1 z2 ) + nπ 2 for some n ∈ Z and then similar argument can show that nπ , 2θ1 = (θ1 + θ2 ) − (θ2 − θ1 ) = Arg (z1 z2 ) − Arg (z1 z2 ) = 2 a contradiction to the fact that θ1 6= nπ 4 . Hence we obtain the results (1.9) which mean that {u2 , v 2 } and {s2 , t2 } are different. This completes the proof of the problem.



Problem 1.10 Bak and Newman Chapter 1 Exercise 10.∗

Proof. We replace z2 by −z2 in the equation (1.2) to obtain

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

(1.10)

Then the sum of the equations (1.2) and (1.10) imply that

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

Geometrically, the sum of the squares of the lengths of the diagonals (left-hand side) is double to the sum of the squares of the lengths of the sides (right-hand side).b This completes the  proof of the problem. Problem 1.11 Bak and Newman Chapter 1 Exercise 11.

Proof. Suppose that z = x + iy 6= 0. By the definition, we notice that π y π − ≤ tan−1 ≤ . 2 x 2 However, recall that Arg z is the angle which the vector (originating from 0) to z makes with the positive x-axis (see the figure on [4, p. 6]), so it takes values [−π, π] (modulo 2π). Thus their connection can be expressed as  y  tan−1 , if x > 0;   x       y   tan−1 + π if x < 0 and y ≥ 0;   x      y tan−1 − π, if x < 0 and y < 0; Arg z = Arg (x + iy) =  x      π   , if x = 0 and y > 0;    2        −π, if x = 0 and y < 0. 2 b

See also [26, Exercise 1.17, p. 9].

7 This completes the proof of the problem.



Problem 1.12 Bak and Newman Chapter 1 Exercise 12.

Proof. Let z = rcis θ. Here we mainly follow the steps in the example on [4, pp. 8, 9].c (a) Since r 6 cis 6θ = 1cis 0, we have r = 1 and 6θ = 0 (modulo 2π). Then we have 6θ = 0, 2π, 4π, 6π, 8π, 10π. Hence the six solutions are given by z1 = 1,

z2 = cis

π , 3

z3 = cis

2π , 3

z4 = cis π = −1,

z5 = cis

4π 3

and z6 = cis

5π . 3

(b) Since r 4 cis 4θ = 1cis π, we have r = 1 and 4θ = π (modulo 2π). Then we obtain 4θ = π, 3π, 5π, 7π. Hence the four solutions are given by z1 = cis

π , 4

z2 = cis

(c) Since r 4 cis 4θ = 2cis 2π 3 , we have r =

3π , 4 √ 4

4θ =

5π 4

z3 = cis

2 and 4θ =

2π 3

and

z4 = cis

7π . 4

(modulo 2π). Then we have

2π 8π 14π 20π , , , . 3 3 3 3

Hence the four solutions are given by z1 =

√ 4

2cis

π , 6

z2 =

√ 2π 4 2cis , 3

z3 =

√ 4

2cis

7π 6

and z4 =

√ 4

2cis

5π . 3

The locations of the roots in parts (a) to (c) have been shown in Figure 1.2 below. This ends  the proof of the problem. Problem 1.13 Bak and Newman Chapter 1 Exercise 13.

Proof. Let ω be an nth root of unity other than 1. Since 0 = ω n − 1 = (ω − 1)(ω n−1 + ω n−2 + · · · + 1), we conclude that ω satisfies the equation z n + z n−1 + · · · + 1 = 0, completing the proof of the problem. c

In fact, there is a general formula of finding the nth roots of the equation z n = r, see [1, p. 16].



Chapter 1. The Complex Numbers

8

(a) The locations of the roots in part (a).

(b) The locations of the roots in part (b).

(c) The locations of the roots in part (c).

Figure 1.2: The locations of the roots.

Problem 1.14 Bak and Newman Chapter 1 Exercise 14.

Proof. By the comments on [4, p. 9] or [14, pp. 13 - 16], the n-th roots of 1 are located at the vertices of the regular n-gon inscribed in the unit circle. Let z1 , z2 , . . . , zn be the n-th roots of zn = 1 with z1 = 1. Without loss of generality, we may assume that the vertices all be connected to z1 . (For example, see Figure 1.3 for the regular 5-gon.) Then the lengths of the diagonals are exactly |z1 − z2 |, |z1 − z3 |, . . . , |z1 − zn |. By Problem 1.13, we know that z n−1 + z n−2 + · · · + 1 = (z − z2 )(z − z3 ) · · · (z − zn )

9 which certainly implies that |1 − z2 | × |1 − z3 | × · · · × |1 − zn | = 1n−1 + 1n−2 + · · · + 1 = n.

Figure 1.3: The product of the 4 diagonals of a regular 5-gon. This ends the proof of the problem.



Problem 1.15 Bak and Newman Chapter 1 Exercise 15.

Proof. In the following, we suppose that z = x + iy. (a) This is a closed disc centred at i with radius 1. By Definition 1.6, it is not a region. (b) The equation becomes |(x − 1) + iy| = |(x + 1) + iy|. By the definition of the modulus and after squaring, we get (x − 1)2 + y 2 = (x + 1)2 + y 2 . (1.11) Thus we must have x = 0, i.e., the set of solutions of the equation is exactly the imaginary axis. Since squaring an equation may create new solutions of the original equation, we have to check that every solution of the equation (1.11) is also a solution of the original equation.d In fact, if z = iy, then it is easy to see that z − 1 | − 1 + iy| = 1. = z+1 |1 + iy|

√ 2 For √ example, if x = 2 − x, then x + x − 2 = 0 whose solutions are x = 1 or x = −2. However, note that −2 = 2 + 2 = 2 which is impossible, so x = −2 is not a solution of the original solution. d

Chapter 1. The Complex Numbers

10

In other words, all solutions of the equation (1.11) are also solutions of the original equation. By Definition 1.6 again, it is not a region. (c) We have (x − 2)2 + y 2 > (x − 3)2 + y 2 which gives x > 25 . Similar to part (b), if x > 25 , then z = x + iy satisfies the original inequality. Hence the solution set of the inequality is Re z > 25 . By Definition 1.6, it is a region. (d) Note that |z| < 1 is the open unit disc centred at 0 and Im z > 0 is the upper half plane, so the solution set is the upper half of the open unit disc centred at 0. By Definition 1.6, it is a region. (e) The equation is equivalent to z · z = 1 which is x2 + y 2 = 1. This is a unit circle centred at 0, so it is not a region by Definition 1.6. (f) We have x2 + y 2 = y which is x2 + (y − 21 )2 = 212 . Hence it is a circle centred at radius is 21 . Similar to part (e), it is not a region.

i 2

and the

(g) By the definition, we have |(x2 − y 2 − 1) + 2xyi| < 1 so that p (x2 − y 2 − 1)2 + 4x2 y 2 < 1 which is equivalent to

(x2 − y 2 )2 − 2(x2 − y 2 ) + 1 + 4x2 y 2 < 1. After simplification, we obtain (x2 + y 2 )2 − 2(x2 − y 2 ) < 0 which is known as a lemniscate of Bernoulli [15, pp. 121 - 123], see Figure 1.4, which is generated by “Desmos” (https://www.desmos.com/), below:

Figure 1.4: The graph of lemniscate of Bernoulli. We notice that this set is open but not connected. By Definition 1.6, it is not a region.

11 We have completed the proof of the problem.



Problem 1.16 Bak and Newman Chapter 1 Exercise 16.∗

Proof. (a) Let z = x + iy. Then the equation becomes p x2 + y 2 = x + 1

(1.12)

which, by squaring both sides, reduces to

y 2 = 2x + 1.

(1.13)

Clearly, this is a parabola which opens rightward. Similar to Problem 1.15(b), we have to check that every solution of the equation (1.13) is also a solution of the equation (1.12). To see this, let (x, y) satisfy the equation (1.13). Since y 2 ≥ 0, we have x ≥ − 12 so that x + 1 > 0. Next, we substitute the equation (1.13) into the left-hand side of the equation (1.12) to get p p p x2 + y 2 = x2 + 2x + 1 = (x + 1)2 = |x + 1| = x + 1. This shows that the solutions of the equations (1.12) and (1.13) are identical.

(b) Geometrically, it says that the sum of the distances from z to 1 and from z to −1 is always 4, so the locus of the points is the ellipse with foci (±1, 0) and the semi-major axis 2, i.e., x2 y 2 + 2 = 1, 22 b where b is the semi-minor axis. By the definition, we have b2 = 22 − 12 = 3 so that the set of points satisfying the given equation is x2 y 2 + = 1. 4 3 (c) Notice that n ∈ Z. There are several cases for consideration. – Case (i): n = 0. The equation becomes

1 z

= z which is Problem 1.15(e).

– Case (ii): n = 1. In this case, z = 1. If z = x + iy, then it implies that x − iy = 1 so that x = 1 and y = 0. – Case (iii): n = 2. Now we have z = z. If z = x + iy, then x + iy = x − iy so that y = 0. In other words, the set of solutions in this case is the real axis. – Case (iv): n ≥ 3. Obviously, z = 0 is a solution of the equation. Suppose that z 6= 0. Then |z n−1 | = |z| implies that |z| = 1. Next, z n−1 = z also gives z n = z · z = |z|2 = 1, i.e., z is an n-th root of unity. Hence the set of solutions in this case consists of the n-th roots of unity and z = 0.

Chapter 1. The Complex Numbers

12

– Case (v): n ≤ −1. In this case, z 6= 0. Similar to Case (iv), |z n−1 | = |z| implies that |z| = 1 and z n−1 = z gives z n = 1, (1.14) where n ≤ −1. If we rewrite the equation (1.14) as z −|n| = 1, then we get z |n| = 1 so that z is an |n|-th root of unity. This completes the proof of the problem.



Problem 1.17 Bak and Newman Chapter 1 Exercise 17.

Proof. Let z = x + iy, where x2 + y 2 = 1. Then it is easy to check that z−1 x − 1 + iy = z+1 x + 1 + iy x − 1 + iy x + 1 − iy · = x + 1 + iy x + 1 − iy x2 + y 2 − 1 2y = +i 2 2 (x + 1) + y (x + 1)2 + y 2 2y . =i (x + 1)2 + y 2 z−1 Note that if y > 0, then z−1 z+1 lies on the positive imaginary axis; if y < 0, then z+1 lies on the negative imaginary axis. By the definition of Arg w given in the problem, we obtain immediately that  π  , if Im z > 0;  2 z − 1  Arg =  z+1   − π , if Im z < 0. 2 We have completed the proof of the problem. 

Problem 1.18 Bak and Newman Chapter 1 Exercise 18.∗

Proof. By [4, Eqn. (4), p. 10], the solutions of√x3 − 6x = 4 are the three real-valued possibilities √ √ √ 3 3 3 3 2 + 2i + √2 − 2i. p Suppose that √ a = 2 + 2i and b = 2 − 2i. Then we have a2 = for p 3 3 3 2 2 2 (2 + 2i) = 2 i, b = (2 − 2i) = 2 3 −i and ab = 2. We note that √ √ √ 3 3 i i 3 i= + , − + , −i 2 2 2 2 and

Thus we have

√ 3

√

i 3 −i = − , 2 2

√ a2 − ab + b2 = 2 3 − 2,

√

i 3 − , 2 2

i.

√ −2 3 − 2

or

−

− 2.

13 By the hint, we conclude that the three real-valued possibilities for by x =a+b=

2 a 3 + b3 =√ , 2 2 a − ab + b 3−1

2 − 3−1 √

or

√ 3

−2=1+

2 + 2i +

√

3,

√ 3

1−

2 − 2i are given

√

completing the proof of the problem.

3 or

− 2, 

Remark 1.1 By direct substitution, we see that x = −2 is a root of the equation x3 − 6x − 4 = 0. Therefore, we have x3 − 6x − 4 = (x + 2)(x2 − 2x − 2) √ so that the other real roots of the equation x3 − 6x − 4 = 0 are exactly 1 ± 3. Problem 1.19 Bak and Newman Chapter 1 Exercise 19.∗

Proof. Suppose that α, β, γ are the roots of the equation x3 + px − q = 0. Then we have

(1.15)

  α + β + γ = 0, αβ + βγ + γα = p,  αβγ = q.

(1.16)

√ If γ = 0, then q = 0 and thus the equation x3 + px = 0 has three roots 0, ± −p. Now the roots are real if and only if p < 0 if and only if 4p3 + 27q 2 = 4p3 < 0. Therefore, we may suppose that all the roots of the equation (1.15) are nonzero. Thus it follows from the equations (1.16) that (α − β)2 = α2 + β 2 + γ 2 − γ 2 −

2αβγ γ

= (α + β + γ)2 − 2(αβ + βγ + γα) − γ 2 − = −2p − γ 2 −

2αβγ γ

2q . γ

(1.17)

Similarly, we also have (β − γ)2 = −2p − α2 −

2q α

and (γ − α)2 = −2p − β 2 −

2q . β

(1.18)

Suppose that (α − β)2 , (β − γ)2 and (γ − α)2 are roots of the new equation y 3 + Ay 2 + By + C = 0, where A, B and C will be determined very soon. By the formulas (1.17) and (1.18), it yields that 2q y = −2p − x2 − x which is equivalent to xy = −2px − x3 − 2q

Chapter 1. The Complex Numbers

14

(x3 + px − q) + xy = −px − 3q xy + px = −3q 3q x=− . y+p

Substituting the expression (1.19) into the original equation (1.15), we arrive at   3q  3q 3 −q =0 +p − − y+p y+p −27q 3 − 3pq(y + p)2 − q(y + p)3 = 0 −qy 3 − 6qpy 2 − 9qp2 y − 27q 3 − 4qp3 = 0.

(1.19)

(1.20)

Recall that q 6= 0, so the equation (1.20) reduces to

y 3 + 6py 2 + 9p2 y + (27q 2 + 4p3 ) = 0

so that the product of its roots is given by (α − β)2 (β − γ)2 (γ − α)2 = −(4p3 + 27q 2 ).

(1.21)

Hence the formula (1.21) implies the following results: • Case (i): The equation (1.15) has three distinct real roots if and only if 4p3 + 27q 2 < 0. • Case (ii): The equation (1.15) has a repeated (real) root if and only if 4p3 + 27q 2 = 0. • Case (iii): The equation (1.15) has two complex roots if and only if 4p3 + 27q 2 > 0. Hence we complete the proof of the problem.



Remark 1.2 (a) By Case (ii) in the proof of Problem 1.19, it is believed that “three real roots” should be replaced by “three distinct real roots”. (b) If we assume that the equation (1.15) has a real root, then the algebra will become much simpler. In fact, let α be a real zero of the function f (x) = x3 + px − q. Then we can write f (x) = (x − α)(x2 + ax + b), where a, b ∈ R. Direct expansion gives f (x) = x3 + (a − α)x2 + (b − aα)x − bα, so we have α = a and then p = b − a2 and q = ba. Now 4p3 + 27q 2 < 0 if and only if

4(b − a2 )3 + 27b2 a2 < 0

4b3 − 12b2 a2 + 12ba4 − 4a6 + 27b2 a2 < 0

4b3 + 15b2 a2 + 12ba4 − 4a6 < 0

(−a2 b2 − 4ba4 − 4a6 ) + (4b3 + 16a2 b2 + 16ba4 ) < 0 −a2 (2a2 + b)2 + 4b(b + 2a2 )2 < 0

(b + 2a2 )2 (4b − a2 ) < 0

if and only if a2 − 4b > 0 which is equivalent to saying that the equation x2 + ax + b = 0 has two (distinct) real roots. Hence, we have completed the proof of the problem.

15 Problem 1.20 Bak and Newman Chapter 1 Exercise 20.∗

Proof. (a) It is clear that (1 − z)P (z) = 1 + 2z + 3z 2 + · · · + nz n−1 − z − 2z 2 − · · · − (n − 1)z n−1 − nz n = 1 + z + z 2 + · · · + z n−1 − nz n .

(1.22)

Assume that z0 was a zero of P (z) with |z0 | = r > 1. Then the expression (1.22) implies that 1 + z0 + z02 + · · · + z0n−1 = nz0n and then nr n ≤ 1 + r + · · · + r n−1 .

(1.23)

However, since r > 1, it is easy to see that r k − 1 > 0 for every k ∈ N so that nr n − (1 + r + · · · + r n−1 ) = (r n − 1) + (r n − r) + · · · + (r n − r n−1 )

= (r n − 1) + r(r n−1 − 1) + · · · + r n−1 (r − 1)

>0

which contradicts the inequality (1.23). Hence no such z0 exists and all the zeros of P (z) lie in |z| ≤ 1. (b) Let P (z) = a0 + a1 z + · · · + an z n . If a0 = a1 = · · · = an = 0, then P (z) ≡ 0 which is meaningless. Therefore, we may assume that ak > 0 for some k ∈ {0, 1, 2, . . . , n}. We consider (1 − z)P (z) = a0 + a1 z + · · · + an z n − a0 z − a1 z 2 − · · · − an−1 z n − an z n+1

= a0 + (a1 − a0 )z + (a2 − a1 )z 2 + · · · + (an − an−1 )z n − an z n+1 .

(1.24)

Assume that w was a zero of P (z) with |w| = r > 1. On the one hand, the expression (1.24) implies that a0 + (a1 − a0 )w + (a2 − a1 )w2 + · · · + (an − an−1 )wn = an wn+1 so that an r n+1 ≤ a0 + (a1 − a0 )r + (a2 − a1 )r 2 + · · · + (an − an−1 )r n .

(1.25)

On the other hand, we know that r k − 1 > 0 for every k ∈ N and since ak > 0 for some k, we see that an r n+1 − [a0 + (a1 − a0 )r + (a2 − a1 )r 2 + · · · + (an − an−1 )r n ]

= an r n (r − 1) + an−1 r n−1 (r − 1) + · · · + a1 r(r − 1) + a0 (r − 1)

= (r − 1)(an r n + an−1 r n−1 + · · · + a1 r + a0 )

> 0,

but this contradicts the inequality (1.25). Hence no such w exists and all the zeros of P (z) lie in |z| ≤ 1.  Consequently, we complete the proof of the problem.

Chapter 1. The Complex Numbers

16

Remark 1.3 Problem 1.20(b) is the well-known Enestr¨om and Kakeya Theorem. See, for examples, [17, pp. 136, 137] and [18].

Problem 1.21 Bak and Newman Chapter 1 Exercise 21.

Proof. (a) Let z0 be a fixed point such that |z0 | < 1. Then there exists a r > 0 such that |z0 | < r < 1. It is clear that |kz k | ≤ kr k for every |z| ≤ r and k = 0, 1, 2, . . .. Let Mk = kr k . Since √ √ k k α = lim sup kr k = lim kr = r < 1, k→∞

k→∞

the Root Test [27, Theorem 6.7, p. 76] ensures that the series orem 1.9 (The Weierstrass M -test), the series

∞ X

∞ X

kr k converges. By The-

k=0

kz k converges uniformly to a continuous

k=0

function f (z) on D(0; r). Since z0 is arbitrary, we conclude that f (z) is continuous on D(0; 1). (b) Let z = x + iy with x > 0. Obviously, we know that 1 1 1 ≤ 2. = p 2 2 2 2 k +z k (k + x) + y

∞ X 1 Since converges, Theorem 1.9 (The Weierstrass M -test) guarantees that the series k2 k=1

∞ X k=1

1 k2 + z

converges uniformly to a continuous function g(z) in the right half-plane Re z > 0. We complete the proof of the problem.



Problem 1.22 Bak and Newman Chapter 1 Exercise 22.

Proof. Suppose that S is a polygonally connected set. Assume that S was disconnected. Then there are disjoint open sets U and V of C such that S ⊆ U ∪ V,

S ∩ U 6= ∅

and S ∩ V 6= ∅.

(1.26)

Select a ∈ S ∩ U and b ∈ S ∩ V . Since U and V are disjoint, a 6= b. Since S is polygonally connected, there exists a polygonally line L : [0, 1] → S connecting a and b, i.e., L(0) = a and L(1) = b. By the definition, we have L = [a, z1 ] ∪ [z1 , z2 ] ∪ · · · ∪ [zn , b]

17 for some n ∈ N. Since each line segment is continuous, L is continuous and then L([0, 1]) is connected by [27, Theorem 7.12, p. 100]. However, the set relations (1.26) imply that L([0, 1]) ⊆ U ∩ V,

L([0, 1]) ∩ U 6= ∅ and

L([0, 1]) ∩ V 6= ∅.

In other words, L([0, 1]) is disconnected, a contradiction. Hence we end the proof of the problem.  Problem 1.23 Bak and Newman Chapter 1 Exercise 23.

Proof. First of all, the graph of S is shown in Figure 1.5.

Figure 1.5: The graph of S in the plane C. Define T = {x + iy | x > 0 and y = sin x1 } and I = {iy | y ∈ R}. Then we have S = I ∪ T . We assume that S was disconnected, i.e., there are disjoint open sets U and V such that S ⊆ U ∪ V, S ∩ U 6= ∅ and S ∩ V 6= ∅. (1.27) Since (0, 0) ∈ S, we may suppose that (0, 0) ∈ U so that I ∩ U 6= ∅. Since I is obviously polygonally connected, it is connected by Problem 1.22. If I ∩ V 6= ∅, then the definition shows that I is disconnected, a contradiction. Hence we know that I ⊆ U. Next, the openness of U implies that there exists a δ > 0 such that {z = a + ib | |z| < δ} ⊆ U.

(1.28)

Chapter 1. The Complex Numbers

18

1 We note that the points xn = 2nπ with n ∈ N satisfy sin x1n = sin 2nπ = 0, so we can select N large enough such that the points zn = xn + i sin x1n satisfy

|zn | =

r

 2 1 (xn − 0)2 + sin − 0 = xn < δ xn

for all n ≥ N .e In other words, it means that zn ∈ U for all n ≥ N or equivalently, T ∩ U 6= ∅. Recall that sin x1 is continuous on R \ {0}, so the function f : (0, ∞) → C defined by f (x) = x + i sin

1 x

is continuous on (0, ∞) and its image f ((0, ∞)) is connected by [27, Theorem 7.12, p. 100]. Since T = f ((0, ∞)), T is also connected and the argument in the previous paragraph shows that T ⊆ U. (1.29) Combining the set relations (1.28) and (1.29), we conclude that S ∩ V = ∅ which contradicts  the assumption (1.27). Hence S is connected which completes the proof of the problem. Remark 1.4 In Problem 1.23, the union of the set T with its limit point (0, 0) is called the Topologist’s sine curve, see [19, Example 7, pp. 156, 157]. Furthermore, it is obvious that the point 2i cannot be connected by any curve in S because | sin x1 | ≤ 1. Problem 1.24 Bak and Newman Chapter 1 Exercise 24.

Proof. Recall from [4, Eqn. (3), p. 17] that ξ=

x , 2 x + y2 + 1

η=

y 2 x + y2 + 1

so that x2 + y 2 = Since ζ ≥ ζ0 , we have

1 1−ζ

≥

1 1−ζ0 .

and ζ =

ζ . 1−ζ

(1.30)

Hence it follows from the equation (1.30) that x2 + y 2 ≥

ζ0 , 1 − ζ0

i.e., T is the exterior of the circle centred at 0 with radius problem.

ζ0 1−ζ0 .

Problem 1.25 Bak and Newman Chapter 1 Exercise 25.

Proof. e

x2 + y 2 x2 + y 2 + 1

In fact, this argument also shows that (0, 0) is a limit point of S.

This completes the proof of the 

19 (a) Suppose that T is a circle on C. Then ζ 6= 1 and T has the form x2 + y 2 + Dx + Ey + F = 0.

(1.31)

Using the expression (1.30) and [4, Eqn. (2), p. 17], the equation (1.31) reduces to ζ Dξ Eη + + +F =0 1−ζ 1−ζ 1−ζ Dξ + Eη + (1 − F )ζ = −F. (1.32) P P Recall that a circle on is the intersection of with a plane Aξ + Bη + Cζ = G, where A, B, C, G ∈ R. As the equation (1.32) is in this form, the corresponding set S is a circle P on that doesn’t contain (0, 0, 1) as required. See Figure 1.6 for an illustration.

Figure 1.6: S is a circle when T is a circle. (b) If T is a line in the form ax + by + c = 0 for some a, b, c ∈ R, then we obtain

Thus S is again a circle on

P

aξ bη + +c=0 1−ζ 1−ζ aξ + bη − cζ = −c. that minus the pole (0, 0, 1), see Figure 1.7.

Figure 1.7: S is a circle minus (0, 0, 1) when T is a line. This completes the proof of the problem.



Chapter 1. The Complex Numbers

20

Problem 1.26 Bak and Newman Chapter 1 Exercise 26.

Proof. Let P (z) = an z n + an−1 z n−1 + · · · + a1 z + a0 , where an 6= 0. Then we have  an−1 a0  a0 an−1 |P (z)| = z n an + + · · · + n = |z|n × an + + · · · + n . (1.33) z z z z By [1, Eqn. (12), p.10], we know that |z1 | − |z2 | ≤ |z1 − z2 | holds for every complex z1 and z2 . Apply this result to the right-hand side of the expression (1.33), we deduce that a a0 n−1 n (1.34) + · · · + n . |P (z)| ≥ |z| × |an | − z z

Let M = max(|a0 |, |a1 |, . . . , |an−1 |). If |z| > 1, then it follows from the triangle inequality that a a0 |an−1 | |an−2 | |a0 | Mn n−1 + ··· + n ≤ + + ··· + n ≤ . 2 z z |z| |z| |z| |z|

≤ |a2n | and the inequality (1.34) becomes a  a0  |an | n−1 |P (z)| ≥ |z|n × |an | − + · · · + n ≥ |z|n × z z 2 which tends to ∞ as |z| → ∞. Consequently, we have P (z) → ∞ as z → ∞ which completes the proof of the problem.  n Therefore, if |z| ≥ max(1, 2M |an | ), then

Mn |z|

Problem 1.27 Bak and Newman Chapter 1 Exercise 27.

Proof. Let z = x + iy. Then we have

1 z

=

1 x+iy

=

x x2 +y 2

y − i x2 +y 2.

(a) Using [4, Eqn. (3), p. 17], we have ξ=

x , 2 x + y2 + 1

η=

y 2 x + y2 + 1

and ζ =

x2 + y 2 . x2 + y 2 + 1

By [4, Eqn. (3), p. 17] again, we see that ξ′ = and ′

η =

x x2 +y 2 −y x 2 2 ( x2 +y 2 ) + ( x2 +y 2 )

−y x2 +y 2 −y x 2 2 ( x2 +y 2 ) + ( x2 +y 2 )

+1

+1

=

=

x2

x2

x =ξ + y2 + 1

−y = −η. + y2 + 1

Finally, we get ′

ζ =

−y x 2 2 ( x2 +y 2 ) + ( x2 +y 2 ) x 2 ( x2 +y 2) +

( x2−y )2 +y 2

+1

=

1 x2 + y 2 = 2 = 1 − ζ. x2 + y 2 + (x2 + y 2 )2 x + y2 + 1

P 1 1 1 1 (b) Obviously, (−P . By part (a), if the point 2 , 0, 2 ) and ( 2 , 0, 2 ) are on P z corresponds to 1 (ξ1 , η1 , ζ1 ) on , then the point z corresponds to (ξ1 , −η1 , 1 − ζ1 ) on . The form of the corresponding points P show that they lie on the circle C which is the intersection of the plane ξ = ξ1 and . It is clear that the centre of circle C is (ξ1 , 0, 21 ). See Figure 1.8 for an illustration.

21

Figure 1.8: The effect of f (z) =

1 z

on

P

.

By considering the projections of the points (ξ1 , η1 , ζ1 ), (ξ1 , −η1 , 1 − ζ1 ) and the circle (ξ1 , 0, 21 ) on the η − ζ plane, we observe that (η1 , ζ1 ) and (−η1 , 1 − ζ1 ) are points on a circle centred at (0, 12 ). Therefore, the equation of the straight line L passing through (0, 12 ) and (η1 , ζ1 ) is (2ζ1 − 1)η − 2η1 ζ + η1 = 0. (1.35) By substituting the point (−η1 , 1 − ζ1 ) into the equation (1.35), it is easy to check that it lies on L. In other words, the point (−η1 , 1 − ζ1 ) is the image of the 180◦ rotation of the point (η1 , ζ1 ) about the centre (0, 21 ). Going back to the three dimensional situation, it implies that the point (ξ1 , −η1 , 1 − ζ1 ) is the image of the 180◦ rotation of the point (ξ1 , η1 , ζ1 ) about the diameter with endpoints (− 21 , 0, 12 ) and ( 21 , 0, 12 ). We complete the proof of the problem.



Remark 1.5 Another way to intercept Problem 1.27(b) can be found in [20, p. 143].

Problem 1.28 Bak and Newman Chapter 1 Exercise 28.

Proof. Let T be a circle or a line in C. By Problem 1.25, its corresponding S ⊂

P

is also a

Chapter 1. The Complex Numbers

22

circle. By Definition 1.11, S has the form Aξ + Bη + Cζ = D,

(1.36)

where A, B, C, D ∈ R. Substituting the result of Problem 1.27(a) into the equation (1.36), it leads Aξ ′ + B(−η ′ ) + C(1 − ζ ′ ) = D

Aξ ′ − Bη ′ − Cζ ′ = D − C.

(1.37)

P By Definition 1.11 again, the set S ′ with the form (1.37) is also a circle on . By Proposition 1.12, its corresponding projection T ′ is a circle or a line in C. In conclusion, we have shown that f (z) = 1z maps circles and lines in C onto other circles and lines. This ends the proof of  the problem. Remark 1.6 The map in Problem 1.28 is sometimes called the inversion through the unit circle, see [14, Exercise I.3.1, p. 17].

CHAPTER

2

Functions of the Complex Variable z

Problem 2.1 Bak and Newman Chapter 2 Exercise 1.

Proof. We consider the monomial P (x, y) = (x + iy)n first, where n ∈ N ∪ {0}. It is clear that Px = n(x + iy)n−1 and Py = in(x + iy)n−1 which imply Py = iPx . Thus the necessity of Proposition 2.3 holds for a monomial. Since an analytic polynomial is a finite linear combination of monomials, the necessity of Proposition 2.3 also holds for an analytic  polynomial. This completes the proof of the problem. Problem 2.2 Bak and Newman Chapter 2 Exercise 2.∗

Proof. (a) For every real z, it follows from Definition 2.4 that f (z + h) − f (z) h→0 h

f ′ (z) = lim

(2.1)

h∈C

exists. In particular, we may take h to be real in the limit (2.1) and the resulting limit still equals to f ′ (z), i.e., f (z + h) − f (z) . (2.2) f ′ (z) = lim h→0 h h∈R

Since both z and h are real, f (z + h) and f (z) are also real so that the quotient in the limit (2.2) is real-valued. Hence f ′ (z) is real-valued for real z. (b) The limit (2.1) exists for all imaginary points z. If z = iy, where y ∈ R, then we have 1 f (iy + ir) − f (iy) f (iy + h) − f (iy) = · lim . h→0 h i r→0 r

f ′ (iy) = lim

r∈R

h∈C

23

(2.3)

Chapter 2. Functions of the Complex Variable z

24

By the hypothesis, both f (iy + ir) and f (iy) are real-valued so that the quotient in the limit (2.3) is also real-valued. Hence f ′ (iy) is purely imaginary. We have completed the proof of the problem.  Problem 2.3 Bak and Newman Chapter 2 Exercise 3.

Proof. (a) Since Px = 3x2 − 3y 2 − 1 + 6ixy and Py = −6xy + i(3x2 − 3y 2 − 1), we have Py = iPx . By Proposition 2.3, P is analytic. (b) Since Px = 2x and Py = 2iy, we have Py 6= iPx . By Proposition 2.3, P is not analytic. (c) Since Px = 2y − 2ix and Py = 2x + 2iy, we have Py = iPx . By Proposition 2.3, P is analytic. This completes the proof of the problem.



Problem 2.4 Bak and Newman Chapter 2 Exercise 4.

Proof. Assume that P was a nonconstant analytic polynomial which takes imaginary values only. Then Q = iP takes only real values. Since Qy = iPy = i(iPx ) = iQx , Q is a nonconstant analytic polynomial. By the Example 1 on p. 24, no such Q exists. Hence no such P exists and  we complete the proof of the problem. Problem 2.5 Bak and Newman Chapter 2 Exercise 5.

Proof. For Problem 2.3(a), we have P (z) = z 3 − z so that P ′ (z) = 3z 2 − 1 = 3(x + iy)2 − 1 = 3x2 − 3y 2 − 1 + 6ixy = Px . Similarly, for Problem 2.3(c), we have P (z) = −iz 2 so that P ′ (z) = −2iz = −2i(x + iy) = 2y − 2ix = Px . We end the proof of the problem. Problem 2.6 Bak and Newman Chapter 2 Exercise 6.

Proof. By Definition 2.4, we have f (z + t) − f (z) g(z + t) − g(z) h1 (z + t) − h1 (z) = lim + lim = f ′ (z) + g ′ (z) t→0 t→0 t→0 t t t

h′1 (z) = lim



25 and h2 (z + t) − h2 (z) t→0 t f (z + t)g(z + t) − f (z)g(z) = lim t→0 t f (z + t)g(z + t) − f (z)g(z + t) + f (z)g(z + t) − f (z)g(z) = lim t→0 t g(z + t) − g(z) f (z + t) − f (z) · g(z + t) + lim · f (z) = lim t→0 t→0 t t = f ′ (z)g(z) + f (z)g ′ (z).

h′2 (z) = lim

(2.4)

If g(z) 6= 0, then we have [g−1 (z)]′ = lim

1 g(z+t)

−

1 g(z)

t g(z) − g(z + t) = lim t→0 tg(z)g(z + t) 1 g(z + t) − g(z) = − lim · lim t→0 g(z)g(z + t) t→0 t g′ (z) =− 2 . g (z) t→0

(2.5)

Using the results (2.4) and (2.5), we see that h′3 (z) = [f (z)g −1 (z)]′ = f ′ (z)g−1 (z) + f (z)([−1 (z)]′ f ′ (z) f (z)g′ (z) − g(z) g2 (z) f ′ (z)g(z) − f (z)g ′ (z) = , g2 (z) =

completing the proof of the problem.



Problem 2.7 Bak and Newman Chapter 2 Exercise 7.

Proof. By Definition 2.4 and the identity an − bn = (a − b)(an−1 + an−2 b + · · · + bn−1 ), we know that (z + h)n − z n = lim [(z + h)n−1 + (z + h)n−2 z + · · · + z n−1 ] = nz n−1 . h→0 h→0 h

(z n )′ = lim

(2.6)

Thus Proposition 2.6 holds for a monomial. If P (z) = α0 + α1 z + · · · + αN z N , then the formula (2.6) and repeated applications of Proposition 2.5 imply that P ′ (z) = α1 + 2α2 z + · · · + N αN z N −1 . This completes the proof of the problem.



Chapter 2. Functions of the Complex Variable z

26

Problem 2.8 Bak and Newman Chapter 2 Exercise 8.

1

Proof. Let Sn = n n , where n ≥ 2. Then log Sn = logn n > 0 which tends to 0 as n → ∞. Since Sn = elog Sn and ex is continuous on R, it has led that   lim Sn = lim elog Sn = exp lim log Sn = e0 = 1. n→∞

n→∞

n→∞

This ends the proof of the problem.



Problem 2.9 Bak and Newman Chapter 2 Exercise 9.

Proof. 1

(a) Now Cn! = 1 so that |Cn! | n! = 1. Therefore, we get 1

1

L = lim sup |Ck | k = lim |Cn! | n! = 1 n→∞

k→∞

and so R = 1 by Theorem 2.8. 1

1

(b) Now Cn = (n + 2n ) so that |Cn | n = (n + 2n ) n = 2 · (1 + we have  n  n1 =1 lim sup 1 + n 2 n→∞ and then

1 n n 2n ) .

Since

n 2n

→ 0 as n → ∞,

 1 n  n1 L = lim sup |Cn | n = lim sup 2 · 1 + n = 2. 2 n→∞ n→∞

By Theorem 2.8, we conclude that R = 12 . We have completed the proof of the problem.



Problem 2.10 Bak and Newman Chapter 2 Exercise 10.

Proof. By the hypothesis, we have 1

L = lim sup |ck | k

and R =

k→∞

1 , L

where 0 < L < ∞. (a) By Problem 2.8, we have p

1

p

1

1

lim sup |kp ck | k = lim sup k k · |ck | k = lim sup k k · lim sup |ck | k = L. k→∞

k→∞

k→∞

k→∞

Therefore, the radius of convergence of the power series is also R.

27 (b) Now it is clear that 1

1

lim sup ||ck || k = lim sup |ck | k = L, k→∞

k→∞

so the radius of convergence of the power series is also R. (c) Obviously, we have 1

1

lim sup |c2k | k = lim sup(|ck | k )2 = L2 k→∞

k→∞

so that the radius of convergence of the power series is R2 . This has completed the proof of the problem.



Problem 2.11 Bak and Newman Chapter 2 Exercise 11.

Proof. Let R be the radius of convergence of the sum of the two power series. Now we claim that R ≥ min(R1 , R2 ). (2.7) To this end, we suppose first that R1 < R2 . Thus for every z with |z| ≤ R1 , the hypotheses imply that both ∞ ∞ X X bn z n (2.8) an z n and n=0

n=0

converge so that their sum

∞ X

n

(an + bn )z =

∞ X

n

an z +

bn z n

(2.9)

n=0

n=0

n=0

∞ X

converges for such z. In other words, R ≥ R1 . However, R1 < R < R2 is impossible. Otherwise, if w satisfies R1 < |w| = R < R2 , then ∞ X

(an + bn )wn −

n=1

∞ X

bn w n =

∞ X

an w n

n=0

n=0

converges which means that the radius of convergence of the first power series (2.8) is greater than R1 , a contradiction. Consequently, we obtain R = R1 which satisfies the inequality (2.7) in this case. Next, if R1 = R2 , then the formula (2.9) is still valid for any z with |z| ≤ R1 = R2 . By the definition, it implies the inequality (2.7) in this case. To show that the strict inequality can hold in the estimate (2.7), we consider the power series ∞ X

zn

and

∞ X

(−z n )

n=0

n=0

both have radius of convergence 1, but their sum ∞ X

(1 − 1)z n = 0

n=0

has radius of convergence ∞. This has completed the proof of the problem.



Chapter 2. Functions of the Complex Variable z

28

Problem 2.12 Bak and Newman Chapter 2 Exercise 12.

Proof. Let |z| = 1. If z = 1, then the series ∞ ∞ X X 1 z = n n n=1 n=1

is divergent. Without loss of generality, we may assume that z 6= 1. Now we want to apply Dirichlet’s Test [27, Theorem 6.14, p. 78]: Lemma 2.1 (Dirichlet’s Test) P Suppose that an is a series of complex numbers whose partial sums {An } is bounded. If {bn } is a monotonically decreasing sequence and lim bn = 0, then the n→∞ P series an bn converges. Let an = z n and bn = n1 . Then the sequence {bn } satisfies the conditions of Lemma 2.1 (Dirichlet’s Test). Next, it follows from the triangle inequality that n n X X zn − 1 2 2|z| |An | = ak = z k = z · = , ≤ z−1 |z − 1| |z − 1| k=1

k=1

so it is bounded for z 6= 1. Hence Lemma 2.1 (Dirichlet’s Test) asserts that the power series ∞ X zn

n=1

n

converges at all points on the unit circle except z = 1. This completes the proof of the problem.  Problem 2.13 Bak and Newman Chapter 2 Exercise 13.

Proof. (a) Suppose that L is finite. For every n ≥ 1, we know that a  a  a  2 3 n+1 an+1 = a1 × × × ··· × . a1 a2 an

(2.10)

By the definitiona , for every ǫ > 0, there is an N ∈ N such that k ≥ N implies a k+1 − L < ǫ ak or equivalently,

L−ǫ< a

For example, [27, p. 49].

ak+1 0, there exists an N ∈ N such that k ≥ N implies ak+1 > M. ak Using similar argument as above, it gives an+1 > aN · M n+1−N 1

1

N

n+1 n+1 an+1 · M 1− n+1 > aN

which shows that

1 n+1 > M. lim an+1

n→∞

Since M is arbitrary large, we conclude that 1

lim ann = ∞.

n→∞

(b) Let an =

1 n!

> 0. Then we have lim

n→∞

an+1 1 = 0. = lim n→∞ n + 1 an

Hence the required result follows immediately from part (a). This ends the proof of the problem.



Problem 2.14 Bak and Newman Chapter 2 Exercise 14.

Proof. (a) Take an =

(−1)n n! .

Then we have lim

n→∞

−1 an+1 = lim = 0. n→∞ an n+1

By Problem 2.13(a) and Theorem 2.8, it can be seen easily that R =

1 L

= ∞.

Chapter 2. Functions of the Complex Variable z

30

(b) Suppose that f (ω) =

∞ X

ω n+1 (2n + 1)! n=0

and

Then it is obvious that g(z) = f (z 2 ). Take an = lim

n→∞

g(z) =

∞ X

z 2n+1 . (2n + 1)! n=0

1 (2n+1)! .

Since

(2n + 1)! an+1 = 0, = lim n→∞ an (2n + 3)! 1

Problem 2.13(a) asserts that lim ann = 0 and then Theorem 2.8 ensures that n→∞

∞ X

n=0

ω n+1 (2n + 1)!

converges for |ω| < ∞. Therefore, the relation g(z) = f (z 2 ) implies that the radius of convergence of the power series of g is also ∞. (c) Let an =

n! nn .

Using [12, §215, Eqn. (1)] or [27, Eqn. (5.24), p. 61], we find that  n n an+1 nn (n + 1)! = e−1 . = lim × = lim n→∞ an n→∞ (n + 1)n+1 n→∞ n + 1 n! lim

By Problem 2.13(a) and Theorem 2.8, we conclude that R = e. (d) Let an =

2n n! .

Then an+1 n! 2n+1 2 × n = lim = 0. = lim n→∞ an n→∞ (n + 1)! n→∞ n + 1 2 lim

Thus Problem 2.13(a) and Theorem 2.8 yield that R = ∞.

This completes the proof of the problem.



Problem 2.15 Bak and Newman Chapter 2 Exercise 15.∗

Proof. (a) Let x ∈ R be such that | sin x| ≤ 21 . We claim that | sin(x + 2)| > condition | sin x| ≤ 12 implies that there is a k ∈ Z such that − Since 3 < π < 4, we have

π 2

. 2

(2.13)

31 Next, since 0 ≤ | sin n| ≤ 1 for every n ∈ N, we certainly have 1

| sin n| n ≤ 1

(2.14)

for every n ∈ N. Now we claim that given ǫ > 0, one can find a positive integer N such that 1 | sin N | N > 1 − ǫ. (2.15) 1

To prove this claim, we first notice that lim 2− n = 1, so there exists an N ′ ∈ N such that n→∞

n ≥ N ′ implies

1

2− n > 1 − ǫ.

(2.16)

On the one hand, if | sin N ′ | > 21 , then we get from the inequality (2.16) that 1

1

| sin N ′ | N ′ > 2− N ′ > 1 − ǫ. On the other hand, if | sin N ′ | ≤ 21 , then the previous claim (2.13) and the inequality (2.16) imply that 1 − 1 | sin(N ′ + 2)| N ′ +2 > 2 N ′ +2 > 1 − ǫ. Hence they mean that our claim (2.15) is true or equivalently, there exists a sequence {nk } of positive integers such that 1

lim | sin nk | nk ≥ 1.

(2.17)

k→∞

Finally, the estimates (2.14) and (2.17) guarantee that 1

lim sup | sin n| n = 1. n→∞

By Theorem 2.8, it concludes that R = 1. (b) By Theorem 2.8, we have 2

1

L = lim sup(e−n ) n = lim e−n = 0 n→∞

n→∞

so that R = ∞.

We complete the proof of the problem.



Problem 2.16 Bak and Newman Chapter 2 Exercise 16.∗

Proof. By the definition, we have 1

1

2k lim c2k = lim (2k ) 2k = lim

k→∞

and

k→∞

k→∞

√

2=

√

2

1 k h h √ 1 k2 i 2k−1 1 k i 2k−1 1+ 1+ = lim = e. k→∞ k→∞ k→∞ k k Since e > 2, we establish from Theorem 2.8 that 1

2k−1 lim c2k−1 = lim

1 R= √ . e This completes the analysis of the problem.



Chapter 2. Functions of the Complex Variable z

32

Problem 2.17 Bak and Newman Chapter 2 Exercise 17.

Proof. The following proof is basically adopted from [22, Theorem 3.50, pp. 74, 75]. Suppose that βn = Bn − B for all n = 0, 1, 2, . . .. It is clear that Cn =

n X

ck

k=0

= a0 b0 + (a0 b1 + a1 b0 ) + · · · + (a0 bn + a1 bn−1 + · · · + an b0 )

= a0 Bn + a1 Bn−1 + · · · + an B0

= a0 (B + βn ) + a1 (B + βn−1 ) + · · · + an (B + β0 )

= (a0 + a1 + · · · + an )B + a0 βn + a1 βn−1 + · · · + an β0

= An B + a0 βn + a1 βn−1 + · · · + an β0 .

(2.18)

Since An B → AB as n → ∞, it follows from the expression (2.18) that Cn → AB if and only if lim (a0 βn + a1 βn−1 + · · · + an β0 ) = 0.

n→∞

Since

∞ X

ak converges absolutely,

k=0

∞ X k=0

(2.19)

|ak | is finite and we denote this number by α. Given ǫ > 0.

Since βn → 0 as n → ∞, there exists an N ∈ N such that n ≥ N implies |βn | < ǫ. Therefore, the triangle inequality shows that |a0 βn + a1 βn−1 + · · · + an β0 | ≤ |a0 βn + · · · + an−(N +1) βN +1 | + |an−N βN + · · · + an β0 |

≤ |an−N βN + · · · + an β0 | + (|a0 | · |βn | + · · · + |an−(N +1) | · |βN +1 |)

< |an−N βN + · · · + an β0 | + ǫ(|a0 | + |a1 | + · · · + |an−(N +1) |)

≤ |an−N βN + · · · + an β0 | + ǫα.

(2.20)

Since an → 0 as n → ∞, we fix N in the inequality (2.20) and then take n → ∞ to establish lim sup |a0 βn + a1 βn−1 + · · · + an β0 | ≤ ǫα. n→∞

Since ǫ is arbitrary, this proves the desired result (2.19) and thus lim Cn = AB.

n→∞

This completes the proof of the problem.



Problem 2.18 Bak and Newman Chapter 2 Exercise 18.

Proof. If the complex number z satisfies |z| < R1 , then the power series ∞ X

n=0

|an | · |z n |

33 converges. Similarly, if |z| < R2 , then the power series ∞ X

n=0

|bn | · |z n |

converges. Therefore, both series converges within |z| < min(R1 , R2 ).

Clearly, by replacing an and bn by an z n and bn z n respectively in Problem 2.17, we get k k X X aj bk−j · z k = ck z k (aj z j )(bk−j z k−j ) = j=0

j=0

and then Problem 2.17 ensures that

∞ X

cn z n

n=0

converges within |z| < min(R1 , R2 ). We have completed the proof of the problem.



Problem 2.19 Bak and Newman Chapter 2 Exercise 19.

Proof. (a) Suppose that |z| < 1. Apply the given identity, we have lim (1 − z)(1 + z + z 2 + · · · + z N ) = lim (1 − z N +1 )

N →∞

(1 − z)

∞ X

n=0 ∞ X

n=0

(b) Let

∞ X

n

cn z be the Cauchy product of

∞ X

N →∞

zn = 1 zn =

1 . 1−z

z n with itself. By the definitionb , we have

n=0

n=0

cn =

n X k=0

(1 × 1) = n + 1.

Combining this fact and part (a), we see that ∞ ∞ ∞ ∞  X  X X X 1 1 n n n nz n + (n + 1)z = = z = z 2 (1 − z) 1−z n=0 n=0 n=0 n=0

which asserts that

∞ X

n=0

nz n =

1 z 1 − = . (1 − z)2 1 − z (1 − z)2

Hence we complete the proof of the problem. b

On p. 28.



Chapter 2. Functions of the Complex Variable z

34

Problem 2.20 Bak and Newman Chapter 2 Exercise 20.

Proof. Let the power series under consideration is ∞ X

Cn z n .

(2.21)

n=0

Recall from [1, p. 53] that x is an accumulation point of the set S if and only if every neighborhood of x contains infinitely many points of S. Particularly, since S has an accumulation point at 0, each neighborhood D(0; n1 ) contains a point zn 6= 0 of S. Thus we have |zn |
0 is impossible. Hence we have completed the proof of the problem.



Problem 2.22 Bak and Newman Chapter 2 Exercise 22.

Proof. We write g(z) = f (z + α) =

∞ X

Cn z n .

n=0

Since lim sup |Cn | n→∞

1 n

< ∞, Theorem 2.8 ensures that the radius of convergence of g is nonzero.

Hence it follows from Corollary 2.11 that Cn =

g(n) (0) f (n) (α) = n! n!

for all n ∈ N ∪ {0}, completing the proof of the problem.



35 Problem 2.23 Bak and Newman Chapter 2 Exercise 23.

Proof. 1

(a) By Problem 2.8, we have lim sup n n = 1 so that the comment following Corollary 2.14 on n→∞

p. 32 implies that the series converges throughout |z − 1| < 1. (b) As we have shown in Problem 2.14(a) that the radius of convergence of the series there is ∞, so the comment following Corollary 2.14 on p. 32 implies that the radius of convergence of ∞ X (−1)n (z + 1)n n! n=0

is also ∞.

(c) Note that

∞ X

n=0

2

n

n (2z − 1) =

By Problem 2.8, we see that

∞ X

n=0

 1 n 2n n2 z − . 2

(2.22)

2

1

lim sup(2n n2 ) n = lim 2n n = 2. n→∞

n→∞

Again the comment following Corollary 2.14 on p. 32 implies that the series (2.22) converges throughout |z − 21 | < 12 . This completes the proof of the problem. 

Chapter 2. Functions of the Complex Variable z

36

CHAPTER

3

Analytic Functions

Problem 3.1 Bak and Newman Chapter 3 Exercise 1.

Proof. Notice that f (z) = f (x, y) and z = x + iy, so it is easy to see that fx = lim

h→0 h∈R

f (x + h, y) − f (x, y) f (z + h) − f (z) = lim h→0 h h h∈R

and fy = lim

h→0 h∈R

f (x, y + h) − f (x, y) f (z + ih) − f (z) = lim . h→0 h h h∈R

This completes the proof of the problem.



Problem 3.2 Bak and Newman Chapter 3 Exercise 2.

Proof. (a) Obviously, fx = 2x and fy = 2iy exist in a neighborhood of any point on the line y = x and furthermore, they are continuous at all points on the line y = x. Since fy = ifx if and only if y = x, Proposition 3.2 asserts that f is differentiable at all points on the line y = x. (b) By Definition 3.1, if f was analytic at z = (x, y), then f is differentiable in a neighborhood of z = (x, y) which is prohibited by part (a).  Hence we have completed the proof of the problem. Problem 3.3 Bak and Newman Chapter 3 Exercise 3.

37

Chapter 3. Analytic Functions

38

Proof. If g is differentiable at w, then Definition 2.4 implies g(s) − g(w) = (s − w)[g ′ (w) + ǫ(s)], where ǫ(s) → 0 as s → w. Therefore, we get g(f (z + h)) − g(f (z)) = [f (z + h) − f (z)] · [g ′ (f (z)) + ǫ(h)],

(3.1)

where ǫ(h) → 0 as h → 0. By dividing both sides of the expression (3.1) by h, we see that g(f (z + h)) − g(f (z)) f (z + h) − f (z) = [g′ (f (z)) + ǫ(h)] × h h and thus g(f (z + h)) − g(f (z)) h→0 h

(g ◦ f )′ (z) = lim

f (z + h) − f (z) h→0 h

= lim [g′ (f (z)) + ǫ(h)] × lim h→0

= g′ (f (z)) · f ′ (z). This completes the analysis of the problem.



Remark 3.1 Problem 3.3 is the Chain Rule for the complex variable case.

Problem 3.4 Bak and Newman Chapter 3 Exercise 4.

Proof. Since g 2 (z) = z, if z0 ∈ C, then we have g2 (z) − g 2 (z0 ) =1 z − z0 which gives

Since g is continuous at g′ (z0 ) = lim

z→z0

√

1 g(z) − g(z0 ) = . z − z0 g(z) + g(z0 )

(3.2)

z0 , it deduces from the expression (3.2) that

g(z) − g(z0 ) 1 1 1 1 = = lim p = √ . = lim z→z0 z→z0 g(z) + g(z0 ) z − z0 2g(z0 ) 2 z0 [g(z)]2 + g(z0 )

Since z0 is arbitrary, we obtain our expected result and we have completed the proof of the  problem. Problem 3.5 Bak and Newman Chapter 3 Exercise 5.

39 Proof. Suppose that f is analytic in the region D ⊆ C. By Definition 3.3, f is differentiable in f D. Since f ′ ≡ 0 in D, Proposition 3.1 implies that iy = fx = f ′ ≡ 0 in D, i.e., fx ≡ fy ≡ 0

(3.3)

in D. If f = u + iv, then the identities (3.3) mean that ux ≡ uy ≡ vx ≡ vy ≡ 0 in D. Therefore, we follow from Theorem 1.10 that both u and v are constant in D. In particular,  f is constant in D. This completes the proof of the problem. Problem 3.6 Bak and Newman Chapter 3 Exercise 6.

Proof. Suppose that f is analytic in the region D ⊆ C. By Definition 3.3, f is differentiable in D. By Problem 3.3, we establish that f 2 is also differentiable in D so that Definition 3.3 again implies that f 2 is analytic in D. Furthermore, we get from Problem 3.3 that [f 2 (z)]′ = 2f (z)f ′ (z) ≡ 0 in D. By Problem 3.5, [f (z)]2 is constant. Let [f (z)]2 = A. Then we take the modulus to both sides to get |f (z)|2 = |A|

in D. In other words, |f | is constant in D. By Proposition 3.7, f is constant in D, completing  the proof of the problem. Problem 3.7 Bak and Newman Chapter 3 Exercise 7.

Proof. Let f : D → C be a nonconstant analytic function in the region D. Let f = u + iv. If f mapped D into a straight line y = ax + b with a, b ∈ R, then the point f (x + iy) as a point in C (or in R2 ), we read v(x, y) = au(x, y) + b in D. Since f is analytic in D, it is differentiable there and Proposition 3.1 says that ux = vy = auy

and uy = −vx = −aux

(3.4)

in D. These two equations (3.4) imply that (1 + a2 )ux = 0 in D. Since a is real, we have ux = 0 and then the second equation (3.4) gives uy = 0 in D. By Theorem 1.10, u is constant so that Proposition 3.6 implies that f is constant which contradicts to the hypothesis. Next, assume that f mapped D into a circular arc of the circle |w − w0 | = r centred at w0 ∈ C and radius r > 0, i.e., |f (z) − w0 | = r. Consider the map g(z) = f (z) − w0 which is also nonconstant analytic in D. Now |g(z)| = |f (z) − w0 | = r in D. Hence Proposition 3.7 ensures that g and then f are both constants, a contradiction again.  This completes the proof of the problem.

Chapter 3. Analytic Functions

40

Problem 3.8 Bak and Newman Chapter 3 Exercise 8.

Proof. Suppose that f (x, y) = (x2 − y 2 ) + iv is analytic at z. Then it must be differentiable in a neighborhood D(z; r) of z and the Cauchy-Riemann equations give ux = vy and uy = −vx in D(z; r). Since ux = 2x and uy = −2y, we have vx = 2y and vy = 2x so that v(x, y) = 2xy + g(y)

and v(x, y) = 2xy + h(x)

(3.5)

in D(z; r). Therefore, the equations (3.5) imply that h(x) = g(y) = C for some constant C in D(z; r). In conclusion, f must take the form f (z) = f (x, y) = (x2 − y 2 ) + i(2xy + C) = (x + iy)2 + iC = z 2 + iC for some constant C. This completes the proof of the problem.



Problem 3.9 Bak and Newman Chapter 3 Exercise 9.

Proof. Assume that f was an analytic function satisfying the requirement. By Proposition 3.1, it satisfies the Cauchy-Riemann equations, i.e., vy = ux = 2x and vx = −uy = −2y. Therefore, we have v(x, y) = 2xy + F (x) and v(x, y) = −2xy + G(y) for some functions F (x) and G(y). Clearly, they imply that F (x) − G(y) = −4xy.

(3.6)

Put y = 0 in the equation (3.6) to get F (x) = G(0) for all x in the domain of F . In other words, F (x) is a constant function. Similarly, it can be shown that G(y) is also a constant function. However, the equation (3.6) means that −4xy is a constant for all x and y, a contradiction and then no such analytic function exits. This  completes the proof of the problem. Problem 3.10 Bak and Newman Chapter 3 Exercise 10.

Proof. Since f is entire, Proposition 3.1 implies that u x = vy .

(3.7)

By the given form, ux and vy are functions of x and y respectively. These facts and the equation (3.7) force that ux = vy = A for some real constant A. Then simple integration gives u(x) = Ax + B

and v(y) = Ay + C

for some B, C ∈ R. Consequently, we have f (z) = Ax + B + i(Ay + C) = A(x + iy) + B + iC = Az + B + iC. This completes the proof of the problem.



41 Problem 3.11 Bak and Newman Chapter 3 Exercise 11.

Proof. By the definition, we have ez = ex cos y + iex sin y so that u = ex cos y and v = ex sin y. (a) Direct computation gives ux = ex cos y, uy = −ex sin y, vx = ex sin y and vy = ex cos y. It is clear that all ux , uy , vx and vy exist and continuous everywhere. Furthermore, since ux = vy and uy = −vx hold everywhere, Proposition 3.2 ensures that ez is differentiable everywhere, i.e., ez is entire. (b) If z1 = a + ib and z2 = c + id, then we see that ez1 +z2 = e(a+c)+i(b+d) = ea+c [cos(b + d) + i sin(b + d)] = ea+c [cos b cos d − sin b sin d + i(sin b cos d + cos b sin d)]

= ea+c (cos b + i sin b) × (cos d + i sin d)

= ez1 ez2 .

This completes the proof of the problem.



Problem 3.12 Bak and Newman Chapter 3 Exercise 12.

Proof. Let z = x + iy. Then it follows from the definition of the modulus of z (see p. 5) that p |ez | = |ex+iy | = |ex | · |eiy | = ex · | cos y + i sin y| = ex · cos2 t + sin2 t = ex , completing the proof of the problem.



Problem 3.13 Bak and Newman Chapter 3 Exercise 13.

Proof. When z = x + iy, we notice that ez = ex eiy .

(3.8)

Suppose that z is a ray lying in the right half-plane {z ∈ C | Re (z) = x > 0}, i.e., x → +∞ and y → L for some finite L or L = ±∞. Then it follows from the expression (3.8) that |ez | = ex → ∞.

Similarly, if z is a ray lying in the left half-plane {z ∈ C | Re (z) = x < 0}, i.e., x → −∞ and y → L for some finite L or L = ±∞. Then the expression (3.8) implies that |ez | = ex → 0.

Finally, if the ray lies on the imaginary axis, then x = 0 and z = iy so that |ez | = |eiy | = 1. Since y → +∞ or y → −∞, ez moves along the unit circle anticlockwise and clockwise infinitely  many times respectively. We have completed the proof of the problem.

Chapter 3. Analytic Functions

42

Problem 3.14 Bak and Newman Chapter 3 Exercise 14.

Proof. (a) Since e0 = 1, ez = e0 implies that z = 2kπi for all k ∈ Z. π

π

(b) Since e 2 i = i, ez = e 2 i implies that z =

πi 2

+ 2kπi = ( π2 + 2kπ)i for all k ∈ Z.

(c) If we write z = x + iy, then ez = −3 will give ex eiy = 3eπi . Then its modulus implies that ex = 3 or equivalently, x = ln 3. In addition, eiy = eπi leads to us that y = π + 2kπ for all k ∈ Z. Hence the solutions of the equation have the form z = x + iy = ln 3 + (2k + 1)πi, where k ∈ Z. (d) Similar to part (c), if we write z = x + iy, then ez = 1 + i implies that √  1 i  √ π ex eiy = 2 · √ + √ = 2e 4 i . 2 2 √ √ Thus ex = 2 and then x = ln 2 = 12 ln 2. Besides, it is easy to see that y = all k ∈ Z. In conclusion, the solutions of the equation take the form  1 1 πi, z = x + iy = ln 2 + 2k + 2 4

π 4

+ 2kπ for

where k ∈ Z.

This completes the proof of the problem.



Problem 3.15 Bak and Newman Chapter 3 Exercise 15.

Proof. (a) Recall the formulas of sin z and cos z, we obtain immediately that 2 sin z cos z = 2 ×

1 1 1 iz (e − e−iz ) × (eiz + e−iz ) = (e2iz − e−2iz ) = sin 2z. 2i 2 2i

(b) Direct computation implies i2 h 1 i2 (eiz + e−iz ) + (eiz − e−iz ) 2 2i 1 2iz 1 −2iz = (e + 2 + e ) − (e2iz − 2 + e−2iz ) 4 4 = 1.

cos2 z + sin2 z =

h1

(c) Using the property (ez )′ = ez and Problem 3.3, we have i 1 1 d h 1 iz (e − e−iz ) = (ieiz + ie−z ) = (eiz + e−iz ) = cos z. (sin z)′ = dz 2i 2i 2 We end the proof of the problem.



43 Problem 3.16 Bak and Newman Chapter 3 Exercise 16.∗

Proof. Recall the definitions of hyperbolic sine and cosine as follows: sinh x =

ex − e−x 2

and

cosh x =

ex + e−x , 2

(3.9)

where x ∈ R. (a) By the definition of sin z, we know that sin

π

 π π π π 1 1 1 + iy = [ei( 2 +iy) − e−i( 2 +iy) ] = (e−y+i 2 − ey− 2 i ) = (ey + e−y ) = cosh y. 2 2i 2i 2

(b) Using the result of Problem 3.20 in advance, we have p | sin z| = (sin x cosh y)2 + (cos x sinh y)2 r  e2y + 2 + e−2y   e2y − 2 + e−2y  + cos2 x · = sin2 x · 4 4 r e2y + 2 + e−2y − 4 cos2 x = 4 r  ey + e−y 2 − cos2 x = 2 q = cosh2 y − cos2 x.

(3.10)

On the left side or the right side of the square, we have z = ±(N + 12 )π + iy, where −(N + 21 )π ≤ y ≤ (N + 12 )π so that cos ±(N + 21 )π = 0 and then the expression (3.10) implies | sin z| = cosh y.

By the A.M. ≥ G.M., we know that cosh y ≥ 1. Hence we have | sin z| ≥ 1 in this case.

Next, on the top side or the bottom side of the square, we see that z = x ± i(N + 12 )π, where −(N + 21 )π ≤ x ≤ (N + 12 )π. By the definition (3.9), cosh y is clearly an even increasing function of y, so we have cosh ±(N + 12 )π = cosh(N + 21 )π ≥ cosh π2 ≥ 2 and then cosh2 y − cos2 x ≥ 4 − 1 = 3. √ Hence we deduce from the expression (3.10) that | sin z| ≥ 3 ≥ 1. (c) If y → ±∞, then it follows from the definition (3.9) that cosh y → ∞. Since 0 ≤ cos2 x ≤ 1 for all x ∈ R, we conclude from the expression (3.10) that | sin z| → ∞ as Im z = y → ±∞. This completes the proof of the problem. Problem 3.17 Bak and Newman Chapter 3 Exercise 17.



Chapter 3. Analytic Functions

44

Proof. Since cos z = 21 (eiz + e−iz ) and sin z =

1 iz 2i (e

− e−iz ), we obtain from the Chain Rule that

d −iz i 1 h d iz (e ) + (e ) 2 dz dz i 1h d d = i (eiz ) − i (e−iz ) 2 d(iz) d(−iz) −1 iz = (e − e−iz ) 2i = − sin z,

(cos z)′ =

completing the proof of the problem.



Problem 3.18 Bak and Newman Chapter 3 Exercise 18.

Proof. By the definition of sin z, we have eiz − e−iz =2 2i eiz − e−iz = 4i.

(3.11)

Set w = eiz . Then the equation (3.11) becomes w2 − 4iw − 1 = 0 and then the quadratic formula [4, Eqn. (1), p. 4] gives p √ √ 4i ± (4i)2 − 4 × (−1) 4i ± 2 3i = = (2 ± 3)i. w= 2 2 √ Thus eiz = (2 ± 3)i and then we get  π √ iz = ln(2 ± 3) + + 2kπ i 2 π  √ z= + 2kπ − i ln(2 ± 3), 2 where k ∈ Z. This ends the proof of the problem.



Problem 3.19 Bak and Newman Chapter 3 Exercise 19.∗

z

Proof. Since ee = 1 = e0 , we have ez = 2kπi,

(3.12)

where k ∈ Z. If k = 0, then ez = 0 which is impossible. Thus we suppose that k 6= 0. π If k > 0, then 2kπi = eln(2kπ)+i 2 and the equation (3.12) gives π

ez = eln(2kπ)+i 2 . Consequently, we obtain z = ln(2kπ) + i where n ∈ Z.

 π π , + 2nπi = ln(2kπ) + i 2nπ + 2 2

45 Next, if k < 0, then we have 2kπi = (−2kπi)(−i) = eln(−2kπ)+i (3.12) becomes 3π ez = eln(−2kπ)+i 2

3π 2

. Similarly, the equation

which implies

 3π 3π  , + 2nπi = ln(−2kπ) + i 2nπ + 2 2 where n ∈ Z. Thus we complete the proof of the problem. z = ln(−2kπ) + i



Problem 3.20 Bak and Newman Chapter 3 Exercise 20.

Proof. By the definition of sin z, we have ei(x+iy) − e−i(x+iy) 2i ix−y e − e−ix+y = 2 −y e (cos x + i sin x) − ey (cos x − i sin x) = 2i y −y e +e e−y − ey = × (i sin x) + cos x 2i 2i ey + e−y ey − e−y = × (sin x) + i · cos x 2 2 = sin x cosh y + i cos x sinh y,

sin(x + iy) =

completing the proof of the problem.



Problem 3.21 Bak and Newman Chapter 3 Exercise 21.

Proof. Using similar attack as in the proof of Problem 2.14(a), the radius of convergence of the series of f (z) is ∞. Furthermore, it is clear that the series converges absolutely, so Problem 2.17 says that f (z)f (w) =

∞ ∞ X z n X wn

n=0

n!

n=0

n!

=

∞ hX n X zk n=0

k=0

k!

·

∞ n  wn−k i X 1  X n k = Ck z · wn−k . (n − k)! n! n=0

(3.13)

k=0

By the binomial theorem, the bracket in the formula (3.13) is exactly (3.13) becomes ∞ X (z + w)n f (z)f (w) = = f (z + w). n!

(z+w)n n! ,

so the formula (3.14)

n=0

Let x ∈ R. By the power series representation of ex [25, Definition 4.5.1, p. 90], it is easy to see that f (x) = ex . (3.15) Recall that the series converges absolutely so that rearrangement is permitted. Thus we have     y3 y2 y4 y3 y2 − i + ··· = 1 − + − ··· + i y − + ··· . (3.16) f (iy) = 1 + iy − 2 3! 2 4 3!

Chapter 3. Analytic Functions

46

By the power series representation of sin y and cos y [25, p. 102]), the equation (3.16) can be written as f (iy) = cos y + i sin y. (3.17) Hence we deduce immediately from the formulas (3.14), (3.15) and (3.17) that f (z) = f (x + iy) = f (x)f (iy) = ex (cos y + i sin y) = ex+iy = ez as desired. This completes the proof of the problem.



Problem 3.22 Bak and Newman Chapter 3 Exercise 22.

Proof. By Problem 3.21, we have iz

e =

∞ X (iz)n

n=0

and e−iz =

(3.18)

n!

∞ X (−iz)n

n=0

n!

.

(3.19)

The difference of the two expressions (3.18) and (3.19) establishes   z3 z5 + − · · · = 2ig(z). eiz − e−iz = 2i z − 3! 5!

Hence it follows from the definition of sin z that g(z) = sin z, completing the proof of the  problem. Problem 3.23 Bak and Newman Chapter 3 Exercise 23.

Proof. Since sin z is entire, it follows from Theorem 2.9 and Problem 3.22 that cos z = (sin z)′ = g′ (z) = 1 − We end the proof of the problem.

z2 z4 + − ··· . 2! 4! 

CHAPTER

4

Line Integrals and Entire Functions

Problem 4.1 Bak and Newman Chapter 4 Exercise 1.

Proof. We say that z ∼ ω if there exists an one-to-one C 1 mapping λ satisfying Definition 4.4. Now we are going to check the definition of an equivalence relation directly. • Reflexivity: The identity i : [a, b] → [a, b] is clearly an one-to-one C 1 mapping, i′ (t) > 0 and z(i(t)) = z(t) for all t ∈ [a, b]. Thus we have z ∼ z. • Symmetry: Suppose that z ∼ ω. By the definition, there exists an one-to-one C 1 mapping λ : [c, d] → [a, b] such that λ(c) = a, λ(d) = b, λ′ (t) > 0 and ω(t) = z(λ(t))

(4.1)

for all t ∈ [c, d].

We need a modified exercise from Rudin as follows:a Lemma 4.1 Suppose that f ′ (x) > 0 on [a, b]. Then f is strictly increasing in [a, b] and if g denotes its inverse function, then g is differentiable on [f (a), f (b)] and g′ (f (x)) =

1 f ′ (x)

on [a, b].

Since λ′ (t) > 0 on [c, d], Lemma 4.1 implies that its inverse function λ−1 : [a, b] → [c, d] is differentiable and 1 >0 (4.2) (λ−1 )′ (λ(t)) = ′ λ (t) for all t ∈ [c, d]. Furthermore, the formula (4.2) ensures that λ−1 is also an one-to-one C 1 mapping on [a, b]. a

In fact, Lemma 4.1 is [22, Exercise 2, p. 114] with the open interval (a, b) replaced by closed interval [a, b]. A proof of this exercise can be found in [26, pp. 85, 86] and a proof of Lemma 4.1 can be modified from it once one-sided limits are considered.

47

Chapter 4. Line Integrals and Entire Functions

48

To continue our proof, if s = λ(t), then t = λ−1 (s) and the expression (4.1) gives immediately that z(s) = ω(λ−1 (s)) which means that ω ∼ z. • Transitivity: Suppose that z ∼ ω and ω ∼ ζ. Then there are one-to-one C 1 mappings λ1 : [c, d] → [a, b] and λ2 : [e, f ] → [c, d] such that λ1 (c) = a, λ1 (d) = b, λ2 (e) = c, λ2 (f ) = d, λ′1 (t) > 0 for all t ∈ [c, d], λ′2 (s) > 0 for all s ∈ [e, f ] and ω(t) = z(λ1 (t))

and ζ(s) = ω(λ2 (s)).

(4.3)

Now the composition function λ = λ1 ◦ λ2 : [e, f ] → [a, b] satisfies λ(e) = a and λ(f ) = b. Furthermore, Problem 3.3 guarantees that λ′ (s) = (λ1 ◦ λ2 )′ (s) = λ′1 (λ2 (s)) · λ′2 (s) > 0 for all s ∈ [e, f ]. Finally, the two equations (4.3) show that z(λ(s)) = z(λ1 (λ2 (s))) = ω(λ2 (s)) = ζ(s) for all s ∈ [e, f ]. By Definition 4.4, z ∼ ζ. Hence these prove that the concept of “smooth curves” is an equivalence relation and this completes the proof of the problem.  Problem 4.2 Bak and Newman Chapter 4 Exercise 2.

Proof. Note that z(t) ˙ = 2t + 2it. By Definition 4.3, we see that Z 1 Z f (t2 + it2 )(2t + 2it) dt f (z) dz = 0 C Z 1 (t4 + it4 )(2t + 2it) dt = 0 Z 1 2 = 2(1 + i) t5 dt 0

2i = , 3

completing the proof of the problem.



Problem 4.3 Bak and Newman Chapter 4 Exercise 3.

Proof. In this case, we have z(t) ˙ = cos t − i sin t so that Definition 4.3 asserts Z 2π Z 2π Z dt = −2πi. (sin t − i cos t)(cos t − i sin t) dt = −i f (z) dz = C

0

(4.4)

0

Our result is different from that in Example 2 because our curve is in the opposite direction of that in Example 2.b We have ended the proof of the problem.  b

We remark that the result (4.4) is an immediate consequence of Proposition 4.7.

49 Problem 4.4 Bak and Newman Chapter 4 Exercise 4.

Proof. We follow the given hint. Suppose that f = u1 + iv1 , g = u2 + iv2 and α = A + iB, where u1 , u2 , v1 , v2 are real-valued functions and A, B ∈ R. Suppose, further, that C is a smooth curve given by z(t), where a ≤ t ≤ b. Then we follow from Definitions 4.1 and 4.3 that Z b Z [f (z(t)) + g(z(t))] · z(t) ˙ dt [f (z) + g(z)] dz = a C Z b  = [u1 (z(t)) + u2 (z(t))] · z(t) ˙ + i[v1 (z(t)) + v2 (z(t))] · z(t) ˙ dt a Z b Z b [v1 (z(t)) + v2 (z(t))] · z(t) ˙ dt [u1 (z(t)) + u2 (z(t))] · z(t) ˙ dt + i = a a Z b hZ b i = v1 (z(t)) · z(t) ˙ dt u1 (z(t)) · z(t) ˙ dt + i a a Z b hZ b i + v2 (z(t)) · z(t) ˙ dt u2 (z(t)) · z(t) ˙ dt + i a a Z b Z b = [u1 (z(t)) + iv1 (z(t))] · z(t) ˙ dt + [u2 (z(t)) + iv2 (z(t))] · z(t) ˙ dt a Z Za g(z) dz. f (z) dz + = C

C

Similarly, we have Z b Z (A + iB)[u1 (z(t)) + iv1 (z(t))] · z(t) ˙ dt αf (z) dz = a C Z b  = [Au1 (z(t)) − Bv1 (z(t))] + i[Av1 (z(t)) + Bu1 (z(t))] · z(t) ˙ dt a Z b Z b [Av1 (z(t)) + Bu1 (z(t))] · z(t) ˙ dt [Au1 (z(t)) − Bv1 (z(t))] · z(t) ˙ dt + i = a a Z b i hZ b v1 (z(t)) · z(t) ˙ dt = (A + iB) u1 (z(t)) · z(t) ˙ dt + i a a Z b [u1 (z(t)) + iv1 (z(t))] · z(t) ˙ dt = (A + iB) a Z f (z) dz. =α C

This completes the proof of the problem.



Problem 4.5 Bak and Newman Chapter 4 Exercise 5.

Proof. Fix a ∈ C. Let b ∈ C with a 6= b and C be a smooth curve connecting a and b. Furthermore, we let z : [0, 1] → C be a parametrization such that z(0) = a and z(1) = b. Since F is analytic on C, Proposition 4.12 implies that Z F ′ (z) dz = 0. F (b) − F (a) = F (z(0)) − F (z(1)) = C

Chapter 4. Line Integrals and Entire Functions

50

In other words, F (b) = F (a). Since b is arbitrary, F is a constant, completing the proof of the problem.  Problem 4.6 Bak and Newman Chapter 4 Exercise 6.

Proof. It is clear that z(t) = eit is a parametrization of the smooth curve |z| = 1. By Definition 4.3, we know that Z 2π Z f (eit )ieit dt. f (z) dz = 0

|z|=1

By the hypothesis, f (eit )ieit is a continuous complex-valued function of t so that the proof of Lemma 4.9 gives Z f (z) dz = Reiθ |z|=1

[0, 2π].c

for some R ≥ 0 and θ ∈ Since R and f (z) are real for all |z| = 1, we have Z f (z) dz =R |z|=1 Z −iθ =e f (z) dz Z

2π

|z|=1

f (eit )iei(t−θ) dt 0 n Z 2π o = Re f (eit )[− sin(t − θ) + i cos(t − θ)] dt 0 Z 2π f (eit ) sin(t − θ) dt. =− =

(4.5)

0

Since −f (eit ) sin(t − θ) ≤ |f (eit ) sin(t − θ)| for all t ∈ [0, 2π], the equation (4.5) reduces to Z

|z|=1

Z f (z) dz ≤

2π 0

Direct computation shows Z

0

|f (eit ) sin(t − θ)| dt ≤

2π

| sin t| dt =

Z

Z

2π

0

π 0

sin t dt −

| sin(t − θ)| dt =

Z

Z

0

2π

| sin t| dt.

(4.6)

2π

sin t dt = 4. π

We conclude from the inequality (4.6) that Z f (z) dz ≤ 4, |z|=1

ending the proof of the problem. Problem 4.7

Bak and Newman Chapter 4 Exercise 7. c

Since the integral is a complex number, we can express it in the polar form.



51 Proof. Suppose that f (z) = α + βz, a ≤ b and c ≤ d. Let z1 = a + ic, z2 = b + ic, z3 = b + id and z4 = a + id. Now the line segment Γ1 from z1 to z2 can be parameterized by γ1 (t) = z1 + t(z2 − z1 ) for t ∈ [0, 1]. Then Definition 4.3 shows that Z 1 Z 1 · (z2 − z1 ) dt = z2 − z1 dz = Γ1

and

Z

z dz = Γ1

Z

1 0

(4.7)

0

[z1 + t(z2 − z1 )](z2 − z1 ) dt = z1 (z2 − z1 ) +

(z2 − z1 )2 z 2 − z12 = 2 . 2 2

(4.8)

Similarly, if Γ2 , Γ3 and Γ4 are the line segments from z2 to z3 , z3 to z4 and z4 to z1 respectively, then we have Z Z Z dz = z1 − z4 (4.9) dz = z4 − z3 and dz = z3 − z2 , as well as

Z

Γ4

Γ3

Γ2

Z

z 2 − z22 z dz = 3 , 2 Γ2

Since

z 2 − z32 z dz = 4 2 Γ3

Z

f (z) dz = Γ

4 Z X

and

Z

z dz =

Γ4

z12 − z42 . 2

(4.10)

f (z) dz,

k=1 Γk

the integrals (4.7), (4.8), (4.9), (4.10) and Proposition 4.8 together imply that Z

Γ

4 h Z X f (z) dz = α k=1

dz + β Γk

Z

4 Z X

i

z dz = α

Γk

k=1 Γk

dz + β

4 Z X

z dz = 0

k=1 Γk

which completes the proof of the problem.



Problem 4.8 Bak and Newman Chapter 4 Exercise 8.

Proof. (a) Clearly, z(θ) ˙ = Rieiθ 6= 0 for all t ∈ [0, 2π], so C is a smooth curve by Definition 4.2. Since k 6= −1, it must be true that  z k+1 ′ = zk . k+1 Since

z k+1 k+1

is analytic on the smooth curve C, Proposition 4.12 implies that Z (Re2πi )k+1 (Re0 )k+1 z k dz = − = 0. k+1 k+1 C

(b) By Definition 4.2, we get Z 2π Z Z f (z) dz = (Reiθ )k · Rieiθ dθ = iRk+1 C

0

This completes the analysis of the problem.

0

2π

ei(k+1)θ dθ =

Rk+1 i(k+1)θ 2π e = 0. k+1 0



Chapter 4. Line Integrals and Entire Functions

52

Problem 4.9 Bak and Newman Chapter 4 Exercise 9.

Proof. 2

(a) Let f (z) = z − i and F (z) = z2 − iz. Since z(t) ˙ = 1 + 2it 6= 0, C is a smooth curve. In ′ addition, F is analytic on C and F (z) = f (z), so we apply Proposition 4.12 to conclude that Z h (1 + i)2 i h (−1 + i)2 i (z − i) dz = F (z(1)) − F (z(−1)) = − i(1 + i) − − i(−1 + i) = 0. 2 2 C (b) If Γ is the line segment from −1 + i to 1 + i which is parameterized by γ(t) = (−1 + i) + t[1 + i − (−1 + i)] = (2t − 1) + i for t ∈ [0, 1], then we have

Z

Γ

(z − i) dz =

Z

1 0

(2t − 1) · 2 dt = 0.

(4.11)

Since the sum C + Γ is a smooth closed curve and f (z) = z − i is entire, Theorem 4.16 (The Closed Curve Theorem) implies that Z Z Z (z − i) dz = 0. (4.12) (z − i) dz + (z − i) dz = C+Γ

Γ

C

Combining the two results (4.11) and (4.12), we conclude that Z (z − i) dz = 0. C

We have completed the proof of the problem.



Problem 4.10 Bak and Newman Chapter 4 Exercise 10.∗

Proof. (a) Recall that (ez )′ = ez . If z(t) = it for t ∈ [0, 1] which is the parametrization of a smooth curve connecting 0 and i, then it follows from Proposition 4.12 that Z i ez dz = ez(1) − ez(0) = ei − 1. 0

(b) Recall that ( 21 sin 2z)′ = cos 2z. If z(t) = π2 + it for t ∈ [0, 1] which is the parametrization of a smooth curve connecting π2 and π2 + i, then it follows from Proposition 4.12 that Z π +i 2 sin(π + 2i) sin π sin(π + 2i) sin 2z(1) sin 2z(0) − = − = . cos 2z dz = π 2 2 2 2 2 2

By Problem 3.20, we see that sin(π + 2i) = −i sinh 2, so we conclude that Z π +i 2 −i sinh 2 . cos 2z dz = π 2 2

This completes the proof of the problem.



53 Problem 4.11 Bak and Newman Chapter 4 Exercise 11.∗

Proof. Let a, b ∈ D. Define the curve C connecting a and b by z(t) = (1 − t)a + tb for t ∈ [0, 1]. Since D is convex, we see that C ⊆ D. Thus Proposition 4.12 is applicable to f ′ along the curve C to get Z f (b) − f (a) =

f ′ (z) dz.

(4.13)

C

Since |f ′ (z)| ≤ 1 on D and C is a line segment of length b− a, Theorem 4.10 (The M -L Formula) reduces the expression (4.13) to Z f ′ (z) dz ≤ 1 · |b − a| = |b − a| |f (b) − f (a)| = C

which is our desired result. This ends the proof of the problem.



Problem 4.12 Bak and Newman Chapter 4 Exercise 12.∗

Proof. If a = b, then we have ea − eb = 0 = a − b. Therefore, we suppose that a 6= b in the following discussion. Let f (z) = ez and D = {z ∈ C | Re z < 0}. Then D is a convex region containing a and b and f is analytic in D. Furthermore, if z = x + iy ∈ D, then x < 0 so that |f ′ (z)| = |ez | = |ex · eiy | = ex ≤ 1.

(4.14)

Hence we may apply Problem 4.11 to conclude that |ea − eb | ≤ |a − b|.

(4.15)

We note that the inequality in Problem 4.11 can be strict provided that we have |f ′ (z)| < 1 in D. Since x < 0, we know that the inequality (4.14) is in fact strict and this implies that the inequality (4.15) there is strict. We complete the proof of the problem. 

Chapter 4. Line Integrals and Entire Functions

54

CHAPTER

5

Properties of Entire Functions

Problem 5.1 Bak and Newman Chapter 5 Exercise 1.

Proof. It is clear that f is entire. Since f ′ (z) = 2z, f ′′ (z) = 2 and f (k) (z) = 0 for all k ≥ 3, it follows from Corollary 5.7 that f (z) = 4 + 4(z − 2) + 2(z − 2)2 . This completes the proof of the problem.



Problem 5.2 Bak and Newman Chapter 5 Exercise 2.

Proof. It is obvious that f is entire. Since f (k) (z) = ez for all k ≥ 1, it follows from Corollary 5.7 that ∞ X (z − a)k ea 2 a a a , f (z) = e + e (z − a) + (z − a) + · · · = e 2! k! k=0

completing the proof of the problem.



Problem 5.3 Bak and Newman Chapter 5 Exercise 3.

Proof. (a) Using Problem 3.3 repeatedly, we have f (k) (z) = (−1)k+1 f (k)(−z) for all k ∈ N. In other words, the derivative of an odd (resp. even) function is an even (resp.) function. Furthermore, f (z) = −f (−z) implies that f (0) = 0. These two facts combine to give f (2n) (0) = 0 55

Chapter 5. Properties of Entire Functions

56

for all n = 0, 1, 2, . . .. Hence we obtain from Corollary 5.7 that ∞

X f (2k−1) (0) f (3) (0) 3 f (z) = f (0)z + z + ··· = z 2k−1 . 3! (2k − 1)! ′

k=1

(b) By the analysis in part (a), we have f (2k−1) (0) = 0 for all k ∈ N, so we follow again from Corollary 5.7 that f (z) = f (0) +

∞

X f (2k) (0) f ′′ (0) 2 z + ··· = z 2k . 2! (2k)! k=1

We have ended the proof of the problem.



Problem 5.4 Bak and Newman Chapter 5 Exercise 4.

Proof. Suppose that f (z) =

∞ X

Ck z k .

k=0

On the one hand, using [4, Eqn. (1), p. 63], we have 1 Ck = 2πi

Z

C

f (ω) dω, ω k+1

(5.1)

where k = 0, 1, 2, . . .. On the other hand, Corollary 2.11 implies that Ck =

f (k) (0) k!

(5.2)

for all k = 0, 1, 2, . . .. Hence our desired formula follows immediately by comparing the different  expressions (5.1) and (5.2). We complete the proof of the problem. Problem 5.5 Bak and Newman Chapter 5 Exercise 5.

Proof. Let g(z) = f (z + a). Then g is entire because f is entire. It is easy to check that g(k) (z) = f (k) (z + a) so that g(k) (0) = f (k) (a) for all k = 0, 1, 2, . . .. Now it follows from Problem 5.4 that Z Z g(ζ) f (ζ + a) k! k! (k) (k) dζ = dζ, (5.3) f (a) = g (0) = k+1 2πi C ζ 2πi C ζ k+1 where C is any circle surrounding the origin. If ζ(t) = Reit for some R > 0 and 0 ≤ t ≤ 2π, then Definition 4.3 shows that the integral in the expression (5.3) becomes Z

C

f (ζ + a) dζ = ζ k+1

Z

0

2π

f (Reit + a) · Rieit dt. (Reit )k+1

(5.4)

57 Let ω = ζ + a. Then the locus of ω is the circle centred at a with radius R. Besides, we have Reit = ζ(t) = ω(t) − a and ω(t) ˙ = Rieit . Therefore, we observe from the integral (5.4) that Z

C

Z

2π

f (Reit + a) · Rieit dt (Reit )k+1 0 Z 2π f (ω(t)) = ω(t) ˙ dt [ω(t) − a]k+1 Z0 f (ω) dω, = k+1 C ′ (ω − a)

f (ζ + a) dζ = ζ k+1

(5.5)

where C ′ is the circle surrounding the point a. Combining the two formulas (5.3) and (5.5), we conclude that Z f (ω) k! (k) f (a) = dω. 2πi C ′ (ω − a)k+1 This completes the proof of the problem.



Problem 5.6 Bak and Newman Chapter 5 Exercise 6.

Proof. (a) By the expression (5.1), we have 1 f (k) (0) = Ck = k! 2πi

Z

C

f (ω) dω, ω k+1

where C is the circle centred at the origin with radius R. Since f is entire, continuous on the smooth curve C. Now the length of C is 2πR and

f (ω) ω k+1

is

f (ω) |f (ω)| M k+1 = k+1 ≤ k+1 ω R R

on C, so Theorem 4.10 (The M -L Formula) implies that Z f (ω) 1 M M 1 dω · k+1 · (2π) = k+1 , |Ck | = ≤ k+1 2π C ω 2π R R

as desired.

(b) Apply part (a) with R = M = 1 to get |Ck | ≤ 1 for all k = 0, 1, 2, . . .. We have completed the proof of the problem.



Problem 5.7 Bak and Newman Chapter 5 Exercise 7.

Proof. Let R > 0. On the circle |z| = R, we have |f (z)| ≤ A + BRk . By Problem 5.6(a), we know that A + BRk , (5.6) |Cj | ≤ Rj

Chapter 5. Properties of Entire Functions

58

where j = 0, 1, 2, . . .. If j > k, then j − k > 0 and the inequality (5.6) implies that A B  + = 0. R→+∞ Rj Rj−k

lim |Cj | ≤ lim

R→+∞

Hence Cj = 0 for all j = k + 1, k + 2, . . . and Theorem 5.11 (The Extended Liouville Theorem)  follows, completing the proof of the problem. Problem 5.8 Bak and Newman Chapter 5 Exercise 8.

Proof. On the circle |z| = R > 0, the hypothesis implies that 3

|f (z)| ≤ A + BR 2 .

(5.7)

Thus we deduce from the bound (5.7), Problem 5.6(a) and Theorem 5.5 (The Taylor Expansion of an Entire Function) that 3

|f

(k)

k!(A + BR 2 ) . (0)| = |k!Ck | ≤ Rk

(5.8)

Notice that the inequality (5.8) holds for every R > 0, so if k ≥ 2, then we conclude that 3

lim |f

R→+∞

(k)

k!(A + BR 2 ) = 0. (0)| ≤ lim R→+∞ Rk

Hence f (k) (0) = 0 for all k = 2, 3, . . . and then f is a linear polynomial. We have completed the  proof of the problem. Problem 5.9 Bak and Newman Chapter 5 Exercise 9.

Proof. Since f is entire, f ′ is everywhere differentiable by Corollary 5.6. By Definition 3.3, f ′ is also entire. By Theorem 5.11 (The Extended Liouville Theorem), f ′ (z) is a polynomial of degree at most 1 and then we apply Theorem 5.5 (The Taylor Expansion of an Entire Function) to f ′ to get f (k) (0) = 0 (5.9) for k = 3, 4, . . .. Furthermore, we have f ′ (0) = 0 so that Theorem 5.5 (The Taylor Expansion of an Entire Function) and the values (5.9) ensure that f (z) = f (0) + f ′ (0)z +

f ′′ (0) 2 z = a + bz 2 , 2!

′′

where a = f (0) and b = f 2(0) . Next, we recall from Problem 5.4 that 1 f (0) = 2πi ′′

Z

C

f ′ (ω) dω, ω2

(5.10)

59 where C is a circle surrounding the origin. In particular, we take C to be the unit circle |ω| = 1 ′ so that f ω(ω) ≪ 1 on C. Therefore, we apply Theorem 4.10 (The M -L Formula) to the integral 2 (5.10) to yield 1 |f ′′ (0)| ≤ · 1 · 2π = 1 2π and hence 1 |f ′′ (0)| ≤ . |b| = 2 2  This completes the proof of the problem. Problem 5.10 Bak and Newman Chapter 5 Exercise 10.

Proof. Assume that f was a nonconstant entire function satisfying the two equations. By repeated applications of the two equations, we can show that f satisfies the following equations f (z) = f (z + n) and f (z) = f (z + in)

(5.11)

for all n ∈ Z.

Consider the unit square D = {a + ib | a, b ∈ [0, 1]} which is compact. Since f is entire, it modulus is continuous on C and thus |f (D)| is a compact subset of R. By the Heine-Borel Theorem [27, p. 30], |f (z)| is bounded on D. Let z = x + iy ∈ C. Recall that x can be written as the sum of [x] and {x}, where [x] and {x} are the greatest integer function and the fractional part of the real number x respectively. We note that [x] ∈ Z and 0 ≤ {x} < 1. Similarly, we have y = [y] + {y}. Therefore, we obtain z = x + iy = ([x] + i[y]) + ω, where ω = {x} + i{y}. It is easily seen that ω ∈ D. By the equations (5.11), we see that f (z) = f ((ω + i[y]) + [x]) = f (ω + i[y]) = f (ω). In other words, f is bounded on C and Theorem 5.10 (Liouville’s Theorem) implies that f is constant, a contradiction. Hence no such function exists and we have completed the proof of the problem.  Problem 5.11 Bak and Newman Chapter 5 Exercise 11.

Proof. Let P (z) be a real polynomial of odd degree. By Problem 1.5, we know that the complex zeros of a real polynomial appear in conjugate pairs. Consequently, the number of complex zeros of P (z) must be even. Since deg P (z) is odd, Theorem 5.12 (The Fundamental Theorem of Algebra) asserts that the number of zeros (counting multiplicity) of P (z) must be odd. Hence P (z) has at least one real zero. This ends the analysis of the problem.  Problem 5.12 Bak and Newman Chapter 5 Exercise 12.

Chapter 5. Properties of Entire Functions

60

Proof. Let P (z) be a real polynomial of degree n. By Theorem 5.12 (The Fundamental Theorem of Algebra), we can express P (z) as P (z) = an (z − z1 )(z − z2 ) · · · (z − zn ) for some an ∈ C and z1 , z2 , . . . , zn are the zeros of P (z). By Problem 1.5, zi is a zero of P (z) if and only if zi is a zero of P (z). There are two cases: • Case (i): zi is real. Then zi is also real and so P (z) has z − zi as its factor. • Case (ii): zi is non-real. Then we have (z − zi )(z − zi ) = z 2 − 2Re (zi )z + |zi |2 which is a real quadratic polynomial. Hence P (z) is a product of real linear and quadratic polynomials, completing the proof of the problem.  Problem 5.13 Bak and Newman Chapter 5 Exercise 13.

Proof. We prove the result in several steps. • Step 1: v(x, y) ≡ 0 if and only if y = 0. If we suppose that P (x+iy) = u(x, y)+iv(x, y), where u and v are real-valued functions, then the hypothesis says that v(x, y) = 0 if and only if y = 0. • Step 2: Either v(x, y) > 0 or v(x, y) < 0 in the upper half-plane. Since P is a polynomial, it is differentiable everywhere and thus v is continuous on C. Fix a point x first, if v(x, y1 ) > 0 and v(x, y2 ) < 0 with 0 < y1 < y2 , then the Intermediate Value Theorem [27, p. 101] asserts that there exists a y0 ∈ (y1 , y2 ) such that v(x, y0 ) = 0 which contradicts Step 1. This means that either v(x, y) > 0 or v(x, y) < 0 for all y > 0. Next, we suppose that x1 is a number such that v(x1 , y) > 0 for all y > 0. Fix a y > 0. If v(x2 , y) < 0, where x2 6= x1 , then we apply the Intermediate Value Theorem to conclude that v(x0 , y) = 0 for some x0 ∈ (x1 , x2 ) or x0 ∈ (x2 , x1 ) which is a contradiction again. Thus we have the desired result. • Step 3: Either vy (x, 0) ≥ 0 for all x ∈ R or vy (x, 0) ≤ 0 for all x ∈ R. Assume that vy (x, 0) changed sign on R, i.e., there exist x1 < x2 such that vy (x1 , 0) > 0 and vy (x2 , 0) < 0. Define G1 , G2 : R → R by G1 (y) = v(x1 , y)

and G2 (y) = v(x2 , y).

The assumption vy (x1 , 0) > 0 implies that G1 is strictly increasing in (−δ1 , δ1 ) for some δ1 > 0. Similarly, G2 is strictly decreasing in (−δ2 , δ2 ) for some δ2 > 0. Therefore, by using Step 1, one can find a ǫ > 0 such that v(x1 , ǫ) > v(x1 , 0) = 0

and v(x2 , ǫ) < v(x2 , 0) = 0.

(5.12)

61 Now if we define F (x) = v(x, ǫ) which is a continuous function with respect to x, then we follow from the Intermediate Value Theorem and the inequalities (5.12) that F (x3 ) = v(x3 , ǫ) = 0 for some x3 ∈ (x1 , x2 ), a contradiction to Step 1. • Step 4: u(x, 0) is monotonic on R. Since P (z) is entire, Proposition 2.3 or 3.1 implies that ux (x, 0) = vy (x, 0) so that either ux (x, 0) ≥ 0 or ux (x, 0) ≤ 0 for all x ∈ R by Step 3. Hence u(x, 0) is monotonic on R. • Step 5: P (z) = α has only one solution for all α ∈ R. Let deg P (z) = n. By Theorem 5.12 (The Fundamental Theorem of Algebra), the polynomial P (z) has exactly n roots z1 , z2 , . . . , zn ∈ C, counted with multiplicity. Let zk = xk +iyk , where k = 1, 2, . . . , n. Then we get P (z) = A(z − z1 )(z − z2 ) · · · (z − zn ) (5.13) for some A ∈ C.

Recall that P = u + iv, so P (zk ) = 0 implies that u(xk , yk ) = v(xk , yk ) = 0. By Step 1, we must have yk = 0 for all k = 1, 2, . . . , n. Hence we obtain from the expression (5.13) that P (z) = A(z − x1 )(z − x2 ) · · · (z − xn ). (5.14) Without loss of generality, we may assume that x1 ≤ x2 ≤ · · · ≤ xn . Now our hypothesis forces that A must be real. Since P (x, 0) = u(x, 0), we get from the expression (5.14) that u(x, 0) = A(x − x1 )(x − x2 ) · · · (x − xn ). Assume that P (x, 0) had at least two distinct zeros, say x1 < x2 . Since P (x, 0) is a continuous function of x, P (x, 0) = u(x, 0) cannot be monotonic, but this contradicts Step 4. Therefore, we conclude that x1 = x2 = · · · = xn and the representation (5.14) reduces to P (z) = A(z − x1 )n . (5.15) πi

Next, if n > 1, then z1 = x1 + e n is a non-real complex number and P (z1 ) = A(z1 − x1 )n = Aeπi = −A.

(5.16)

Recall that A is real, so the result (5.16) contradicts to our hypothesis that P is real if and only if z is real. Consequently, we have n = 1 and so the expression (5.15) gives P (z) = Az − Az1 , as desired. We complete the analysis of the problem. Problem 5.14 Bak and Newman Chapter 5 Exercise 14.



Chapter 5. Properties of Entire Functions

62

Proof. If P (α) = P ′ (α) = · · · = P (k−1) (α) = 0 and P (k) (α) 6= 0, then Corollary 5.7 implies that P (z) =

P (k) (α) P (k+1) (α) (z − α)k + (z − α)k+1 + · · · = (z − α)k Q(z), k! (k + 1)!

where Q(z) = and Q(α) =

P (k) (α) k!

P (k) (α) P (k+1) (α) + (z − α) + · · · k! (k + 1)!

6= 0.

Conversely, suppose that P (z) = (z − α)k Q0 (z), where Q0 (α) 6= 0. Then we have P ′ (z) = (z − α)k−1 Q1 (z), where Q1 (z) = (z − α)Q′0 (z) + kQ0 (z). Therefore, it is easily seen that P (n) (z) = (z − α)k−n Qn (z),

(5.17)

where Qn (z) = (z − α)Q′n−1 (z)+ kQn−1 (z) which satisfies Qn (α) = kQn−1 (α) for n = 1, 2, . . . , k. Thus the expression (5.17) implies immediately that  if 1 ≤ n ≤ k − 1;  0, P (n) (α) =  k!Q0 (α), if k = n. Hence we have completed the analysis of the problem.



Problem 5.15 Bak and Newman Chapter 5 Exercise 15.

Proof. Fix a point z ∈ C. Consider the line segment connecting 0 and z which is parameterized by {tz | t ∈ [0, 1]}. Suppose that Kz = {t ∈ [0, 1] | |f (tz)| ≤ 1} which is a bounded set. Define   sup Kz , if Kz 6= ∅; d= (5.18)  0, otherwise.

Obviously, we have 0 ≤ d ≤ 1. Let zd = dz. We claim that

|f (zd )| ≤ max{1, |f (0)|}.

(5.19)

On the one hand, if d = 0, then zd = 0 so that |f (zd )| = |f (0)|.

(5.20)

On the other hand, suppose that d > 0 and {tn } is a sequence of Kz with limit point t0 . Since f is continuous and |f (tn z)| ≤ 1 for all n ∈ N, we have |f (t0 z)| = lim |f (tn z)| ≤ 1. n→∞

In other words, t0 ∈ Kz and thus Kz is closed in [0, 1]. By the Heine-Borel Theorem, Kz is compact. Furthermore, we recall from [27, Problem 4.27, pp. 44, 45] that d ∈ Kz which implies |f (zd )| = |f (dz)| ≤ 1.

(5.21)

63 Hence our claim (5.19) follows immediately from the results (5.20) and (5.21). By the definition (5.18), we have |f (tz)| > 1 for all t ∈ [d, 1] and the hypothesis implies that |f ′ (tz)| ≤ 1 for all t ∈ [d, 1], i.e., |f ′ (ω)| ≤ 1 along the line segment connecting zd and z. Combining this fact and Proposition 4.12, we havea Z z Z z ′ f ′ (ω) dω = |f (z) − f (zd )| ≥ |f (z)| − |f (zd )| |f (ω)| dω ≥ |z| ≥ |z − zd | ≥ zd

zd

and so

|f (z)| ≤ |f (zd )| + |z| ≤ max{1, |f (0)|} + |z|, where z ∈ C. Consequently, it follows from Theorem 5.11 (The Extended Liouville Theorem) that f is a linear polynomial. This completes the proof of the problem.  Problem 5.16 Bak and Newman Chapter 5 Exercise 16.∗

Proof. Let P (z) = an z n + an−1 z n−1 + · · · + an be a polynomial having zeros z1 , z2 , . . . , zn , where an 6= 0. By [4, Eqn. (4), p. 67], the centroid of zeros of P (z) is z1 + z2 + · · · + zn an−1 =− . n nan

(5.22)

Now we let ω1 , ω2 , . . . , ωn−1 be the zeros of P ′ (z) = nan z n−1 + (n − 1)an−1 z n−2 + · · · + a2 z + a1 . In other words, we have nan z n−1 + (n − 1)an−1 z n−2 + · · · + a2 z + a1 = nan (z − ω1 )(z − ω2 ) · · · (z − ωn−1 ).

(5.23)

By comparing coefficients of z n−2 on both sides of the expression (5.23), we see immediately that (n − 1)an−1 = −nan (ω1 + ω2 + · · · + ωn−1 ) an−1 ω1 + ω2 + · · · + ωn−1 =− . n−1 nan

(5.24)

Hence the result follows if we combine the two expressions (5.22) and (5.24). This completes  the proof of the problem. Problem 5.17 Bak and Newman Chapter 5 Exercise 17.∗

Proof. Let S be a convex set. The statement for n = 1 and n = 2 are trivial. Assume that the statement is true for n = k ∈ N, i.e., if z1 , z2 , . . . , zk ∈ S, then a1 z1 + a2 z2 + · · · + ak zk ∈ S, where aj ≥ 0 for all 1 ≤ j ≤ k and 1 ≤ j ≤ k + 1 and a

k+1 X

k X j=1

(5.25)

aj = 1. Now let z1 , z2 , . . . , zk , zk+1 ∈ S, aj ≥ 0 for all

aj = 1.

j=1

In fact, we also apply Definition 4.3 and Lemma 4.9 to show the third inequality holds.

Chapter 5. Properties of Entire Functions

Define A =

k X j=1

64

ai . Then the definition gives 1 − A = ak+1 which implies that

a a2 ak  1 (a1 z1 + a2 z2 + · · · + ak zk ) + ak+1 zk+1 = A z1 + z2 + · · · + zk + (1 − A)zk+1 . (5.26) A A A k X aj

Since

j=1

A

= 1, the induction assumption (5.25) ensures that the bracket on the right-hand

side of the expression (5.26) is an element of S. Therefore, the convexity of S further implies that the right-hand side of the expression (5.26) belongs to S. Hence our required result follows from induction and it ends the proof of the problem.  Problem 5.18 Bak and Newman Chapter 5 Exercise 18.∗

Proof. (a) Let z1,k , z2,k , . . . , zk,k be the zeros of Pk (z), where k ∈ N. Applying [4, Eqn. (4), p. 67], we have 1 z1,k + z2,k + · · · + zk,k 1 h (k−1)! i = −1. = × − 1 k k k! (b) Without loss of generality, we may assume that zk,k is a zero of Pk (z) with maximal possible modulus. We note that ′ Pk+1 (z) = 1 + z + · · · +

zk = Pk (z) k!

for every k = 0, 1, 2, . . .. By the hypothesis, we know that |zj,k+1 | ≤ |zk+1,k+1 | for all j = 1, 2 . . . , k. In other words, we get zj,k+1 ∈ D(0; |zk+1,k+1 |) for all j = 1, 2 . . . , k + 1. Since D(0; |zk+1,k+1 |) is convex, it contains the convex hull of z1,k+1 , z2,k+1 , . . . , zk+1,k+1 by Definition 5.13. Finally, Theorem 5.14 (The Gauss-Lucas ′ Theorem) implies that all the zeros of Pk+1 (z) = Pk (z) lie within the convex hull of the zeros of Pk+1 (z). Particularly, we obtain |zk,k | ≤ |zk+1,k+1 | and hence {|zk,k |} is an increasing sequence.

We end the proof of the problem. Problem 5.19

Bak and Newman Chapter 5 Exercise 19.∗



65 Proof. Let Q(z) = 1 + z + z 2 + · · · + z n . Then it is easy to check that Q′ (z) = P (z). Note that Q(z) =

z n+1 − 1 , z−1

so the zeros of Q(z) lie on the unit circle. Let S = {z1 , z2 , . . . , zn+1 } be set of the zeros of Q(z) and Conv (S) the convex hull of S. Since S ⊆ D(0; 1) and D(0; 1) is a convex set, Definition 5.13 ensures that Conv (S) ⊆ D(0; 1). Hence we deduce from Theorem 5.14 (The Gauss-Lucas Theorem) that the zeros of P (z) lie within Conv (S) ⊆ D(0; 1). This completes the proof of the problem.  Remark 5.1 In fact, the convex hull of a finite set S = {z1 , z2 , . . . , zn+1 } is given by n+1 n X o aj . Conv (S) = a1 z1 + a2 z2 + · · · + an+1 zn+1 aj ≥ 0 for 1 ≤ j ≤ n + 1 and j=1

See, for example, [26, Eqn. (10.33), p. 268].

Problem 5.20 Bak and Newman Chapter 5 Exercise 20.∗

Proof. Using the method of Chapter 1, we see that the solutions of the polynomial equation z 2 = i are 1 z = ± √ (1 + i). 2 √ Therefore, we have i = √12 (1 + i). √ To find the estimates of i by Newton’s method, we first let f (z) = z 2 − i

and g(z) = z −

f (z) z2 − i = z − . f ′ (z) 2z

Take z0 = 1. Then we have 1+i 1−i = , 2 2 − 2i 1+i 1+i 1+i 3 z2 = g(z1 ) = − = + = (1 + i), 2 1+i 2 4 4 17 (1 + i), z3 = g(z2 ) = 24 865 z4 = g(z3 ) = (1 + i). 1224 z1 = g(z0 ) = 1 −

This completes the proof of the problem.



Chapter 5. Properties of Entire Functions

66

CHAPTER

6

Properties of Analytic Functions

Problem 6.1 Bak and Newman Chapter 6 Exercise 1.

Proof. It is easily seen that f (z) = √ z ∈ D(1 + i; 2), then we obtain

1 z

is analytic everywhere except at z = 0. Therefore, if

1 1 = z (1 + i) + [z − (1 − i)] 1 1 · = z−(1+i) 1+i 1+ 1+i h i z − (1 + i) [z − (1 + i)]2 1 · 1− + − · · · = 1+i 1+i (1 + i)2 ∞ X (−1)k (z − 1 − i)k = . (1 + i)k+1 k=0

This completes the proof of the problem.



Problem 6.2 Bak and Newman Chapter 6 Exercise 2.∗

Proof. By partial fractions, we have f (z) = We know that

1 1 2 1 1 1 = = · + · . 1 − z − 2z 2 (1 + z)(1 − 2z) 3 1 + z 3 1 − 2z ∞

X 1 (−z)n = 1 + z n=0

and

∞

X 1 (2z)n = 1 − 2z n=0

(6.1)

(6.2)

for |z| < 1 and for |z| < 12 respectively. Hence, for |z| < 21 , by putting the identities (6.2) into the expression (6.1), we see that f (z) =

∞ h X 2(−1)n

n=0

ending the proof of the problem.

3

67

+

2n i n z , 3 

Chapter 6. Properties of Analytic Functions

68

Problem 6.3 Bak and Newman Chapter 6 Exercise 3.

Proof. We note that the series

∞ X

n=0

z n converges for |z| < 1, so Theorem 2.9 implies that ∞ X

z

n

n=0

′

=

∞ X

nz n−1

(6.3)

n=0

throughout |z| < 1. Therefore, we observe from the formula (6.3) that ∞ X

nz n = z

∞ X

nz n−1 = z

zn

n=0

n=0

n=0

∞ X

′

=z·

 1 ′ z = 1−z (1 − z)2

(6.4)

throughout |z| < 1.

Similarly, further application of Theorem 2.9 gives ∞ X

zn

n=0

′′

∞ X

=

n=0

n(n − 1)z n−2 .

(6.5)

throughout |z| < 1. Hence we deduce from the formulas (6.4) and (6.5) that ∞ X

n2 z n =

n=0

∞ X

n(n − 1)z n +

n=0 ∞ X 2

=z

n=0

= z2

∞ X

nz n

n=0

z (1 − z)2

n(n − 1)z n−2 +

∞ X

zn

n=0

′′

+

z (1 − z)2

 1 ′′ z + =z 1−z (1 − z)2 2z 2 z = + 3 (1 − z) (1 − z)2 z(1 + z) . = (1 − z)3 2

We have completed the proof of the problem.



Problem 6.4 Bak and Newman Chapter 6 Exercise 4.

Proof. Assume that f ( n1 ) = g

1 n+1

1 n

for all n ∈ N. Consider g(z) = f (z) −

=f

1 n

−

1 n 1 n

+1

=

z 1+z .

Then we have

1 1 − =0 n+1 n+1

for all n ∈ N. Since { n1 } is a sequence of distinct points and Uniqueness Theorem) implies that z f (z) = 1+z

1 n

→ 0 ∈ D(0; 1), Theorem 6.9 (The (6.6)

69 in D(0; 1). Since f is analytic in D(0; 1), it must be continuous there. This means that lim f (zn ) = f (−1)

n→∞

(6.7)

holds for every sequence {zn } ⊆ D(0; 1) such that zn → −1 as n → ∞ and zn 6= −1 for all n ∈ N. However, the representation (6.6) guarantees that the limit (6.7) is impossible. Hence  no such analytic function exists and we complete the proof of the problem. Problem 6.5 Bak and Newman Chapter 6 Exercise 5.

Proof. We fix z2 ∈ R. Let f (z) = sin(z + z2 ) and g(z) = sin z cos z2 + cos z sin z2 . Then f and g are entire with respect to z. Since 1 1 =g f n n

for all n ∈ N, it follows from Corollary 6.10 that f ≡ g in C (with respect to z). Next, we fix z1 ∈ C and consider F (ω) = sin(z1 + ω) and G(ω) = sin z1 cos ω + cos z1 sin ω. Obviously, they are entire with respect to ω. The previous paragraph ensures that F

1

 1 1 1 1 = sin z1 + = f (z1 ) = g(z1 ) = sin z1 cos + cos z1 sin =G n n n n n

for all n ∈ N. Thus Corollary 6.10 shows that F ≡ G in C (with respect to ω) and hence the formula sin(z1 + z2 ) = sin z1 cos z2 + cos z1 sin z2 holds for all z1 , z2 ∈ C. We complete the proof of the problem.



Problem 6.6 Bak and Newman Chapter 6 Exercise 6.

Proof. Let D = C \ {(k + 21 )π | k ∈ Z}. Since π is irrational, we have n1 ∈ D for all n ∈ N. Since f (z) agrees with the analytic function tan z on the set { n1 } and n1 → 0 in D as n → ∞, Corollary 6.10 implies that f (z) = tan z in D. Thus f (z) = i if and only if tan z = i. Using the identities for sin z and cos z on [4, p. 41], we have eiz − e−iz = −1 eiz + e−iz eiz − e−iz = −eiz − e−iz eiz = 0

which is impossible. Furthermore, if f is entire, then f is continuous in C. Thus using similar argument as in the proof of Problem 6.4, it is impossible that f is not entire. This completes the proof of the problem. 

Chapter 6. Properties of Analytic Functions

70

Problem 6.7 Bak and Newman Chapter 6 Exercise 7.

Proof. We notice that N ∈ N. Since f is entire and |f (z)| ≥ |z|N , f (z) → ∞ as z → ∞. By Theorem 6.11, f is a polynomial. Let f (z) = an z n + an−1 z n−1 + · · · + a1 z + a0 , where an is a nonzero constant. Then the triangle inequality implies that |f (z)| ≤ |an | · |z|n + |an−1 | · |z|n−1 + · · · + |a1 | · |z| + |a0 |  |an−1 | |a0 |  ≤ |z|n · |an | + + ··· + n |z| |z| n ≤ 2|an | · |z|

(6.8)

for sufficiently large z. Assume that n < N . Now the hypothesis and the inequality (6.8) give |z|N ≤ 2|an | · |z|n for sufficiently large enough z, but this means that an = ∞, a contradiction. Hence deg f (z) ≥ N  and we complete the proof of the problem. Problem 6.8 Bak and Newman Chapter 6 Exercise 8.

Proof. Suppose that f is constant. Since |f (z)| ≤ 2 for√all |z| = 1 with Im z ≥ 0, it is actually true for all |z| ≤ 1. In particular, we have |f (0)| ≤ 2 < 6.

Without loss of generality, we may assume that f is nonconstant. Define F (z) = f (z)f (−z). By the definition on [4, p. 86], F is analytic in D(0; 1) and continuous on the compact set D(0; 1) so that max |F (z)| exits in D(0; 1) by the Extreme Value Theorem. Now Theorem 6.13 (The Maximum Modulus Theorem) implies that this maximum must lie on |z| = 1. Hence our hypotheses assert that |F (z)| ≤ max F (z) = max(|f (z)| · |f (−z)|) ≤ 2 × 3 = 6 |z|=1

|z|=1

for all |z| ≤ 1. Particularly, if z = 0, we have which implies |f (0)| ≤

√

|f (0)|2 = |F (0)| ≤ 6 6, completing the proof of the problem.



Problem 6.9 Bak and Newman Chapter 6 Exercise 9.

Proof. Let z = x + iy and K ⊆ C be compact. Since ez is continuous on K, the Extreme Value Theorem ensures that the function |ez | attains its maximum and minimum in K. Let K ◦ be the interior of K and z0 = x0 + iy0 ∈ K ◦ . Assume that z0 was an extreme of the |ez |. We know that K ◦ is open, so there exists a δ > 0 such that D(z0 ; δ) ⊆ K ◦ . It is obvious that z± = z0 ± 2δ ∈ D(z0 ; δ) so that δ

δ

|ez+ | = ex0 + 2 > ex0 = |ez0 | and |ez− | = ex0 − 2 < ex0 = |ez0 |

71 which imply contradictions. Hence the modulus of ez can only have an extreme on ∂K which completes the proof of the problem.  Problem 6.10 Bak and Newman Chapter 6 Exercise 10.

Proof. Let f (z) = z 2 − z = z(z − 1) which is clearly nonconstant and analytic in D(0; 1). On the one hand, since f (0) = 0, we have min |f (z)| = 0.

|z|≤1

On the other hand, the Extreme Value Theorem and Theorem 6.13 (The Maximum Modulus Theorem) ensure that max |f (z)| exists on |z| = 1. To find this maximum, if |z| = 1, then z = reiθ so that

|z|≤1

|f (z)| = |z − 1| =

q

(cos θ − 1)2 + sin2 θ =

√

2 − 2 cos θ.

Therefore, f takes its maximum when θ = π or equivalently at z = −1 and furthermore, we have max |f (z)| = 2. |z|≤1

This completes the proof of the problem.



Problem 6.11 Bak and Newman Chapter 6 Exercise 11.∗

Proof. Let C = ∂D(α; r) for some α ∈ C and r > 0. Since f is analytic in the compact set D(α; r), it is analytic in an open set D containing D(α; r) by Definition 3.3. Consequently, there exists a R > r such that D(α; r) ⊆ D(α; R) ⊆ D. (6.9)

/ D. Since {ωn } ⊆ D(α; r+1) Otherwise, for each n ∈ N, one can find a ωn ∈ D(α; r+ n1 ) but ωn ∈ and |ωn | ≤ |ωn − α| + |α|, {|ωn |} is a bounded sequence. Thus the Bolzano-Weierstrass Theorem [27, Problem 5.25] ensures that there is a convergent subsequence {ωnk }. Let ωnk → ω0 as k → ∞. Next, we note that {ωnk } ⊆ C \ D. Since D is open in C, C \ D is closed in C so that ω0 ∈ C \ D. However, since |ωnk − α| < r +

1 nk

(6.10)

for all k ∈ N, we have |ω0 − α| ≤ r,

i.e., ω0 ∈ D(α; r) which is contrary to the set relation (6.10). Hence the set relation (6.9) holds for some R > r. Let z0 ∈ D(α; r) and n ∈ N. Using Theorem 6.4 (The Cauchy Integral Formula) to the analytic function f n to get Z f n (z) 1 n dz. (6.11) f (z0 ) = 2πi C z − z0

Chapter 6. Properties of Analytic Functions

72

Now we want to apply Theorem 4.10 (The M -L Formula) to the expression (6.11), so we check n (z) the hypotheses. Since z0 ∈ / C, the function fz−z is continuous on C. Let ρC (z0 ) = inf |z − z0 |. 0 z∈C

Recall that ρC (z0 ) = 0 if and only if z0 ∈ C. Therefore, ρC (z0 ) > 0. Since ρC (z0 ) ≤ |z − z0 | for all z ∈ C, it must be true that f n (z) Mn . (6.12) ≤ z − z0 ρC (z0 )

Consequently, it follows from Theorem 4.10 (The M -L Formula) and the estimate (6.12) that 1 Z f n (z) 1 Mn M nr n |f (z0 )| = dz ≤ · · (2πr) = 2πi C z − z0 2π ρC (z0 ) ρC (z0 ) which gives

|f (z0 )| ≤



r  n1 · M. ρC (z0 )

(6.13)

Take n → ∞, the inequality (6.13) implies that |f (z0 )| ≤ M for all z0 ∈ D(α; r). This proves Theorem 6.13 (The Maximum-Modulus Theorem) and we complete the proof of the problem.  Remark 6.1 For Problem 6.11, please also see the book of P´ olya and Szeg˝ o [21, p. 29].

Problem 6.12 Bak and Newman Chapter 6 Exercise 12.

Proof. If both f and g are constant, then there is nothing to prove. Thus, without loss of generality, we may assume that at least one of f and g is nonconstant. Let z0 be a maximum of |f (z)| + |g(z)| in D. Furthermore, we let f (z0 ) = R1 e−iα and g(z0 ) = R2 e−iβ for some α, β ∈ [0, 2π]. We consider the function h : C → C defined by h(z) = f (z)eiα + g(z)eiβ . Since f and g are analytic in D and at least one of them is nonconstant, h is nonconstant and analytic in D. Therefore, the Extreme Value Theorem and Theorem 6.13 (The Maximum Modulus Theorem) show that the modulus |h(z)| attains its maximum on ∂D. By the definition, we have

|h(z)| ≤ |eiα | · |f (z)| + |eiβ | · |g(z)| = |f (z)| + |g(z)|

(6.14)

for all z ∈ D. Besides, we also have |h(z0 )| = |f (z0 )eiα + g(z0 )eiβ | = R1 + R2 = |f (z0 )| + |g(z0 )|.

(6.15)

Now these two facts (6.14) and (6.15) combine to say that the modulus |h(z)| attains its maximum, which is exactly |f (z0 )| + |g(z0 )|, at z0 . Hence we conclude from the previous paragraph  that |f (z)| + |g(z)| takes its maximum on ∂D. This completes the proof of the problem. Problem 6.13 Bak and Newman Chapter 6 Exercise 13.

73 Proof. Suppose that P (z) is a nonconstant polynomial and R > 0. Assume that P (z) 6= 0 in the region D(0; R). Now Theorem 6.14 (The Minimum Modulus Theorem) says that the minimum modulus of P (z) in D(0; R) exists on |z| = R, i.e., there exists a z0 such that |z0 | = R and |P (z0 )| < |P (z)| holds for all |z| < R. Particularly, we have |P (z0 )| < |P (0)|

(6.16)

However, we recall from Problem 1.26 that |P (z)| ≥ M |z|n for large |z|, where M is a positive constant and n = deg P . So if R is large enough, then we have |P (z0 )| > 2|P (0)| which definitely contradicts the inequality (6.16). Hence P (ω) = 0 for some ω ∈ C which proves the Fundamental Theorem of Algebra. This ends the analysis of the problem.  Problem 6.14 Bak and Newman Chapter 6 Exercise 14.

Proof. If P (z) is constant, then P (z) = a0 with |a0 | ≤ 1 and thus there is nothing to prove. We suppose that P (z) is nonconstant. By Problem 5.6(b), we have |ak | ≤ 1

(6.17)

for all k = 0, 1, 2, . . . , n. Take R > 1 and consider f (z) =

P (z) a0 a1 = n + n−1 + · · · + an n z z z

(6.18)

defined in the compact annulus D = {z ∈ C | 1 ≤ |z| ≤ R}. Since f is C-analytic in the region {z ∈ C | 1 < |z| < R}, the Extreme Value Theorem and Theorem 6.13 (The Maximum Modulus Theorem) guarantee that the maximum modulus of f in D occurs on ∂D. Thus if |z| = 1, since |P (z)| ≤ 1 for all |z| ≤ 1, then |f (z)| ≤ 1. Next, if |z| = R, then we obtain from the inequality (6.17) and the expression (6.18) that |f (z)| ≤

 1 |a1 | 1 1  R |a0 | + · · · + →1 + + · · · + |a | ≤ 1 + = 1 − · n n n−1 n n+1 |z| |z| R R R R−1

as R → ∞. In other words, this means that |f (z)| ≤ 1 for all |z| ≥ 1 and hence |P (z)| ≤ |z|n for all |z| ≥ 1. This completes the proof of the problem. Problem 6.15 Bak and Newman Chapter 6 Exercise 15.∗



Chapter 6. Properties of Analytic Functions

74

Proof. We notice that any line passing through z = 2 can be parameterized as z(t) = 2 + teiθ ,

(6.19)

where t ∈ R and θ ∈ [0, 2π] is the angle between the line and the positive real axis. Therefore, we observe that f (z(t)) = (1 + teiθ ) · (−2 + teiθ )2 = 4 − 3t2 e2iθ + t3 e3iθ

= (4 − 3t2 cos 2θ + t3 cos 3θ) + i(−3t2 sin 2θ + t3 sin 3θ). Direct computation gives g(t) = |f (z(t))|2

= (4 − 3t2 cos 2θ + t3 cos 3θ)2 + (−3t2 sin 2θ + t3 sin 3θ)2

= t6 + 9t4 + 16 − 6t5 cos θ − 24t2 cos 2θ + 8t3 cos 3θ so that

g′ (t) = 6t5 + 36t3 − 30t4 cos θ − 48t cos 2θ + 24t2 cos 3θ = 6t(t4 − 5t3 cos θ + 6t2 + 4t cos 3θ − 8 cos 2θ)

(6.20)

and g′′ (t) = 30t4 + 108t2 − 120t3 cos θ − 48 cos 2θ + 48t cos 3θ. Since the problem requires that |f (z)| has a relative extremum at z = 2, it follows from Fermat’s Theorem [27, Theorem 8.6, p. 129] that g′ (0) = 0 which is obvious by the expression (6.20). It is evident that g ′′ (0) = −48 cos 2θ, 5π 7π ′′ so we have g′′ (0) > 0 if and only if π4 < θ < 3π 4 and 4 < θ < 4 . Similarly, g (0) < 0 if and 5π 7π only if 0 ≤ θ < π4 , 3π 4 < θ < 4 and 4 < θ ≤ 2π. In other words, it follows from the Second Derivative Test [2, §4.17, pp. 188, 189] that |f (z)| has a local minimum at z = 2 on any line 5π 7π (6.19) with θ ∈ ( π4 , 3π 4 ) ∪ ( 4 , 4 ) and |f (z)| has a relative maximum at z = 2 through any line π 3π 5π (6.19) with θ ∈ [0, 4 ) ∪ ( 4 , 4 ) ∪ ( 7π 4 , 2π].

Now it remains to test the behaviour of |f | along the line (6.19) for θ=

π 3π 5π , , 4 4 4

and

7π . 4

In fact, it follows easily from the First Derivative Test [2, Theorem 4.8, p. 188] that |f | attains its local maximum at z = 2 when θ = π4 , 7π 4 and |f | has a local minimum at z = 2 when 5π θ = 3π  , . This completes the proof of the problem. 4 4 Problem 6.16 Bak and Newman Chapter 6 Exercise 16.∗

Proof. It is clear that f is analytic in C \ {0}. Note that f ′ (z) =

(z − 1)(z + 1) z2 − 1 = 2 z z2

75 so that f ′ (z) = 0 if and only if z = ±1. Since f (1) 6= 0, Theorem 6.17 ensues that z = 1 is a saddle point of f . Since f (−1) = 0, the modulus |f | definitely has the absolute minimum at z = −1 and thus z = −1 is not a saddle point of f by Definition 6.16. In conclusion, z = 1 is the only saddle point of f . Suppose that z(t) = 1 + teiθ (6.21) where t ∈ R and θ ∈ [0, 2π] is the angle between the line and the positive real axis. Therefore, we observe that (2 + teiθ )2 1 + teiθ (4 + 4teiθ + t2 e2iθ )(1 + te−iθ ) = 1 + 2t cos θ + t2 4(1 + t2 ) + t(4 + t2 )eiθ + 4te−iθ + t2 e2iθ = 1 + 2t cos θ + t2 t3 sin θ + t2 sin 2θ 4(1 + t2 ) + t(8 + t2 ) cos θ + t2 cos 2θ +i = 2 1 + 2t cos θ + t 1 + 2t cos θ + t2

f (z(t)) =

so that g(t) = |f (z(t))|2 h 4(1 + t2 ) + t(8 + t2 ) cos θ + t2 cos 2θ i2  t3 sin θ + t2 sin 2θ 2 = + 1 + 2t cos θ + t2 1 + 2t cos θ + t2 h2 (t) + h2 (t) = 1 2 2 , h3 (t)

(6.22)

where h1 (t) = 4(1 + t2 ) + t(8 + t2 ) cos θ + t2 cos 2θ, h2 (t) = t3 sin θ + t2 sin 2θ, h3 (t) = 1 + 2t cos θ + t2 . Direct computation shows h1 (0) = 4,

h′1 (0) = 8 cos θ,

h2 (0) = 0,

h′2 (0) = 0,

h3 (0) = 1,

h′3 (0)

h′′1 (0) = 8 + 2 cos 2θ,

h′′2 (0) = 2 sin 2θ,

= 2 cos θ,

h′′3 (0)

(6.23)

= 2.

Now it yields from the expression (6.22) that g′ = 2 ·

h3 (h1 h′1 + h2 h′2 ) − (h21 + h22 )h′3 h33

and then the values (6.23) imply that g′ (0) = 0 for every θ ∈ [0, 2π]. Next, further differentiation gives h33 [h3 (h1 h′1 + h2 h′2 ) − (h21 + h22 )h′3 ]′ − [h3 (h1 h′1 + h2 h′2 ) − (h21 + h22 )h′3 ] · (3h23 h′3 ) h63 h3 [h3 (h1 h′1 + h2 h′2 ) − (h21 + h22 )h′3 ]′ − [h3 (h1 h′1 + h2 h′2 ) − (h21 + h22 )h′3 ] · (3h′3 ) =2· h43

g′′ = 2 ·

Chapter 6. Properties of Analytic Functions and then g′′ (0) = 2 ·

[h3 (h1 h′1 + h2 h′2 )]′ − [(h21 + h22 )h′3 ]′ . t=0 h33

76

(6.24)

On the one hand, [h3 (h1 h′1 + h2 h′2 )]′ = h′3 (h1 h′1 + h2 h′2 ) + h3 [h1 h′′1 + (h′1 )2 + h2 h′′2 + (h′2 )2 ], so the values (6.23) give d ′ ′ [h3 (h1 h1 + h2 h2 )] = 2 cos θ(32 cos θ) + [4(8 + 2 cos 2θ) + 64 cos2 θ] dt t=0 = 128 cos2 θ + 8(4 + cos 2θ).

(6.25)

On the other hand, since [(h21 + h22 )h′3 ]′ = (2h1 h′1 + 2h2 h′2 )h′3 + (h21 + h22 )h′′3 , we get from the values (6.23) that d [(h21 + h22 )h′3 ] = 32 + 128 cos 2 θ. (6.26) dt t=0 Putting the results (6.25) and (6.26) into the expression (6.24), we observe that g′′ (0) = 2[128 cos 2 θ + 8(4 + cos 2θ) − 32 − 128 cos2 θ] = 16 cos 2θ Hence the argument of the proof of Problem 6.15 can be applied here to yield that |f | has a 5π 7π local minimum at z = 1 on any line (6.21) with θ ∈ [0, π4 ) ∪ ( 3π 4 , 4 ) ∪ ( 4 , 2π]. Similarly, |f | 7π π 3π has a local maximum at z = 1 through any line (6.21) with θ ∈ ( 4 , 4 ) ∪ ( 5π 4 , 4 ). Now the remaining unclear points are 7π π 3π 5π and . (6.27) θ= , , 4 4 4 4 However, we can check from the graphs of |f | that the point z = 1 is in fact an inflection pointa along the lines corresponding to these values (6.27). Figure 6.1 to Figure 6.3 below are some typical examples for the above discussion. In fact, the p orange line, the ′light red ′′line and the blue line correspond to the graphs of the function g(t) = |f (z(t))|, g (t) and g (t) respectively. They show us that |f | has a local maximum, a local minimum and an inflection point at z = 1 on the line 1 + t exp( π2 ), 1 + t exp( 33π 18 ) and 1 + t exp( 3π ) respectively. 4

Figure 6.1: The function |f (z(t))| has a local maximum at z = 1 on the line 1 + t exp( π2 ). a

Points where the concavity changes direction, see [2, p. 191].

77

Figure 6.2: The function |f (z(t))| has a local minimum at z = 1 on the line 1 + t exp( 33π 18 ).

Figure 6.3: The function |f (z(t))| has an inflection point at z = 1 on the line 1 + t exp( 3π 4 ). This completes the proof of the problem.



Problem 6.17 Bak and Newman Chapter 6 Exercise 17.∗

Proof. (a) By direction computation, we have f ′ (z) =

(z − 1)(z 2 + z + 2) z3 + z − 2 = . z3 z3

So f ′ (z) = 0 if and only if z = 1, −1±i 2

√

7

. By Theorem 6.17, the saddle points of f (z) are

Chapter 6. Properties of Analytic Functions exactly

√ −1 + i 7 z1 = 2

78

√ −1 − i 7 and z2 = . 2

(b) We divide the proof into several steps: √ – Step 1: √ Finding possible relative extrema of |f (z)|. Note that |z1 | = |z2 | = 2. Let z = 2eiθ , where θ ∈ [0, 2π]. Then we have √ |f (z)| = |f ( 2eiθ )| p √ √ 3 − 2 2 cos θ · 5 − 4 cos 2θ (6.28) = 2 p √ √ 15 − 12 cos 2θ − 10 2 cos θ + 8 2 cos θ cos 2θ = . (6.29) 2 Suppose that g(θ) is the function inside the square root of the expression (6.29). It is clear from the expression (6.29) that it suffices to find the extrema of g(θ). Simple differentiation shows that √ √ √ g′ (θ) = 24 sin 2θ + 10 2 sin θ − 8 2 sin θ cos 2θ − 16 2 cos θ sin 2θ √ √ √ = 24 sin 2θ + 10 2 sin θ − 8 2 sin θ(1 − 2 sin2 θ) − 32 2(1 − sin2 θ) sin θ √ √ = 48 2 sin3 θ − 30 2 sin θ + 24 sin 2θ. Therefore, g′ (θ) = 0 if and only if √ √ 48 2 sin3 θ − 30 2 sin θ + 24 sin 2θ = 0 √ √ sin θ(8 2 sin2 θ − 5 2 + 8 cos θ) = 0 √ √ sin θ(−8 2 cos2 θ + 8 cos θ + 3 2) = 0 if and only if sin θ = 0 or

√ √ 3 2 2 ,− . cos θ = 4 4

√

Note that cos θ = 3 4 2 is impossible because of the definition. Consequently, Fermat’s Theorem ensues that g(θ) takes its possible relative maxima or minima when sin θ = 0 √ 2 or cos θ = − 4 .

– Step 2: Determination of the relative minima of |f (z)|. We know that √ √ g′′ (θ) = 144 2 sin2 θ cos θ − 30 2 cos θ + 48 cos 2θ. If sin θ = 0, then cos θ = ±1 and cos 2θ = 1 so that √ g ′′ (θ) = 48 ∓ 30 2 > 0.

By the Second Derivative Test, g(θ) takes its relative minima when sin θ = 0 and we follow from the expression (6.28) that √ |f ( 2)| =

p √ 3−2 2 2

are relative minima of |f (z)|.

√

or |f (− 2)| =

p

√ 3+2 2 2

79 – Step 3: Determination of the relative maxima of |f (z)|. If cos θ = − sin θ = ±

√

14 4

and cos 2θ =

− 34 .

√

2 4 ,

then

In this case, we have g′′ (θ) = −75 < 0,

so the Second Derivative Test implies that g(θ) and thus |f (z)| takes its relative √ 2 maxima when cos θ = − 4 . Next, the pair of values cos θ = − z=

√

√ 2 4

and sin θ =

z=

14 4

corresponds to

√ √ 1 −1 + i 7 28 2e = − + i = = z1 . 2 4 2 iθ

Similarly, the pair of values cos θ = − √

√

√ 2 4

and sin θ = −

√ 14 4

corresponds to

√ √ −1 − i 7 28 1 = = z2 . 2e = − − i 2 4 2 iθ

Hence we conclude immediately that

on |z| =

|f (zi )| = max |f (z)|

√ 2, where i = 1, 2.

(c) Rewrite f as f (z) =

 1−z (z − 1)2 (z + 1)  z 2 z  z  = 1− . (6.30) 1− = z −1+ (1+z) = z − 2 2 2 2 z |z| |z| |z| |z|2

We apply the idea of the proof of Problem 6.16. – Case (i): Lines through z1 . Consider z1 (t) = z1 + teiθ , where t ∈ R and θ ∈ [0, 2π] is √the angle between the line and the positive real axis. Since z1 (t) = − 12 + t cos θ + i( 27 + t sin θ), we have √ |z1 (t)|2 = 2 + ( 7 sin θ − cos θ)t + t2 .

Since z1 = − 21 + t cos θ − i(

√

7 2

+ t sin θ), the expression (6.30) becomes √

 √7  3 − t cos θ + i( 7 + t sin θ) 3 2 √ + t sin θ + 2 f (z1 (t)) = − + t cos θ + i 2 2 2 + ( 7 sin θ − cos θ)t + t2 3 − t cos θ 3 √ 2 = − + t cos θ + 2 2 + ( 7 sin θ − cos θ)t + t2 √ √ 7 h 7 i + t sin θ √ 2 i + t sin θ + 2 2 + ( 7 sin θ − cos θ)t + t2  −1 − (√7 sin θ − cos θ)t − t2 3 √ − t cos θ · = 2 2 + ( 7 sin θ − cos θ)t + t2  √7  3 + (√7 sin θ − cos θ)t + t2 √ +i + t sin θ · 2 2 + ( 7 sin θ − cos θ)t + t2

Chapter 6. Properties of Analytic Functions

80

and then g(t) = |f (z1 (t))|2 =

h21 (t) + h22 (t) , h23 (t)

where h1 (t) =

3

 √ − t cos θ [1 + ( 7 sin θ − cos θ)t + t2 ],

2  √7

 √ + t sin θ [3 + ( 7 sin θ − cos θ)t + t2 ], 2 √ h3 (t) = 2 + ( 7 sin θ − cos θ)t + t2 .

h2 (t) =

Direct computation shows that 3 1 √ h1 (0) = , h′1 (0) = (3 7 sin θ − 5 cos θ), 2√ 2 √ 3 7 1 h2 (0) = , h′2 (0) = (13 sin θ − 7 cos θ), 2 √ 2 h3 (0) = 2, h′3 (0) = 7 sin θ − cos θ. Therefore, we obtain h3 (h1 h′1 + h2 h′2 ) − (h21 + h22 )h′3 t=0 h33 √ √ √ 3 3 7 2[ 4 (3 7 sin θ − 5 cos θ) + 4 (13 sin θ − 7 cos θ)] − ( 49 + = 4 √ √ 6(4 7 sin θ − 3 cos θ) − 18( 7 sin θ − cos θ) = 4 √ 3 7 sin θ = 2

g ′ (0) = 2 ·

√

63 4 )(

7 sin θ − cos θ)

so that g′ (0) = 0 if and only if θ = 0, π. By the First Derivative Test, it is easily seen that |f | has local minima at z = z1 through the lines z1 ± t.

– Case (ii): Lines through z2 . Next, we consider

z2 (t) = z2 + teiθ . Since z2 (t) = − 12 + t cos θ + i(−

√

7 2

+ t sin θ), we have √ |z2 (t)|2 = 2 − ( 7 sin θ + cos θ)t + t2 .

Since z2 = − 21 + t cos θ − i(−

√

7 2

+ t sin θ), the expression (6.30) gives

√  √7  3 − t cos θ + i( − 7 + t sin θ) 3 2 √ + t sin θ + 2 f (z2 (t)) = − + t cos θ + i − 2 2 2 − ( 7 sin θ + cos θ)t + t2 3 − t cos θ 3 √ 2 = − + t cos θ + 2 2 − ( 7 sin θ + cos θ)t + t2 √ √ − 7 i h + t sin θ 7 √ 2 + t sin θ + i − 2 2 − ( 7 sin θ + cos θ)t + t2  −1 + (√7 sin θ + cos θ)t − t2 3 √ − t cos θ · = 2 2 − ( 7 sin θ + cos θ)t + t2

81  3 − (√7 sin θ + cos θ)t + t2  √7 √ + t sin θ · . +i − 2 2 − ( 7 sin θ + cos θ)t + t2 Thus we get G(t) = |f (z2 (t))|2 =

H12 (t) + H22 (t) , H32 (t)

where 3

 √ − t cos θ [−1 + ( 7 sin θ + cos θ)t − t2 ], 2  √7  √ H2 (t) = − + t sin θ [3 − ( 7 sin θ + cos θ)t + t2 ], 2 √ H3 (t) = 2 − ( 7 sin θ + cos θ)t + t2 .

H1 (t) =

Simple differentiation yields that 1 √ 3 H1 (0) = − , H1′ (0) = (3 7 sin θ + 5 cos θ), 2√ 2 √ 3 7 1 H2 (0) = − , H2′ (0) = (13 sin θ + 7 cos θ), 2 √ 2 H3 (0) = 2, H3′ (0) = −( 7 sin θ + cos θ). Clearly, these values imply that H3 (H1 H1′ + H2 H2′ ) − (H12 + H22 )H3′ t=0 H33 √ √ √ √ 2[− 43 (3 7 sin θ + 5 cos θ) − 3 4 7 (13 sin θ + 7 cos θ)] + 18( 7 sin θ + cos θ) = 4 √ √ 2(−12 7 sin θ − 9 cos θ) + 18( 7 sin θ + cos θ) = 4 √ 3 7 =− sin θ. 2

G′ (0) = 2 ·

Hence G′ (0) = 0 if and only if θ = 0, π. Similar to Case (i), |f | has local maxima at z = z2 through the lines z2 ± t. We complete the analysis of the problem.



Chapter 6. Properties of Analytic Functions

82

CHAPTER

7

Further Properties of Analytic Functions

Problem 7.1 Bak and Newman Chapter 7 Exercise 1.

Proof. Let K be the compact domain of f . Consider ef (z) and e−if (z) . Since f nonconstant analytic on K, both ef (z) and e−if (z) are nonconstant analytic on K. Furthermore, we know that ef (z) 6= 0 and e−if (z) 6= 0 on K. Thus the Extreme Value Theorem, Theorems 6.13 (The Maximum Modulus Theorem) and 6.14 (The Minimum Modulus Theorem) together assert that max |ef (z) |, max |eif (z) |, min |ef (z) | and min |e−if (z) | assume on the boundary ∂K. Recall that ex is an increasing function on R, so the facts |ef (z) | = eRe f (z)

and |e−if (z) | = eIm f (z)

imply that maxima/minima of eRe f (z) correspond to maxima/minima of Re f (z) and similarly, maxima/minima of eIm f (z) correspond to maxima/minima of Im f (z). Hence Re f and Im f assume their maxima and minima on ∂K, completing the proof of the problem.  Problem 7.2 Bak and Newman Chapter 7 Exercise 2.

Proof. Let f be a nonconstant analytic function on the region D. By Theorem 7.1 (The Open Mapping Theorem), f (D) is open in C. Since f is continuous and D is connected, f (D) is also connected. By Definition 1.6, f (D) is also a region. This completes the proof of the problem.  Problem 7.3 Bak and Newman Chapter 7 Exercise 3.

Proof. (a) Assume that z was an interior point of S. Then there exists a δ > 0 such that D(z; δ) ⊆ S. By Theorem 7.1 (The Open Mapping Theorem), f (D(z; δ)) is an open set in C containing f (z), i.e., f (z) is an interior point of f (S) = T which contradicts the hypothesis. 83

Chapter 7. Further Properties of Analytic Functions

84

(b) The graph of the set S is shown in Figure 7.1 below. By the definition of f (z) = z 2 ,

Figure 7.1: The graph of the set S. we see that f (S) ⊆ D(0; 4). We claim that f (S) = D(0; 4). Let u + iv ∈ D(0; 4) and f (x + iy) = u + iv. Then we have u = x2 − y 2 and v = 2xy. Solving these equations give 4x4 − 4ux2 − v 2 = 0 and it implies that x =

u±

s

u+

2

If we take x=−

√

√

u2 + v 2 . 2

u2 + v 2 < 0, 2

then we have √ z = x + iy ∈ S1 ⊂ S and we have the claim. Next, the points z1 = 1 and √ 1 z2 = 2 + i 23 are clearly boundary points on S. Since f (z1 ) = 1 and f (z2 ) = − 21 + i 23 , they are clearly interior points of D(0; 4). We complete the proof of the problem.



Problem 7.4 Bak and Newman Chapter 7 Exercise 4.

Proof. Suppose that f is nonconstant and D = D(0; 1). We want to verify that f (D) = D. If ζ ∈ D \ f (D), then the map g : D → D defined by g(z) =

1 f (z) − ζ

(7.1)

85 is nonconstant C-analytic in D. Now the Extreme Value Theorem and Theorem 6.13 (The Maximum Modulus Theorem) assert that |g(z)| ≤ max |g(w)| and then w∈C(0;1)



1 f (z) − ζ

≤ max w∈C(0;1)

1 f (w) − ζ

≤ max w∈C(0;1)

1 1 = |f (w)| − |ζ| 1 − |ζ|

for all z ∈ D. Thus we have 0 < 1 − |ζ| ≤ |f (z) − ζ| which means that the distance between the point ζ and the set f (D)a is at least 1 − |ζ|. Therefore, we have  1 − |ζ|  D ζ, ⊆ D \ f (D) 2

and so the set D \ f (D) is open in C. Next, we follow from Theorem 7.1 (The Open Mapping Theorem) that f (D) is open. Recall the set relation D = (D \ f (D)) ∪ f (D).

(7.2)

Since D \ f (D) and f (D) are disjoint open subsets of D, the relation (7.2) implies that D is disconnected by the definition, but this is definitely a contradiction. In other words, we have either f (D) = D or f (D) = ∅. However, the latter is impossible by the hypothesis and then we have the result (7.1), completing the analysis of the problem.  Problem 7.5 Bak and Newman Chapter 7 Exercise 5.

Proof. Let D = D(0; 1). If f (z) = 1, then we are done, so we suppose that f is nonconstant. We claim that the number of zeros of f inside D is finite. To this end, let {αn } be the sequence of zeros (counted with their multiplicities) of f inside D. Since |f (z)| = 1 on |z| = 1, we actually have {αn } ⊆ D. As a consequence of the Bolzano-Weierstrass Theorem, {αn } contains a convergent subsequence {αnk }. Let αnk → α0 . Since f is continuous in D, we have 0 = lim f (αnk ) = f (α0 ) k→∞

which implies that α0 ∈ / C(0; 1), i.e., α0 ∈ D. Since D is a region, Theorem 6.9 (The Uniqueness Theorem) says that f ≡ 0 in D but it contradicts the continuity of f in D. This proves our claim. Next, we define the function g(z) =

f (z) , Bα1 (z)Bα2 (z) · · · Bαn (z)

(7.3)

where Bα (z) is the M¨obius transformation considered on [4, p. 95]. There are two cases. • Case (i): g is constant. Then it means that f (z) = cBα1 (z)Bα2 (z) · · · Bαn (z)

(7.4)

for some c ∈ C. If αj 6= 0, then f is undefined at z = α1j which contradicts the hypothesis that f is entire. Consequently, we have αj = 0 for all j = 1, 2, . . . , n so that the expression (7.4) reduces to f (z) = cz n as required. a

That is ρf (D) (ζ) = inf |f (z) − ζ|. z∈D

Chapter 7. Further Properties of Analytic Functions

86

• Case (ii): g is nonconstant. We claim that g is analytic in D. In fact, it is clear for z ∈ D \ {α1 , α2 , . . . , αn }. In the case z = αj , we need the following lemma:b Lemma 7.1 If f is analytic in a region G and f 6≡ 0 on G, then for each α ∈ G with f (α) = 0 there exists an integer n ≥ 1 and an analytic function g : G → C such that f (z) = (z − α)n g(z)

and

g(α) 6= 0.

Apply Lemma 7.1 to the function f (z), the expression (7.3) can be rewritten as n

g(z) =

Y (z − α1 )(z − α2 ) · · · (z − αn )h(z) (1 − αj z), = h(z) n Y z − αj

(7.5)

j=1

j=1

1 − αj z

where h is analytic in D and h(αj ) = 6 0 for all j = 1, 2, . . . , n. Thus the representation (7.5) implies that g(z) is analytic at z = αj for all j = 1, 2, . . . , n. Since g is nonconstant analytic in D, we apply the Extreme Value Theorem and Theorem 6.13 (The Maximum Modulus Theorem) to the expression (7.3) to get |f (ζ)| = max |f (ζ)| = 1 |g(z)| ≤ max |g(ζ)| = max n |ζ|=1 |ζ|=1 |ζ|=1 Y ζ − αj 1 − αj ζ

(7.6)

j=1

for all z ∈ D. Recall that |αj | < 1 so that |αj |−1 > 1. By the representation (7.5) and the property of h, we see that g(z) 6= 0 for all z ∈ D. Now it follows from Theorem 6.14 (The Minimum Modulus Theorem) that 1 = min |f (ζ)| = min |g(ζ)| ≤ |g(z)| |ζ|=1

|ζ|=1

(7.7)

for all z ∈ D. Combining the inequalities (7.6) and (7.7), we have |g(z)| = 1 for all z ∈ D, i.e., g(z) = eiθ for some θ so that f has the form (7.4) and then our result follows in this case. We have completed the proof of the problem.



Problem 7.6 Bak and Newman Chapter 7 Exercise 6.∗

Proof. Suppose that D = D(0; 1) and f (z) has poles at {α1 , α2 , . . . , αn } ⊆ D. Let f (z) = where P (z) and Q(z) are polynomials. Then we have f (z) =

P (z) n Y

.

(z − αj )

j=1 b

Read [7, Corollary 3.9, p. 79] for a proof of it. The case for polynomials has been given on [4, p. 67].

P (z) Q(z) ,

87 Define g(z) = f (z)Bαj (z) =

n Y P (z) z − αj = n n Y Y 1 − αj z (z − αj ) j=1 (1 − αj z)

P (z)

·

j=1

(7.8)

j=1

which is rational in D. Since |αj | < 1, we have |αj |−1 > 1 and then the expression (7.8) ensures that g(z) has no poles in D. Furthermore, the representation (7.8) also implies that, on |z| = 1, |g(z)| = |f (z)| · |Bαj (z)| = |f (z)|. This completes the proof of the problem.



Problem 7.7 Bak and Newman Chapter 7 Exercise 7.∗

Proof. (a) Let D = D(0; 1) and |a| < 1. Recall that Ba : D → D given by Ba (ω) =

ω−a 1−a·ω

(7.9)

α is analytic and bounded in D by 1. Put a = R and ω = Rz into the representation (7.9) to get z z −α R(z − α) B Rα = R α Rz = 2 R R −α·z 1− R · R

which is analytic in D(0; R). Note that |α| < R. Following the idea on [4, p. 95], on |z| = R, we have   R(z − α) R(z − α) α z 2 · = 2 B R R R − α · z R2 − α · z R2 (|z|2 − αz − αz + |α|2 ) = 4 R − R2 αz − R2 αz + |α|2 · |z|2 R2 − αz − αz + |α|2 = 2 R − αz − αz + |α|2 = 1. In other words, the function B Rα ( Rz ) maps the circle |z| = R into C(0; 1). n 1 Y 2 (b) Let f (z) = (z − α1 )(z − α2 ) · · · (z − αn ) and g(z) = n (R − αk z). It is clear that both R k=1

f and g are analytic throughout D(0; R) and part (a) ensures that they have the same magnitude on |z| = R, i.e., |f (z)| = |g(z)| (7.10)

on |z| = R. Since we have

1 · R2n = Rn 6= 0 Rn Theorem 6.13 (The Maximum Module Theorem) and Theorem 6.14 (The Minimum Module Theorem) imply that there exist points |z1 | = |z2 | = R such that g(0) =

|g(z1 )| > Rn

and |g(z2 )| < Rn .

(7.11)

Chapter 7. Further Properties of Analytic Functions

88

Consequently, we obtain from the results (7.10) and (7.11) that p p n |z1 − α1 | · |z1 − α2 | · · · |z1 − αn | = n |f (z1 )| > R

and

p n

|z2 − α1 | · |z2 − α2 | · · · |z2 − αn | =

We end the proof of the problem.

p n |f (z2 )| < R.



Problem 7.8 Bak and Newman Chapter 7 Exercise 8.

Proof. If f is constant, then there is nothing to prove. Thus we may assume that f is nonconstant. Since f is analytic in the annulus 1 ≤ |z| ≤ 2, if we define g(z) =

f (z) , z2

then g is also nonconstant analytic in the annulus. By Theorem 6.13 (The Maximum Modulus Theorem), max |g(ω)| occurs on |ω| = 1 or |ω| = 2 so that we obtain |f (z)| |f (ω)| = |g(z)| ≤ max |g(ω)| = max ≤1 |z|2 |ω|2 throughout the annulus. Hence we have |f (z)| ≤ |z|2 throughout the annulus which ends the  proof of the problem. Problem 7.9 Bak and Newman Chapter 7 Exercise 9.

1 f (2z). Then F is analytic in D(0; 1). Furthermore, we have |F (z)| ≤ 1 Proof. Define F (z) = 10 1 in D(0; 1) and F ( 2 ) = 0. It is easily seen that our F is exactly the function f considered in [4, Example 1, p. 95]. Therefore, we have

or equivalently,

Put z =

1 4

z−1 2 |F (z)| ≤ 1 1 − 2z

z−1 2 |f (2z)| ≤ 10 1 . 1 − 2z

(7.12)

into the inequality (7.12), we get

completing the proof of the problem.

 1  20 , ≤ f 2 7

Problem 7.10 Bak and Newman Chapter 7 Exercise 10.



89 Proof. Let D = D(0; 1). We define g : D → C by g = Bf (α) ◦ f where Bα is the M¨obius transformation Bα (z) =

z−α . 1 − αz

Since f and Bf (α) are analytic in D, g is also analytic in D. Furthermore, since f and Bf (α) are bounded by 1 in D, g is also bounded by 1 in D. By direct computation, we know that Bα′ (α) =

1 . 1 − |α|2

(7.13)

Using Problem 3.3 and the fact (7.13), we obtain g′ (α) = Bf′ (α) (f (α)) × f ′ (α) =

f ′ (α) 1 − |f (α)|2

(7.14)

Since f (α) 6= 0, we have 1 − |f (α)|2 < 1. Furthermore, it is impossible that |f (α)| = 1 by Theorem 6.13 (The Maximum Modulus Theorem). Hence the expression (7.14) implies that |f ′ (α)| < |g ′ (α)|. This completes the proof of the problem.



Problem 7.11 Bak and Newman Chapter 7 Exercise 11.

Proof. By Problem 7.10, we may assume that f (α) = 0 so that f ′ (α) = lim

z→α

f (z) f (z) − f (α) = lim . z→α z−α z−α

(7.15)

Again the assumption and [4, Example 1, p. 95] give |f (z)| ≤ |Bα (z)|

(7.16)

throughout D(0; 1). Therefore, the limit (7.15) and the estimate (7.16) combine to imply Bα (z) Bα (z) − Bα (α) = lim = Bα′ (α). z→α z − α z→α z−α

f ′ (α) ≪ lim

Since f is arbitrary, we conclude that max |f ′ (α)| = Bα′ (α). f

This ends the proof of the problem. Problem 7.12 Bak and Newman Chapter 7 Exercise 12.



Chapter 7. Further Properties of Analytic Functions

90

Proof. We imitate the proof of Proposition 7.3. In fact, we introduce the auxiliary function g(z) = (z 2 + R2 )2 f (z). For any z with |z| = R and Im z ≥ 0, we have |(z − iR)2 f (z)| ≤ for some θ ∈ [0, π4 ] so that

 |z − iR| 2 |Re z|

= sec2 θ

|(z − iR)2 f (z)| ≤ 2.

(7.17)

Similarly, if |z| = R and Im z ≤ 0, then we obtain |(z + iR)2 f (z)| ≤ 2.

(7.18)

Combining the estimates (7.17) and (7.18), for all z with |z| = R, we get |g(z)| = |(z + iR)2 | · |(z − iR)2 | · |f (z)| ≤ 2 × (|z| + |R|)2 = 8R2 . Since g is nonconstant analytic in D(0; R), Theorem 6.13 (The Maximum Modulus Theorem) implies that |g(z)| ≤ 8R2 in D(0; R) and so |f (z)| ≤

8R2 |z 2 + R2 |2

(7.19)

for all z ∈ D(0; R). If we take R → ∞, then we deduce from the inequality (7.19) that f (z) = 0 for all z ∈ C, completing the proof of the problem.  Problem 7.13 Bak and Newman Chapter 7 Exercise 13.

Proof. (a) Let Γ be the boundary of a rectangle R in C and z ∈ C. We want to apply Theorem 7.4 (Morera’s Theorem), so we check its hypotheses into four steps. – Step 1: The continuity of an auxiliary function. Define ϕ : C × [0, 1] → C by  sin zt  , if t 6= 0;  t ϕ(z, t) = (7.20)   z, if t = 0.

We claim that ϕ is continuous on C×[0, 1]. To this end, we need the following lemma: Lemma 7.2 Suppose that ζ is complex. Then we have lim

ζ→0

sin ζ = 1. ζ

91 Proof of Lemma 7.2. Notice that lim

ζ→0

sin ζ sin ζ − sin 0 = lim = cos 0 = 1 ζ→0 ζ ζ−0

as required.



It is clear from the definition (7.20) that ϕ(z, t) is continuous for every z ∈ C and t ∈ (0, 1], so it suffices to check its continuity at t = 0. In fact, for every fixed z ∈ C, we get from Lemma 7.2 that lim ϕ(z, t) = z lim

zt→0 t6=0

t→0 t6=0

sin zt = z = ϕ(z, 0). zt

This shows that ϕ is continuous on C × [0, 1] as desired.

– Step 2: f is continuous on C. With the aid of the auxiliary function (7.20), we can write Z 1 f (z) = ϕ(z, t) dt 0

so that

|f (z) − f (ω)| ≤

Z

0

1

|ϕ(z, t) − ϕ(ω, t)| dt.

(7.21)

Given ǫ > 0. By Step 1, there exists a δ > 0 such that |(z, t) − (ω, s)| < δ implies |ϕ(z, t) − ϕ(ω, s)| < ǫ. Particularly, if we take t = s, then |z − ω| < δ implies |ϕ(z, t) − ϕ(ω, t)| < ǫ and then the inequality (7.21) can be reduced to |f (z) − f (ω)| < ǫ. In other words, f is continuous on C. – Step 3: Absolute Convergence of the Double Integral. To evaluate the integral Z Z 1 Z sin zt f (z) dz = dt dz, (7.22) t Γ 0 Γ we need to interchange the order of integration of the double integral (7.22). As suggested by the example given in [4, pp. 98, 99], we have to establish the absolute convergence of the double integral (7.22). To this end, we fix z ∈ C. Now Lemma 7.2 ensures that there exists a M > 0 such that sin ζ (7.23) ≤M ζ for all 0 < |ζ| ≤ |z|. If ζ = tz, where t ∈ (0, 1], then the inequality (7.23) implies that sin zt ≤ M |z| t which gives

Z

0

sin zt dt ≤ M |z|. t

1

(7.24)

Chapter 7. Further Properties of Analytic Functions

92

Since Γ is a rectangle, it must be of finite length. Since the function F (z) = |z| is continuous on the compact set Γ, F is bounded there. Hence we follow from the inequality (7.24) and Theorem 4.10 (The M -L Formula) that Z Z Z 1 sin zt |z| dz < ∞. dt dz ≤ M t Γ Γ 0

– Step 4: The Application of Theorem 7.4 (Morera’s Theorem). For each fixed t ∈ [0, 1], since sin zt is entire, we follow from Theorem 4.14 (The Rectangle Theorem) that Z sin zt dz = 0.

Γ

Thus we establish from this and Step 3 that Z Z 1 Z Z 1Z Z 1 Z  sin zt sin zt 1 f (z) dz = sin zt dz dt = 0. dt dz = dz dt = t t Γ 0 Γ Γ 0 Γ 0 t Finally, we conclude that f is entire.

(b) Fix z ∈ C and let R = |z|. By Problem 3.22, we have ∞

sin zt X (−1)n z 2n+1 2n = t , t (2n + 1)!

(7.25)

n=0

where t ∈ (0, 1]. It is clear that tz ∈ D(0; R), so we see that (−1)n z 2n+1 R2n+1 t2n ≤ . (2n + 1)! (2n + 1)!

Since

R2 (2n + 1)! R2n+3 × = 0, = lim sup R2n+1 n→∞ (2n + 3)(2n + 2) n→∞ (2n + 3)! the Ratio Test [27, Theorem 6.8, p. 77] implies that the series lim sup

∞ X R2n+1 (2n + 1)! n=0

converges. Thus the Weierstrass M -test [28, Theorem 10.5, p. 3] asserts that the series (7.25) converges uniformly for all t ∈ (0, 1]. Consequently, this fact allows us to interchange the integration and summation in the following calculation Z 1X ∞ (−1)n z 2n+1 2n t dt f (z) = 0 n=0 (2n + 1)! Z ∞ X (−1)n z 2n+1 1 2n = t dt (2n + 1)! 0 =

n=0 ∞ X

(−1)n z 2n+1 . (2n + 1)(2n + 1)! n=0

(7.26)

Since z is arbitrary, the power series representation (7.25) of f holds throughout its domain of convergence. To determine the radius of convergence, since lim

n→∞

(2n + 1)(2n + 1)! 2n + 1 = lim = 0, n→∞ (2n + 3)(2n + 3)! (2n + 3)2 (2n + 2)

we apply Problem 2.13 and Theorem 2.8 to conclude that the power series (7.26) converges for all z ∈ C. In other words, f is entire.  We have completed the proof of the problem.

93 Remark 7.1 (a) In Step 3 above, the reason why absolute convergence of the double integral implies the interchange of order of integration depends on the application of an advanced tool called Fubini Theorem, see [23, Theorem 8.8, pp. 164, 165]. (b) We also remark that Problem 7.13(a) is actually a special case of the complex version of the so-called Leibniz’s Rule: Suppose that ϕ(z, t) is defined on Ω × [0, 1], where Ω is open in C. If ϕ(z, t) is analytic in z for each t and ϕ is continuous on Ω × [0, 1], then the function f : Ω → C given by Z 1 ϕ(z, t) dt f (z) = 0

is analytic. This statement can be found in Conway [7, Exercise 2, pp. 73, 74] or Stein and Shakarchi [24, Theorem 5.4, p. 56]. The tools Stein and Shakarchi applied there are Riemann sums and Theorem 7.6. The corresponding real version can be found in [22, Theorem 9.42, pp. 236, 237].

Problem 7.14 Bak and Newman Chapter 7 Exercise 14.

Proof. (a) Clearly, we have f (z) =

Z

0

1

sin zt dt = t

Z

0

1Z z

We let g(ζ) =

cos ζt dζ dt =

0

Z

0

1

cos ζt dt =

Z

z 0

Z

1



cos ζt dt dζ.

0

sin ζ ζ

which is entire. Now by the proof of Theorem 4.15 (The Integral Theorem), we conclude immediately that Z 1 Z z d ′ cos zt dt. g(ζ) dζ = g(z) = f (z) = dz 0 0 (b) Applying Theorem 2.9 to the power series (7.26) and then using Problem 3.22, we conclude that Z 1 ∞ X (−1)n z 2n sin z ′ cos zt dt. f (z) = = = (2n + 1)! z 0 n=0

We end the proof of the problem. Problem 7.15 Bak and Newman Chapter 7 Exercise 15.

Proof. It is reasonable to assume that z0 6= z1 . For every real x, we have g(x) = z0 + xeiθ



Chapter 7. Further Properties of Analytic Functions

94

x|z1 − z0 | · eiArg (z1 −z0 ) |z1 − z0 | x(z1 − z0 ) = z0 + |z1 − z0 |   x x z0 + z1 . = 1− |z1 − z0 | |z1 − z0 | = z0 +

(7.27)

Therefore, the formula (7.27) implies exactly that g(R) = L. This completes the proof of the problem.  Problem 7.16 Bak and Newman Chapter 7 Exercise 16.

Proof. Let Ω+ = {z ∈ C | Im z > 0} and Ω− = {z ∈ C | Im z < 0}. Since f is analytic in Ω+ ∪ R and f is real for real z, Theorem 7.8 (The Schwarz Reflection Principle)c implies that f can be extended to an entire function F such that F (z) = f (z) on Ω+ . Since f is bounded in Ω+ ∪ R, F is bounded in C. By Theorem 5.10 (Liouville’s Theorem), F and then f is constant which  completes the proof of the problem. Problem 7.17 Bak and Newman Chapter 7 Exercise 17.

Proof. Since f is entire, it deduces from Theorem 5.5 (The Taylor Expansion of an Entire Function) that ∞ X f (k) (0) k z (7.28) f (z) = k! k=0

for all z ∈ C. Since C is certainly a region symmetric with respect to R and f is real on R, Corollary 7.9 shows that f (z) = f (z) (7.29)

for all z ∈ C. Combining the two expressions (7.28) and (7.29), we obtain ∞ X f (k) (0) k=0

k!

k

z =

∞ X f (k)(0) k=0

k!

zk

=

∞ X f (k) (0) k=0

k!

zk

for all z ∈ C. By Theorem 6.9 (The Uniqueness Theorem), we have f (k) (0) f (k) (0) = k! k! for all k = 0, 1, 2, . . .. Next, we consider g(z) = if (iz) =

∞ X k=0

c

ik+1 ·

(7.30)

f (k) (0) k z . k!

Here we use the stronger form of the Schwarz Reflection Principle given in [7, §1, pp. 210 - 213] or [24, p. 60] because the form of the principle in our book is only applicable to “C-analytic function f in a region D” whose definition assumes that D is a bounded region.

95 Since f is imaginary on the imaginary axis, g is real on the real axis. Thus the preceding paragraph shows that f (k) (0) ik+1 f (k) (0) f (k) (0) ik+1 = = ik+1 · . (7.31) k! k! k! If k is even, then k = 2m so that ik+1 = (−1)m i = −(−1)m i = −i2m+1 = −ik+1 and the expression (7.31) reduces to f (k) (0) f (k) (0) =− . (7.32) k! k! Combining the results (7.30) and (7.32), we conclude that f (k) (0) =0 k! for even k. Hence the representation (7.28) can be written as ∞ X f (2m+1) (0) 2m+1 f (z) = z (2m + 1)! m=0

which is indeed an odd function. This completes the proof of the problem.



Problem 7.18 Bak and Newman Chapter 7 Exercise 18.∗

Proof. Let z = x + iy, ω = γ(z) = u + iv and ω ∗ = γ(z). Since the line u = v can be parameterized by γ(t) = t + it, where t ∈ R, we have z + iz = u + iv so that (x + iy) + i(x + iy) = u + iv (x − y) + i(x + y) = u + iv. Thus we have x =

u+v 2

ω ∗ = γ(z) = γ

and y =

u + v 2

+i

v−u 2

so that

u + v v − u u + v v−u v − u = = v + iu −i +i −i 2 2 2 2 2

as required, completing the proof of the problem.



Problem 7.19 Bak and Newman Chapter 7 Exercise 19.

Proof. Let D = {z ∈ C | |z| < 1 and Im z > 0} and D ∗ = {z ∈ C | |z| > 1 and Im z > 0}. By Problem 1.17, we know that the mapping ω=i

z−1 z+1

(7.33)

maps the upper unit circle C + = {z ∈ C | |z| = 1 and Im z > 0} into the negative real axis R− . In fact, it is easy to show that this mapping is bijective. Therefore, its inverse z=

i+ω i−ω

(7.34)

Chapter 7. Further Properties of Analytic Functions

96

maps R− one-one and onto C + . We claim that the mapping (7.34) maps the region H = {ω ∈ C | Re ω < 0 and Im ω < 0} onto D.d Let ω = a + ib, where a < 0 and b < 0. Then direct computation gives z=

1 − a2 − b2 2a a + (b + 1)i = 2 −i 2 2 −a − (b − 1)i a + (b − 1) a + (b − 1)2

and so |z|2 =

(1 − a2 − b2 )2 + 4a2 1 + 2a2 + a4 − 2b2 + 2a2 b2 + b4 = . [a2 + (b − 1)2 ]2 [a2 + (b − 1)2 ]2

(7.35)

Further simplification of the expression (7.35) implies that |z| < 1 if and only if b[a2 + (b − 1)2 ] < 0 or equivalently, b < 0. This means that the mapping (7.34) maps H into D. Next, given z = x + iy ∈ D, where y > 0 and x2 + y 2 < 1. Now the expression (7.33) implies that ω=i

x − 1 + iy 2y 1 − x2 − y 2 x + iy − 1 =i =− − i . x + iy + 1 x + 1 + iy (x + 1)2 + y 2 (x + 1)2 + y 2

(7.36)

Since y > 0 and x2 + y 2 < 1, the expression (7.36) ensures that Re ω < 0 and Im ω < 0. Consequently, the mapping is onto and our claim follows. Define F : H → D by i + ω  F (ω) = f i−ω which is obviously analytic in H and is real for ω < 0. Let H ∗ = {ω ∈ C | Re ω < 0 and Im ω > 0}, the left upper half-plane. By Theorem 7.8 (The Schwarz Reflection Principle), F has an analytic extension G : H ∪ R− ∪ H ∗ → C which is defined by   F (ω), if ω ∈ H ∪ R− ; (7.37) G(ω) =  ∗ F (ω), if ω ∈ H . By the expression (7.34), we know thate

i+ω 1 = . z i−ω

Now the analysis of the above claim also shows that if ω ∈ H ∗ , then Im z > 0 and |z| > 1. Hence the analytic function (7.37) can be expressed as   f (z), if z ∈ D ∪ C + ;   z − 1  g(z) = G i =  1 z+1   f , if z ∈ D ∗ . z

This concludes the proof of the problem.



Problem 7.20 Bak and Newman Chapter 7 Exercise 20. Actually, it is easily seen that H is the left lower half-plane. For example, if ω = −1 − i, then z = − 15 + 52 i which belongs to D. e By the example on [4, pp. 102, 103], z1 is the reflection of the point z across the curve C + . d

97 Proof. Assume that f : D(0; 1) → C was a nonconstant analytic function in D(0; 1) and f (z) is real for all z ∈ C(0; 1). By Definition 3.3, f is continuous on the closed unit disc D(0; 1). Similar to the proof of Problem 7.19, we can show that the mapping T : C → C defined by T (z) =

z−i z+i

maps the sets U = {z ∈ C | Im z > 0} and R onto D(0; 1) and C(0; 1) respectively. Therefore, the composition F = f ◦ T : U → C is also nonconstant and continuous. Furthermore, F is analytic in U and F (z) takes real values for real z. Thus Theorem 7.8 (The Schwarz Reflection Principle)f implies that F can be extended to a nonconstant entire function G defined by

G(z) =

  F (z), if z ∈ U ∪ R; 

F (z), if z ∈

(7.38)

U ∗,

where U ∗ = {z ∈ C | z ∈ U }. Since D(0; 1) is compact, we know that there exists a positive constant M such that |F (z)| = |f (T (z))| ≤ M for all z ∈ U ∪ R. Therefore, this fact and the expression (7.38) imply that |G(z)| = |f (T (z))| ≤ M for all z ∈ C. By Theorem 5.10 (Liouville’s Theorem), G is constant and hence f is also constant,  a contradiction. We complete the proof of the problem. Problem 7.21 Bak and Newman Chapter 7 Exercise 21.

Proof. Assume that f (x) = |x| for all x ∈ R. Now, since f is analytic in the bounded region Ω+ = {z ∈ C | |z| < 1 and Im z > 0} and f is continuous on the upper semi-disc, we observe from Theorem 7.8 (The Schwarz Reflection Principle) that f can be extended to an analytic function g in D(0, 1). Since g is analytic in D(0; 1), we see that lim

h→0 h>0

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

h>0

but f (h) − f (0) −h − 0 g(h) − g(0) = lim = lim = −1. h→0 h→0 h→0 h h h lim

h 0. We claim that the annulus B = {ω ∈ C | r − δ ≤ |ω − α| ≤ r + δ} is contained in S. To see this, we let ω ′ be the projection of the point ω ∈ B on C. Now the set relation (8.1) ensures that ω ′ ∈ D(zk ; δzk ) for some 1 ≤ k ≤ N , see Figure 8.1 below.

Figure 8.1: The geometry of the circle C and the annulus B. Then the triangle inequality implies that |ω − zk | ≤ |ω − ω ′ | + |ω ′ − zk | < δ + δzk < 2δzk . In other words, we have established that ω ∈ D(z; 2δzk ) ⊆ S and thus B ⊆ S. We have completed the proof of the problem. 

103 Problem 8.4 Bak and Newman Chapter 8 Exercise 4.

Proof. Rewrite Se = T ∪ I, where

n o 1 T = x + i sin 0 < x ≤ 1 x

and

I = {iy | − 1 ≤ y < ∞}.

We note that the set T is part of the Topologist’s sine curve studied in Problem 1.23. Since Se contains all its limit pointsa , it is closed in C and therefore, S is open in C. Given x ∈ (0, 1] and ǫ > 0. Let n ∈ N be large enough such that lim sin

1 t→ nπ

1 nπ

< x. Since

1 = sin nπ = 0, t

the point x + i sin x1 ∈ T ⊆ Se is connected to the point 1 γ1 : [ nπ , x] → T given by

1 nπ

∈ T ⊆ Se by the continuous curve

  1 1 − t + i sin . γ1 (t) = x + 1 nπ x + nπ −t

Recall the definition from [22, Exercise 20, p. 20] that e = inf d(γ1 (t), z). d(γ1 (t), S) z∈Se

If we choose n >

1 πǫ ,

1 then since d(( nπ , 0), (0, 0)) =

1 , ∞) → Se by Next, if we define γ : [ nπ

1 nπ

< ǫ, we obtain

e < ǫ. d(γ1 (x), S)

 h 1 i   γ (t), if t ∈ ,x ; 1   nπ     1 2−t γ(t) = · , if t ∈ (x, 2];   nπ 2 − x       i(t − 2), if t > 2,

1 , ∞). Furthermore, if t ∈ (x, 2], then γ(t) < then γ is a continuous curve on [ nπ

1 1 ) = γ1 ( nπ ) = x + i sin x1 and Since γ( nπ

(8.2)

1 nπ

so that

e < ǫ. d(γ(t), S)

lim γ(t) = lim i(t − 2) = ∞,

t→∞

t→∞

we see that our curve (8.2) satisfies the requirements of Definition 8.1 and hence D is simply connected, completing the proof of the problem.  a

Recall that (0, 0) is a limit point of the set {x + i sin

1 x

| 0 < x ≤ 1}.

Chapter 8. Simply Connected Domains

104

Problem 8.5 Bak and Newman Chapter 8 Exercise 5.∗

Proof. Let R = {x + iy | a < x < b and c < y < d} be a rectangle containing z, Γ = ∂R and γ be a continuous curve connecting z to ∞. We further set E = {t ∈ [0, ∞) | γ(t) ∈ R} and t0 = sup E. Since γ(0) = z ∈ R, E is not empty so that t0 exists in R. If γ(t0 ) ∈ Γ, then there is nothing to prove. Therefore, without loss of generality, we may assume that γ(t0 ) ∈ / Γ in the following discussion. Suppose that t0 ∈ E, i.e., γ(t0 ) ∈ R. Since R is open in C, there exists a ǫ > 0 such that D(γ(t0 ); ǫ) ⊆ R. Since γ is continuous on [0, ∞), there exists a δ > 0 such that |t − t0 | < δ implies γ(t) ∈ D(γ(t0 ); ǫ).

Particularly, we have γ(t0 + 2δ ) ∈ R which means that t0 + that t0 = sup E.

δ 2

∈ E but this contradicts the fact

Now the observation in the previous paragraph forces that t0 ∈ C \ E. The assumption γ(t0 ) ∈ / Γ means that {γ(t0 )} ∩ Γ = ∅. Since Γ is compact and {γ(t0 )} is closed in C, it follows from the well-known fact [22, Exercise 21, p. 101] that one can find a ǫ1 > 0 such that d(γ(t0 ), Γ) = 2ǫ1 > 0. Consequently, we have D(γ(t0 ); ǫ1 ) ∩ R = ∅.

(8.3)

Again, the continuity of γ implies that there is a δ1 > 0 such that |t − t0 | < 2δ1 must imply γ(t) ∈ D(γ(t0 ); ǫ1 ). In particular, γ(t0 − δ1 ) ∈ D(γ(t0 ); ǫ1 ) or equivalently, γ(t0 − δ1 ) ∈ / R by the result (8.3). Since t0 − δ1 < t0 , the definition of t0 shows that there is a p ∈ E such that t0 − δ1 < p < t0 . Obviously, we have |p − t0 | < δ1 < 2δ1 , so γ(p) ∈ D(γ(t0 ); ǫ1 ) and then the set relation (8.3) ensures that γ(p) ∈ / R. However, p ∈ E certainly implies that γ(p) ∈ R, a contradiction. Hence we conclude  that γ(t0 ) ∈ Γ which completes the proof of the problem. Remark 8.2 We note that only the continuity of γ is required in the proof of Problem 8.5.

Problem 8.6 Bak and Newman Chapter 8 Exercise 6.

Proof. Suppose that Γ is a simple closed polygonal path contained in a simply connected domain D. Recall from the proof of Lemma 8.3 that if Γ has only two levels, then it is the boundary of a single rectangle R. Thus it is natural to define the “inside” of Γ to be the points of the rectangle R. Assume that the “inside” of a simple closed polygonal path Γ ⊆ D can be defined up to k(≥ 2) levels and the “inside” of Γ is contained in D. Let Γ′ be a simple closed polygonal path of (k + 1) levels, see Figure 8.2 below. Then we can write Γ′ = Γ ∪ ∂R, (8.4)

105 where Γ and ∂R are simple closed polygonal paths and boundaries of rectangles R respectively. Thus the “inside” of Γ′ is just the union of the “inside” of Γ and the points of R.

Figure 8.2: The “inside” of a simple closed polygonal path. Therefore, if Γ′ is contained in the simply connected domain D, then so are Γ and ∂R. By the assumption step, the “inside” of Γ is contained in D. Next, if z ∈ R, then we must have z ∈ D. Otherwise, we have z ∈ D c . On the one hand, we know that D c is closed in C, ∂R is compact and D c ∩ (∂R) = ∅, so there exists a δ > 0 such that d(∂R, D c ) = δ > 0.

(8.5)

On the other hand, we recall from Definition 8.1 that z is connected within ǫ to ∞, i.e., there exists a continuous curve γ : [0, ∞) → ∞ such that d(γ(t), D c ) < δ for all t ∈ [0, ∞). However, Problem 8.5 means that we can find a t0 ∈ [0, ∞) such that γ(t0 ) ∈ ∂R and d(γ(t0 ), D c ) < δ, contradicting the result (8.5). Therefore, R lies in D and we observe from the set relation (8.4) that the “inside” of Γ′ must lie in D too. By induction, our desired result follows and we  complete the proof of the problem. Problem 8.7 Bak and Newman Chapter 8 Exercise 7.

Proof. By Definition 4.13, let z : [0, 1] → C be a non-simple closed polygonal path. Let p11 , p12 be a pair of points such that 0 < p11 < p12 < 1 and z(s) = z(t) for all s, t ∈ [p11 , p12 ], see Figure 8.3 for an example. In this example, we may define z1 : [0, 1] → C and γ1 : [0, 1] → C by  if t ∈ [0, 12 ];  z(2p11 t), z1 (t) =  z(p12 + 2(1 − p12 )(t − 12 )), if t ∈ [ 21 , 1] and

γ1 (t) = z(p11 + (p12 − p11 )t). Then it is easy to see that both z1 and γ1 are continuous on [0, 1] so that we can write z = z1 ∪ γ1 ,

Chapter 8. Simply Connected Domains

106

where z1 is a simple closed polygonal path and γ1 is a line segment traversed twice in opposite directions.b

Figure 8.3: A decomposition of a closed polygonal path. Generally speaking, if there are k pairs of points {p11 , p12 , p21 , p22 , . . . , pk1 , pk2 } such that z(sj ) = z(tj ) for all sj , tj ∈ [pj1 , pj2 ] and 1 ≤ j ≤ k. Then the above argument can be repeated to decompose z into z = (z1 ∪ z2 ∪ · · · ∪ zk ) ∪ (γ1 ∪ γ2 ∪ · · · ∪ γk ),

where each pair of zj and γj composes of simple closed polygonal paths and line segments traversed twice in opposite directions. Hence we have completed the proof of the problem.  Problem 8.8 Bak and Newman Chapter 8 Exercise 8.

Proof. Let D be the whole plane minus the non-negative real axis. Then D is simply connected and 0 ∈ / D. Since −1 ∈ D and log(−1) = log | − 1| + iArg (−1) = πi, Theorem 8.8 says that the function Z z dζ (8.6) f (z) = πi + −1 ζ is an analytic branch of log z in D. Next, since z ∈ / R+ , then we have Arg z ∈ (0, 2π). If z lies on the negative axis, then z = −|z| and Arg z = π so thatc Z −|z| −|z| dζ = log |z| + πi. (8.7) = πi + log |ζ| f (z) = πi + ζ −1 −1

Thus we have Im (log z) = Im (f (z)) = Arg z = π. If z ∈ D \ R− , then the line integral (8.6) goes from −1 to −|z| and then from −|z| to z to get Z −|z| Z Z dζ dζ dζ f (z) = πi + + = log |z| + πi + , (8.8) ζ −1 C ζ C ζ b We remark that there are examples consisting of several simple closed polygonal paths (e.g. two rectangles intersecting at one of their vertices only) or line segments (see the horizontal and vertical line segments at z(pk1 ) = z(pk2 ) in Figure 8.3). c The integral in the expression (8.7) is the usual Riemann integral.

107 where C is the circular arc from connecting −|z| and z clockwise, see Figure 8.4. If C is

Figure 8.4: The circular arc C connecting −|z| and z clockwise. parameterized as γ(t) = |z|ei(π−t) , where 0 ≤ t ≤ π − Arg z, then we deduce from the expression (8.8) that Z π−Arg z 1 · [−i|z|ei(π−t) ] dt f (z) = log |z| + πi + i(π−t) |z|e 0 Z π−Arg z dt = log |z| + πi − i 0

= log |z| + πi − i(π − Arg z) = log |z| + iArg z

which implies that Im (log z) = Arg z ∈ (0, 2π). This completes the proof of the problem.



Problem 8.9 Bak and Newman Chapter 8 Exercise 9.∗

Proof. Let D be the plane minus the non-positive real axis. By the analysis on [4, p. 115], we can define an analytic branch of log z in D by log z = log |z| + iArg z,

(8.9)

where Arg z ∈ (−π, π). Next, we define f (z) = ez log z .

(8.10)

Since log z is analytic in D, f is also analytic in D. Besides, if x > 0, then f (x) = ex log x = xx . In other words, the function f defined by the expression (8.10) satisfies our requirements. Using the expressions (8.9) and (8.10) directly, we see that π

f (i) = ei log i = ei(log |i|+iArg (i)) = e− 2 and

π

f (−i) = e−i log −i = e−i(log |−i|+iArg (−i)) = e− 2 . Finally, let z = x + iy ∈ D. If y = 0, then we have f (x) = f (x), so we assume that y 6= 0 so that z = x − iy ∈ D. In addition, we have log z = log |z| + iArg z = log |z| − iArg z = log |z| + iArg z = log z

Chapter 8. Simply Connected Domains

108

which gives f (z) = exp(z · log z) = exp(z log z) = exp(z log z) = f (z). This ends the proof of the problem.



CHAPTER

9

Isolated Singularities of an Analytic Function

Problem 9.1 Bak and Newman Chapter 9 Exercise 1.

Proof. Assume that z0 was a removable singularity of f . By Definition 9.2, there exists an analytic function g defined in a neighborhood of z0 such that f (z) = g(z) for all z 6= z0 . Now the continuity of g yields lim |f (z)| = lim |g(z)| = |g(z0 )| < ∞

z→z0

z→z0

which contradicts our hypothesis. Consequently, z0 is not a removable singularity of f . Next, we assume that z0 was an essential singularity of f . Since f (z) → ∞ as z → z0 , for every ǫ > 0, there exists a M > 0 such that |f (z)| > M for all z ∈ D(z0 ; ǫ). Therefore, it implies that f (D(z0 ; ǫ)) ∩ D(0; M ) = ∅. In other words, f (D(z0 ; ǫ)) is not dense in C which contradicts Theorem 9.6 (The Casorati-Weierstrass Theorem). Hence z0 cannot be an essential singularity of f and Definition 9.2 forces that f has a pole at z0 . This completes the analysis of  the problem. Problem 9.2 Bak and Newman Chapter 9 Exercise 2.

Proof. The answer is negative. By the hypothesis, we get 1

lim |f (z)| = lim e |z| = ∞,

z→0

z→0

so Problem 9.1 implies that f has a pole at 0 and then f has the form f (z) =

∞ X

an z n

(9.1)

n=−k

for some k ∈ N. Therefore, the form (9.1) implies that |f (z)| ∼

M |z|k

as z → 0 for some positive constant M . This is certainly a contradiction and we complete the proof of the problem.  109

Chapter 9. Isolated Singularities of an Analytic Function

110

Problem 9.3 Bak and Newman Chapter 9 Exercise 3.∗

Proof. Suppose that f is entire and injective. Then f is nonconstant. Consider the function g : C \ {0} → C defined by g(z) = f (z −1 ) which has a singularity at 0. Since f is entire, Theorem 5.5 (The Taylor Expansion of an Entire Function) says that ∞ X f (ω) = ak ω k k=0

for some ak ∈ C and then g(z) =

0 X

ak z k .

k=−∞

If g has a removable singularity at 0, then a−1 = a−2 = · · · = 0 which implies that g(z) = a0 . Consequently, f is also a constant which contradicts to our hypothesis. Therefore, g has either a pole or an essential singularity at 0. Assume that the singularity was essential. By Theorem 9.6 (The Casorati-Weierstrass Theorem), the set g(D(0; 1) \ {0}) is dense in C. Since g is analytic in the region D(2; 12 ), Problem 7.2 implies that g(D(2, 12 )) is also a region. In particular, g(D(2, 21 )) is open in C. Thus we have   1  6= ∅. g(D(0; 1) \ {0}) ∩ g D 2; 2 In other words, there exists z1 ∈ D(0; 1) \ {0} and z2 ∈ D(2; 12 ) such that z1 6= z2 but f (z1−1 ) = g(z1 ) = g(z2 ) = f (z2−1 ). This contradicts the fact that f is injective and hence the singularity is not essential. The previous analysis ensures that the singularity of g at 0 is a pole, so f must be a polynomial. Suppose that f (z) = a0 + a1 z + · · · + an z n . By Theorem 5.12 (The Fundamental Theorem of Algebra), f has n roots counting multiplicity. Since f is injective, it cannot have any two distinct roots. In other words, f must have the form f (z) = a(z − z0 )n for some a, z0 ∈ C and a 6= 0. If n ≥ 2, then the fact f (z0 + 1) = a = f (z0 + e

2πi n

)

shows that f is not injective, a contradiction. Now we have n = 1 and so f (z) = a(z − z0 ) which  is a linear polynomial as desired. We end the proof of the problem. Problem 9.4 Bak and Newman Chapter 9 Exercise 4.

111 Proof. If f is analytic at the origin, then f is entire. For |z| > 1, we have p |f (z)| < 2 |z| < 2|z|

1 |z|

< |z| so that

there. Next, since |f (z)| is continuous on the compact set D(0; 1), |f (z)| is bounded by a positive constant M there. These two facts imply that one can find a positive constant M ′ such that |f (z)| ≤ M ′ |z| holds for all z ∈ C. Hence Theorem 5.11 (The Extended Liouville Theorem) guarantees that f (z) = A + Bz for some A, B ∈ C. Now the triangle inequality gives p 2 |z| > |f (z)| = |Bz + A| ≥ |B| · |z| − |A|

for large z and this forces that B = 0. In other words, f (z) = A in this case. Suppose that f has an isolated singularity at 0. Then we consider its Laurent expansion at 0: ∞ X f (z) = ak z k . k=−∞

Let R > 0. Now if k 6= 0, then we consider the circle C = C(0; R) and we apply Theorem 4.10 (The M -L Formula) to obtain 1 Z f (ω) 1 1 √ 1 √ 1  1  √ √ |ak | = dω ≤ × 2πR × = . (9.2) R + R + 2πi C(0;R) ω k+1 2π Rk+1 Rk R R If k > 0, then the inequality (9.2) shows that

1  1 √ √ =0 R + R→∞ Rk R

lim |ak | ≤ lim

R→∞

which means a1 = a2 = · · · = 0. If k < 0, then the inequality (9.2) gives 1 √ 1  √ R + = 0. R→0 Rk R

lim |ak | ≤ lim

R→0

In other words, a−1 = a−2 = · · · = 0. Hence we have shown that f (z) = a0 in this case, completing the proof of the problem.  Problem 9.5 Bak and Newman Chapter 9 Exercise 5.∗

Proof. If g ≡ 0, then there is nothing to prove. Thus all the zeros of g are isolated. Furthermore, (z) we have | fg(z) | ≤ 1 in a deleted neighborhood of a zero of g, so Corollary 9.4 asserts that all the

singularities of fg are removable. As a result, fg can be extended to an entire function. By the continuity of the extended function, it is also true that f (z) ≤1 g(z)

for all z ∈ C. Now Theorem 5.10 (Liouville’s Theorem) leads to us that completes the proof of the problem.

f g

is a constant which 

Chapter 9. Isolated Singularities of an Analytic Function

112

Problem 9.6 Bak and Newman Chapter 9 Exercise 6.

1

Proof. It is easy to see that e z 6= 0 for all z ∈ C. Then 0 is a missing value. Let ω ∈ C \ {0}. Set z1 = log ω = log |ω| + i(Arg ω + 2πk), where Arg ω ∈ (−π, π] and k ∈ Z. Certainly, we have 1

e z = elog ω = ω. In addition, it is evident that 1 log |ω| + i(Arg ω + 2πk) log |ω| − i(Arg ω + 2πk) 1 · = log |ω| + i(Arg ω + 2πk) log |ω| − i(Arg ω + 2πk) log |ω| − i(Arg ω + 2πk) = (log |ω|)2 + (Arg ω + 2πk)2

z=

which gives |z| = p

1 (log |ω|)2

+ (Arg ω + 2πk)2

.

(9.3)

Hence the modulus (9.3) will ensure that 0 < |z| < 1 for all large enough k. This completes the  proof of the problem. Problem 9.7 Bak and Newman Chapter 9 Exercise 7.

Proof. For simplicity, we assume that z0 = 0 in the following discussion. Suppose that we have f (z) =

∞ X

ak z

k

and g(z) =

∞ X

bp z p ,

(9.4)

p=−n

k=−m

where ak , bp ∈ C for all k = −m, −m + 1, . . . and p = −n, −n + 1, . . .. If m 6= n, then it is easy to see from the forms (9.4) that f + g has a pole of order max{m, n} at 0. If m = n, then the order of the pole at 0 depends on the coefficients ak and bp . For example, if an = −bn , then the order of pole of f + g at 0 is less than n. Anyhow the order of pole of f + g at 0 is less than or equal to n. By Definition 9.2, let’s write f (z) =

A(z) z m B(z)

and

g(z) =

P (z) , z n Q(z)

(9.5)

where A, B, P and Q are analytic at 0, A(0), B(0), P (0), Q(0) 6= 0. It follows from the forms (9.5) that A(z)P (z) f (z) · g(z) = m+n , z B(z)Q(z) where A(0)P (0) 6= 0 and B(0)Q(0) 6= 0. Hence f · g has a pole of order m + n at 0. Furthermore, the form (9.5) also implies that A(z) z n Q(z) A(z)Q(z) f (z) = m × = m−n g(z) z B(z) P (z) z B(z)P (z)

(9.6)

113 which shows that fg has a zero of order n − m at 0 if n > m and similarly, it has a pole of order m − n at 0 if m > n. Finally, if m = n, then the representation (9.6) reduces to A(z)Q(z) f (z) = g(z) B(z)P (z) so that lim z ·

z→0

f (z) A(z)Q(z) = lim z · = 0. z→0 g(z) B(z)P (z)

By Theorem 9.3 (The Riemann’s Principle of Removable Singularities), singularity at 0. This completes the proof of the problem.

f g

has a removable 

Problem 9.8 Bak and Newman Chapter 9 Exercise 8.∗

Proof. Suppose that z0 is an essential singularity of f . By Theorem 9.6 (The Casorati-Weierstrass Theorem), f (D(z0 ; n1 ) \ {z0 }) is dense in C for every n ∈ N. In other words, it is true that   1   1 \ {z0 } ∩ D 0; 6= ∅ f D z0 ; n n

(9.7)

for every n ∈ N. Equivalently, the set relation (9.7) asserts that there is an αn ∈ D(z0 ; n1 ) \ {z0 } such that f (αn ) ∈ D(0; n1 ) for each n ∈ N. Hence we obtain the result that both αn → z0 and f (αn ) → 0 hold. Similarly, we also have   1   1 \ {z0 } ∩ D n; 6= ∅ f D z0 ; n n

for every n ∈ N. Consequently, one can find a βn ∈ D(z0 ; n1 )\{z0 } such that f (βn ) ∈ D(n, n1 ) for each n ∈ N. Since f (βn ) > n − n1 holds for all n ∈ N, we conclude that βn → z0 and f (βn ) → ∞ as n → ∞. Conversely, we suppose that there exist sequences {αn } and {βn } tending to z0 such that f (αn ) → 0 and f (βn ) → ∞. If z0 was a removable singularity of f , then lim f (z) exists by z→z0

Definition 9.2 or by its Laurent expansion about z0 . However, this means that lim f (αn ) = lim f (βn )

n→∞

n→∞

which is impossible. If it was a pole of f , then Definition 9.2 implies that lim f (z) = ∞

z→z0

which contradicts to our hypothesis. Hence they assert that z0 is an essential singularity of f ,  completing the proof of the problem. Problem 9.9 Bak and Newman Chapter 9 Exercise 9.

Proof.

Chapter 9. Isolated Singularities of an Analytic Function (a) We know that z4

1 1 1 = 2 2 = 2 . 2 +z z (z + 1) z (z + i)(z − i)

114

(9.8)

By Definition 9.2, it has a pole of order 2 at 0 and poles of order 1 at ±i. z (b) We note that cot z = cos sin z and sin z is zero if and only if z = nπ, where n ∈ Z. Thus cot z is analytic in a deleted neighborhood of nπ. By Lemma 7.2, we establish

lim (z − nπ) ·

z→nπ

cos(ζ + nπ) cos ζ cos z = lim ζ · = lim ζ · = 1. ζ→0 sin z sin(ζ + nπ) ζ→0 sin ζ

(9.9)

Furthermore, we have lim (z − nπ)2 ·

z→nπ

 cos ζ  cos z = 0. = lim ζ × ζ · ζ→0 sin z sin ζ

Therefore, it follows from Theorem 9.5 that cot z has a pole of order 1 at nπ, where n ∈ Z. (c) By similar analysis as part (b), csc z has a pole of order 1 at nπ, where n ∈ Z. (d) The function exp( z12 ) z−1 has singularities at 0 and 1. It is easily seen that f is analytic in a deleted neighborhood of 1. Since 1 (9.10) lim (z − 1)f (z) = lim exp 2 = e and lim (z − 1)2 f (z) = 0, z→1 z→1 z→1 z f (z) =

Theorem 9.5 ensures that f has a simple pole at 1.

For the singularities at 0, we consider the Laurent expansion of f at 0. In fact, we know that 1 1 1 1 exp 2 = 1 + 2 + 4 + · · · and = −1 − z − z 2 − · · · . z z z z−1 Thus we have   1 1 (9.11) f (z) = −(1 + z + z 2 + · · · ) × 1 + 2 + 4 + · · · . z z Since every term in the brackets of the expression (9.11) has positive coefficient, the resulting expansion of f will have infinitely many nonzero terms in its principal part. By the notes on [4, p. 125], we conclude that f has an essential singularity at 0. Hence we end the proof of the problem.  Problem 9.10 Bak and Newman Chapter 9 Exercise 10.∗

Proof. We note that z 2 + 1 = (z − i)(z + i) so that f (z) =

(z 2

1 1 1 = · . 2 2 + 1) (z − i) (z + i)2

(9.12)

1 Let F (z) = (z+i) 2 . Then F is clearly analytic in a neighbourhood of z = i. By Theorem 6.5 (The Power Series Representation for Functions Analytic in a Disc), we know that

F (z) =

∞ X F (k) (i) k=0

k!

(z − i)k .

(9.13)

115 Combining the two expressions (9.12) and (9.13), we get f (z) =

∞ X F (k) (i) k=0

Since F (i) = given by

1 (2i)2

=

− 14

k!

k−2

(z − i)

∞

F (i) F ′ (i) X F (k+2) (i) = + + (z − i)k . (z − i)2 z−i (k + 2)! k=0

− 4i ,

and F ′ (i) =

the principal part of the Laurent expansion of f is

1 i . − 2 4(z − i) 4(z − i) This completes the proof of the problem. −



Problem 9.11 Bak and Newman Chapter 9 Exercise 11.

Proof. (a) We have ∞ ∞ X 1 1 1 X 1 2 4 k 2k = 2 = 2 · (1 − z + z − · · · ) = 2 (−1) z = (−1)k+1 z 2k . z4 + z2 z (1 + z 2 ) z z k=0

(b) Suppose that

k=−1

∞ X exp( z12 ) ak z k . = z−1 k=−∞

Since

we have

1 1 1 + + ··· exp 2 = 1 + z 1!z 2 2!z 4

and

1 = −(1 + z + z 2 + · · · ), z−1

∞   X 1 1 2 − 1+ + + · · · (1 + z + z + · · · ) = ak z k . 1!z 2 2!z 4 k=−∞

For k ≥ 0, we obtain

  1 1 ak = − 1 + + + · · · = −e. 1! 2! Let n ∈ N. If k = −2n, then we get h1 i 1 1 1 a−2n = − + + · · · = −e + 1 + + · · · + . n! (n + 1)! 2! (n − 1)!

Similarly, if k = −2n + 1, then we conclude i h1 1 1 1 + + · · · = −e + 1 + + · · · + . a−2n+1 = − n! (n + 1)! 2! (n − 1)!

(c) We note that z2 which implies that

1 1 1 1 = = · −4 (z − 2)(z + 2) z − 2 4(1 + z−2 4 )

∞ ∞ k  X X (−1)k+1 1 1 k z−2 = = (z − 2)k . (−1) 2 z −4 4(z − 2) 4 4k+2 k=0

We have completed the analysis of the problem.

k=−1



Chapter 9. Isolated Singularities of an Analytic Function

116

Problem 9.12 Bak and Newman Chapter 9 Exercise 12.∗

Proof. We know that f (z) = If |z| < 1, then we have If |z| > 1, then

1 1 1  − . + z 2−z 1−z

1 = 1 + z + z2 + · · · . 1−z

(9.14)

 1 1 1 1 1 1 1 = − 1 + =− + + · · · = − − 2 − ··· . 1 2 1−z z z z z z z(1 − z )

(9.15)

1 |z|

< 1 and thus

Combining the formulas (9.14) and (9.15), we see that  ∞ X    zk ,     k=0

if |z| < 1;

1 = 1−z  −1  X    z k , if |z| > 1.   −

(9.16)

k=−∞

Similarly, we have

1 1 1 = × 2−z 2 1−

z 2

=

 ∞ 1 X  z k    ,     2 k=0 2

if |z| < 2;

 −1   1 X  z k   , if |z| > 2.   −2 2

(9.17)

k=−∞

(a) Now the formulas (9.16) and (9.17) imply that f (z) =

∞ ∞  ∞ ∞ X 1 h 1 X  z k X k i 1 X  1  1  − + z = 1 − k+2 z k . 1 − k+1 z k = z 2 2 z 2 2 k=0

k=0

k=−1

k=0

(b) In this case, the formulas (9.16) and (9.17) give f (z) =

−1 −2 ∞ ∞ i X X X 1 k 1 h 1 X  z k − − z . zk = − zk − z 2 2 2k+2 k=0

k=−∞

k=−∞

k=−1

(c) Finally, we have −1 −2  −1 −1  i 1 X   X X 1 h 1 X  z k 1 1 k k z = − f (z) = − 1 z = − 1 zk . z 2 2 z 2k+1 2k+2 k=−∞

k=−∞

We have completed the proof of the problem.

k=−∞

k=−∞



117 Problem 9.13 Bak and Newman Chapter 9 Exercise 13.∗

Proof. For n = 1, 2, . . . , k, we notice that z an − 1 = (z − 1)Pn (z), where Pn (z) = z an −1 + z an −2 + · · · + z + 1. Thus we have R(z) =

1 (z −

1)k P1 (z)P2 (z) · · · Pk (z)

.

Now the product P (z) = P1 (z)P2 (z) · · · Pk (z) is a polynomial in z, so Theorem 5.12 (The Fundamental Theorem of Algebra) ensures that there exists a δ > 0 such that C(1; δ) does not contain any zero of P (z). Thus we observe easily from Corollary 9.10 that if C = C(1; δ), then c−k

1 = 2πi

Z

C

R(z) 1 dz = −k+1 (z − 1) 2πi

Z

C

1 P (z)

z−1

dz.

(9.18)

Since P 1(z) is analytic inside and on C, we may apply Theorem 6.4 (The Cauchy Integral Formula) to the last integral (9.18) to obtain c−k =

1 1 = . P (1) ak ak−1 · · · a1

This ends the proof of the problem.



Problem 9.14 Bak and Newman Chapter 9 Exercise 14.

Proof. Suppose that f (z) =

∞ X

ak z k .

(9.19)

k=−∞

Then we have f (−z) =

∞ X

(−1)k ak z k .

(9.20)

k=−∞

Since f is odd, we have f (z) = and (9.20) that ∞ X

k=−∞

ak z k =

f (z)−f (−z) . 2

Thus it follows from the two representations (9.19)

∞ ∞ X X ak + (−1)k+1 ak k z = a2k+1 z 2k+1 . 2

k=−∞

k=−∞

Consequently, it means that a2k = 0 for all k ∈ Z. This completes the proof of the problem.  Problem 9.15 Bak and Newman Chapter 9 Exercise 15.

Chapter 9. Isolated Singularities of an Analytic Function

118

Proof. (a) We observe from the expression (9.8) that 1 A1 A2 B C A1 z(z 2 + 1) + A2 (z 2 + 1) + Bz 2 (z + i) + Cz 2 (z − i) = + + + = z4 + z2 z z2 z − i z + i z 2 (z − i)(z + i) which means that A1 z(z 2 + 1) + A2 (z 2 + 1) + Bz 2 (z + i) + Cz 2 (z − i) = 1 1 . If z = −i, then C = for all z ∈ C. If z = 0, then A2 = 1. If z = i, then B = − 2i Finally, if z = 1, then A1 = 0. Consequently, we conclude that

z4

1 2i .

1 1 1 1 = 2− + . 2 +z z 2i(z − i) 2i(z + i)

(b) Similar as part (a), we have z2

1 1 1 = − . +1 2i(z − i) 2i(z + i)

This completes the proof of the problem.



Problem 9.16 Bak and Newman Chapter 9 Exercise 16.

Proof. We follow the given hint. Assume that there were ω ∈ C and δ > 0 such that |f (z)−ω| > δ for all z ∈ D. Then we have 1 1 < f (z) − ω δ throughout D. By Corollary 9.4, the function

g(z) =

1 f (z) − ω

has a removable singularity at z0 . Therefore, g is analytic at z0 . Now g is analytic in the region D ∪ {z0 } and g(zn ) = 0 for all n ∈ N, where zn → z0 . It follows from Theorem 6.9 (The Uniqueness Theorem) that g ≡ 0 in D ∪ {z0 }, i.e., f (z) = ∞ throughout D which is impossible.  We have completed the analysis of the problem. Problem 9.17 Bak and Newman Chapter 9 Exercise 17.

Proof. Suppose that D0 (0; 1) = D(0; 1) \ {0} and g(z) = sin 1z . (a) We know that sin

1 1 1 1 = − + − ··· , 3 z z 3!z 5!z 5

(9.21)

119 so it has an essential singularity at 0 and Theorem 9.6 (The Casorati-Weierstrass Theorem) ensures that g(D0 (0; 1)) is dense in C. In other words, given ω ∈ C, there exists a sequence {zn } ⊆ D0 (0; 1) such that g(zn ) → ω1 as n → ∞. Since f = g1 , we conclude that 1 =ω n→∞ g(zn )

lim f (zn ) = lim

n→∞

which means that f (D0 (0; 1)) is also dense in C. (b) Let zn =

1 nπ ,

where n ∈ Z. Clearly, we follow from Lemma 7.2 that lim (z − zn ) csc

z→zn

and lim (z − zn )2 csc

z→zn

z − zn 1 = lim z z→zn sin 1z ζ − nπ 1 · lim =− nπ ζ→nπ sin ζ 1 ω =− · lim n nπ ω→0 (−1) sin ω (−1)n+1 =− nπ 6= 0

1 z − zn = lim (z − zn ) · = 0. z→z z n sin 1z

It follows from Theorem 9.5 that csc z1 has a simple pole at every zn . Since zn → 0 as n → ±∞. By Problem 9.16, we see that f (D0 (0; 1)) is dense in C. Hence we complete the proof of the problem.



Problem 9.18 Bak and Newman Chapter 9 Exercise 18.

Proof. If f is a polynomial, then Theorem 5.12 (The Fundamental Theorem of Algebra) ensures that f takes every value in the plane C. Suppose that f is not a polynomial and we consider g(z) = f ( 1z ). Similar to the proof of Problem 9.3, it can be shown that g has either a pole or an essential singularity at 0. If 0 is a pole of g, then we have g(z) =

∞ X

an z n

n=−k

for some k ∈ N so that f (z) =

k X

an z n .

(9.22)

n=−∞

Since f is analytic at 0, the principal part of the Laurent expansion (9.22) disappears, i.e., a−1 = a−2 = · · · = 0. In other words, f reduces to a polynomial, a contradiction. Therefore, 0 is an essential singularity of g and Theorem 9.6 (The Casorati-Weierstrass Theorem) implies that g(C \ {0}) is dense in C. Since g(C \ {0}) ⊆ f (C), the set f (C) is also dense in C and we  complete the analysis of the problem.

Chapter 9. Isolated Singularities of an Analytic Function

120

CHAPTER

10

The Residue Theorem

Problem 10.1 Bak and Newman Chapter 10 Exercise 1.

Proof. 1 (a) By Problem 9.9(a), z 4 +z 2 has a pole of order 2 at 0 and poles of order 1 at ±i. Furthermore, we apply [4, Eqn. (1), p. 129] to the expression (9.8) to get   1 1 1 1 i ; i = lim(z − i) × 4 = lim =− = Res 4 z→i z + z2 z + z 2 z→i z 2 (z + i) 2i 2

and

Res



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

Since we know z4

1 1 1 1 1 = 2× = 2 (1 − z 2 + z 4 − · · · ) = 2 − 1 + z 2 − · · · , 2 2 +z z 1+z z z

Definition 10.1 implies that Res



 1 ; 0 = 0. z4 + z2

(b) Recall from Problem 9.9(b) that cot z has a simple pole at nπ for every n ∈ Z. Applying [4, Eqn. (1), p. 129] to the limit (9.9), we conclude that Res (cot z; nπ) = 1 for all n ∈ Z. (c) By Problem 9.9(c), csc z has a simple pole at nπ for every n ∈ Z. Then we have lim (z − nπ) csc z = lim

z→nπ

z→nπ

z − nπ = lim sin z − sin nπ z→nπ exp(

1

1 sin z−sin nπ z−nπ

)

=

1 = (−1)n . cos nπ

(d) By Problem 9.9(d), the function z−1z2 has a simple pole at 1 and an essential singularity at 0. By the limit (9.10), we conclude at once that Res

 exp( 1 )  z2 ; 1 = e. z−1 121

Chapter 10. The Residue Theorem

122

Next, we get from Problem 9.11(b) (with n = 1) that Res (e) We note that z2

 exp( 1 )  z2 ; 0 = −e + 1. z−1

1 1 = , + 3z + 2 (z + 1)(z + 2)

so it has simple poles at −1 and −2. According to [4, Eqn. (1), p. 129], we see that   1 1 1 Res 2 ; −1 = lim (z + 1) × 2 = lim =1 z→−1 z + 3z + 2 z + 3z + 2 z→−1 z + 2

and

Res



 1 1 1 ; −2 = lim (z + 2) × 2 = lim = −1. z→−2 z 2 + 3z + 2 z + 3z + 2 z→−2 z + 1

(f) We know from the condition of Problem 9.17(a) that sin 1z has an essential singularity at 0. Thus we follow from the expression (9.21) that  1  Res sin ; 0 = 1. z 3

(g) It is clear that the function ze z has an essential singularity at 0. By [4, Example (iv), p. 124], we have 3 9 1 Res (ze z ; 0) = 1 2 = 2. 2 · (3) (h) By the quadratic formula, the function az 2 √

1 + bz + c

2

b b −4ac if b2 − 4ac 6= 0 and a pole of order 2 at z0 = − 2a if has simple poles at z± = −b± 2a 2 b − 4ac = 0. Therefore, we have   1 1 Res ; z + = lim (z − z+ ) 2 z→z+ az + bz + c a(z − z+ )(z − z− ) 1 = lim z→z+ a(z − z− ) 1 = a(z+ − z− ) 1 =√ 2 b − 4ac

and Res



 1 1 1 . ; z = −√ − = lim (z − z− ) 2 z→z− az + bz + c a(z − z+ )(z − z− ) b2 − 4ac

Finally, we get   i 1 1 dh d  1  2 Res (z − z ) × = = 0. ; z = 0 0 az 2 + bz + c dz a(z − z0 )2 z=z0 dz a z=z0

Hence we complete the proof of the problem.



123 Problem 10.2 Bak and Newman Chapter 10 Exercise 2.

Proof. (a) Obviously, cot z is analytic in the simply connected domain D(0; 1 + δ) except 0 for some small δ > 0 and C(0; 1) ⊆ D(0; 1 + δ) is a closed curve such that 0 lies inside C(0; 1), it follows from Theorem 10.5 (The Cauchy’s Residue Theorem) and Problem 10.1(b) that Z cot z dz = 2πiRes (cot z; 0) = 2πi. |z|=1

2πki

1 (b) The function f (z) = (z−4)(z 3 −1) has simple poles at 4 and zk = e 3 , where k = 1, 2, 3. Now f is analytic in the simply connected domain D(0; 2 + δ) except z1 , z2 , z3 for some small δ > 0 and C(0; 2) ⊆ D(0; 2 + δ) is a closed curve such that zk lie inside C(0; 2) and 4 lies outside C(0; 2). If B(z) = (z − 4)(z 3 − 1), then B ′ (z) = (z 3 − 1) + 3z 2 (z − 4). Thus we obtain from [4, Eqn. (1), p. 129] that

Res (f (z); 4) =

43

1 1 = −1 63

and Res (f (z); zk ) =

where k = 1, 2, 3. Since zk3 = 1, we have zk2 =

1 zk

Res (f (z); zk ) =

1 3zk2 (zk

− 4)

,

so that zk , 3(zk − 4)

where k = 1, 2, 3. Next, recall that z1 + z2 + z3 = z1 z2 + z1 z3 + z2 z3 = 0 and z1 z2 z3 = 1, so direct computation gives z2 z3 z1 + + z1 − 4 z2 − 4 z3 − 4 z1 (z2 − 4)(z3 − 4) + z2 (z1 − 4)(z3 − 4) + z3 (z1 − 4)(z2 − 4) = (z1 − 4)(z2 − 4)(z3 − 4) 3z1 z2 z3 − 8(z1 z2 + z1 z3 + z2 z3 ) + 16(z1 + z2 + z3 ) = z1 z2 z3 − 4(z1 z2 + z1 z3 + z2 z3 ) + 16(z1 + z2 + z3 ) − 64 1 =− . 21 Hence Theorem 10.5 (The Cauchy’s Residue Theorem) leads to us that Z   dz 1 z1 z2 z3 = 2πi 0 × + 1 × + 1 × + 1 × 3 63 3(z1 − 4) 3(z2 − 4) 3(z3 − 4) |z|=2 (z − 4)(z − 1) 2πi . =− 63 (c) Now sin z1 has an essential singularity at 0 and it is analytic in the simply connected domain D(0; 1 + δ) except 0 for some small δ > 0 and C(0; 1) ⊆ D(0; 1 + δ) is a closed curve such that 0 lies inside C(0; 1). Combining Problem 10.1(f) and Theorem 10.5 (The Cauchy’s Residue Theorem), we establish Z  1  1 sin dz = 2πiRes sin ; 0 = 2πi. z z |z|=1

Chapter 10. The Residue Theorem

124

(d) By using Problem 10.1(g) and Theorem 10.5 (The Cauchy’s Residue Theorem) directly, we see that Z 3 3 ze z dz = 2πiRes (ze z ; 0) = 9πi. |z|=2

Hence we end the proof of the problem.



Problem 10.3 Bak and Newman Chapter 10 Exercise 3.

1 Proof. It is clear that the function f (z) = [1−exp(−z)] n is analytic everywhere except zk = 2πki, where k ∈ Z. If C is a regular closed curve surrounding only the singularity z0 = 0, then it deduces from Theorem 10.5 (The Cauchy’s Residue Theorem) that Z   dz 1 1 ; 0 = . (10.1) Res −z n (1 − e ) 2πi C (1 − e−z )n

By the substitution ω = ω(z) = 1 − e−z , the integral in the formula (10.1) becomes Z Z dz dω , = −z )n n (1 − ω) (1 − e ω ′ C C

(10.2)

where C ′ = ω(C). 1 has (simple) poles at 0 and 1. Now we need to construct a suitable C The function ωn (1−ω) ′ so that the curve C surrounds only the pole 0 but not 1. To this end, we consider the rectangle R = {z = x + iy | −1 ≤ x ≤ 1 and − π2 ≤ y ≤ π2 }. Now its boundary C = ∂R is a regular closed curve. Obviously, we have ω = ω(z) = 1 − e−x−iy = (1 − e−x cos y) + ie−x sin y which implies that |ω(z) − 1| = e−x , where −1 ≤ x ≤ 1, so we have actually 1 ≤ |ω(z) − 1| ≤ e e for all z ∈ R. Furthermore, Re ω(z) = 1 − e−x cos y implies that Re ω(z) > 1 if and only if cos y < 0 if and only if π2 < y < 3π 2 . In other words, we have Re ω(z) ≤ 1 for all z ∈ R which assures us that ω(R) is a subset of the left half of the annulus A centred at 1 with inner and outer radii e−1 and e respectively. Finally, it is easy to see from the definition that the curves π n πo −1 −1 γ1 = ω(1 + iy) = (1 − e cos y) + ie sin y − ≤ y ≤ 2 2 and

π n πo γ2 = ω(−1 + iy) = (1 − e cos y) + ie sin y − ≤ y ≤ 2 2 are exactly the inner arc and the outer arc of A respectively. Similarly, the curves o n  π = 1 + ie−x − 1 ≤ x ≤ 1 γ3 = ω x + i 2

125 and

o n  π = 1 − ie−x − 1 ≤ x ≤ 1 γ4 = ω x − i 2 are the remaining boundary of the annulus A. Consequently, we obtain the desired result that the regular closed curve C ′ = ω(C) surrounds only the origin. Hence it deduces from Theorem 10.5 (The Cauchy’s Residue Theorem) that Z   dω 1 = Res ; 0 = 1. (10.3) n ω n (1 − ω) C ′ ω (1 − ω)

Now our desired result follows immediately from the comparison of the expressions (10.1), (10.2) and (10.3), completing the proof of the problem.  Problem 10.4 Bak and Newman Chapter 10 Exercise 4.∗

Proof. We know that f (z) = (z + z1 )2m+1 has a pole of order 2m+1 at 0 for every m = 0, 1, 2, . . .. By Theorem 10.5 (The Cauchy’s Residue Theorem), we establish Z   1 2m+1  1 2m+1 dz = 2πiRes z + ;0 . z+ z z |z|=1 By the binomial theorem, we have 2m+1 2m+1  X X 1 2m+1 2m+1 2m+1−k −k z+ = Ck z z = Ck2m+1 z 2m+1−2k . z k=0

k=0

2m+1 If k = m + 1, then Cm+1 is the coefficient of z −1 so that Z  1 2m+1 2m+1 2m+1 dz = 2πiCm+1 = 2πiCm , z+ z |z|=1

completing the proof of the problem.



Problem 10.5 Bak and Newman Chapter 10 Exercise 5.∗

Proof. Suppose that f (ω) is analytic in the region D, D contains the regular closed curve C and F (ω) =

f (ω) p(ω) − p(z) · , p(ω) ω−z

where ω ∈ D\{ω1 , ω2 , . . . , ωn } and z ∈ C. Intuitively, one may apply Corollary 10.6 and Theorem 10.11 (The Generalized Cauchy Integral Formula) directly but we can’t do that because they require the domain of the analyticity of f should be simply connected. Thus we have to search another versions of the results that fit our case. In fact, such variations can be found in [3, Theorem 3.8.6, p. 213; Theorem 5.1.2, p. 294]. Since C encloses the distinct points ω1 , ω2 , . . . , ωn , it follows from [3, Theorem 5.1.2, p. 294] that Z n n   f (ω) p(ω) − p(z) X X · ; ωk . (10.4) F (ω) dω = 2πi Res (F (ω); ωk ) = 2πi Res p(ω) ω−z C k=1

k=1

Chapter 10. The Residue Theorem

126

Since p is a polynomial of degree n, direct computation shows that the quotient polynomial of degree n − 1. Thus if we may let

p(ω)−p(z) ω−z

is a

n−1 X p(ω) − p(z) Qm (ω)z m , = ω−z m=0

where Qk (ω) is a polynomial in ω, then we get from [4, Eqn. (1), p. 129] that Res

    f (ω) p(ω) − p(z) f (ω) · p(ω)−p(z) ω−z ; ωk = Res · ; ωk p(ω) ω−z (ω − ω1 )(ω − ω2 ) · · · (ω − ωn ) p(ωk ) − p(z) f (ωk ) · ωk − z Y = (ω − ωj ) j6=k

=

n−1 X

f (ωk ) · Qm (ωk ) m Y z . (ω − ωj ) m=0

(10.5)

j6=k

Using the expressions (10.4) and (10.5), we conclude that 1 P (z) = 2πi

Z

F (ω) dω =

C

n n−1 X X f (ωk ) · Qm (ωk ) Y zm (ω − ωj ) k=1 m=0 j6=k

which is a polynomial in z of degree n − 1. Besides, we follow from the definition and [3, Theorem 3.8.6, p. 213] that Z Z f (ω) p(ω) f (ω) 1 1 · dω = dω = f (ωk ) P (ωk ) = 2πi C p(ω) ω − ωk 2πi C ω − ωk for every k = 1, 2, . . . , n. We complete the analysis of the problem.



Problem 10.6 Bak and Newman Chapter 10 Exercise 6.

Proof. For small h, we observe that Z Z φ(ω) φ(ω)  f (z + h) − f (z) 1 1  dω = = − dω. h h ω − z − h ω − z (ω − z − h)(ω − z) γ γ Then we have f (z + h) − f (z) f (z) = lim = lim h→0 h→0 h ′

Z

γ

φ(ω) dω. (ω − z − h)(ω − z)

If γ : ω = ω(t) with a ≤ t ≤ b, then it follows from Definition 4.3 that Z b Z φ(ω) F (t, h) dt, dω = G(h) = a γ (ω − z − h)(ω − z) where F (t, h) =

φ(ω(t)) ω(t). ˙ [ω(t) − z − h][ω(t) − z]

(10.6)

(10.7)

127 We claim that G is a continuous function of h. To this end, since z ∈ / γ, we obtain from [22, Exercise 20, p. 101] that d(z, γ) = inf |ω − z| > 0. Furthermore, we know that d(z, γ) is ω∈γ

a continuous function in z, so we also have d(z + h, γ) > 0 for sufficiently small h > 0. These facts imply that 1 [ω(t) − z − h][ω(t) − z]

is a continuous function of t and h. Finally, the continuity of φ on γ and the piecewise continuity of ω(t) ˙ on [a, b] ensure that F (t, h) is a piecewise continuous function of t and h. By the definition (10.7), we have Z b |F (t, h) − F (t, h′ )| dt, |G(h) − G(h′ )| ≤ a

so G is a continuous function of h, as desired. Consequently, it is legitimate by the above claim that we can interchange the limit and the integral in the expression (10.6) and get Z Z φ(ω) φ(ω) lim f ′ (z) = dω = dω h→0 (ω − z − h)(ω − z) (ω − z)2 γ γ which proves the first assertion. For the second assertion, suppose that f is analytic in a simply connected domain D and γ is a regular closed curve contained in D. By Definition 4.3, γ is compact. Since f is obviously continuous on γ, the first assertion can be applied here to conclude Z f (ω) 1 ′ dω, f (z) = 2πi γ (ω − z)2 where z lies inside γ. Suppose that we have f

(k)

k! (z) = 2πi

Z

γ

f (ω) dω. (ω − z)k+1

Using similar attack as proving the first assertion, Z i k! 1h 1 f (k) (z + h) − f (k) (z) 1 dω. = f (ω) · − h 2πi γ h (ω − z − h)k+1 (ω − z)k+1 Recall that ak+1 − bk+1 = (a − b)(ak + ak−1 b + · · · + abk−1 + bk ), so we see that 1 h 1 × − = k+1 k+1 (ω − z − h) (ω − z) (ω − z − h)(ω − z) i h 1 1 1 + + · · · + (ω − z − h)k (ω − z − h)k−1 (ω − z) (ω − z)k and thus f (k) (z + h) − f (k) (z) h→0 h Z i 1 1h 1 k! f (ω) · − dω = lim h→0 2πi γ h (ω − z − h)k+1 (ω − z)k+1 Z i 1h 1 1 k! dω f (ω) · lim − = h→0 h (ω − z − h)k+1 2πi γ (ω − z)k+1

f (k+1) (z) = lim

Chapter 10. The Residue Theorem Z

128

1 × (ω − z − h)(ω − z) γ i 1 1 1 dω + + · · · + (ω − z − h)k (ω − z − h)k−1 (ω − z) (ω − z)k Z k! k+1 dω f (ω) · = 2πi γ (ω − z)k+2 Z (k + 1)! f (ω) = dω. k+2 2πi γ (ω − z) k! = 2πi h

f (ω) · lim

h→0

Hence Theorem 10.11 (The Generalized Cauchy Integral Formula) follows immediately from the induction. It completes the proof of the problem.  Problem 10.7 Bak and Newman Chapter 10 Exercise 7.

Proof. It is impossible that f (z) ≡ 0 on C because it contradicts our hypothesis. Assume that f had more than one zero, say z1 and z2 . Let r be so large that z1 and z2 lie inside C(0; r). By the hypothesis again, we know that both z1 and z2 must be real. Suppose that z1 > z2 which gives r > z1 > z2 > −r. Furthermore, we know from the hypothesis that f (±r) are the only real values on C(0; r). Thus if f (r) = 0, then we can select another r > r1 so that D(0; r1 ) contains z1 , z2 , i.e., r > r1 > z1 > z2 > −r1 > −r. This process will stop in a finite number of steps, otherwise, one can find a decreasing sequence {rn } such that r > r1 > r2 > · · · > z1 > z2 > · · · > −r2 > −r1 > −r. Since the sequence {rn } is obviously bounded, it follows from [22, Theorem 3.14, p. 55] that it converges. Since f (rn ) = 0 for all n ∈ N, Theorem 6.9 (The Uniqueness Theorem) ensures that f ≡ 0 in C, a contradiction. Hence we may suppose that our chosen r satisfies f (r) 6= 0. Similarly, we also assume that f (−r) 6= 0. Next, we define g : (0, π) → R by

g(θ) = f (reiθ ). Since f is entire, g is a continuous function of θ and the previous paragraph shows that its imaginary part is always positive or negative, i.e., {f (reiθ ) | 0 < θ < π} lies entirely in the upper half-plane or the lower half-plane. Similarly, the set {f (reiθ ) | π < θ < 2π} lies entirely in the upper half-plane or the lower half-plane. Since f is analytic in D(0; r) and f (z) 6= 0 on C(0; r), we obtain from Corollary 10.9 (The Argument Principle) that Z(f ) =

1 ∆Arg f (z) 2π

(10.8)

as z travels around the circle C(0; r) from re0·i to re2πi . Since ∆Arg f (z) ≤ π in the upper half-plane or the lower half-plane, we establish from the formula (10.8) that Z(f ) ≤ 1, completing the proof of the problem.



129 Problem 10.8 Bak and Newman Chapter 10 Exercise 8.∗

Proof. (a) If z is a zero of f , then |f (z)| ≥ |g(z)| implies that g(z) = 0 and so f (z) + g(z) = 0, a contradiction. In other words, f has no zero on γ so that 1 + fg is well-defined on γ. Now we can follow the proof of Theorem 10.10 (Rouch´e’s Theorem) to get Z(f + g) = Z(f ) +

1 2πi

Z d(1 + g ) f γ

1+

g f

= Z(f ) + n

  g (γ); 0 . 1+ f

By the hypothesis, we know that |(1 + fg ) − 1| ≤ 1 on γ, so (1 + fg )(γ) remains within or on

the disc of radius 1 centred at 1. In addition, we observe that 1 + fg(z) (z) = 0 if and only if g f (z) + g(z) = 0, so the analytic function 1 + f will not touch the origin. Finally, we note that γ is a closed curve. Hence these observations ensure that (1 + fg )(γ) is also a closed curve lying entirely in the right half-plane without surrounding 0. As a consequence, we have   g (γ); 0 = 0, n 1+ f so Rouch´e’s Theorem holds in this case.

(b) Let f (z) = 2z 4 and g(z) = z 5 + 1. If e5iθ + 2e4iθ + 1 = 0 for some θ ∈ [0, 2π], then |eiθ + 2| = 1 or (2 + cos θ)2 + sin2 θ = 1. (10.9) After simplification, the equation (10.9) reduces to cos θ = −2 which is impossible. Hence f (z) + g(z) = z 5 + 2z 4 + 1 6= 0 on C(0; 1). Since |g(z)| ≤ 2 = |f (z)|, part (a) implies that Z(z 5 + 2z 4 + 1) = Z(2z 4 ) = 4 in the unit disc. We complete the proof of the problem.



Problem 10.9 Bak and Newman Chapter 10 Exercise 9.

Proof. (a) Now the functions f (z) = 3ez and g(z) = −z which are analytic inside and on the regular closed curve C(0; 1). Since |f (z)| = 3|ez | ≥ 3e−1 > |z| for all z ∈ C(0; 1), Theorem 10.10 (Rouch´e’s Theorem) implies that Z(f1 ) = Z(f ) = Z(3ez ) = 0 inside C(0; 1). (b) The functions f (z) = −z and g(z) = 31 ez are analytic inside and on the regular closed curve C(0; 1). Since |f (z)| = |z| = 1 > 3e ≥ |g(z)| on C(0; 1), it follows from Theorem 10.10 (Rouch´e’s Theorem) that Z(f2 ) = Z(f ) = Z(−z) = 1 inside C(0; 1).

Chapter 10. The Residue Theorem

130

(c) The functions f (z) = z 4 and g(z) = −5z + 1 are analytic inside and on the regular closed curve C(0; 2). Since |f (z)| = |z 4 | = 32 > |5z − 1| = |g(z)| on C(0; 2), it follows from Theorem 10.10 (Rouch´e’s Theorem) that Z(f3 ) = Z(f ) = Z(z 4 ) = 4

(10.10)

inside C(0; 2). Similarly, the functions F (z) = −5z and G(z) = z 4 + 1 are analytic inside and on the regular closed curve C(0; 1). Since |F (z)| = |5z| = 5 > |z 4 + 1| = |G(z)|, it follows from Theorem 10.10 (Rouch´e’s Theorem) that Z(f3 ) = Z(F ) = Z(5z) = 1

(10.11)

inside C(0; 1). Combining the results (10.10) and (10.11), Z(f3 ) = 4 − 1 = 3 inside 1 ≤ |z| ≤ 2. (d) Suppose that f (z) = −5z 4 + 3z 2 and g(z) = z 6 − 1. Both f and g are analytic inside and on the regular closed curve C(0; 1). The triangle inequality implies that |f (z)| = | − z 4 + 3z 2 | ≥ |3z 2 | − |z 4 | = 2

and |g(z)| ≤ 2

on |z| = 1. Put z = eiθ into |f (z)| = 2 to get | − 5e2iθ + 3| = 2. Then we derive from this that cos 2θ = 1 so that θ = 0, π. Consequently, |f (z)| = 2 if and only if z = ±1. Since |g(±1)| = 0 < 2, it means that we always have |f (z)| > |g(z)| on C(0; 1). By Theorem 10.10 (Rouch´e’s Theorem), we see that Z(f4 ) = Z(f )

(10.12)

q inside C(0; 1). Notice that f (z) = 0 if and only if z = 0 (repeated roots) and z = ± 35 , so all the zeros of f lie inside C(0; 1) and hence the formula (10.12) gives Z(f3 ) = 4 inside C(0; 1). We complete the analysis of the problem.  Problem 10.10 Bak and Newman Chapter 10 Exercise 10.∗

Proof. We take f (z) = λ − z and g(z) = −e−z . It is clear that z = λ is the only zero of f (z) and it lies in the right half-plane. Let R > 2λ and we consider the rectangle in the right half-plane SR = {z = x + iy | 0 ≤ x ≤ R and −R ≤ y ≤ R}. Now the boundary γR = ∂SR is easily seen to be a regular closed curve and SR becomes the right half-plane as R → ∞. Furthermore, our functions f and g are entire. Let z ∈ γR . If x = 0, then z = iy and so we observe that p |λ − iy| = λ2 + y 2 ≥ λ > 1 = |e−iy |.

Next, we suppose that x > 0. Since z = x + iR for 0 < x ≤ R, we have |z| = that the triangle inequality gives |λ − z| ≥ |z| − λ > λ > 1 ≥ |e−z |.

(10.13) √

x2 + R2 > R so

(10.14)

131 p Finally, if z = R + iy, then we also have |z| = R2 + y 2 ≥ R and thus the inequality (10.14) remains valid in this case. Consequently, the inequalities (10.13) and (10.14) together imply that |f (z)| > |g(z)| for all z ∈ γR . Hence it yields from Theorem 10.10 (Rouch´e’s Theorem) that Z(λ − z − e−z ) = Z(f + g) = Z(f ) = 1 inside γR and the desired result follows if we let R → ∞. This completes the proof of the  problem. Problem 10.11 Bak and Newman Chapter 10 Exercise 11.

Proof. If f is a nonzero constant, then there is nothing to prove. Without loss of generality, we ′ (z) may assume that f is nonconstant. Let F (z) = z m ff (z) . Since f 6= 0 on γ, F is analytic inside and on γ except for isolated singularities inside γ. According to [3, Theorem 5.1.2, p. 294], we see immediately that Z Z ′ X 1 1 m f (z) z F (z) dz = Res (F ; zk ), dz = 2πi γ f (z) 2πi γ k

where the zk denote the isolated singularities of F inside γ which are exactly the zeros of f inside γ. By Lemma 7.1, there exists a function g analytic inside and on the regular closed curve γ such that f (z) = (z − zk )pk g(z) and g(zk ) 6= 0. Therefore, we have F (z) = z m

pk z m z m g′ (z) f ′ (z) = + . f (z) z − zk g(z)

(10.15)

If zk = 0, then F is actually analytic at 0 so that Res (F ; zk ) = 0. Otherwise, F has a simple pole at each zero of f . Thus it yields from [4, Eqn. (1), p. 129] and the expression (10.15) that Res (F ; zk ) = lim (z − zk )F (z) = pk zkm . z→zk

Consequently, we obtain 1 2πi

Z

z γ

mf

′ (z)

1 dz = f (z) 2πi

Z

γ

F (z) dz =

X

pk zkm ,

k

where the sum takes over all zeros of f inside γ. This completes the proof of the problem.



Problem 10.12 Bak and Newman Chapter 10 Exercise 12.

Proof. On the one hand, we have ez 6= 0 for all z ∈ D(0; R) with R > 0, so there exists a ǫR > 0 such that |ez | > ǫR (10.16)

Chapter 10. The Residue Theorem

132 k

for all z ∈ D(0; R). On the other hand, since | zk! | ≤

Rk k!

and the series

∞ X zk k=0

k!

is obviously convergent, we deduce from Theorem 1.9 (The Weierstrass M -Test) that Pn (z) → ez uniformly in D(0; R). In other words, there exists an N ∈ N such that n ≥ N implies |Pn (z) − ez | < ǫR

(10.17)

for all z ∈ D(0; R). By combining the inequalities (10.16) and (10.17), it is easy to see that |Pn (z) − ez | < |ez | holds for all n ≥ N and z ∈ D(0; R). As both Pn (z)−ez and ez are analytic in D(0; R), Theorem 10.10 (Rouch´e’s Theorem) asserts that Z(Pn (z)) = Z(Pn (z) − ez + ez ) = Z(ez ) = 0 inside C(0; R), completing the proof of the problem.



Problem 10.13 Bak and Newman Chapter 10 Exercise 13.∗

Proof. (a) Direct computation gives (1 − z)P (z) = a0 + a1 z + a2 z 2 + · · · + an z n − a0 z − a1 z 2 − a2 z 3 − · · · − an z n+1 n X = a0 + (ak − ak−1 )z k − an z n+1 . (10.18) k=1

Let R > 1. If |z| = R, then we have |z|n+1 > |z|k for all k = 0, 1, 2, . . . , n. Therefore, it follows from the expression (10.18) and the fact an ≥ an−1 ≥ · · · ≥ a0 ≥ 0 that n X (ak − ak−1 )z k |(1 − z)P (z) − (−an z n+1 )| = a0 +

h

< a0 +

k=1 n X k=1

i (ak − ak−1 ) · |z n+1 |

= | − an z n+1 |.

By Theorem 10.10 (Rouch´e’s Theorem), we conclude that Z((1 − z)P (z)) = Z(−an z n+1 ) = n + 1 inside C(0; R). By letting R → 1, all zeros of (1 − z)P (z) (except z = 1) and then the zeros of P (z) lie in C(0; 1).

133 (b) We consider P (z) =

1 (1 − z)2

(10.19)

which is analytic in D(0; 1). Since P (k) (z) = (k + 1)!(1 − z)−k−2 for every k = 0, 1, 2, . . ., we observe from Theorem 6.5 (The Power Series Representation for Functions Analytic in a Disc) that P (k) (0) =k+1 Ck = k! and so ∞ X (k + 1)z k (10.20) P (z) = k=1

in D(0; 1). Now if |z| < 1, then |1 − z| < 2 and thus it follows from the definition (10.19) that 1 |P (z)| > . 4

(10.21)

Next, since the power series (10.20) has radius of convergence 1, we recall from [4, p. 27] that the series Pn (z) → P (z) uniformly in the smallest disc D(0; ρ). Consequently, there exists an N ∈ N such that n ≥ N implies |Pn (z) − P (z)| ≤

1 4

(10.22)

for all z ∈ D(0; ρ). Combining the inequalities (10.21) and (10.22), we know that |Pn (z) − P (z)| < |P (z)| on C(0; ρ). Hence we apply Theorem 10.10 (Rouch´e’s Theorem) that if n ≥ N , then we have Z(Pn ) = Z(P ) inside C(0; ρ). Since P has no zeros in D(0; 1), we establish that Z(Pn ) = 0 inside C(0; ρ) for all n ≥ N .

This completes the proof of the problem.



Problem 10.14 Bak and Newman Chapter 10 Exercise 14.

Proof. Let P (z) = z n + an−1 z n−1 + · · · + a1 z + a0 , where a0 , a1 , . . . , an−1 ∈ C. Let f (z) = z n and g(z) = an−1 z n−1 + · · · + a1 z + a0 . On the regular closed curve C(0; R) for large enough R, we know that |g(z)| ≤ |an−1 | · |z|n−1 + |an−2 | · |z|n−2 + · · · + |a0 | ≤ M Rn−1 < Rn = |f (z)|, where M is a positive constant. Then Theorem 10.10 (Rouch´e’s Theorem) applies in this case to conclude that Z(P ) = Z(f + g) = Z(f ) = n inside C(0; R). It completes the proof of the problem.



Chapter 10. The Residue Theorem

134

Problem 10.15 Bak and Newman Chapter 10 Exercise 15.

Proof. Since |f | > |g| on γ, we get from the triangle inequality that |f (z) + λg(z)| ≥ |f (z)| − λ|g(z)| ≥ |f (z)| − |g(z)| > 0

(10.23)

for all z ∈ γ and all λ ∈ [0, 1]. Thus J(λ) is well-defined on [0, 1]. If s, t ∈ [0, 1], then we have Z Z 1  f ′ + sg ′ f ′ + tg ′  |s − t| f g′ − f ′ g |J(s) − J(t)| = dz = · − · dz . (10.24) 2π f + sg f + tg 2π γ γ (f + tg)(f + sg)

Using the inequality (10.23), we know that

|(f + tg)(f + sg)| = |f + tg| · |f − sg| ≥ (|f | − |g|)2 on γ and then



|f g′ − f ′ g| f g′ − f ′ g ≤ (f + tg)(f + sg) (|f | − |g|)2

(10.25)

on γ. By Definition 4.2, γ : [a, b] → C is continuous, so γ([a, b]) is compact and hence it is of finite length L. Since f, g, f ′ and g ′ are analytic on γ, they are continuous functions on γ([a, b]) so that all the sets f (γ([a, b])), g(γ([a, b])), f ′ (γ([a, b])) and g′ (γ([a, b])) are also compact. Thus there exists a M > 0 such that f, g, f ′ , g ′ ≪ M throughout γ which shows that |f g′ − f ′ g| ≤ 2M 2

(10.26)

throughout γ. By the inequality (10.23) again, there exists a ǫ > 0 such that |f (z)| − |g(z)| ≥

√

ǫ>0

(10.27)

on γ. Substutiting the inequalities (10.26) and (10.27) into the inequality (10.25), we conclude that 2M 2 f g′ − f ′ g ≤ (f + tg)(f + sg) ǫ on γ. Now we apply Theorem 4.10 (The M -L Formula) to the inequality (10.24) to obtain |J(s) − J(t)| ≤

|s − t| 2M 2 × × L = A|s − t| 2π ǫ

for some constant A. Hence J(λ) is a continuous function of λ. Next, direct computation shows f ′ + λg ′ f′ (f + λg)′ = = + f + λg f + λg f and then J(λ) =

1 2πi

Z

γ

f (λg)′ −f ′ (λg) f2 1 + λg f

(f + λg)′ 1 dz = f + λg 2πi

Z

γ

=

λg ′ f ′ (1 + f ) + f 1 + λg f

f′ 1 dz + f 2πi

Z (1 + γ

1

λg ′ f ) + λg f

dz.

(10.28)

135 Our hypothesis definitely implies that |f (z)| > 0 on γ, so we are able to employ Theorem 10.8 to conclude that the second integral in the expression (10.28) is an integer N . Next, the inequality (10.23) implies that |1 + λg f | > 0 on γ, so the third integral in the expression (10.28) is also an integer Nλ by Theorem 10.8. In other words, J(λ) ∈ Z for every λ ∈ [0, 1]. Recall from the previous paragraph that J(λ) is a continuous function of λ, so it must be a constant. In particular, J(0) = J(1) and Theorem 10.8 establishes that Z ′ Z f (f + g)′ 1 1 dz = dz = Z(f + g) Z(f ) = 2πi γ f 2πi γ f + g as required. We end the proof of the problem.



Problem 10.16 Bak and Newman Chapter 10 Exercise 16.

Proof. By the discussion on [4, p. 115], we know that the function   1 Z z 2ζ  1 p 2 2 F (z) = z − 1 = exp log(z − 1) = exp dζ 2 2 √2 ζ 2 − 1

(10.29)

√ is analytic in C \ (−∞, 1] because f (z) = z 2 − 1 is analytic there and f ( 2) 6= 0. Suppose that D = C \ [−1, 1]. Then D is open in C. • Step 1: F is continuous on D. If Im z > 0, then the path of integration in the integral (10.29) is taken to be in the upper half-plane. Similarly, if Im z < 0, then the path of integration in the integral (10.29) is taken to be in the lower half-plane. Since F is analytic in C \ (−∞, 1], F is clearly continuous there. Therefore, it remains to show that F is continuous on (−∞, −1), i.e., for every x0 ∈ (−∞, −1), we have F (x0 ) = z→x lim F (z). 0

(10.30)

z∈D

√ To this end, suppose that Im z > 0. Let γ1 and γ2 be the√ paths from z to x0 and 2 in the upper half-plane respectively. Let γ3 be the path from 2 to x0 in the lower half-plane. See Figure 10.1 for an illustration.

Figure 10.1: The regular closed curve C surrounding ±1.

Chapter 10. The Residue Theorem

136

√ Let C = γ1 − γ2 − γ3 be the regular closed curve connecting 2 and x0 in D. Then C must surround ±1. Since the function g(z) = z 2 − 1 is analytic and nonzero on C, it observes from Corollary 10.9 (The Argument Principle) that Z 2ζ dζ = 4πi 2 C ζ −1 and then

Z

1 2

γ1

2ζ 1 dζ − 2 ζ −1 2

Z

γ2 +γ3

2ζ dζ = 2πi −1

ζ2

which implies that exp

1 Z 2

γ1

Z    2ζ 2ζ 1 dζ = exp 2πi + dζ ζ2 − 1 2 γ2 +γ3 ζ 2 − 1 1 Z  2ζ = exp dζ . 2 γ2 +γ3 ζ 2 − 1

Hence we conclude that lim F (z) =

z→x0 z∈γ1

lim

z→x0 z∈γ2 +γ3

F (z).

In other words, this proves the validity of (10.30) in the case Im z > 0. Now the case Im z < 0 can be verified similarly and so we have established the desired result (10.30). • Step 2: F is analytic in D. Let N be a large positive integer. By Step 1, we see that F is continuous in the open set DN = D(0; N ) \ [−1, 1] ⊆ D and analytic there except on the line segment (−N, −1), so Theorem 7.7 ensures that F is analytic throughout DN . Since N is arbitrary and DN → D as N → ∞, we conclude that F is analytic in D. This completes the proof of the problem.



Problem 10.17 Bak and Newman Chapter 10 Exercise 17.

Proof. Similar to Problem 10.16, we define p F (z) = 3 (z − 1)(z − 2)(z − 3)  1 log[(z − 1)(z − 2)(z − 3)] = exp 3  1 Z z [(ζ − 1)(ζ − 2)(ζ − 3)]′  1 = exp dζ + log 6 3 4 (ζ − 1)(ζ − 2)(ζ − 3) 3  1 Z z [(ζ − 1)(ζ − 2)(ζ − 3)]′  √ 3 dζ = 6 exp 3 4 (ζ − 1)(ζ − 2)(ζ − 3)

(10.31)

is analytic in C \ (−∞, 3] because f (z) = (z − 1)(z − 2)(z − 3) is analytic there and f (4) 6= 0. If we define D = C \ [1, 3], then D is open in C. • Step 1: F is continuous on D. If Im z > 0, then the path of integration in the integral (10.31) is supposed to be in the upper half-plane. Similarly, if Im z < 0, then the path of integration in the integral (10.31) is assumed to be in the lower half-plane. Since F is

137 analytic in C \ (−∞, 3], F is clearly continuous there. Therefore, it remains to show that F is continuous on (−∞, 1), i.e., for every x0 ∈ (−∞, 1), we have F (x0 ) = z→x lim F (z). 0

(10.32)

z∈D

To this end, suppose that Im z > 0. Similar to Figure 10.1, we let γ1 and γ2 be the paths from z to x0 and 4 in the upper half-plane respectively. Furthermore, let γ3 be the path from 4 to x0 in the lower half-plane. Let C = γ1 − γ2 − γ3 be the regular closed curve connecting 4 and x0 in D. Then C must surround 1, 2 and 3. Since the function g(z) = (z − 1)(z − 2)(z − 3) is analytic and nonzero on C, it observes from Corollary 10.9 (The Argument Principle) that Z [(ζ − 1)(ζ − 2)(ζ − 3)]′ dζ = 6πi C (ζ − 1)(ζ − 2)(ζ − 3) and then 1 3

Z

γ1

1 [(ζ − 1)(ζ − 2)(ζ − 3)]′ dζ − (ζ − 1)(ζ − 2)(ζ − 3) 3

Z

γ2 +γ3

[(ζ − 1)(ζ − 2)(ζ − 3)]′ dζ = 2πi (ζ − 1)(ζ − 2)(ζ − 3)

which implies that  1 Z [(ζ − 1)(ζ − 2)(ζ − 3)]′  1 Z [(ζ − 1)(ζ − 2)(ζ − 3)]′  exp dζ = exp dζ . 3 γ1 (ζ − 1)(ζ − 2)(ζ − 3) 3 γ2 +γ3 (ζ − 1)(ζ − 2)(ζ − 3) Hence we conclude that lim F (z) =

z→x0 z∈γ1

lim

z→x0 z∈γ2 +γ3

F (z),

i.e., the limit (10.32) exists in the case Im z > 0. Now the case Im z < 0 can be verified similarly and so we have proved that the limit (10.32) exists. • Step 2: F is analytic in D. Let N be a large positive integer. Now our F is continuous in the open set DN = D(0; N ) \ [1, 3] ⊆ D and analytic there except on the line segment (−N, 1). Hence Theorem 7.7 guarantees that F is analytic throughout DN and then throughout D as desired if we take N → ∞. We complete the proof of the problem.



Chapter 10. The Residue Theorem

138

CHAPTER

11

Applications of the Residue Theorem to the Evaluation of Integrals and Sums

Problem 11.1 Bak and Newman Chapter 11 Exercise 1.

Proof. (a) It is a Type I integral, so we have Z ∞   x2 dx z2 = 2πiRes ; i . 2 2 (1 + z 2 )2 −∞ (1 + x )

(11.1)

Since i is a pole of order 2 of the function f (z) = we see that  Res

z2 , (z + i)2 (z − i)2

i  d h z 2 i dh i z2 2 (z − i) f (z) = ; i = =− . z=i (1 + z 2 )2 dz dz (z + i)2 z=i 4

(11.2)

Combining the expressions (11.1) and (11.2), we conclude immediately that Z ∞ x2 dx π = . 2 )2 (1 + x 2 −∞ (b) Since

x2 (x2 +4)2 (x2 +9)

is an even function, we have Z

0

∞

1 x2 dx = 2 2 2 (x + 4) (x + 9) 2

Z

∞ −∞

x2 dx (x2 + 4)2 (x2 + 9)

which is a Type I integral, so we have Z ∞ h   x2 dx 1 z2 = · 2πi Res ; 3i (x2 + 4)2 (x2 + 9) 2 (z 2 + 4)2 (z 2 + 9) 0 i  z2 ; 2i . + Res (z 2 + 4)2 (z 2 + 9) 139

(11.3)

Chapter 11. Applications of the Residue Theorem to the Evaluation ... Here 3i and 2i are a simple pole and a pole of order 2 of the function f (z) = respectively. z2 , (z+2i)2 (z 2 +9)

Since (z − 2i)2 f (z) =

140 z2 (z 2 +4)2 (z 2 +9)

we have

i  z2 13i z2 dh Res =− ; 2i = . (z 2 + 4)2 (z 2 + 9) dz (z + 2i)2 (z 2 + 9) z=2i 200 

(11.4)

Using [4, Eqn. (1), p. 129], we see that Res



 z2 z2 ; 3i = (z 2 + 4)2 (z 2 + 9) 2(z 2 + 4) · (2z)(z 2 + 9) + (z 2 + 4)2 · (2z) z=3i 3 =− . (11.5) 50i

Therefore, if we put the residues (11.4) and (11.5) back into the expression (11.3), then we obtain Z ∞ 13π 3π π x2 dx − = . 2 2 2 (x + 4) (x + 9) 200 50 200 0 (c) Since

1 x4 +x2 +1

is an even function, we have Z

∞ 0

dx 1 = x4 + x2 + 1 2

Z

∞ −∞

dx x4 + x2 + 1

which is a Type I integral, so we have Z ∞   i h  1 1 1 dx + Res , = · 2πi Res ; z ; z 1 2 x4 + x2 + 1 2 z4 + z2 + 1 z4 + z2 + 1 0 where z1 = cis π3 and z2 = cis 2π 3 . Since both z1 and z2 are simple poles of f (z) = [4, Eqn. (1), p. 129] gives Res



and Res

(11.6) 1 z 4 +z 2 +1 ,

 1 1 1 √ = ; z 1 = z4 + z2 + 1 4z13 + 2z1 −3 + i 3

(11.7)

 1 1 1 √ . = ; z 2 = 3 4 2 z +z +1 4z2 + 2z2 3+i 3

(11.8)



Hence, by substituting the residues (11.7) and (11.8) into the expression (11.6), we obtain Z ∞  √3π  1 1 dx √ + √ = = πi . x4 + x2 + 1 6 −3 + i 3 3 + i 3 0 (d) It is clear that it is a Type II integral. Since the function Z Since that

eiz z

∞ 0

sin x dx 1 = 2 x(1 + x ) 2

Z

∞ −∞

1 sin x dx = Im 2 x(1 + x ) 2

sin x x(1+x2 )

Z

∞ −∞

is even, we have

eix dx. x

has a simple pole at 0, we follow the idea of the example on [4, p. 146], it is true Z

0

∞

1 sin x dx = Im x(1 + x2 ) 2

Z

∞

−∞

 eiz − 1 i eix − 1 1 h dx = Im 2πiRes ;i . x(1 + x2 ) 2 z(1 + z 2 )

141 Using [4, Eqn. (1), p. 129], we know that  eiz − 1  eiz − 1 e−1 − 1 Res ; i = = − z(1 + z 2 ) (1 + z 2 ) + 2z 2 z=i 2 and hence

Z

∞

0

 (1 − e−1 )π sin x dx 1  −1 = Im (1 − e )πi = . x(1 + x2 ) 2 2

(e) Again, it is a Type II integral. Since the integrand is even, we can write Z ∞ Z  eiz i cos x dx 1 ∞ cos x dx 1 h = = Re 2πiRes ;i . 1 + x2 2 −∞ 1 + x2 2 1 + z2 0 By [4, Eqn. (1), p. 129], we have  eiz  eiz e−1 ; i = = Res 1 + z2 2z z=i 2i which implies Z ∞ π cos x dx = . 2 1+x 2e 0

(f) It is a Type III (A) integral, so Z ∞ 2  log z  X dx Res = − ; z , k x3 + 8 z3 + 8 0 k=0

where zk = 2 exp[ (2k+1)πi ] for k = 0, 1, 2. Since they are simple poles of z 31+8 , it follows 3 from [4, Eqn. (1), p. 129] that  log z  log z Res 3 ; zk = z +8 3z 2 z=zk zk log zk =− 24 zk (log |zk | + iArg zk ) =− 24 (2k+1)π (2 + i 3 ) exp (2k+1)πi 3 =− . 12 Hence we conclude that Z ∞ 5πi (2 + i π3 ) exp πi (2 + iπ) exp(πi) (2 + i 5π dx 3 3 ) exp 3 = + + x3 + 8 12 12 12 0 √  √ i h   πi 1 5πi  1 1 3i 3i 2+ − (2 + iπ) + 2 + + − = 12 2 2 3 2 2 √ 3 1 4 3π = × 12 6 √ 3π = . 18 (g) It is a Type III (C) integral. Thus Z ∞ α−1  z α−1  x 2πi(α−1) dx = 2πiRes ; −1 = 2πi(−1)α−1 = 2πie(α−1)πi [1 − e ] 1 + x 1 + z 0 which implies that Z ∞ α−1 x −2πieαπi 2πie(α−1)πi −2πi π = dx = = −απi = . 2απi απi 2πi(α−1) 1 + x 1 − e e − e sin(απ) 1 − e 0

Chapter 11. Applications of the Residue Theorem to the Evaluation ...

142

(h) The integral is of Type IV, so [4, Eqn. (5), p. 150] gives Z

2π

0

dx = (2 + cos x)2

z (z 2 +4z+1)2

Res

1

|z|=1

4 i

Z

(2 +

z+z −1 2 2 )

×

dz iz

z dz + 4z + 1)2 |z|=1   √ z = 8πRes 3 − 2 . ; (z 2 + 4z + 1)2 =

If f (z) =

Z

=

√ z √ , (z+2− 3)2 (z+2+ 3)2

(z 2

then we know that

 i √ 2 √ z dh (z + 2 − ; 3 − 2 = 3) f (z) √ (z 2 + 4z + 1)2 dz z=−2+ 3 i dh z √ = √ 2 dz (z + 2 + 3) z=−2+ 3 √ 3 . = 18



(11.9)

(11.10)

Now it yields from the expressions (11.9) and (11.10) that Z

2π 0

√ √ dx 4 3π 3 = 8π × = . (2 + cos x)2 18 9

(i) It is a Type IV integral, so [4, Eqn. (5), p. 150] implies Z

2π

0

sin2 x dx = 5 + 3 cos x

Z

|z|=1

Z

− 14 (z − z1 )2 dz 3 1 × iz 5 + 2 (z + z )

(z 2 − 1)2 dz 2 2 |z|=1 z (3z + 10z + 3)   X i (z 2 − 1)2 = × 2πi Res 2 2 ; zk , 2 z (3z + 10z + 3) i = 2

(11.11)

k

where zk are the zeros of z 2 (3z 2 + 10z + 3) inside the unit circle |z| = 1. Therefore, the formula (11.11) reduces to Z

2π 0

h    1 i (z 2 − 1)2 (z 2 − 1)2 sin2x dx . = −π Res 2 2 ; 0 + Res 2 2 ;− 5 + 3 cos x z (3z + 10z + 3) z (3z + 10z + 3) 3

Notice that Res and Res



Thus we get



 (z 2 − 1)2 10 d h (z 2 − 1)2 i =− ; 0 = 2 2 2 z (3z + 10z + 3) dz 3z + 10z + 3 z=0 9

8 (z 2 − 1)2 (z 2 − 1)2 1 = . = ; − z 2 (3z 2 + 10z + 3) 3 2z(3z 2 + 10z + 3) + z 2 (6z + 10) z=− 31 9 Z

0

2π

 10 8  2π sin2 x dx = = −π − + . 5 + 3 cos x 9 9 9

143 (j) The integral is of Type IV and then Z

2π 0

Z

1 dz 1 1 × iz a + (z + ) |z|=1 2 z Z 2 dz = i |z|=1 z 2 + 2az + 1   X 1 ; zk , = 4π Res 2 z + 2az + 1

dx = a + cos x

(11.12)

k

where zk are zeros of z 2 + 2az + 1 lying inside the unit circle. We note that z 2 + 2az + 1 = 0 if and only if p z = −a ± a2 − 1. √ √ If a > 1,√then | − a + a2 − 1| < 1√but | − a − a2 − 1| > 1. Similarly, if a < −1, then | − a − a2 − 1| < 1 but | − a + a2 − 1| > 1. Therefore, we get from the expression (11.12) that    p 1  2 − 1 , if a > 1;  4πRes a ; −a +  Z 2π  z 2 + 2az + 1 dx =   a + cos x  p 0  1  4πRes 2 − 1 , if a < −1  ; −a − a z 2 + 2az + 1  1     4π × 2z + 2a z=−a+√a2 −1 , if a > 1; =   1   4π × , if a < −1 √ 2z + 2a z=−a− a2 −1  2π  √ , if a > 1;    a2 − 1 =   2π   −√ , if a < −1. a2 − 1

We have completed the proof of the problem.



Problem 11.2 Bak and Newman Chapter 11 Exercise 2.

Proof. By the power series expansion of e2iz around 0, we have e2iz − 1 − 2iz 4i = −2 − z + · · · . 2 z 3 2iz

has a removable singularity at 0, it is entire. Let CR be the closed contour used Since e −1−2iz z2 on [4, p. 143], we follow from Theorem 6.3 (The Closed Curve Theorem) that Z

CR

Recall that

e2iz − 1 − 2iz dz = 0. z2 Z

CR

=

Z

+ ΓR

Z

R

. −R

Chapter 11. Applications of the Residue Theorem to the Evaluation ... Therefore, we get

Z

R −R

e2ix − 1 − 2ix dx + x2

Z

ΓR

e2iz − 1 − 2iz dz = 0. z2

144

(11.13)

On the one hand, if z ∈ ΓR , then we have e2iz − 1 e−2Im z + 1 2 ≤ 2 ≤ 2 2 z R R

and Theorem 4.10 (The M -L Formula) implies that Z

2 2π e2iz − 1 dz ≤ 2 × πR = . z2 R R

ΓR

Consequently, we obtain

Z

lim

R→∞ ΓR

e2iz − 1 dz = 0. z2

(11.14)

By Definition 4.3, we know that Z

ΓR

dz = z

Z

π 0

iReiθ dθ = Reiθ

Z

π

i dθ = iπ

(11.15)

0

for every R > 0. Now it follows from the results (11.14) and (11.15) that lim

Z

R→∞ ΓR

e2iz − 1 − 2iz dz = −2i · iπ = 2π. z2

(11.16)

On the other hand, since Z

0

−R

e2ix − 1 − 2ix dx = x2

Z

0

R

e−2ix − 1 + 2ix dx, x2

the first integral in the expression (11.13) can be written as Z

R

−R

e2ix − 1 − 2ix dx = x2 =

Z

0

e2ix − 1 − 2ix dx + x2

−R R −2ix e

Z

0

=

Z

R

0

=

Z

R

0

Z

− 1 + 2ix dx + x2 e2ix + e−2ix − 2 dx x2 −4 sin2 x dx. x2

R

0

Z

e2ix − 1 − 2ix dx x2

R 0

e2ix − 1 − 2ix dx x2

(11.17)

Hence, by putting the results (11.16) and (11.17) into the expression (11.13), we conclude that −4 so that

Z

∞

sin2 x dx + 2π = 0 x2

∞

π sin2 x dx = . 2 x 2

0

Z

0

This completes the proof of the problem.



145 Problem 11.3 Bak and Newman Chapter 11 Exercise 3.

′ be Proof. Let CR be the given contour of large enough radius R. Furthermore, let ΓR , ηR and ηR 2π the circular arc, the ray on the real axis and the ray with angle n with the direction indicated in the exercise respectively. Note that if R is large enough, then the point exp( πi n ) lies inside CR . n On the one hand, since exp( πi n ) is a simple zero of the equation z + 1 = 0, Theorem 10.5 (The Cauchy’s Residue Theorem)a implies that Z  1  πi  dz = 2πiRes ; exp n 1 + zn n CR 1 + z 1 = 2πi × n−1 nz z=exp( πi ) n z = −2πi × n z=exp( πi ) n   −2πi πi = . (11.18) exp n n

n n n On the other hand, if z ∈ ΓR , then z = Reiθ with 0 ≤ θ ≤ 2π n . Since |z +1| ≥ |z| −1 = R −1 on ΓR for large R, Theorem 4.10 (The M -L Formula) gives Z 1 dz 2πR × n . (11.19) ≤ n 1 + z n R −1 ΓR

Since n ≥ 2, the inequality (11.19) shows that Z lim R→∞ ΓR

Next, it is obvious that Z

0

∞

dx = lim 1 + xn R→∞

Z

ηR

dz = 0. 1 + zn dx = lim 1 + xn R→∞

(11.20) Z

ηR

dz . 1 + zn

′ , we have z = (R − x) exp( 2πi ) with 0 ≤ x ≤ R so that Finally, on ηR n Z Z R Z 0 2πi  2πi  Z R dx exp( n ) dx exp( 2πi dz n ) dy = − = = − exp . n n ′ 1 + zn 1 + yn n ηR 0 1+x 0 1 + (R − x) R

(11.21)

(11.22)

′ , we substitute the results (11.20), (11.21) and (11.22) into the Recall that CR = ΓR + ηR + ηR expression (11.18) to get Z Z Z  πi  dz dz dz 2πi exp + lim + lim =− lim R→∞ η′ 1 + z n R→∞ ΓR 1 + z n R→∞ ηR 1 + z n n n R Z ∞ Z ∞    dx dx 2πi 2πi πi  − exp = − exp 1 + xn n 1 + xn n n 0 0 Z ∞ exp( πi 2πi dx n) · = − n 1+x n 1 − exp( 2πi 0 n ) π = . n sin πn

This ends the proof of the problem. a

Of course, we apply the version [3, Theorem 5.1.2, p. 294] of Asmar and Grafakos here.



Chapter 11. Applications of the Residue Theorem to the Evaluation ...

146

Problem 11.4 Bak and Newman Chapter 11 Exercise 4.∗

Proof. (a) By the substitution y = ax, we have Z Z ∞ Z ∞ a3 ∞ cos y dy cos y dy cos ax 3 dx = a = (1 + x2 )2 (a2 + y 2 )2 2 −∞ (a2 + y 2 )2 0 0 which is a Type II integral. Since Z ∞ h  i cos y dy eiz = Re 2πiRes ; ai 2 2 2 (a2 + z 2 )2 −∞ (a + y ) n o i d h eiz = Re 2πi dz (z + ai)2 z=ai (a + 1)e−a π , = 2a3 we have Z ∞ cos ax (a + 1)e−a π . dx = (1 + x2 )2 4 0

(b) Using the same strategy of Problem 11.3, we see that Z  3πi   z2  πi  z 2 dz πi = − . exp = 2πiRes ; exp 10 1 + z 10 10 5 10 CR 1 + z

(11.23)

On ΓR , similar to the inequality (11.19), we have Z z 2 dz πR R2 ≤ × 10 5 R10 − 1 ΓR 1 + z so that

lim

Z

z 2 dz = 0. 1 + z 10

Z

x2 dx = lim 1 + x10 R→∞

R→∞ ΓR

On ηR , we have

Z

0

∞

x2 dx = lim 1 + x10 R→∞

ηR

(11.24) Z

ηR

z 2 dz . 1 + z 10

′ , we have On ηR Z R Z  3πi  Z R x2 dx (R − x)2 exp( 3πi z 2 dz 5 ) = − dx = − exp . 10 ′ 1 + z 10 1 + (R − x)2 5 0 1+x 0 ηR

(11.25)

(11.26)

After putting the results (11.24), (11.25) and (11.26) into the expression (11.23), we establish that h  3πi i Z ∞ x2 dx  3πi  πi 1 − exp = − exp 5 1 + x10 5 10 0 Z ∞ 2 πi 1 x dx =− × 3πi 3πi 10 1 + x 5 exp(− ) 0 10 − exp( 10 ) 1 πi =− × 5 −2i sin 3π 10 π = . 10 sin 3π 10

147 (c) We note that Z 2π 0

Z

dz e exp(e ) dθ = = −i iz |z|=1 iθ

z

Z

|z|=1

 ez  ez dz = 2πRes ; 0 = 2π. z z

Hence we have completed the analysis of the problem.



Problem 11.5 Bak and Newman Chapter 11 Exercise 5.∗

Proof. It is clear that Z

2

2m

Since (z + 1)

=

2m X

2π

2m

(cos x)

dx =

Z

|z|=1

0

Z

 1 1 2m dz × z + 4m z iz

(z 2 + 1)2m dz z 2m+1 |z|=1  (z 2 + 1)2m  2π ;0 . = m Res 4 z 2m+1

1 = m i4

Ck2m z 4m−2k , we have

k=0

2m 2m 2m X X Cm (z 2 + 1)2m 2m 2m−2k−1 Ck2m z 2m−2k−1 . C z = + = k z 2m+1 z k=0

(11.27)

k=0 k6=m

Hence we immediately conclude from the expansion (11.27) that Z 2π 2π 2m (cos x)2m dx = m Cm , 4 0 completing the proof of the problem.



Problem 11.6 Bak and Newman Chapter 11 Exercise 6.∗

Proof. This is a Type I integral so that Z ∞   dx 1 = 2πiRes ; i . 2 n+1 (1 + z 2 )n+1 −∞ (1 + x )

(11.28)

Since f (z) = (z + i)−n−1 (z − i)−n−1 has a pole of order n + 1 at i, we know from the formula on [4, p. 130] that   1 dn 1 −n−1 ;i = · (z + i) Res (1 + z 2 )n+1 n! dz n z=i 1 = · (−n − 1)(−n − 2) · · · (−2n)(z + i)−2n−1 n! z=i 1 (−1)n (n + 1)(n + 2) · · · (2n) = × n! (2i)2n+1

Chapter 11. Applications of the Residue Theorem to the Evaluation ... (2n)! (n!)2 · 22n+1 1 · 3 · 5 · · · (2n − 1) = −i · . 2 · 4 · 6 · · · (2n) · 2

148

= −i ·

(11.29)

Hence our desired result follows immediately if we combine the expressions (11.28) and (11.29).  This ends the proof of the problem. Problem 11.7 Bak and Newman Chapter 11 Exercise 7.

′ as described in Problem 11.3 with the central Proof. Consider the contour CR = ΓR + ηR + ηR π angle 4 , see Figure 11.1 below.

Figure 11.1: The contour CR with central angle

π 4.

2

Since f (z) = eiz is entire and CR is a closed curve, Theorem 4.16 (The Closed Curve Theorem) ensures that Z Z Z Z 2 iz 2 iz 2 iz 2 eiz dz = 0. (11.30) e dz = e dz + e dz + ′ ηR

ηR

ΓR

CR

(a) On ΓR , if z = Rcis θ, where 0 ≤ θ ≤ π4 , then 2

2

|eiz | = e−R

sin 2θ

.

(11.31)

π Recall from [22, Exercise 7, p. 197] that 2x π ≤ sin x ≤ x ≤ π for all x ∈ [0, 2 ], so if we put x = 2θ, then we have 4θ ≤ sin 2θ ≤ π (11.32) π

for every θ ∈ [0, π4 ]. By applying the result (11.32) to the expression (11.31) we conclude that   iz 2 2 4θ |e | ≤ exp − R · π on ΓR and thus Lemma 4.9 gives Z

iz 2

e ΓR

Z dz =

π 4

0

i[z(θ)]2

e |

· R(− sin θ + i cos θ) dθ {z } This is G(θ).

149

≤

Z

π 4

0

 4θ  dθ R exp − R2 · π } | {z This is |G(θ)|.

 π  π 2 4θ 4 = −R · exp − R · 4R2 π 0 π 2 = (1 − e−R ) 4R which obviously tends to 0 as R → ∞. (b) On ηR , we know that Z Z iz 2 e dz = ηR

R

ix2

e

dx =

Z

R

2

cos x dx + i

0

0

Z

R

sin x2 dx.

(11.33)

0

π

′ , we have z = (R − x)ei 4 with 0 ≤ x ≤ R so that Next, on ηR Z R Z Z 2 i π4 i π4 iz 2 exp[−(R − x) ] dx = −e e dz = −e ′ −ηR

0

R

2

e−y dy.

(11.34)

0

Hence we put the result of part (a), the expressions (11.33) and (11.34) into the equation (11.30) to obtain Z i hZ 2 2 eiz dz = 0 eiz dz + lim R→∞

Z

∞

ηR

cos x2 dx + i

0

Z

′ ηR ∞

π

sin x2 dx = ei 4

0

Z

∞

2

e−x dx.

(11.35)

0

Recall the fact [11, §3.321, Eqn. (3), p. 336] that √ Z ∞ π −x2 , dx = e 2 0 so the equation (11.35) becomes Z Z ∞ 2 cos x dx = 0

∞

0

√ √ π π 2π sin x dx = · cos = . 2 4 4 2

This completes the analysis of the problem.



Remark 11.1 The integrals considered in Problem 11.7(b) are called Fresnel integrals.

Problem 11.8 Bak and Newman Chapter 11 Exercise 8.

Proof. Let z1 , z2 , . . . , zm be the poles of f (i.e., the zeros of Q). Let R be large enough so that all poles of f lie inside the circle CR = C(0; R). Then we follow from Theorem 10.5 (The Cauchy’s Residue Theorem) that Z m  P (z)  X P (z) dz = 2πi ; zk . (11.36) Res Q(z) CR Q(z) k=1

Chapter 11. Applications of the Residue Theorem to the Evaluation ...

150

Let P (z) = an z n + an−1 z n−1 + · · · + a0 and Q(z) = bm z m + bm−1 z m−1 + · · · + b0 , where m − n ≥ 2 and an bm 6= 0. If M = max{|a0 |, |a1 |, . . . , |an |}, then it is clear that M > 0 and |P (z)| ≤ M |z|n

(11.37)

for large |z|. Furthermore, we recall from Problem 1.26 that |Q(z)| ≥

|bm | m |z| 2

(11.38)

for large enough |z|. Hence if |z| = R is large enough, then we yield from the inequalities (11.37) and (11.38) that P (z) 2M ≤ Q(z) |bm |Rm−n on C(0; R). Thus we deduce from Theorem 4.10 (The M -L Formula) that Z P (z) 2M 4M π dz ≤ × 2πR = →0 m−n |bm |R |bm |Rm−n−1 CR Q(z)

as R → ∞ because m − n − 1 ≥ 1, and hence the equation (11.36) becomes Z m  P (z)  X P (z) 1 Res ; zk = lim dz = 0. Q(z) 2πi R→∞ CR Q(z) k=1

This ends the proof of the problem.



Problem 11.9 Bak and Newman Chapter 11 Exercise 9.

Proof. (a) This is a Type I sum. Let f (z) =

1 1+z 2 .

Then we have

h  π cot πz   π cot πz   π cot πz i f (n) = − Res ; 0 + Res ; i + Res ; −i . 1 + z2 1 + z2 1 + z2 n=−∞ ∞ X

(11.39)

n6=0

Since 0 and ±i are simple poles of

π cot πz , 1+z 2

we establish from [4, Eqn. (1), p. 129] that

 π ; 0 = = 1, 2 2 2 1+z 2z tan πz + π(1 + z ) sec πz z=0  π cot πz   π cot πz  ; i = Res ; −i Res 1 + z2 1 + z2 π cot πi . = 2i

Res

 π cot πz

By the identities of sin z and cos z on [4, p. 41], we have cot πi =

i(e−π + eπ ) , e−π − eπ

so we get from the formula (11.39) that 2

∞ X

n=1

 π(e−π + eπ ) π cot πi  π(e2π + 1) 1 = −1 + = − 1 + = −1 + . n2 + 1 i eπ − e−π e2π − 1

151 As a consequence, we have 1 π(e2π + 1) i 1h − 1 + . = n2 + 1 2 e2π − 1 n=1 ∞ X

(b) Again it is a Type I sum, so if f (z) = ∞ X

n=−∞ n6=0

1 z4

which has a pole of order 4 at 0, then we have

f (n) = −Res

 π cot πz z4

 ;0 .

Since the Laurent series of cot πz around 0 is given byb cot πz =

2 5 5 1 πz π 3 z 3 − − − π z + ··· , π 3 45 945

we obtain at once that

∞ X 1 π4 = . n4 90

n=1

(c) This is a Type II sum. Suppose that f (z) = Eqn. (1), p. 129], we have

1 1+z 2

which has simple poles at ±i. By [4,

∞  π csc πz   π csc πz  X (−1)n = −Res ; i − Res ; −i 1 + n2 1 + z2 1 + z2 n=−∞ n6=0

π csc πz π csc πz − 2z 2z z=i z=−i π =− i sin πi 2π = π e − e−π =−

which implies that

∞ X (−1)n e2π + 2πeπ − 1 = . 1 + n2 2(e2π − 1)

n=1

Hence we have completed the analysis of the problem.



Problem 11.10 Bak and Newman Chapter 11 Exercise 10.∗

Proof. (a) Figure 11.2 shows the square CN = γ1 + γ2 + γ3 + γ4 considered in the problem. Let A = N + 12 . On γ1 , we have z = A + it with t ∈ [−A, A]. Therefore, we obtain from Definition 4.3 that Z A Z dz i dt N = (−1) . −A (A + it) cos itπ γ1 z sin πz b

See [16, Exercise A(5)(a), p. 730].

Chapter 11. Applications of the Residue Theorem to the Evaluation ...

152

Figure 11.2: The square CN with vertices ±(N + 12 ) ± i(N + 12 ). Similarly, on −γ3 , we have z = −A + it with t ∈ [−A, A] so that Z Z Z A dz dz i dt N =− = −(−1) . −γ3 z sin πz γ3 z sin πz −A (A − it) cos itπ

(11.40)

Using the substitution t = −u to the last integral (11.40), we see that Z Z A dz i du N = −(−1) . γ3 z sin πz −A (A + iu) cos iuπ Consequently, we have

and then

Z Z

CN

γ1

dz + z sin πz

dz = z sin πz

Z

γ2

Z

γ3

dz =0 z sin πz

dz + z sin πz

Z

γ4

dz . z sin πz

(11.41)

On γ2 or γ4 , we have z = t ± iA, where t ∈ [−A, A], so 1 2 2 ≤ ≤ √ 2 Aπ − e−Aπ ) 2 −Aπ Aπ z sin πz A(e A + t · |e −e |

for large enough A and so for large N . By Theorem 4.10 (The M -L Formula), we conclude that Z Z dz dz 4 4 and ≤ ≤ Aπ Aπ − e−Aπ z sin πz e z sin πz e − e−Aπ γ2 γ4 which tend to 0 as A → ∞ or equivalently as N → ∞. By the relation (11.41), we see at once that Z dz = 0. lim N →∞ CN z sin πz

(b) Similar to Type II series, if f (z) = Z

CN

h π dz = 2πi (2z − 1) sin πz

N X

1 2z−1

n=−(N −1)

which has a simple pole at 12 , then we have

 1 i (−1)n π + Res ; 2n − 1 (2z − 1) sin πz 2

N  h X (−1)N i 1 π (−1)n − + Res ; = 2πi 2n − 1 (2z − 1) sin πz 2 2N − 1 n=−N

153 N h X  (−1)n (−1)N i 1 π = 2πi 2 − , + Res ; 2n − 1 (2z − 1) sin πz 2 2N − 1

(11.42)

n=1

where CN is the rectangle with vertices (N + 12 ) ± i(N + 12 ) and −(N − 21 ) ± i(N + 12 ) for large positive integer N , see Figure 11.3.

Figure 11.3: The rectangle CN with vertices (N + 12 ) ± i(N + 12 ) and −(N − 21 ) ± i(N + 12 ). Let A = N + 21 , if z = A + it with |t| ≤ A, then we get Z

γ1

(−1)N i π dz = (2z − 1) sin πz 2

Z

A

−A

dt (N + it) cos iπt

and if z = −(N − 21 ) + it with |t| ≤ A, then we have Z

γ3

π dz (−1)N i =− (2z − 1) sin πz 2

Z

A

−A

dt . (N + it) cos iπt

Hence, by similar strategy applied in part (a), we can show that Z π dz lim = 0, N →∞ CN (2z − 1) sin πz so the equation (11.42) reduces to 2

1 π ; 2n − 1 (2z − 1) sin πz 2 π = 2 sin πz + π(2z − 1) cos πz z= 21 π = 2

∞ X (−1)n+1

n=1

= Res



which definitely implies that 1−

∞ X (−1)n+1 1 1 1 π + − + ··· = = . 3 5 7 2n − 1 4 n=1

We end the proof of the problem.



Chapter 11. Applications of the Residue Theorem to the Evaluation ...

154

Problem 11.11 Bak and Newman Chapter 11 Exercise 11.∗

kz

e Proof. Suppose that f (z) = 1+e z and consider the rectangle ΓR with vertices at ±R and ±R+2πi as shown in Figure 11.4. It is clear that 1 + ez vanishes inside the rectangle if and only if z = πi.

Figure 11.4: The rectangle ΓR with vertices at ±R and ±R + 2πi. Thus we represent (z − πi)f (z) = ekz ·

z − πi z − πi = ekz · z = ekz · z 1+e e − eπi

1 ez −eπi z−πi

which implies lim (z − πi)f (z) = lim ekz ·

z→πi

z→πi

1 ez −eπi z−πi

= −ekπi .

By Theorem 10.5 (The Cauchy’s Residue Theorem), we see that Z

ΓR

 ekz  ekz dz = 2πiRes ; πi = −2πiekπi . 1 + ez 1 + ez

(11.43)

Now it is obvious that, on γ4 , we have Z

γ4

ekz dz = 1 + ez

Z

R −R

ekx dx. 1 + ex

(11.44)

Similarly, we know that points on −γ2 have the form z = x + 2πi with −R ≤ x ≤ R, so it follows from Proposition 4.7 that Z

γ2

ekz dz = − 1 + ez

Z

−γ2

ekz dz = − 1 + ez

Z

R

−R

ek(x+2πi) dx = −e2kπi 1 + ex+2πi

Z

R

−R

ekx dx. 1 + ex

(11.45)

Next, suppose that z = R + it ∈ γ1 with t ∈ [0, 2π]. For large enough R, the triangle inequality implies that eR |1 + eR+it | ≥ |eR+it | − 1 ≥ 2 so that ekz ek(R+it) = ≤ 2e(k−1)R . z R+it 1+e 1+e

155 Hence Theorem 4.10 (The M -L Formula) gives Z ekz dz ≤ 4πe(k−1)R z γ1 1 + e

which ensures that the integral along γ1 tends to 0 as R → ∞ because k < 1. By similar analysis, the integral along γ3 tends to 0 as R → ∞ because k > 0. Combining these facts and the expressions (11.44) and (11.45), this assures us that Z ∞ ekx 2kπi (1 − e ) dx = −2πiekπi x 1 + e −∞ Z ∞ ekπi ekx dx = −2πi · x 1 − e2kπi −∞ 1 + e 1 = 2πi · kπi e − e−kπi π . = sin kπ Hence we complete the proof of the problem.



Problem 11.12 Bak and Newman Chapter 11 Exercise 12.∗

Proof. We consider the keyhole contour Kǫ,M used on [4, p. 147], but this time we let M → 1. By Theorem 10.5 (The Cauchy’s Residue Theorem), we have Z X f (z) log z dz = 2πi Res (f (z) log z; zk ), Kǫ,M

k

where the sum is taken over all the poles of f inside Kǫ,M . Since f is analytic for |z| ≤ 1, f has no poles so that Z f (z) log z dz = 0 (11.46) Kǫ,M

for any 0 < ǫ < M < 1. By Theorem 4.10 (The M -L Formula), we have Z f (z) log z dz ≤ πǫ max |f (z) log z|. Cǫ

Cǫ

(11.47)

Since f is continuous at 0 and | log z| < | log |z||+2π = | log ǫ|+2π, we obtain from the inequality (11.47) that Z (11.48) f (z) log z dz ≤ Aπǫ(| log ǫ| + 2π) → 0 Cǫ

as ǫ → 0, where A is a positive constant. Next, we√recall that CM is the circular arc of radius √ M traced counterclockwise from M 2 − ǫ2 + iǫ to M 2 − ǫ2 − iǫ, so it is true that Z Z f (z) log z dz. (11.49) f (z) log z dz = lim ǫ→0 M →1 CM

Finally, since lim

Z

ǫ→0 M →1 I1

|z|=1

f (z) log z dz =

Z

0

1

f (x) log x dx

(11.50)

Chapter 11. Applications of the Residue Theorem to the Evaluation ... and similarly, lim

Z

ǫ→0 M →1 I2

f (z) log z dz = −

Z

156

1

f (x)(log x + 2πi) dx.

(11.51)

0

Since Kǫ,M = CM + I2 + Cǫ + I1 , we put all the results (11.48), (11.49), (11.50) and (11.51) into the equation (11.46) to conclude that Z f (z) log z dz = 0 lim lim

Z

ǫ→0 M →1 CM

Z

f (z) log z dz + lim

Z

ǫ→0 M →1 I1 Z 1

f (z) log z dz +

0

|z|=1

ǫ→0 M →1 Kǫ,M

f (z) log z dz + lim

f (x) log x dx −

Z

1

Z

ǫ→0 M →1 I1

f (z) log z dz = 0

f (x)(log x + 2πi) dx = 0

0

1 2πi

Z

f (z) log z dz =

|z|=1

Z

1

f (x) dx. 0

We end the proof of the problem.



Problem 11.13 Bak and Newman Chapter 11 Exercise 13.

Proof. We note that Cn3n is the coefficient of z n in (1 + z)3n . Therefore, we have Z 1 (1 + z)3n 3n Cn = dz 2πi C z n+1 where C is a simple closed contour surrounding the origin. Similar to the analysis of [4, Example 1, p. 155], we choose C to be the circle C(0; 21 ) because (1 + z)3 (1 + |z|)3 27 = 0 as well as 1+

√

1 − 2x 2 − 2x < z+ < 2x 2x

and

√ 1 − 2x 1 − 2 − 2x < z− < 2x 2x

if x < 0. Direct computation shows that z+ > 1 if x > 0 and z+ < −1 if x < 0. Hence it follows from Theorem 10.5 (The Cauchy’s Residue Theorem) that ∞ X

n=0



 1 ; z − xz 2 + (2x − 1)z + x 1 =− 2xz + (2x − 1) z=z− 1 =√ 1 − 4x

Cn2n xn = −Res

which completes the proof of the problem.



Problem 11.15 Bak and Newman Chapter 11 Exercise 15.

Proof. To find the maximum of a2 b, where a2 + b2 = 4 and 0 ≤ a, b ≤ 2. We define the function f (b) = (4 − b2 )b = 4b − b3 . Then we have max a2 b = max f (b).

0≤a,b≤2

0≤b≤2

Note that f ′ (b) = 4 − 3b2 = 0 if and only if b =

attains its maximum at b =

√ 2 3 3

√ 2 3 3

√

and f ′′ ( 2 3 3 ) = −6 ·

and its maximum value is given by

max f (b) = f

0≤b≤2

 2√3  3

√ 16 3 = . 9

√ 2 3 3

√ = −4 3 < 0, so f

(11.54)

If |z| = 1, then we see from the figure provided in the problem that |z − 1|2 + |z + 1|2 = 4. Thus we may let a = |z − 1| and b = |z + 1| so that our a and b satisfy the hypotheses in the previous paragraph. Consequently, we conclude from the value (11.54) that √ 16 3 , |(z − 1) (z + 1)| ≤ 9 2

completing the analysis of the problem.



Problem 11.16 Bak and Newman Chapter 11 Exercise 16.

Proof. Since part (b) considers √ points on the circle of radius R > 0, we do this general case in part (a) and then take R = 2 to obtain the desired result. Let z ∈ C(0; R), a = |z + 1| and b = |z − 1|. Now the points z, −z, 1 and −1 form a parallelogram in the plane C.

159 (a) It is well-known that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.c In other words, we have (2R)2 + 22 = 2(a2 + b2 ) which reduces to a2 + b2 = 2(1 + R2 ), (11.55) p √ √ where 0 ≤ a, b ≤ 2(1 + R2 ). If R = 2, then we have a2 = 6 − b2 and 0 ≤ a, b ≤ 6. To 2 find the maximum of a2 b , we define 6b − b3 , 2

f (b) = where b ∈ [0,

√

6]. Then their maximums are the same, i.e., a2 b = max √ f (b). 6] 2 b∈[0, 6]

max√

a,b∈[0,

√ √ 2 Now f ′ (b) = 6−3b = 0 if and only if b = 2 and f ′′ (b) = −3b so that f ′′ ( 2) < 0. 2 √ Consequently, f attains its maximum at b = 2 and the maximum value is √ √ √ 2(6 − 2) max√ f (b) = f ( 2) = =2 2 2 0≤b≤ 6 which gives the desired result. (b) In the general case, we note that R > 0 and f (b) = where 0 ≤ b ≤

p

[2(1 + R2 ) − b2 ]b , R2

2(1 + R2 ). As usual, we have f ′ (b) =

2(1 + R2 ) − 3b2 =0 R2

if and only if b=

r

2(1 + R2 ) . 3

Since f ′′ (b) = − R6b2 , we must have

r 6 2(1 + R2 )  2(1 + R2 ) =− 2 · < 0. f 3 R 3 q 2) Hence f attains its maximum at b = 2(1+R and the maximum value is given by 3 ′′

0≤b≤

max √



r

2(1+R2 )

f (b) = f

r  2(1 + R2 )  3

√ 3 4 6 (1 + R2 ) 2 = · . 9 R2

To find the minimum of the expression (11.56), we define F : (0, ∞) → (0, ∞) by 3

F (R) = c

Read [22, Exercise 17, p. 23].

(1 + R2 ) 2 . R2

(11.56)

Chapter 11. Applications of the Residue Theorem to the Evaluation ...

160

Direct computation gives

√ (R2 − 2) 1 + R2 F (R) = R3 √ and then F ′ (R) = 0 if and only if R = 2. √ It can be easily seen from the First Derivative Test that F attains its minimum at R = 2 and actually, √ 3 3 3 (1 + 2) 2 = . (11.57) min F (R) = 0 3. Since az = ez log a , if z = x + iy for x ≤ 1, then we have

|az | = |ez log a | = |ex log a | = ex log a ≤ max(1, elog a ) = max(1, a)

in the left half-plane as R → ∞. As a result, Theorem 4.10 (The M -L Formula) implies that Z az lim dz = 0 R→∞ CR z 2

and hence the equation (12.3) reduces to Z z  az   ez log a  a dz = 2πiRes ; 0 = 2πiRes ; 0 = 2πi log a 2 z2 z2 I z

because

ez log a = 1 +

z log a (z log a)2 + + ··· . 1! 2!

We end the proof of the problem.

 163

Chapter 12. Further Contour Integral Techniques

164

Remark 12.1 There is a general formula for the integral Z x0 +i∞

az F (z) dz

x0 −i∞

in [16, Eqn. (4.13.4), p. 952] for the case a ≥ 1 and x0 ∈ R. Problem 12.2 Bak and Newman Chapter 12 Exercise 2.∗ √ Proof. The function 4z 2 − 8z + 3 has zeros at z = 12 and z = 32 , so 4z 2 − 8z + 3 is analytic in √ the plane minus [ 12 , 32 ].a Since 4z 2 − 8z + 3 ∼ 2z for large z, it follows that Z Z dz 1 dz √ lim = πi. (12.4) = lim 2 R→∞ R→∞ |z|=R 2 4z − 8z + 3 |z|=R z Now Figure 12.1 shows the formations of the contours C(0; R) and C(0; 2).

Figure 12.1: The formations of the contours C(0; R) and C(0; 2). Note that C(0; R) − C(0; 2) = Γ1 + Γ2 + (γ1 + γ2 ) = (Γ1 + I1 + γ1 + I2 ) + (Γ2 − I2 + γ2 − I1 ). 1

Since (4z 2 − 8z + 3)− 2 is analytic in a simply connected domain containing the smooth closed curve Γ1 + I1 + γ1 + I2 , Theorem 8.6 (The General Closed Curve Theorem) ensures that Z dz √ = 0. (12.5) 2 4z − 8z + 3 Γ1 +I1 +γ1 +I2 a

Refer to Problem 10.16.

165 Similarly, we also have

Z

Γ2 −I2 +γ2 −I1

√

dz = 0. 4z 2 − 8z + 3

(12.6)

Therefore, we follow from the results (12.5) and (12.6) that Z Z dz dz √ √ = 4z 2 − 8z + 3 4z 2 − 8z + 3 C(0;R) C(0;2) and hence the limit (12.4) implies Z

C(0;2)

√

4z 2

dz = πi. − 8z + 3

This completes the proof of the problem.



Problem 12.3 Bak and Newman Chapter 12 Exercise 3.

Proof. Let R > 0 be large and ǫ > 0 be small. We consider the closed contour ΓR formed by the following curves and lines: • γ1 : [−R, R] → C is the parabola defined by γ1 (t) = 1 − t2 + it; • the straight line γ2 connecting γ1 (R) = 1 − R2 + iR and the point ǫ exp(iθR ), where θR = Arg γ1 (R); • γ3 : [θR , θ−R ] → C defined by

γ3 (t) = ǫeit ,

where θ−R = Arg γ1 (−R); • the straight line γ4 connecting ǫ exp(iθ−R ) to γ1 (−R) = 1 − R2 − iR. This contour is shown in Figure 12.2 below:

Figure 12.2: The closed contour formed by γ1 , γ2 , γ3 and γ4 .

Chapter 12. Further Contour Integral Techniques

166

It is clear that f (z) = ez log z is analytic inside and on the closed contour ΓR and f has no singularities inside ΓR , Theorem 10.5 (The Cauchy’s Residue Theorem), we find Z ez log z dz = 0 ΓR

and thus

Z

z

γ1

e log z dz = −

Z

z

γ2

e log z dz −

Z

z

γ3

e log z dz −

Z

ez log z dz.

(12.7)

γ4

√ Notice that both −γ2 and γ4 have the same form γ : [ǫ, R4 − R2 + 1] → C given by γ(t) = teiθ , R R where θ satisfies either tan θ = 1−R 2 or tan θ = − 1−R2 . When R → ∞ and ǫ → 0, in the case of −γ2 , we have θ → −π and for γ4 , we find θ → π. By Proposition 4.7 and then Definition 4.3, we have Z Z f (z) dz f (z) dz = − γ2

=− =−

Z

Z

−γ2 √ R4 −R2 +1 ǫ

ǫ

= −eiθ iθ

= −e so that

Z

lim

R→∞ γ2 ǫ→0

√

R4 −R2 +1

Z

ǫ

Z

f (−γ2 (t))(−γ2 )′ (t) dt eiθ f (teiθ ) dt

√ R4 −R2 +1 √ R4 −R2 +1

exp(teiθ ) log(teiθ ) dt exp(teiθ )(log t + iθ) dt

ǫ

f (z) dz =

Z

∞

0

e−t (log t − iπ) dt.

(12.8)

Similarly, we also have Z

f (z) dz = γ4

Z

√

R4 −R2 +1

ǫ

which gives lim

Z

f (γ4 (t))γ4′ (t) dt

R→∞ γ4 ǫ→0

f (z) dz = −

iθ

=e

Z

√

R4 −R2 +1

exp(teiθ )(log t + iθ) dt

ǫ

Z

∞

e−t (log t + iπ) dt.

(12.9)

0

By [11, §4.331, Eqn. (1), p. 571], we know that Z ∞ e−t log t dt = −C, 0

where C is the Euler’s constant, so the integrals (12.8) and (12.9) can be split into two convergent integrals and then Z Z ∞ Z lim e−t dt = 2πi. f (z) dz + lim f (z) dz = 2πi R→∞ γ2 ǫ→0

R→∞ γ4 ǫ→0

0

167 Substitute this result into the equation (12.7), we see that Z Z z lim e log z dz = 2πi − lim ez log z dz. R→∞ ǫ→0

R→∞ ǫ→0

γ1

(12.10)

γ3

Finally, if z = γ3 (t) = ǫeit , where t ∈ [θR , θ−R ] ⊆ [−π, π], then we know that |γ3 (t)| ≤ 2πǫ and |f (z)| = | exp(ǫeit ) log(ǫeit )| = eǫ cos t | log ǫ + it| ≤ e1 (| log ǫ| + 2π). Since f is obviously continuous on γ3 , Theorem 4.10 (The M -L Formula) implies that Z f (z) dz ≤ 2πe1 (| log ǫ| + 2π)ǫ → 0 γ3

as ǫ → 0. Hence we conclude from the expression (12.10) that Z ez log z dz = 2πi γ

which ends the proof of the problem.



Problem 12.4 Bak and Newman Chapter 12 Exercise 4.

Proof. Using exactly the same argument as the proof of [4, Example 3, pp. 163, 164], we have Z n X 1 1 π n n 13 [f (z)] 3 · dz, (−1) (Ck ) = 2πi C sin πz

(12.11)

k=0

where C is any contour in −1 < Re z < n + 1 and f (z) =

sin πz . πz(1 − z)(1 − 2z ) · · · (1 − nz )

Furthermore, we can split the integral (12.11) as Z 1 Z n+ 1 +i∞ n i X 2 1 1 h − 2 −i∞ π n n 13 (−1) (Ck ) = dz , + [f (z)] 3 · 2πi − 1 +i∞ sin πz n+ 1 −i∞ k=0

2

2

If z = n − ω, then we have n! sin π(n − ω) π(n − ω)[(ω + 1 − n)(ω + 2 − n) · · · (ω − 1)(ω − 0)] (−1)n+1 n! sin πω = n π(n − ω)(−1) [−ω(1 − ω) · · · (n − 2 − ω)(n − 1 − ω)] n! sin πω = πω[(1 − ω)(2 − ω) · · · (n − 2 − ω)(n − 1 − ω)(n − ω)] sin πω = (1−ω) 2−ω (n−2−ω) (n−1−ω) (n−ω) πω[ 1 · 2 · · · n−2 · n−1 · n ] sin πω = πω(1 − ω)(1 − ω2 ) · · · (1 − ωn )

f (n − ω) =

(12.12)

Chapter 12. Further Contour Integral Techniques

168

= f (ω). Therefore, we see that Z

n+ 12 +i∞ n+ 21 −i∞

1

[f (z)] 3 ·

Z − 21 −i∞ 1 π π dz = dω [f (ω)] 3 · 1 sin πz sin πω − +i∞ 2

and when Re z = − 21 , we deduce from the expression (12.12) that

∞ 1 Z − 12 −i∞ X 1 1 π (−1)n (Ckn ) 3 ≤ dz [f (z)] 3 · π − 1 +i∞ sin πz 2 k=0  2 1 Z 2 dz 1 3 ≤ × π3 × √ 2 π n+1 Re z=− 21 (sin πz) 3 √ Z 3 dz 2 √ = √ 2 . 3 π · 6 n + 1 Re z=− 21 (sin πz) 3

(12.13)

Note that if z = − 12 + iy, then the definition of sin z gives

1 1 πi πi | sin πz| = (e− 2 −πy − e 2 +πy ) = (e−πy + eπy ). 2i 2

Consequently, if y ∈ [0, ∞), then | sin πz| ≥ 12 eπy . If y ∈ (−∞, 0), then we have | sin πz| ≥ 21 e−πy . Now these imply that Z Z ∞ h Z 0 2π i dz √ 3 − 2π y y 3 3 = 4 · e e dy + dy 2 −∞ Re z=− 12 (sin πz) 3 0 i h 3 2π 0 √ 2π ∞ 3 3 e 3 y e− 3 y − = 4· 2π 2π −∞ 0 √ 3 3 = 4· . π Hence it follows from the inequality (12.13) that √ ∞ X 3 √ 2 3 6 3 n n 31 √ √ 4· = √ (−1) (Ck ) ≤ √ · 6 3 3 π π· n+1 π π· 6 n+1 k=0 which guarantees

∞ X 1 (−1)n (Ckn ) 3 → 0 k=0

as n → ∞, completing the proof of the problem.



Problem 12.5 Bak and Newman Chapter 12 Exercise 5.

Proof. (a) Using the same argument as in the proof of [4, Example 3, pp. 163, 164], we know that Z 3 Z n+ 3 +i∞ p ∞ i X p 4 1 h − 4 −i∞ π f (z) · + dz . (−1)k Ckn = 2πi − 3 +i∞ sin πz n+ 3 −i∞ k=0

4

4

(12.14)

169 p As we have shown in Problem 12.4, the integrand f (z) · sinππz is invariant under the substitution z 7→ n − z, it suffices to consider the first integral in the expression (12.14). On the line Re z = − 43 , we have z = − 34 + iy and   z z  z z ··· 1 − z(1 − z) 1 − = |z| · |1 − z| · 1 − · · · 1 − 2 n 2 n  3  3  3  3  · 1+ ··· 1 + ≥ · 1+ 4 4 4·2 4n n   Y 3 3 = · . 1+ 4 4k

(12.15)

k=1

Recall the binomial theorem ([22, Exercise 22, p. 201]) that α

(1 + x) =

∞ X

Cnα xn

=

1 k

3

Cnα =

n=0

n=0

If we put x =

∞ X

α(α − 1) · · · (α − n + 1) n x . n!

and α = 34 , then we obtain 

1+

3 5 3 3 1  34 − + − ··· ≤ 1 + =1+ 2 3 k 4k 32k 128k 4k

(12.16)

because |Cn4 | is a decreasing function of n. Combining the inequalities (12.15) and (12.16), we see immediately that n   z  3 Y  1  43 z ··· 1 − 1+ ≥ · z(1 − z) 1 − 2 n 4 k k=1

n 3 h Y  k + 1 i 34 = · 4 k k=1

3 3 = · (n + 1) 4 . 4

(12.17)

Hence it concludes that Z

− 34 −i∞ p

− 34 +i∞

p π dz | sin πz| p · z z 3 sin πz ) · · · (1 − )| |πz(1 − z)(1 − − 4 +i∞ 2 n √ Z − 3 −i∞ dz 4 2 π 1 ≤ · · √ √ 3 π − 3 +i∞ sin πz 8 3(n + 1) 4 Z − 3 −i∞ 4 2 dz √ =√ 3 sin πz 3π(n + 1) 8 − 34 +i∞

1 π dz ≤ · f (z) · sin πz π

Z

− 34 −i∞

3

≤ An− 8 for some constant A > 0.

(b) Let δ > 0 be small. Suppose that −t = −1 + δ so that t ∈ (0, 1). Now our integration becomes along Re z = −t and Re z = n + t. On Re z = −t, since t ∈ (0, 1), instead of the inequality (12.17), we establish n   Y z  z  t − z) 1 − · · · 1 − ≥ t 1 + z(1 2 n k k=1

Chapter 12. Further Contour Integral Techniques

≥t

170 n  hY 1 it 1+ k k=1

= (1 − δ)(n + 1)1−δ

≥ (1 − δ)n1−δ .

Recall the facts from the proof of Problem 12.4 that | sin πz| ≥ | sin πz| ≥ 12 e−πy if y ∈ (−∞, 0), where y = Im z.

1 πy 2e

(12.18) if y ∈ [0, ∞) and

Now instead of (12.14), we have

Z n+1−δ+i∞ p Z ∞ i X p n π 1 h −1+δ−i∞ k f (z) · + dz . (−1) Ck = 2πi −1+δ+i∞ sin πz n+1−δ−i∞ k=0

Applying similar argument we have employed in part (a), we follow from the inequality (12.18) that Z ∞ X p n 1 −1+δ−i∞ p π dz k (−1) Ck ≤ f (z) · π −1+δ+i∞ sin πz k=0 Z −1+δ−i∞ 1 dz =√ p √ π −1+δ+i∞ sin πz × z(1 − z)(1 − 2z ) · · · (1 − nz ) Z dz 1 ≤p √ sin πz π(1 − δ)n1−δ Re z=−1+δ A ≤p (1 − δ)n1−δ

for some positive constant A. We have completed the proof of the problem.



Problem 12.6 Bak and Newman Chapter 12 Exercise 6.∗

1 > 0 on [0, ∞). Since ext ≥ 1 + xt, we observe from Proof. Suppose that x > 0 and h(t) = tt−1 the definition that Z ∞ xt Z ∞ Z ∞ hZ 1 i e dt g(x + 2πi) = f (x) = dt ≥ x =x h(t) dt . (12.19) h(t) dt + tt tt−1 0 0 1 0

Since t1−t is bounded on [0, 1], the integral Z Z 1 h(t) dt = 0

1

t1−t dt

0

is convergent and then the convergence of the integral Z ∞ dt t−1 t 0 depends on the convergence of the second integral inside the brackets (12.19). To this end, we study the convergence of the series ∞ X

n=1

h(n) =

∞ X

n=1

1 . nn−1

(12.20)

171 Let an =

1 . nn−1

Then it is easy to see that α = lim sup n→∞

√ n

an = lim

n→∞

1 1

n1− n

= 0,

so [27, Theorem 6.7, p. 76] ensures that the series (12.20) converges. Since h(t) ≥ 0 and h decreases monotonically on [1, ∞), it follows from [28, Theorem 1.10, p. 37] that the second integral in the brackets (12.19) is convergent. Therefore, there is a positive constant M such that g(x + 2πi) ≥ M x which shows certainly that g(x + 2πi) → ∞ as x → ∞. This completes the analysis of the  problem.

Chapter 12. Further Contour Integral Techniques

172

CHAPTER

13

Introduction to Conformal Mapping

Problem 13.1 Bak and Newman Chapter 13 Exercise 1.

Proof. Suppose that z0 6= 0 and z1 , z2 ∈ C are distinct numbers such that f (z1 ) = f (z2 ). If z2 = 0, then it forces that z1 = 0. Therefore, without loss of generality, we may assume that z2 6= 0. Then we can write h z k i 1 f (z1 ) − f (z2 ) = z2k −1 z2 which means that f (z1 ) = f (z2 ) if and only if z1 = z1 (n) = z2 exp( 2nπi k ) if and only if Arg z1 − Arg z2 =

2nπ k

(13.1)

for some n ∈ {1, 2, . . . , k − 1}. Hence the equation (13.1) insures that if δ > 0 is chosen such that D(z0 ; δ) lies in the sector S = {z ∈ C | α < Arg z < β}, where 0 < β − α < 2π k (see Figure 13.1 below), then f is locally 1-1 at z0 by Definition 13.3.

Figure 13.1: The sector S and the disc D(z0 ; δ). More precisely, if θ = Arg z0 , then we may take α = θ − πk and β = θ + πk so that the ray connecting the origin and the point z0 bisects the sector S. This completes the proof of the problem.  173

Chapter 13. Introduction to Conformal Mapping

174

Problem 13.2 Bak and Newman Chapter 13 Exercise 2.

Proof. If x = c, then we have |ω| = | exp(c + iy)| = ec

which is a circle centred at 0 with radius ec . Next, if y = c, then ω = ex · eic which is a ray with angle c and ex > 0. (Note that when x runs through R, ex runs through (0, ∞).) We have  completed the proof of the problem. Problem 13.3 Bak and Newman Chapter 13 Exercise 3.

Proof. We remark that throughout this chapter, we denote H to be the upper half-plane. (i) Note that f1 (z) = z+2 maps the strip S onto the strip S1 = {z | 0 < Re z < 3} conformally. Next, the mapping f2 (z) = π3 f1 (z) maps S1 onto the strip S2 = {z | 0 < Re z < π} conformally. Furthermore, the mapping f3 (z) = if2 (z) sends S2 onto the vertical strip S4 = {z | 0 < Im z < π} and the mapping f4 (z) = ef3 (z) maps S4 onto H conformally. (z)−i conformally maps H onto T . Hence a By Theorem 13.16, the mapping f5 (z) = ff44 (z)+i required conformal mapping f is given by f (z) =

exp( πi 3 (z + 2)) − i

exp( πi 3 (z + 2)) + i

.

(ii) Theorem 13.23 implies that the (unique) bilinear transformation f mapping −2, 0, 2 into −1, 0, 2 is given by (f (z) − 0)(2 + 1) (z − 0)(2 + 2) = (f (z) + 1)(2 − 0) (z + 2)(2 − 0) which reduces to

f (z) =

4z . 6−z

(iii) The mapping f1 (z) = z 4 maps S onto H conformally. Next, the mapping f2 (z) = log f1 (z) maps H onto the strip {z | 0 < Im z < π} conformally. Finally, f3 (z) = π1 f2 (z) conformally maps the strip {z | 0 < Im z < π} onto the strip {z | 0 < Im z < 1}. Hence a required conformal mapping f is given by f (z) =

4 log z. π

1

(iv) The mapping f1 (z) = iz 2 sends S onto the upper semi-disc S1 = {z | |z| = 1 and Im z > 0} (z−1)2 conformally. By [4, Example 1, p. 180], the mapping f2 (z) = − 4(z+1) 2 sends S1 conforz−i mally onto H. According to Theorem 13.6, f3 (z) = z+i is a conformal mapping of H onto the T . Therefore, a required conformal mapping f is given by √ √ i h i−1 h (i z − 1)2 (i z − 1)2 √ √ −i × − +i . f (z) = − 4(i z + 1)2 4(i z + 1)2

We complete the analysis of the problem.



175 Problem 13.4 Bak and Newman Chapter 13 Exercise 4.∗

Proof. Let the region between the two circles be G. The two circles touch internally at 2, 1 so [4, Example 1, p. 181] indicates that the mapping f1 (z) = z−2 sends G conformally onto 1 1 two parallel lines. Since f1 (0) = − 2 and f1 (−2) = − 4 , the two parallel lines are Re z = − 14 1 z and Re z = − 21 . Then the mapping f2 (z) = z−2 + 12 = 2(z−2) maps G onto the infinite strip

S = {z ∈ C | 0 < Re z < 14 } conformally. Next, the mapping f3 (z) = 4πif2 (z) = conformally onto the strip T = {z ∈ C | 0 < Im z < π} and so the mapping f4 (z) = exp f3 (z) = exp

2πiz z−2

sends G

 2πiz  z−2

sends G onto H conformally. Finally, Theorem 13.16 shows that the mapping f given by f (z) =

2πi )−i exp( z−2 2πi exp( z−2 )+i

maps G onto the unit disc D(0; 1) conformally, completing the proof of the problem.



Problem 13.5 Bak and Newman Chapter 13 Exercise 5.∗

Proof. Let S = {z = x + iy ∈ C | x > 0 and 0 < y < 1}. Then f1 (z) = πiz sends S conformally onto S1 = {z = x + iy, | −π < x < 0 and y > 0}. Next, f2 (z) = πiz + π2 will map S onto the semi-infinite strip o n S2 = z ∈ C − π2 < Re z < π2 and Im z > 0

conformally. We know that the Schwarz-Christoffel transformation f (z) = sin z maps S2 onto H, so the composition f3 = f ◦ f2 maps S onto H conformally. Since π  f3 (z) = f (f2 (z)) = sin + πiz = cos(πiz), 2

we observe from Theorem 13.16 that the mapping g(z) =

cos(πiz) − i cos(πiz) + i

sends S conformally onto the unit disc D(0; 1). This ends the proof of the problem. Problem 13.6 Bak and Newman Chapter 13 Exercise 6.∗

Proof. By [4, Example 1, p. 180], the mapping f defined by f (z) = −

(z − 1)2 4(z + 1)2



Chapter 13. Introduction to Conformal Mapping

176

sends S onto H conformally. Now Theorem 13.16 ensures that the mapping g given by g(z) =

i h i−1 f (z) − i h (z − 1)2 (z − 1)2 = − − i × − + i f (z) + i 4(z + 1)2 4(z + 1)2

maps S conformally onto the unit disc D(0; 1) and we complete the proof of the problem.



Problem 13.7 Bak and Newman Chapter 13 Exercise 7.∗

Proof. Suppose that U, V, W are regions and there exist conformal mappings f : U → V , g : V → W such that f (U ) = V and g(V ) = W . In other words, we have U ∼ V and V ∼ W. • Reflexive Property: The identity map id : U → U is certainly bijective. Furthermore, it satisfies id ′ (z) = 1 for all z ∈ U , so Theorem 13.4 implies that id is conformal, i.e., U ∼ U. • Symmetric Property: Since f : U → V is a conformal mapping, Theorem 13.8 shows that F = f −1 : V → U is also a conformal mapping. Since f is bijective, F is also bijective. Thus we conclude that V ∼ U . • Transitive Property: We know that h = g ◦ f : U → W is bijective because f and g are bijective. Since f and g are analytic, Problem 3.3 ensures that h is also analytic. By Definition 13.9, h is a conformal mapping which means that U ∼ W . According to the definition, “conformal equivalence” is an equivalence relation which completes  the proof of the problem. Problem 13.8 Bak and Newman Chapter 13 Exercise 8.

Proof. (a) Suppose that R = {z = x+iy | a ≤ x ≤ b and c ≤ y ≤ d} and ∂R = γ1 +γ2 +γ3 +γ4 , where γ1 and γ3 are the horizontal lines and γ2 and γ4 are the vertical sides.a Let f (z) = az + b, where a 6= 0. It is clear that f : ∂R → f (∂R) is bijective. By Theorem 13.11, f (γ1 ), f (γ2 ), f (γ3 ) and f (γ4 ) are lines. Since f is conformal at the four vertices, f (γ1 ) and f (γ3 ) are parallel lines such that they are perpendicular to both f (γ2 ) and f (γ4 ). Consequently, f (∂R) is also a polygon. (b) Let R and R′ = f (R) be the said rectangles. The hypotheses imply that f is nonconstant. By a rotation and a translation if necessary, we may assume that points z of R have the form 0 ≤ Re z ≤ a and 0 ≤ Im z ≤ b for some positive constants a and b. Denote 0, v1 , v2 and v3 to be its vertices. Similarly, we may also assume that if z ∈ R′ , then 0 ≤ Re z ≤ c and 0 ≤ Im z ≤ d for some positive constants c and d with vertices 0, v1′ , v2′ and v3′ . Figure 13.2 shows the setting for this problem.

a

If not, then we can consider the rectangle R′ = eiθ R for some θ whose sides are either horizontal or vertical.

177

Figure 13.2: The rectangles R and R′ . Suppose that f (z0 ) ∈ {0, v1′ , v2′ , v3′ }. By Problem 7.3, we have z0 ∈ ∂R. We claim that if z0 ∈ / {0, v1 , v2 , v3 }, then f (z0 ) ∈ / {0, v1′ , v2′ , v3′ }. Otherwise, we assume, without loss of generality, that f (z0 ) = v2′ . Now we want to compare ∠γ1 , γ2 and ∠f (γ1 ), f (γ2 ), see Figure 13.3 below for an illustration:

Figure 13.3: The angles of ∠γ1 , γ2 and ∠f (γ1 ), f (γ2 ). By Theorems 13.4 and 13.7, we have ∠γ1 , γ2 = kπ, where k is the least positive integer such that f (k) (z0 ) 6= 0. However, as the interior angle at v2′ is π2 , so we will have ∠γ1 , γ2 6= ∠f (γ1 ), f (γ2 ), a contradiction. Thus this proves our claim. By Theorem 7.1 (The Open Mapping Theorem), we have the observation that interior points of R are mapped by f to interior points of R′ . Assume that there was a ω ∈ R◦ such that f (ω) ∈ ∂R. For each n ∈ N, the continuity of f ensures that there exists a δn > 0 such that zn ∈ D(ω; δn ) ⊆ R◦ implies |ω − f (zn )|
0.

Thus f1 ◦ f2 is also a bilinear transformation. If f1 , f2 and f3 are bilinear transformations, then it is true that (f1 ◦ f2 ) ◦ f3 = f1 ◦ (f2 ◦ f3 ) as compositions of functions. Obviously, the identity id (z) = z is a bilinear transformation satisfying f ◦ id = id ◦ f = f. Finally, we notice from [4, p. 177] that if f (z) =

az+b cz+d ,

then f −1 (z) =

dz−b −cz+a

and therefore,

f ◦ f −1 = f −1 ◦ f = id . Hence bilinear mappings form a group under composition which completes the analysis0 of the  problem. Problem 13.10 Bak and Newman Chapter 13 Exercise 10.

Proof. The mappings are bilinear, so Theorem 13.11 ensures that they map the unit circle onto a circle or a line. (a) Suppose that z = eiθ . Then we have ω = cis (−θ) which means that |ω| = 1. (b) Since ω(−1) = − 12 and ω has a pole at 1, its image is the vertical line Im z = − 21 .

179 (c) Notice that ω = ω2 ◦ ω1 , where ω1 (z) = z − 2 and ω2 = z1 . It is easily seen that ω1 (C(0; 1)) = C(−2; 1). Since 1 6= | − 2|, it follows from the proof of Lemma 13.10 that ω2 (C(−2; 1)) = C(β; R), where −2 1 1 2 β= and R = = . =− 2 2 2 | − 2| − 1 3 | − 2| − 1 3 Hence we have completed the proof of the problem.



Problem 13.11 Bak and Newman Chapter 13 Exercise 11.

Proof. We deduce from Theorem 13.15 that f has the form f (z) = eiθ

 z−α  , 1 − αz

where |α| < 1. Since 0 = f (0) = eiθ α, we have α = 0 and thus f (z) = eiθ z. Since f ′ (0) = eiθ > 0, θ = 0 which implies that f (z) = z as desired. This completes the proof of the problem.  Problem 13.12 Bak and Newman Chapter 13 Exercise 12.

Proof. Note that f1 , f2 : D → D(0; 1) are bijective conformal mappings. Then the mapping f2−1 : D(0; 1) → D is also bijective and conformal. By Problem 13.7, we see that the mapping f = f1 ◦ f2−1 : D(0; 1) → D(0; 1) is an automorphism. Clearly, f (0) = f1 (f2−1 (0)) = f1 (z0 ) = 0. By Problem 3.3 and Proposition 3.5, we obtain f ′ (0) = f1′ (f2−1 (0)) · (f2−1 )′ (0) = f1′ (f2−1 (0)) ·

1 f2′ (f2−1 (0))

=

f1′ (z0 ) > 0. f2′ (z0 )

Hence f satisfies the hypotheses of Problem 13.11 so that f (z) ≡ z, i.e., f1 ≡ f2 which ends the proof of the problem.  Problem 13.13 Bak and Newman Chapter 13 Exercise 13.

Proof. Suppose that D(z1 ; r1 ), D(z2 ; r2 ) are discs and U1 , U2 are half-planes, where r1 , r2 > 0. Let U be H. As what we have used in Problem 13.7, D1 ∼ D2 means that D1 and D2 are conformally equivalent. • Case (i): D(z1 ; r1 ) ∼ D(z2 ; r2 ). Let f be this conformal mapping. Now the mapping gk (z) = r1k (z − zk ) maps D(zk ; rk ) conformally onto the unit disc D(0; 1), where k = 1, 2. It is obvious that each gk is a bilinear transformation. Since f = g2−1 ◦ g1 , it follows from Problem 13.9 that f is also in the form of a bilinear transformation.

Chapter 13. Introduction to Conformal Mapping

180

• Case (ii): D(z1 ; r1 ) ∼ U1 . Let f be this conformal mapping. It is easily seen that there exist constants θ and b such that the mapping h(z) = eiθ (z + b) sends U1 conformally onto U . It deduces from Theorem 13.16 that the mapping φ(z) = z−i z+i is a conformal mapping −1 −1 of U onto D(0; 1). We know that f = h ◦ φ ◦ g, where g(z) = r11 (z − z1 ). Since the inverses h−1 and φ−1 are bilinear transformations, Problem 13.9 guarantees that our f is in the form of a bilinear transformation. • Case (iii): U1 ∼ U2 . Let f be this conformal mapping. Suppose that h1 (z) = eiθ1 (z + b1 ) and h2 (z) = eiθ2 (z +b2 ) map U1 and U2 conformally onto U respectively, where θ1 , θ2 , b1 , b2 are some constants. Then we have f = h−1 2 ◦ h1 which is a bilinear transformation by Problem 13.9. Consequently, any conformal mapping of a half-plane or disc onto another half-plane or disc must be in the form of a bilinear transformation. This completes the analysis of the problem.  Problem 13.14 Bak and Newman Chapter 13 Exercise 14.

Proof. Let g = −f . Then we have g(z) =

(−a)z + (−b) cz + d

and (−a)d − (−b)c = −(ad − bc) > 0. It follows from Theorem 13.17 that g is an automorphism of H. Therefore, f maps H conformally onto the lower half-plane, completing the proof of the  problem. Problem 13.15 Bak and Newman Chapter 13 Exercise 15.

Proof. Let Π1 be the first quadrant. Now Theorem 13.17 shows that an automorphism of H is of the form az + b h(z) = , cz + d √ where ad − bc > 0. We note that f (z) = z is a conformal mapping of H onto Π1 . Similarly, g(z) = z 2 is a conformal mapping of Π1 onto H. Since the composition of two conformal mappings is also a conformal mapping, the mapping F = f ◦h◦g : Π1 → Π1 is an automorphism. Hence we conclude that r az 2 + b F (z) = f (h(g(z))) = . cz 2 + d We complete the proof of the problem. Problem 13.16 Bak and Newman Chapter 13 Exercise 16.



181 z−i Proof. We follow the given hint. Let h1 = f −1 ◦eiθ ◦f and h2 = f −1 ◦e−iθ g ◦f , where f (z) = z+i z−α and g(z) = eiθ ( 1−αz ) with |α| < 1. By the properties of composition of functions, we obtain

h = f −1 ◦ g ◦ f = f −1 ◦ eiθ ◦ e−iθ g ◦ f = (f −1 ◦ eiθ ◦ f ) ◦ (f −1 ◦ e−iθ g ◦ f ) = h1 ◦ h2 . Since f (z) =

z−i z+i ,

we have f −1 (z) = −iθ

e

i(z+1) −z+1 .

g(f (z)) =

z−i z+i

1−

(13.2)

Therefore, we see that −α

z−i α( z+i )

=

(1 − α)z − (1 + α)i (1 − α)z + (1 + α)i

and direct computation shows h2 (z) = i ·

(1−α)z−(1+α)i (1−α)z+(1+α)i + 1 (1−α)z−(1+α)i +1 − (1−α)z+(1+α)i

(1 − α)z − (1 + α)i + (1 − α)z + (1 + α)i −(1 − α)z + (1 + α)i + (1 − α)z + (1 + α)i [2 − (α + α)]z + (α − α)i =i· (α − α)z + [2 + (α + α)]i (1 − Re α)z + Im α . = (Im α)z + (1 + Re α) =i·

Recall that |α| < 1, so (1 − Re α)(1 + Re α) − (Im α)2 = 1 − [(Re α)2 + (Im α)2 ] = 1 − |α|2 > 0. In other words, h2 is a bilinear transformation. If θ = π, then h1 (z) = −z and we obtain from the formula (13.2) that h = −h2 and thus this insures that h is in the form of a bilinear transformation. Suppose that θ 6= π. Then we have cos θ 6= −1 and cos 2θ 6= 0. Now simple algebra shows that h1 (z) =

eiθ ( z−i z+i ) + 1 −eiθ ( z−i z+i ) + 1

eiθ (z − i) + z + i −eiθ (z − i) + z + i (1 + eiθ )z + (1 − eiθ )i =i· (1 − eiθ )z + (1 + eiθ )i =i·

=i·

(2 cos 2θ )z − (2i sin θ2 )i

(−2i sin 2θ )z + (2 cos θ2 )i (1 + cos θ)z + sin θ . = (− sin θ)z + (1 + cos θ) As (2 + cos θ)2 + sin2 θ = 5 + 4 cos θ > 0, h1 is in the form of a bilinear transformation. By Problem 13.9, h is also a bilinear transformation in this case and we have completed the proof  of the problem. Problem 13.17 Bak and Newman Chapter 13 Exercise 17.

Proof. (a) Now

z−1 z+1

= z if and only if z 2 = −1 if and only if z = ±i.

Chapter 13. Introduction to Conformal Mapping (b) Similarly,

z z+1

182

= z if and only if z 2 = 0 if and only if z = 0.

This completes the proof of the problem.



Problem 13.18 Bak and Newman Chapter 13 Exercise 18.

Proof. By Lemma 13.20, we know that T (z) =

(z − z2 )(z3 − z1 ) (z − z1 )(z3 − z2 )

is the (unique) bilinear mapping sending z1 , z2 and z3 to ∞, 0 and 1 respectively. Next, Theorem 13.11 implies that the image of the circle or line containing z1 , z2 and z3 under T is either a circle or a line. Since ∞, 0 and 1 lie on the real axis, the image must be the real axis. Therefore, T (z4 ) is real if and only if z4 lies on the circle or the line containing z1 , z2 and z3 , completing  the proof of the problem. Problem 13.19 Bak and Newman Chapter 13 Exercise 19.

Proof. (a) By Theorem 13.23, we conclude that 2(ω − i) −2(z − i) = . (ω + 1)(1 − i) (z − 1)(−1 − i)

After simplification, we conclude that ω = − 1z . (b) By Theorem 13.23 again, we get

2i(ω − i) 2iz = iω i(z + i) which gives ω = z + i. (c) We know that ω( 3ω1 1 − 1) (ω − 0)( 13 − ω1 ) = lim 1 ω = 3ω, lim ω1 →∞ ω1 →∞ (ω − ω1 )( 1 − 0) 3 3 ( ω1 − 1)

so Theorem 13.23 implies that

3ω =

(z − i)(2i + i) z−i =3· (z + i)(2i − i) z+i

z−i . which reduces to ω = z+i We have ended the proof of the problem.



Problem 13.20 Bak and Newman Chapter 13 Exercise 20.

Proof. Figure 13.4 shows the conformal mapping f of the region between the two circles (the light blue part) and the annulus (the light orange part).

183

Figure 13.4: The conformal mapping of the region between the two circles and the annulus. We know from Theorem 13.15 that f (z) = eiθ ·

z−α 1 − αz

(13.3)

is an automorphism of D(0; 1), where |α| < 1 and θ ∈ R. Now it remains to determine α so that f maps D( 14 ; 14 ) conformally onto D(0; a). Obviously, α 6= 0; otherwise, f = eiθ · id which is impossible. Now the hypothesis and the fact that f maps C(0; 1) onto itself certainly force that our f must map C( 14 ; 14 ) onto the circle C(0; a). By choosing θ appropriately in the formula (13.3), we may assume that f ( 21 ) = a. Since f is conformal, we have f (0) = −a so that 1 2

−α = e−iθ a = α 1 − α2 4α = 1 + |α|2

√ 2 − 4α + 1 = 0 which implies that α = 2 ± 3. and thus α must be real. Consequently, we have α √ Since |α| < 1, we obtain α = 2 − 3 which gives the desired conformal mapping. This completes the proof of the problem.  Problem 13.21 Bak and Newman Chapter 13 Exercise 21.∗

Proof. If ζ = cosh z, then we can show easily from Proposition 3.5 or from [16, Eqn. (8), p. 201] directly that d (cosh−1 z) = dz

1 d dζ (cosh ζ)

=

Therefore, we conclude that f (z) = cosh

1 1 1 =p =√ . sinh ζ z2 − 1 cosh2 ζ − 1

−1

z=

Z

z 0

p

dζ ζ2 − 1

.

By [16, Eqn. (8), p. 201] again, we know that cosh−1 z = i cos−1 z. Using [3, Example 1.8.8, p. 91; Exercise 48, p. 93] we can derive the identity sin−1 z + cos−1 z = π2 , so we establish f (z) = cosh−1 z =

πi − i sin−1 z. 2

(13.4)

Chapter 13. Introduction to Conformal Mapping

184

We note from [4, pp. 187, 188] that sin−1 z maps H conformally onto the semi-infinite strip {z ∈ C | − π2 < Re z < π2 and Im z > 0}, thus the equation (13.4) ensures that our mapping f sends H conformally onto the semi-infinite strip {z ∈ C | Re z > 0 and 0 < Im z < π}.b We have completed the proof of the problem.  Problem 13.22 Bak and Newman Chapter 13 Exercise 22.∗

Proof. Applying the Schwarz-Christoffel transformation and idea as well as the notations of [4, Sect. III, pp. 191, 192], we choose a1 = −1, a2 = 1 and a3 > 1. Since the exterior angles of an 3π π isosceles right triangle are 3π 4 , 4 and 2 , we obtain α1 π = which give α1 = α2 =

3 4

3π , 4

α2 π =

3π 4

π 2

and α3 π =

and α3 = 12 . Thus the mapping Z z dζ f (z) = 3 3 1 0 (ζ + 1) 4 (ζ − 1) 4 (ζ − a3 ) 2

(13.5)

is a conformal mapping sends H onto an isosceles right triangle. In fact, we can omit the factor 1 (ζ − a3 ) 2 in the formula (13.5), so we can reduce the formula (13.5) to Z z dζ f (z) = 3 0 (ζ 2 − 1) 4 which is our required mapping. This ends the proof of the problem.



Problem 13.23 Bak and Newman Chapter 13 Exercise 23.∗

Proof. Take a = −1, b = 0, c = 1 and replace d by ∞ in [4, Eqn. (3), p. 189] to get Z z dζ p , f (z) = ζ(1 − ζ 2 ) 0

which sends H conformally onto a rectangle, where the branch of one that is positive when 0 < ζ < 1, see Figure 13.5 below:

p

ζ(1 − ζ 2 ) is taken to be the

Figure 13.5: The conformal mapping of H onto a rectangle. b

The reader is suggested to refer to [13, Maps 5.53, 5.54, pp. 173, 174].

185 Recall from [4, p. 191] that the point at infinity is mapped by f onto one of the vertices of the rectangle. We claim that this rectangle is actually a square. To this end, we know that f (0) = 0 and 1 f ′ (z) = p >0 z(1 − z 2 )

on the interval 0 < z < 1, so f maps the interval [0, 1] onto the interval [0, A], i.e., f (1) = A or A=

Z

1 0

p

dx x(1 − x2 )

.

As z crosses over the point 1, the argument of f ′ (z) will increase by π2 . Next, we consider the path of integration [0, ∞]. As z crosses “over” the point ∞ (from ∞ to −∞), the argument of f ′ (z) will also increase by π2 so that f (∞) = A + iB. Since Z ∞ dx p A + iB = f (∞) = f (1) + , x(1 − x2 ) 1 we get

iB =

Z

∞

1

p

dx x(x2 − 1)

.

(13.6)

Applying the substitution x = y1 to the integral (13.6), it can be shown easily that B = A which means that the rectangle is in fact a square. This completes the proof of the problem. 

Chapter 13. Introduction to Conformal Mapping

186

CHAPTER

14

The Riemann Mapping Theorem

Problem 14.1 Bak and Newman Chapter 14 Exercise 1.

Proof. By the discussion on [4, p. 195], we know that Z g(ζ) dζ = 0

(14.1)

C

for any closed curve C in D, so by using similar argument as in the proof of Theorem 8.5, we conclude that our F is well-defined and analytic in D. Let z1 , z2 ∈ C and γ be a (smooth) curve from z1 to z2 . Furthermore, suppose that γ1 and γ2 are curves from z0 to z1 and z2 respectively. Then γ1 + γ − γ2 forms a closed curve and so we observe from the equation (14.1) that Z Z Z g(ζ) dζ − g(ζ) dζ = − g(ζ) dζ γ1

γ2

γ

or equivalently, F (z2 ) − F (z1 ) =

Z

g(ζ) dζ.

(14.2)

γ

If we write g = u + iv and z(t) = x(t) + iy(t), then we apply Definition 4.3 to the expression (14.2) to get F (z2 ) − F (z1 ) = = =

Z

t2

t1 Z t2

t1 Z t2

(u − iv)z(t) ˙ dt (u − iv)( dx + i dy) (u dx + v dy) + i

t1

Z

t2 t1

(u dy − v dx),

(14.3)

where t1 and t2 are the values such that z1 = z(t1 ) and z2 = z(t2 ). If Re F (z) = C1 for some constant C1 on γ, then the representation (14.3) implies that Z

t2

(u dx + v dy) = 0

t1

187

(14.4)

Chapter 14. The Riemann Mapping Theorem

188

and if we represent g = u + iv in the vector form (u, v), then the result (14.4) means that the velocity vector (u, v) is orthogonal to the tangle vector (x′ (t), y ′ (t)), completing the proof of the problem.  Problem 14.2 Bak and Newman Chapter 14 Exercise 2.

Proof. It follows from the representation (14.3) that if Im F (z) = C2 for a constant C2 , then we have Z t2 (u dy − v dx) = 0 t1

which means that the velocity vector (u, v) is orthogonal to the vector (−y ′ (t), x′ (t)). Since (x′ (t), y ′ (t)) · (−y ′ (t), x′ (t)) = 0, the flow g is parallel to the tangent vector (x′ (t), y ′ (t)), i.e., g  is tangent to Im F (z) = C2 . This completes the proof of the problem, Problem 14.3 Bak and Newman Chapter 14 Exercise 3.

Proof. (a) By Problem 14.1, we take z0 = 0 to get F (z) =

Z

z

g(ζ) dζ =

Z

0

0

z

ζ dζ =

ζ 2 z z2 = . 2 0 2

(14.5)

Let c ∈ R. If z = x + iy, then z 2 = x2 − y 2 + 2ixy. By Problem 14.2, the streamlines of g is Im F (z) = c. Therefore, we follow from the expression (14.5) that the streamlines of g are xy = c. (b) For simplicity, we take z0 = 1 so that Z Z z g(ζ) dζ = F (z) = 1

z 1

dζ = log z = log |z| + iArg z. ζ

In this case, Im F (z) = Arg z. Hence the streamlines of g are rays from the origin. We end the proof of the problem.



Problem 14.4 Bak and Newman Chapter 14 Exercise 4.

Proof. We follow the given hint. Let C be a real constant. Let D be the exterior of D(0; 1) and for simplicity, we pick I = [0, 1] to be the interval for consideration. Suppose that F : D → C \ I is a conformal mapping with F (z) ∼ z as z → ∞. Thus we may suppose that F (z) = z + A0 +

A1 A2 + 2 + ··· , z z

189 where Ak ∈ C for all k = 0, 1, 2, . . .. If z = eiθ with θ ∈ [−π, π], then we notice that Ak z −k = (Re Ak + iIm Ak )(cos kθ − i sin kθ)

= (Re Ak cos kθ + Im Ak sin kθ) + i(−Re Ak sin kθ + Im Ak cos kθ).

(14.6)

Now Im F (eiθ ) = C, so the expression (14.6) implies that ∞ X sin θ + Im A0 + (−Re Ak sin kθ + Im Ak cos kθ) = C.

(14.7)

k=1

By the identities sin z =

1 iz 2i (e

− e−iz ) and cos z = 12 (eiz + e−iz ), we can rewrite (14.7) as ∞ X

ck eikθ = C,

(14.8)

−∞

where

We apply the factsa

 Im Ak + iRe Ak   ,   2        Im A1 − (1 − Re A1 )i     2     Im A0 ck =       Im A1 + (1 − Re A1 )i      2         Im A|k| − iRe A|k| , 2 1 2π

Z

π

einx dx =

−π

to the representation (14.8) to get ck =

if k ≥ 2; if k = 1; if k = 0;

(14.9)

if k = −1; if k ≤ −2.

  1, if n = 0; 

0, if n = ±1, ±2, . . .

  C, if k = 0; 

0,

if k ∈ Z \ {0}.

Therefore, the definition (14.9) asserts that Re A1 = 1, Re Ak = 0 for all k ≥ 2 and Im Ak = 0 for all k ≥ 1. In other words, we have established that 1 F (z) = z + A0 + . z This completes the proof of the problem.



Remark 14.1 In fact, the mapping in Problem 14.5 is classically called the Joukowsky transform, see [3, Example 7.1.8, pp. 408-410] or [9, Exercise 15, pp. 66, 67]. a

See [22, Eqn. (61) & (62), pp. 185, 186]

Chapter 14. The Riemann Mapping Theorem

190

Problem 14.5 Bak and Newman Chapter 14 Exercise 5.

Proof. (a) Let f (z) = 2z + z1 and let D and D ′ be the exteriors of D(0; 1) and the said ellipse / D. Thus f is respectively. Since f ′ (z) = 2 − z12 , f ′ (z) = 0 if and only if z = ± √12 ∈ conformal throughout D by Theorem 13.4. Take z = eiθ , so

f (eiθ ) = 2(cos θ + i sin θ) + (cos θ − i sin θ) = 3 cos θ + i sin θ which means that

x2 n o f (C(0; 1)) = (x, y) + y2 = 1 . 9 As a conformal map, f is continuous on the connected set D so that f (D) is connected. In addition, since f (z) → ∞ as z → ∞, f must map D into D ′ . It remains to show that the mapping f : D → D ′ is surjective, but it is easy to see because if (x, y) ∈ D ′ satisfies x2 x 2 9 + y > 1, then the point z = 3 + iy lies in D. Hence the f is actually a surjective conformal mapping. (b) Without loss of generality, we let the real line segment be [−2, 2]. By [4, Example 2, p. 197], g(z) = z + 1z maps D conformally onto C \ [−2, 2]. Thus the composition g ◦ f −1 : D ′ → C \ [−2, 2] is a desired conformal mapping. Consequently, we have completed the proof of the problem.



Problem 14.6 Bak and Newman Chapter 14 Exercise 6.

Proof. Since f (z0 ) ∈ D(0; 1), Theorem 13.15 implies that the mapping of the form h(z) = eiθ

 z − f (z )  0 1 − f (z0 )z

(14.10)

is an automorphism of D(0; 1), where θ ∈ R. By Problem 13.7, the composition g = h ◦ f is a conformal mapping of R onto D(0; 1). Obviously, the expression (14.10) gives g(z0 ) = h(f (z0 )) = 0. Furthermore, direct computation gives g′ (z) = eiθ ·

[1 − f (z0 ) · f (z)]f ′ (z) − [f (z) − f (z0 )] · [−f (z0 ) · f ′ (z)] [1 − f (z0 ) · f (z)]2

so that g′ (z0 ) = eiθ ·

f ′ (z0 ) . 1 − |f (z0 )|2

Now if we take θ = −Arg f ′ (z0 ), then we obtain the desired result that g′ (z0 ) > 0. We end the  proof of the problem.

191 Problem 14.7 Bak and Newman Chapter 14 Exercise 7.∗

Proof. We define g : R → U by g(z) = f (z). We claim that g is analytic in R. To this end, let z = x + iy ∈ R and f (z) = u(x, y) + iv(x, y), where u and v are real-valued functions. Then we have g(z) = f (z) = u(x, −y) + iv(x, −y) = u(x, −y) − iv(x, −y) = U (x, y) + iV (x, y), where U (x, y) = u(x, −y) and V (x, y) = −v(x, −y). Since f is analytic in R, Theorem 6.6 and Corollary 2.10 together imply that u and v have C 1 partial derivatives. In particular, gx = Ux + iVx and gy = Uy + iVy are continuous. By Proposition 3.1, we have ∂u(x, y) ∂v(x, y) = ∂x ∂y

and

∂u(x, y) ∂v(x, y) =− ∂y ∂x

which give ∂U (x, y) ∂u(x, −y) ∂v(x, −y) ∂V (x, y) = = = ∂x ∂x ∂(−y) ∂y and

∂(−v(x, −y)) ∂u(x, −y) ∂U (x, y) ∂V (x, y) = = =− ∂x ∂x ∂(−y) ∂y Hence it follows from Proposition 3.2 that g is analytic at z, so the claim is true. Since R is symmetric respect to R, it is easy to see that the map z 7→ z is bijective in R. Since f : R → U is the Riemann mapping, it is bijective. Finally, since g is the composition of bijective maps, it is also bijective. By Definition 13.9, g is a conformal mapping. Recall that z0 is real, thus we obtain g(z0 ) = f (z0 ) = f (z0 ) = 0. Furthermore, for every ω ∈ R, the definition of analyticity of g guarantees that g(z) − g(ω) z→ω z−ω f (z) − f (ω) = lim z→ω z−ω f (z) − f (ω) = lim z→ω z−ω h f (z) − f (ω) i = lim z→ω z−ω h f (ζ) − f (ω) i = lim ζ→ω ζ −ω

g′ (ω) = lim

= f ′ (ω).

(14.11)

Particularly, if we take ω = z0 , then we deduce from the expression (14.11) and the hypotheses that g ′ (z0 ) = f ′ (z0 ) = f ′ (z0 ) > 0. By the Riemann Mapping Theorem, the mapping f is unique, so we conclude that f (z) = g(z) which means f (z) = f (z) for all z ∈ R, as desired. This completes the proof of the problem.  Problem 14.8 Bak and Newman Chapter 14 Exercise 8.∗

Chapter 14. The Riemann Mapping Theorem

192

Proof. By Theorem 13.16, every conformal mapping f of H onto D(0; 1) is of the form f (z) = eiθ

z − α , z−α

(14.12)

where Im α > 0. (a) Since f (−1) = 1, f (0) = i and f (1) = −1, we deduce from the formula (14.12) that eiθ

1 + α 1+α

= 1,

eiθ ·

1 − α α = −1. = i and eiθ α 1−α

(14.13)

Suppose that α = a + ib. Then the first and the third equations (14.13) yield that a2 + b2 = 1. Besides, the first two equations (14.13) imply that a − b + 1 = 0.

(14.14)

Finally, the last two equations (14.13) establish that a + b − 1 = 0.

(14.15)

Solving the equations (14.14) and (14.15), we get a = 0 and b = 1, i.e., α = i. Using the second expression (14.13), we find that eiθ = −i. Hence the desired conformal mapping is given by z−i iz + 1 f (z) = −i · =− . z+i z+i (b) The condition f (i) = 0 implies that α = i. Next, the condition f (1) = 1 implies that eiθ = i. Hence the required conformal mapping is f (z) = i · We have completed the proof of the problem.

iz + 1 z−i = . z+i z+i 

Problem 14.9 Bak and Newman Chapter 14 Exercise 9.

Proof. If R = C, then the mapping f : C → C defined by f (z) = z − z1 + z2 is conformal at every z ∈ C because f ′ (z) = 1 by Theorem 13.4. Since f (z1 ) = z2 , it satisfies our requirements. Next, we suppose that R 6= C. Since R is simply connected, the Riemann Mapping Theorem ensures that there exist conformal mapping ϕ1 : R → D(0; 1) and ϕ2 : R → D(0; 1) such that ϕ1 (z1 ) = ϕ2 (z2 ) = 0. By Problem 13.7, the mapping ϕ = ϕ−1 2 ◦ ϕ1 : R → R is a conformal mapping and we have −1 ϕ = ϕ−1 2 (ϕ1 (z1 )) = ϕ2 (0) = z2 , completing the analysis of the problem. Problem 14.10 Bak and Newman Chapter 14 Exercise 10.



193 Proof. Assume that f : C → R was a conformal mapping. By the Riemann Mapping Theorem, there exists a conformal mapping g : R → D(0; 1). By Problem 13.7, h = g ◦ f : C → D(0; 1) is conformal, i.e., h is a bounded entire function which certainly contradicts Theorem 5.10 (Liouville’s Theorem). This completes the proof of the problem.  Problem 14.11 Bak and Newman Chapter 14 Exercise 11.

Proof. (a) Note that G = {g : R → U | g is analytic and g ′ (z0 ) > 0}. By the proof of Part (B) of the Riemann Mapping Theorem [4, p. 202], for every g ∈ G, we see easily that 1 g′ (z0 ) = |g ′ (z0 )| < , δ where δ is a positive number independent of g such that D(z0 ; δ) ⊂ R. In conclusion, we find that 1 sup g′ (z0 ) = M ∗ ≤ . δ g∈G (b) Following the same argument as the proof of Part (B), we can verify that there exists a function Φ ∈ G such that Φ′ (z0 ) = M ∗ . Next, if Φ(z0 ) = α with 0 < |α| < 1, then the map gb : R → U defined by Φ(z) − α gb(z) = 1 − αΦ(z) is clearly analytic, i.e., gb ∈ G. Besides, we have gb′ (z0 ) =

Φ′ (z0 ) > Φ′ (z0 ), 1 − |α|2

a contradiction. Therefore, we conclude that Φ(z0 ) = 0. Let ϕ : R → U be the Riemann mapping function with ϕ(z0 ) = 0 and ϕ′ (z0 ) = M > 0. Notice that F ⊆ G, so it happens that Φ′ (z0 ) = M ∗ ≥ M > 0. Now we consider the map f = Φ ◦ ϕ−1 : U → U. As a composition of two analytic functions, f is also analytic in U and |f (z)| < 1 on U . Simple algebra shows that f (0) = Φ(ϕ−1 (0)) = Φ(z0 ) = 0. Consequently, our f satisfies the hypotheses of Theorem 7.2 (Schwarz’s Lemma), so we see that |f (z)| ≤ |z| and |f ′ (0)| ≤ 1.

On the other hand, by combining Proposition 3.5 and Problem 3.3, we know that f ′ (0) =

M∗ Φ′ (z0 ) = ≥ 1. ϕ′ (z0 ) M

Hence according to Theorem 7.2 (Schwarz’s Lemma), it is true that f (z) = z, i.e., f = id . By the definition, we finally get Φ=ϕ so that Φ is injective. We complete the analysis of the problem.



Chapter 14. The Riemann Mapping Theorem

194

Problem 14.12 Bak and Newman Chapter 14 Exercise 12.∗

Proof. (a) By similar idea as the proof of Problem 13.6, the mapping f : S → U defined by f (z) = = =

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

−iz 2 − 2z − i z 2 + 2iz + 1 =i· 2 z − 2iz + 1

(14.16)

is conformal. Since S and U are Jordan regions by Definition 14.2, we follow from Theorem 14.3 (The Carath´eodory-Osgood Theorem) that f can be extended to a homeomorphism between S and U . √ / S, the representation (14.16) (b) Since the roots of z 2 − 2i + 1 = 0 are z = (1 ± 2)i ∈ ensures that f is analytic on S. However, it follows from the representation (14.16) that the inverse f −1 : U → S has the form √ (−1 + iz) + 2 − 2z 2 −1 f (z) = i−z which has a simple pole at −i ∈ U , so f −1 is not analytic on U .

This completes the proof of the problem.



Problem 14.13 Bak and Newman Chapter 14 Exercise 13.∗

Proof. Figure 14.1 shows the shape of a Norman window N . Without loss of generality, we may suppose that 0 lies on ∂N .

Figure 14.1: The shape of a Norman window N .

195 Assume that f : U → N was analytic and surjective. By Theorem 7.1 (The Open Mapping Theorem)b , f will map a z0 ∈ C(0; 1) onto the origin. Now we may further suppose that z0 = 1. If f ′ (1) 6= 0, then Theorem 13.4 implies that f maps the rays I1 = {1 + it | 0 ≤ t ≤ ∞} and I2 = {1 − it | 0 ≤ t ≤ ∞} onto two curves whose tangent lines form a straight angle, ∠f (I1 ), f (I2 ) = π,

(14.17)

but it is impossible because the angle at 0 is just π2 . Next, if f ′ (1) = 0, then Theorem 13.7 will ensure that, instead of the value (14.17), we have ∠f (I1 ), f (I2 ) = kπ, where k is the least positive integer such that f (k) (1) 6= 0. However, this is also impossible. Hence  no such analytic and surjective map f exists. This completes the proof of the problem.

b

See also the proof of Problem 13.8.

Chapter 14. The Riemann Mapping Theorem

196

CHAPTER

15

Maximum-Modulus Theorems for Unbounded Domains

Problem 15.1 Bak and Newman Chapter 15 Exercise 1.∗

Proof. Without loss of generality, we may assume that f is non-constant. We basically follow the proof of Theorem 15.1. Suppose first that D = {z ∈ C | Re z > 0}, f (z) ≪ 1 on ∂D, f (z) ≪ log z for all z ∈ D and z0 ∈ D. Consider the auxiliary function h(z) =

f N (z) , z+1

(15.1)

where N ∈ N. By the hypothesis, we have |h(z)| ≤ 1 along the imaginary axis. In addition, if z = x + iy ∈ D, then x > 0 and so p p |z + 1| = x2 + y 2 + 2x + 1 ≥ x2 + y 2 = |z|.

Therefore, for all z ∈ D with |z| = R > 0, we have

N

|h(z)| ≤

[(log R)2 + (Arg z)2 ] 2 | log z|N ≤ . R R

(15.2)

Since −π ≤ Arg z ≤ π, we can take R large enough so that the bound (15.2) reduces to √ 2N (log R)N |h(z)| ≤ ≤1 R

(15.3)

for all z in the right semi-disc DR = {z ∈ D | |z| ≤ R}. Since h is certainly analytic in DR and DR is compact, Theorem 6.13 (The Maximum Modulus Theorem) implies that |h(z0 )| ≤ 1. By the definition (15.1), we achieve f N (z ) 0 ≤1 z0 + 1

or equivalently,

1

|f (z0 )| ≤ |1 + z0 | N .

Since N is arbitrary, we take N → ∞ to conclude that |f (z0 )| ≤ 1 as desired. 197

Chapter 15. Maximum-Modulus Theorems for Unbounded Domains

198

Next, we suppose that D is an arbitrary region. Then we consider g(z) =

f (z) − f (a) , z−a

where a is a fixed point in D. By Proposition 6.7, g is C-analytic in D. Furthermore, the hypothesis implies that, for large |z|, |g(z)| =

| log z| + | log a| |f (z) − f (a)| ≤ |z − a| |z| − |a|

so that g(z) → 0 as z → ∞. Consequently, there exists a constant M > 0 such that |g(z)| ≤ M on D. Similar to the previous paragraph, we set DR = {z ∈ D | |z| ≤ R} and h(z) = f N (z)g(z), where R > 0 and N ∈ N. Thus it must be true that |h(z)| ≤ M on ∂DR for sufficiently large R. Hence it follows from Theorem 6.13 (The Maximum Modulus Theorem) that |h(z0 )| ≤ M for every z0 ∈ D. If g(z0 ) 6= 0, then we get 1 h(z) 1 MN N |f (z0 )| ≤ ≤ 1 . g(z) |g(z0 )| N

By taking N → ∞, we get |f (z0 )| ≤ 1. According to Theorem 6.9 (The Uniqueness Theorem), since f is nonconstant, the set S = {z ∈ D | g(z) = 0} is discrete, so we must have |f (z)| ≤ 1

(15.4)

on D \ S. Finally, the analyticity of f ensures that the bound (15.4) holds in D. Hence our conclusion of Theorem 15.1 is still valid in this case. We see that our argument works because of the inequality (15.3). Thus if f (z) ≪ P (log z) in D, where P (ω) is a polynomial in ω of degree n ≥ 1, then there exists a positive constant M such that M N (log R)nN ≤1 |h(z)| ≤ R holds in DR = {z ∈ D | |z| ≤ R} for sufficiently large enough R. This completes the proof of the  problem. Problem 15.2 Bak and Newman Chapter 15 Exercise 2.

Proof. We consider D1 = {z ∈ C | −

π π < Arg z < } and F (z) = exp(z 2 ). 4 4

On z = r exp(± π4 i), we see that z 2 = ±ir 2 so that |F (z)| = | exp(±ir 2 )| = 1, i.e., F is bounded on ∂D1 . However, it is easy to check that 2

F (x) = ex → ∞ as x → ∞. Consequently, F is unbounded in D1 .

Suppose that F is the class of all functions f analytic in D1 such that f is bounded on ∂D1 but f (z) is unbounded in D1 . Let f (z) be an analytic function “smaller” than F (z) in D1 in the sense that for every ǫ > 0, there exists a constant Aǫ such that |f (z)| ≤ Aǫ exp(ǫ|z|2 )

199 for all z ∈ D1 . Now Corollary 15.5 with α = π2 implies immediately that this function f must be bounded in the region D1 . Combining this fact and the observation in the previous paragraph, 2 we conclude that F ∈ F but f ∈ / F . Hence F (z) = ez is the “smallest” analytic function in this sense. 3π By replacing z by e−i 4 z, we see at once that the function G(z) = F (e−i

3π 4

z) = exp(e−i

3π 2

z 2 ) = eiz

2

is the desired “smallest” non-constant analytic function in D, completing the proof of the prob lem. Problem 15.3 Bak and Newman Chapter 15 Exercise 3.

Proof. Let F (z) = exp(ez ) and D = {x + iy | x ∈ R and − π2 < y < π2 }. Clearly, F is C-analytic in D. If z = x ± i π2 , then we have ez = ±iex . In other words, ez maps ∂D onto the imaginary axis which implies that |F (z)| = | exp(ez )| = | exp(±iex )| = 1 on ∂D, proving the first assertion. However, we note that F (log x) = exp(elog x ) = ex → ∞ as x → ∞. In other words, F is unbounded in the region D. Suppose that f (z) is an analytic function “smaller” than F (z) in D in the sense that for every ǫ > 0, there exists a constant Aǫ such that |f (z)| ≤ Aǫ exp(ǫ|ez |) for all z ∈ D. In this case, the analytic function g(z) = f (log z) satisfies |g(z)| ≤ Aǫ eǫ|z| in the region D. By Theorem 15.4 (The Phragm´en-Lindel¨of Theorem), g(z) and then f (z) are also bounded throughout D. Suppose that F is the class of all functions f analytic in D such that f is bounded on ∂D but f (log z) is unbounded in D. Then the first assertion says immediately that F ∈ F but the previous paragraph means that any analytic function f “smaller” than F implies that f ∈ / F.  This completes the proof of the problem. Problem 15.4 Bak and Newman Chapter 15 Exercise 4.∗

Proof. Assume that there were constants A and B such that |g(z)| ≤ A exp(|z|B )

(15.5)

Chapter 15. Maximum-Modulus Theorems for Unbounded Domains

200

in C.a As given by the hint, we choose a positive integer N such that N > 2B and then we π divide the plane into N wedges of equal angles α = N . Let W1 , W2 , . . . , WN be these wedges. Since g is entire, it is C-analytic in each Wk . Next, the hypothesis ensures that |g(z)| ≤ Mk

(15.6)

on ∂Wk for some positive constant Mk . Furthermore, it is clear from the assumption (15.5) that π

N

|g(z)| ≤ A exp(|z|B ) ≤ A exp(|z| 2 ) = A exp(|z| 2α ). π

(15.7) 2α

π

Given ǫ > 0. Then the inequality |z| 2α ≤ ǫ|z| α holds if and only if |z| ≥ ǫ− π . Thus the inequality (15.7) gives π (15.8) |g(z)| ≤ A exp(ǫ|z| α ) 2α

2α

if z ∈ Wk and |z| ≥ ǫ− π . If z ∈ Wk and |z| < ǫ− π , then since ex is strictly increasing, it is easy to see from the inequality (15.7) that |g(z)| ≤ A exp(ǫ−1 ).

(15.9)

Let Aǫ = max(A, A exp(ǫ−1 )). We conclude from the inequalities (15.8) and (15.9) that π

|g(z)| ≤ Aǫ exp(ǫ|z| α ) for all z ∈ Wk . In other words, our function g satisfies the hypotheses of Corollary 15.5. Hence the inequality (15.6) holds throughout Wk and then g is bounded in C. Therefore, Theorem 5.10 (Liouville’s Theorem) implies that g is constant which contradicts our hypothesis. This  completes the analysis of the problem.

a

By Definition 16.12, g is of finite order.

CHAPTER

16

Harmonic Functions

Problem 16.1 Bak and Newman Chapter 16 Exercise 1.

Proof. Since f is analytic, Theorem 16.2 says that u and v are harmonic, i.e, uxx + uyy = 0 and vxx + vyy = 0. Since (u + v)xx + (u + v)yy = (uxx + uyy ) + (vxx + vyy ) = 0, u+v is harmonic by Definition 16.1. By direct computation, we have (uv)xx = uxx v+2ux vx +uvxx and (uv)yy = uyy v + 2uy vy + uvyy whose sum is (uv)xx + (uv)yy = 2(ux vx + uy vy ).

(16.1)

By Proposition 3.1, the equation (16.1) reduces to 0, so uv is also harmonic by Definition 16.1,  completing the proof of the problem. Problem 16.2 Bak and Newman Chapter 16 Exercise 2.

Proof. Suppose that u is a harmonic function. By Theorem 16.3, ux is the real part of an analytic function f . As an analytic function, f is infinitely differentiable. Hence u is also infinitely differentiable. Let g = ux . Then we deduce from [28, Theorem 15.12, p. 146] that gxx + gyy =

∂ ∂ ∂ uxx + uyyx = uxx + uxyy = (uxx + uyy ) = 0. ∂x ∂x ∂x

Consequently, ux is harmonic. Similarly, uy is also harmonic and we have completed the proof  of the problem. Problem 16.3 Bak and Newman Chapter 16 Exercise 3.

201

Chapter 16. Harmonic Functions

202

Proof. Let f = u2 . Obviously, we have fx = 2uux and fxx = 2(uuxx + u2x ). Similarly, we also have fyy = 2(uuyy + u2y ). Thus we obtain fxx + fyy = 2u(uxx + uyy ) + 2(u2x + u2y ) = 2(u2x + u2y ) ≥ 0 so that fxx + fyy = 0 if and only if ux = uy = 0 if and only if u is a constant by Theorem 1.10, a contradiction. This ends the proof of the problem.  Problem 16.4 Bak and Newman Chapter 16 Exercise 4.

Proof. Let u = log(x2 + y 2 ). Then we get ux =

2x x2 +y 2

and

uxx =

2(y 2 − x2 ) . (x2 + y 2 )2

(16.2)

uyy =

2(x2 − y 2 ) . (x2 + y 2 )2

(16.3)

Similarly, we get

Thus the sum of the formulas (16.2) and (16.3) certainly imply that uxx + uyy = 0, i.e., u is harmonic in D = C \ {0}.

Assume that u was the real part of an analytic function f = u + iv in D. We note that u = log(x2 + y 2 ) = 2 log |z|, so we write f = 2g, where g = log |z| + i v2 is analytic in D. Therefore, g is an analytic branch of log z up to an imaginary constant in D. Next, recall from Problem 8.8 that an analytic branch of log z can be defined with 0 < Arg z < 2π in the plane C \ [0, ∞). This fact forces that g is not continuous on [0, ∞) which contradicts the assumption that g is analytic in D. Hence no such g and then f exists and we complete the analysis of the  problem. Problem 16.5 Bak and Newman Chapter 16 Exercise 5.

Proof. (a) Note that x = r cos θ and y = r sin θ, so we obtain ∂x = cos θ, ∂r

∂x = −r sin θ, ∂θ

∂y = sin θ ∂r

and

∂y = r cos θ. ∂θ

These imply that

and thus

∂u ∂x ∂u ∂y ∂u ∂u ∂u = · + · = cos θ + sin θ ∂r ∂x ∂r ∂y ∂r ∂x ∂y

(16.4)

2 2 ∂2u ∂2u 2 ∂ u 2 ∂ u = cos θ + 2 cos θ sin θ + sin θ . ∂r 2 ∂x2 ∂x∂y ∂y 2

(16.5)

Similarly, we have ∂u ∂u ∂u = −r sin θ + r cos θ ∂θ ∂x ∂y

203 so that   2 2  ∂2u ∂u ∂u  ∂2u 2 ∂ u 2 2 ∂ u sin θ + r = −r cos θ + sin θ − 2 cos θ sin θ + cos θ . (16.6) ∂θ 2 ∂x ∂y ∂x2 ∂x∂y ∂y 2

Combining the expressions (16.4), (16.5) and (16.6), we conclude that 1 1 urr + ur + 2 uθθ = uxx + uyy . r r

(16.7)

Now if u(r, θ) is a harmonic function depending on r alone, then uθθ = 0 so that Laplace’s equation (16.7) becomes 1 urr + ur = 0 (16.8) r as desired. (b) The differential equation (16.8) can be written in the form (rur )r = 0 which means that rur = a for some constant a, or equivalently ur = ar . By integration, we establish u(r, θ) = a log r + b for some constant b. We complete the proof of the problem.



Problem 16.6 Bak and Newman Chapter 16 Exercise 6.

Proof. We start with the following form of Poisson Formula [1, Theorem 22, p. 168]: Z 1 − |a|2 dζ 1 U (ζ) U (a) = 2 2π |ζ|=1 |ζ − a| iζ

(16.9)

where U is C-harmonic in D(0; 1) and a ∈ D(0; 1). By Theorem 13.16, the map ϕ : H → D(0; 1) given by z−i ϕ(z) = (16.10) z+i is conformal and surjective. Besides, it is easy to check that ϕ maps ∂H = R onto C(0; 1) \ {1}. Now we consider the map U = u ◦ ϕ−1 : D(0; 1) → R. Since H is simply connected, Theorem 16.3 ensures that u = Re f for some analytic function f on H. Write f = u + iv so that f ◦ ϕ−1 = u ◦ ϕ−1 + iv ◦ ϕ−1 . Since both f and ϕ−1 are analytic, its composition f ◦ ϕ−1 : D(0; 1) → C is also analytic in D(0; 1) and we follow from Theorem 16.2 that u ◦ ϕ−1 is harmonic in D(0; 1). Since u is continuous on R and bounded, u ◦ ϕ−1 can be made to be continuous at 1 so that it is actually continuous on C(0; 1). In other words, u◦ϕ−1 is C-harmonic in D(0; 1). Thus we put U = u◦ϕ−1 into the formula (16.9) to get Z 1 − |a|2 dζ 1 −1 u(ϕ−1 (ζ)) . (16.11) u(ϕ (a)) = 2 2π |ζ|=1 |ζ − a| iζ Put ζ = ϕ(t) and a = ϕ(z) into the formula (16.11) and then using the expression (16.10) and 2i the fact ϕ′ (t) = (t+i) 2 to establish −1

u(x + iy) = u(z) = u(ϕ

1 (a)) = 2π

Z

R

1 − |ϕ(z)|2 2 · u(t) · 2 dt. |ϕ(t) − ϕ(z)|2 t +1

(16.12)

Chapter 16. Harmonic Functions 2 Since | z−i z+i | =

x2 +(y−1)2 x2 +(y+1)2

204

and

t − i z − i 2 4[(t − x)2 + y 2 ] |2y + 2(t − x)i|2 − = , |ϕ(t) − ϕ(z)| = = t+i z+i (1 + t2 )[x2 + (y + 1)2 ] (1 + t2 )[x2 + (y + 1)2 ] 2

the expression (16.12) becomes

Z ∞h 1 x2 + (y − 1)2 i (1 + t2 )[x2 + (y + 1)2 ] 2 1− 2 · u(t) · 2 dt · 2π −∞ x + (y + 1)2 4[(t − x)2 + y 2 ] t +1 Z ∞ 1 (y + 1)2 − (y − 1)2 = u(t) dt 4π −∞ (t − x)2 + y 2 Z 1 ∞ y · u(t) = dt π −∞ (t − x)2 + y 2

u(x + iy) =

which is our desired result. We have completed the proof of the problem.



Problem 16.7 Bak and Newman Chapter 16 Exercise 7.

Proof. Following the idea of [4, Example i, p. 232], since z 3 is analytic in D(0; 1), Re (z 3 ) is harmonic in D(0; 1) by Theorem 16.2. Note that Re (z 3 ) = x3 − 3xy 2 .

(16.13)

Since x2 + y 2 = 1, the equation (16.13) reduces to Re (z 3 ) = 4x3 − 3x on the boundary C(0; 1). Hence if we choose 1 1 u(x, y) = [Re (z 3 ) + 3x] = (x3 − 3xy 2 + 3x), 4 4 then it is harmonic in D(0; 1) by Definition 16.1 and satisfies u(x, y) = x3 on C(0; 1). This ends  the proof of the problem. Problem 16.8 Bak and Newman Chapter 16 Exercise 8.

Proof. By [4, Example ii, pp. 232, 233], we know that u(z) =

z − 1 3 1 , − Arg 2 π z+1

where |z| ≤ 1. If u(x, y) = k, where k ∈ [0, 1], then we get θk = Arg

z − 1 z+1

=π

3 2

 −k .

It is clear that as k runs through [0, 1], the formula (16.14) indicates that θk runs from Suppose that z = x + iy. Simple algebra gives x2 + y 2 − 1 2y z−1 = +i . z+1 (x + 1)2 + y 2 (x + 1)2 + y 2

(16.14) 3π 2

to

π 2.

205 By the definition, we know that tan θk =

x2

2y + y2 − 1

and then the value (16.14) implies that x2

1 2y = 2 +y −1 tan kπ

or equivalently x2 + y 2 − (2 tan kπ)y − 1 = 0.

(16.15)

It is trivial that (−1, 0) and (1, 0) always lie on the locus represented by the equation (16.15). Furthermore, if k = 12 , then we conclude from the equation (16.15) that y = 0 and x = ±1. If k = 0 or 1, then the equation (16.15) reduces to the unit circle C(0; 1).

Figure 16.1: The level curves of u(x, y) = k for k ∈ (0, 21 ) ∪ ( 12 , 1). Next, suppose that k ∈ (0, 12 ) ∪ ( 21 , 1). Then the centre and the radius of the circle (16.15) are given by p = (0, tan kπ) and | sec kπ| respectively. In fact, for 0 < k < 21 , we know that tan kπ > 0 and thus the centre lies on the positive imaginary axis which implies that the level curve u(x, y) = k is the lower circular segment of the circle C(p, sec kπ). This is exactly the dark red arc in Figure 16.1 lying inside the unit disc D(0; 1). Similarly, for 21 < k < π, then tan kπ < 0 so that its centre lies on the negative imaginary axis. In this case, the level curve u(x, y) = k is the upper circular segment of the circle C(p, | sec kπ|), see the purple arc lying inside the unit disc D(0; 1) in Figure 16.1.  We have completed the analysis of the problem. Problem 16.9 Bak and Newman Chapter 16 Exercise 9.

Chapter 16. Harmonic Functions

206

Proof. By Problem 16.6 with the function u(t) = we get

  1, if t > 0; 

1 u(x + iy) = π

0, if t < 0,

Z

0

∞

y dt. (t − x)2 + y 2

(16.16)

Using the substitution y tan u = t − x and the identity tan−1 (−x) = − tan−1 x for all x ∈ R, we can reduce the expression (16.16) to the following form u(x + iy) =

1 1 x + tan−1 2 π y

which is our desired harmonic function. This ends the proof of the problem.  Problem 16.10 Bak and Newman Chapter 16 Exercise 10.

Proof. Let S be the semi-infinite strip {z = x + iy | − π2 < x < π2 and y > 0}. The graph of the the temperature problem with the prescribed boundary values is shown in Figure 16.2. We note that this is a Dirichlet problem for the region S with its boundary in the z plane. Our method of solution is to obtain a new Dirichlet problem for the upper half-plane H with its boundary in the ω plane.

Figure 16.2: The temperature problem with the prescribed boundary values. Recall from §13.3 that the function ω = f (z) = sin z maps S conformally onto H and the interval [− π2 , π2 ] is mapped onto [−1, 1]. On z = − π2 + iy with y > 0, we have 1 sin z = − cos(iy) = − (ey + e−y ). 2 Therefore, f maps the vertical line z = − π2 + iy onto (−∞, −1). Similarly, it can be shown easily that sin z maps the vertical line z = π2 + iy with y > 0 onto (1, ∞). Consequently, the mapping f (z) = sin z transforms this boundary value problem into another boundary value problem in H.

207 Stimulating by [4, Example ii, pp. 232, 233], we consider the mapping ζ = log(ω 2 − 1) = log |ω 2 − 1| + iArg (ω 2 − 1) which is analytic in H. By the definition, we know that Arg (ω 2 − 1) = Arg (ω + 1) + Arg (ω − 1) = θ2 + θ1 , where θ1 and θ2 are shown in Figure 16.3 below.

Figure 16.3: The angle Arg (ω 2 − 1). Clearly, the function Im ζ = Arg (ω 2 − 1) is harmonic in H by Theorem 16.2. In addition, since Arg (ω 2 − 1) = Arg (ω + 1) + Arg (ω − 1), we actually obtain  2π, if ω ∈ (−∞, −1);      π, if ω ∈ (−1, 1); (16.17) Arg (ω 2 − 1) =      0, if ω ∈ (1, ∞).

These facts show that, in H, Im ζ = Arg (ω 2 − 1) is harmonic with the prescribed boundary values (16.17). Hence the composition function u(x, y) =

1 Arg (sin2 z − 1) π

is harmonic in S with the prescribed boundary values  π   2, if z = − 2 + iy with y > 0;    u(x, y) = 1, if z ∈ (−1, 1);      0, if z = π2 + iy with y > 0.

This completes the proof of the problem. Problem 16.11

Bak and Newman Chapter 16 Exercise 11.



Chapter 16. Harmonic Functions

208

Proof. If the number of zeros of f (z) = ez − P (z) was finite, then we deduce from Theorem 16.13 that f has the form ez − P (z) = P1 (z)eP2 (z) (16.18) for some polynomials P1 and P2 . If deg P2 ≥ 2, then the equation (16.18) implies that ez − P (z) = 0. z→∞ eP2 (z)

lim P1 (z) = lim

z→∞

By Problem 1.26, P1 (z) ≡ 0. In this case, the expression (16.18) becomes ez = P (z) which is impossible. If deg P2 = 0, then P2 is a constant and so the expression (16.18) yields that ez = Q(z) for some polynomial Q, a contradiction. Hence we must have deg P2 = 1 and furthermore, P2 (z) = z. Now we rewrite the expression (16.18) as [1 − P1 (z)]ez = P (z).

(16.19)

If P1 (z) 6= 1, then we follow from the expression (16.19) that ez is a rational function, a contradiction. Thus P1 (z) ≡ 1 and so P (z) = 0 by the expression (16.19), but this contradicts to our hypothesis. For the entire function g(z) = sin z − P (z), if sin z − P (z) = P1 (z)eP2 (z) for some polynomials P1 and P2 , then we have sin z − P (z) =0 lim P (z) = z→∞ lim z→∞ 1 eP2 (z) z∈R z∈R which contradicts Problem 1.26. Thus we must have P1 (z) ≡ 0 and then sin z − P (z) = 0 which is another contradiction. Now we complete the analysis of the problem.  Problem 16.12 Bak and Newman Chapter 16 Exercise 12.

Proof. Suppose that f does not have infinitely many zeros. By Theorem 16.13, f is in the form f (z) = Q(z)eP (z) ,

(16.20)

where P and Q are polynomials. Since f is non-vanishing, the representation (16.20) and Theorem 5.12 (The Fundamental Theorem of Algebra) force that Q is a non-zero constant and thus we may assume further that Q ≡ 1, i.e., f (z) = eP (z) .  Let S = k ∈ R lim |f (z)| exp(−|z|k ) = 0 and P (z) = an z n + an−1 z n−1 + · · · + a0 , where z→∞ an 6= 0. Then it is easy to see that lim

z→∞

so that lim

z→∞

|P (z)| =1 |an | · |z|n

|f (z)| = 1. exp(|an | · |z|n )

(16.21)

Assume that n > j. Since j = inf S, the definition of infimum means that n is not a lower bound of S, i.e., there exists a k ∈ S such that n > k. Therefore, we have exp(|z|−n ) < exp(|z|−k ) for large z which gives |f (z)| |f (z)| lim = lim =0 z→∞ exp(|z|n ) z→∞ exp(|z|k )

209 and this means that n ∈ S, but it implies the contrary result n > j > n. In other words, we must have n ≤ j. If n < j, then we can find a k0 ∈ (n, j). Combining this fact and the limit (16.21), it is trivial to see that |f (z)| exp(|an | · |z|n ) |f (z)| = lim × = 0. k n z→∞ exp(|z| 0 ) z→∞ exp(|an | · |z| ) exp(|z|k0 ) lim

Consequently, it yields that k0 ∈ S which implies that k0 ≥ j, a contradiction. Hence it must  been that n = j and this completes the analysis of the problem. Problem 16.13 Bak and Newman Chapter 16 Exercise 13.∗

Proof. Assume that sin z − z 6= c for some c ∈ C and for all z ∈ C. On the one hand, we have sin z − z 6= c + 2π

(16.22)

on C. Otherwise, if sin z0 − z0 − 2π = c for some z0 ∈ C, then we have sin(z0 + 2π) − (z0 + 2π) = sin z0 − z0 − 2π = c, a contradiction. On the other hand, since the function f (z) = sin z − z is clearly an entire function of finite order, the Little Picard Theorem ensures that f assumes other values infinitely many times, but this contradicts the observation (16.22). Hence we obtain the desired result  that sin z − z = c has a solution. This completes the proof of the problem. Problem 16.14 Bak and Newman Chapter 16 Exercise 14.∗

Proof. Since g0 (0) = 0, we only consider the function F (z) =

∞ X fk (z) k=1

gk (k)

and Fn (z) =

n X fk (z) k=1

gk (k)

,

where n ∈ N. Since each fk is entire, every Fn is also entire. Let K be a compact set of C and z ∈ K. Then it can be shown by induction easily that k

|fk (z)| ≤ exp(exp(· · · exp(|Re z|)) {z } |

and

.. k. gk (k) = k | {z } .

(16.23)

k + 1 terms

k exponential factors

Since K is compact, the set {|Re z| | z ∈ K} must be bounded by a positive constant M . Suppose that N is a positive integer such that N ≥ 2 max(e, M ). In this case, N must satisfy M 1 ≤ N 2 so that

N

N

eM ≤ e 2 ≤ N 2 . Consequently, we achieve

f (z) 1 1 eM 1 ≤ N ≤ N ≤ g1 (z) N 2 N2

Chapter 16. Harmonic Functions

210

and because N ≥ 2, so 1 N M f (z) exp( 21 N N ) ee 1 N 2N 1 2 √ ≤ 2. ≤ ≤ NN ≤ N N = N N N N g2 (z) 2 N N N N

By induction, it can be shown that k ≥ N implies

f (z) 1 k ≤ k. gk (k) 2

Now Theorem 1.9 (The Weierstrass M -Test) can be applied to conclude that Fn → F uniformly in K. Finally, since K is taken arbitrary, it follows from Theorem 7.6 that F is entire. This proves the first assertion. For the second assertion, we note that if z = x > 0, then F (x) is a series of positive terms which gives fn (x) |F (z)| = F (x) ≥ gn (n) for every n ∈ N. Given that k ∈ N, since {gn (n)} is a strictly increasing sequence of positive integers, the binomial theorem implies that for all large positive integers N , we have egN+k (N +k) = (1 + α)gN+k (N +k) [gN +k (N + k) − 1]α2 · gN +k (N + k) 2 3 > kgN +k (N + k), 2 ≥

(16.24)

where α = e − 1 > 1. Now, for each fixed j ∈ N, we know from the representation (16.23) that gj (j) ≥ 1, so we may take xj = egj (j) > 1. Combining the inequality (16.24) and the fact fj+2 (xN +k ) = exp(exp(exp · · · exp(xN +k ))) = exp(exp(exp · · · exp[gN +k (N + k)])), | {z } {z } | (j + 2) terms

we obtain

F (xN +k ) =

(j + 3) terms

fj+2 (xN +k ) ≥ exp(exp(exp · · · exp[gN +k (N + k)])) | {z } gj+2 (j + 2) (j + 2) terms

for all large positive integers N . Therefore, the result

log(log(log · · · log[F (xN +k )])) ≥ exp(exp[gN +k (N + k)]) {z } | j terms

> exp

h3

2

i kgN +k (N + k)

= exp[kgN +k (N + k)] · exp > exp[kgN +k (N + k)]

hk

2

i gN +k (N + k)

= xkN +k holds for all large positive integers N . By the definition, F is not of j-fold exponential order for any positive integer j. This proves the second assertion and then we complete the proof of the problem. 

CHAPTER

17

Different Forms of Analytic Functions

Problem 17.1 Bak and Newman Chapter 17 Exercise 1.

Proof. By Definition 17.1, we get PN = =

N  Y

k=2 N h Y k=2

1−

1 k2

(k − 1)(k + 1) i k2

1·3 2·4 3·5 (N − 3)(N − 1) (N − 2)N (N − 1)(N + 1) × × × ··· × × × 2·2 3·3 4·4 (N − 2)(N − 2) (N − 1)(N − 1) N ·N 1 N +1 = × 2 N =

so that PN →

1 2

as N → ∞, completing the proof of the problem.



Problem 17.2 Bak and Newman Chapter 17 Exercise 2.

Proof. By Definition 17.1, we see that PN = =

N h Y

k=2 N h Y

k=2

=

1+

(−1)k i k

k + (−1)k i k

3 2 5 4 N − 1 + (−1)N −1 N + (−1)N . × × × × ··· × × N {z N − 1 } |2 3 4 5 (N − 1) terms

211

(17.1)

Chapter 17. Different Forms of Analytic Functions

212

Obviously, we know that  N −1     N , if N is odd;

N + (−1)N =  N    N + 1 , otherwise. N

Hence the expression (17.1) reduces to  if N is odd;   1, PN =   N + 1 , otherwise. N

Now we conclude that PN → 1 as N → ∞. This completes the proof of the problem.



Problem 17.3 Bak and Newman Chapter 17 Exercise 3.∗

Proof. For every N ∈ N, we consider N X

n=1

N   i i i i  Xh log 1 + + iArg 1 + = log 1 + n n n

=

n=1 N X

n=1

N X 1 i arctan . log 1 + + i n n

(17.2)

n=1

Since ex ≥ 1 + x for x ≥ 0, we obtain log(1 + x) ≤ x and then r  i 1 1 1 1 log 1 + = log 1 + 2 = log 1 + 2 ≤ 2 . n n 2 n 2n

(17.3)

Applying the Comparison Test [27, Theorem 6.6, p. 76] to the inequality (17.3), we conclude that ∞ X i (17.4) log 1 + n n=1

∞ Y i converges. By Proposition 17.2, the product 1 + converges. n n=1

Assume that the product

∞  Y

n=1

1+

i

n

was convergent. Then Proposition 17.2 tells us that

 i converges. Applying this fact with the convergence of the series (17.4) log 1 + n n=1 to the sum (17.2), we see that the series the series

∞ X

∞ X

arctan

n=1

is also convergent. However, since arctan n1 ≥

π 4n

1 n

(17.5)

for all n ≥ 1, the Comparison Test implies ∞  Y i is divergent. 1+ that the series (17.5) is divergent, a contradiction. Hence the product n n=1 This ends the proof of the problem. 

213 Problem 17.4 Bak and Newman Chapter 17 Exercise 4.

Proof. Using similar argument as in the proof of Proposition 17.3, we take N ∈ N such that k > N implies |zk | < 21 . Then, for every k > N , we have log(1 + zk ) − zk = − so that

 1 z  zk2 zk3 k + − · · · = zk2 − + − ··· 2 3 2 3

(17.6)

1 z k | log(1 + zk ) − zk | = |zk | · − + − ··· 2 3  1 |z | |z |2  k k ≤ |zk |2 · + + + ··· 2 3 4 1 1 1  ≤ |zk |2 · + + + ··· 2 4 8 = |zk |2 . 2

Hence the series Since

∞ X

∞ X

[log(1 + zk ) − zk ] and then the series

k=N +1

∞ X [log(1 + zk ) − zk ] are convergent. k=1

zk converges, we conclude that the series

k=1

∞ X

log(1 + zk )

k=1

is convergent. By Proposition 17.2, the product

∞ Y

(1 + zk ) converges which completes the proof

k=1

of the problem.



Problem 17.5 Bak and Newman Chapter 17 Exercise 5.

Proof. Let zk =

k (−1) √ , k

where k = 2, 3, . . .. Assume that the product

By Proposition 17.2, the series

∞ Y

(1 + zk ) was convergent.

k=2 ∞ X

log(1 + zk )

k=2

is convergent. Using the formula (17.6), if k ≥ 4, then we havea 1 log(1 + zk ) − zk = k 1 ≤ k a

Notice that all zk and 1 + zk are real numbers.

h

i 1 (−1)k 1 √ · − + + ··· − 2 4k 3 k  1 1  · − + √ 2 3 k

(17.7)

Chapter 17. Different Forms of Analytic Functions

214

1  1 1 · − + k 2 6 1 ≤− 3k

≤

or equivalently, zk − log(1 + zk ) ≥ Thus the convergence of

∞ X

1 . 3k

(17.8)

zk and the series (17.7) imply that the series

k=2

∞ X [zk − log(1 + zk )] k=4

is convergent, so it follows from the inequality (17.8) and the Comparison Test that the series ∞ X 1 k k=4

is convergent, a contradiction. Consequently, the product

∞ Y

(1 + zk ) must be divergent, com-

k=2

pleting the proof of the problem.



Problem 17.6 Bak and Newman Chapter 17 Exercise 6.

Proof. Now we claim that N−1

PN = (1 + z)(1 + z 2 ) · · · (1 + z 2

N −1

) = 1 + z + z2 + · · · + z2

for every N ∈ N. The case for N = 1 is trivial. Suppose that k−1

Pk = (1 + z)(1 + z 2 ) · · · (1 + z 2

k −1

) = 1 + z + z2 + · · · + z2

(17.9)

for some k ∈ N. If N = k + 1, then the assumption step (17.9) implies that k−1

Pk+1 = [(1 + z)(1 + z 2 ) · · · (1 + z 2 k −1

= (1 + z + z 2 + · · · + z 2

k −1

= 1 + z + z2 + · · · + z2

k

)](1 + z 2 ) k

)(1 + z 2 ) k

k +1

+ z2 + z2

k+1 −1

+ · · · + z2

.

Hence our claim follows from induction. If K is a compact subset of D(0; 1), then there exists a 0 < δ < 0 such that K ⊆ D(0; δ). In ∞ X 1 b D(0; δ), we see that the series z k converges uniformly to 1−z , so it yields that k=0

PN →

∞ X k=0

zk =

1 1−z

uniformly on D(0; δ) and in particularly, on K. Since K is arbitrary, we know from Definition ∞ Y k 1 in |z| < 1. This (1 + z 2 ) converges uniformly on compacta to 1−z 7.5 that the product k=0

completes the proof of the problem. b

See also Problem 2.19 and [4, p. 27].



215 Problem 17.7 Bak and Newman Chapter 17 Exercise 7.

Proof. Since λk → ∞, Theorem 17.7 (The Weierstrass Product Theorem) ensures the existence ∞ X 1 converges, we deduce from the discussion on [4, p. of such an entire function g. Since k2 k=1 245] that the function ∞  Y z 1− 2 g(z) = (17.10) k k=1

satisfies our requirements, completing the analysis of the problem.



Problem 17.8 Bak and Newman Chapter 17 Exercise 8.

Proof. Recall from Problem 3.22 that sin z = z − so we obtain

∞

X (−1)k z3 z5 + − ··· = z 2k+1 , 3! 5! (2k + 1)! k=0

√ ∞ sin(π z) X (−1)k π 2k k √ z . = (2k + 1)! π z

(17.11)

k=0

If ak =

(−1)k π 2k (2k+1)! ,

then ak+1 −π 2 = lim = 0. k→∞ ak k→∞ (2k + 3)(2k + 2) lim

By Problem √ 2.13 and Theorem 2.8, the power series (17.11) converges everywhere, i.e., the sin(π √ function π z z) is in fact entire. Since the zeros of sin z are kπ for all k ∈ Z, the zeros of √ sin(π z) are k2 for all k ∈ N. In other words, one solution to Problem 17.7 is given by the entire function (17.11). This ends the proof of the problem.  Problem 17.9 Bak and Newman Chapter 17 Exercise 9.

Proof. By the inspiration of [4, Example 3, p. 247], an entire function with a single zero at every k + 12 with k ∈ Z is given by ∞ h  z io n Y  z io z  z  exp × exp 1 − k + 12 k + 12 −k − 12 −k − 12 k=0 k=0 ∞ h  z   Y z  z i z  exp 1 + exp − = 1− k + 12 k + 12 k + 21 k + 21 k=0

f (z) =

∞ h nY

1−

Chapter 17. Different Forms of Analytic Functions

=

∞ h Y

k=0

1−

216

4z 2 i . (2k + 1)2

Note that cos πz is an entire function having simple zeros at every k + 12 , where k ∈ Z. Therefore, we have ∞ h Y 4z 2 i (17.12) cos πz = C 1− (2k + 1)2 k=0

for some nonzero constant C. Put z = 0 into the expression (17.12), we get immediately that C = 1. Hence we have completed the proof of the problem.  Problem 17.10 Bak and Newman Chapter 17 Exercise 10.

Proof. (a) Applying similar idea of [4, Example 1, p. 246], an entire function g with zeros at every positive integer is given by ∞  Y z z g(z) = 1− ek . k k=1

Next, we define

f (z) = g

∞   1  Y  1 1 1− = e k(1−z) 1−z k(1 − z) k=1

which is clearly analytic in |z| < 1 and f (z) = 0 if and only if z = 1 − k1 , where k ∈ N. (b) Suppose that {zk } is a sequence of distinct numbers such that zk → z0 as k → ∞. Then the 1 } satisfies λk → ∞ as k → ∞. By Theorem 17.7 (The Weierstrass sequence {λk = zk −z 0 Product Theorem), one can find an entire function g such that g(z) = 0 if and only if z = λk for every k ∈ N. Now if we define  1  f (z) = g , z − z0 then f is analytic in |z| < |z0 | and f (z) = 0 if and only if z = z1 , z2 , . . ..

We complete the proof of the problem.



Problem 17.11 Bak and Newman Chapter 17 Exercise 11.

Proof. Since f (z) = ϕ(z, t) satisties the hypotheses of Theorem 17.9, the function F (z) is analytic in a region D. Thus we get Z Z Z b  f (ζ) F (ζ) 1 1 ′ dζ = dt dζ, (17.13) F (z) = 2 2πi C (ζ − z)2 2πi C a (ζ − z)

where C is a disc contained in D with centre z. As suggested by the question, we switch the order of integration in the formula (17.13) to obtain Z Z b  1 f (ζ) ′ (17.14) dζ dt. F (z) = 2πi C (ζ − z)2 a

217 Using the formula at the bottom of [4, p. 80], we see that Z 1 f (ζ) ′ ϕz (z, t) = f (z) = dζ. 2πi C (ζ − z)2 After the substitution of this formula into the expression (17.14), we reduce it to Z b ′ ϕz (z, t) dt F (z) = a

which is our desired formula and we end the proof of the problem.



Problem 17.12 Bak and Newman Chapter 17 Exercise 12.

Proof. Given that ǫ > 0 which will be determined later. We write Z x−ǫ Z x+ǫ Z β h(u)y h(u)y h(u)y du = du + du 2 2 2 2 2 2 (u − x) + y α x−ǫ (u − x) + y α (u − x) + y Z β h(u)y + du, 2 2 x+ǫ (u − x) + y

(17.15)

where h is a continuous function on [α, β] and x ∈ (α, β). Since [α, β] is compact, h is bounded by a positive constant M . Furthermore, on [α, x−ǫ], it is easy to see that (u−x)2 +y 2 ≥ ǫ2 +y 2 ≥ ǫ2 . Therefore, we get Z x−ǫ Z x−ǫ Z x−ǫ M |y| M |y|(β − α) h(u)y h(u)y du ≤ du ≤ . (17.16) du ≤ 2 2 2 2 2 (u − x) + y (u − x) + y ǫ ǫ2 α α α Similarly, it is true that

Z

β

x+ǫ

Finally, since the function

M |y|(β − α) h(u)y du . ≤ (u − x)2 + y 2 ǫ2 g(u) =

(17.17)

y (u − x)2 + y 2

is integrable and does not change sign on [x − ǫ, x + ǫ], we observe from the Weighted Mean Value Theorem for Integrals [5, Exercise 17, p. 215] that Z x+ǫ Z x+ǫ y du h(u)y du = h(ξ) (17.18) 2 2 2 2 x−ǫ (u − x) + y x−ǫ (u − x) + y for some ξ ∈ [x − ǫ, x + ǫ]. We apply the substitution u = x + y tan θ to the integral on the right-hand side of the expression (17.18), we have Z

x+ǫ x−ǫ

Finally, we take ǫ = (17.15) to get

h(u)y du = h(ξ) (u − x)2 + y 2 √ 4

Z

tan−1 (ǫ/y) tan−1 (−ǫ/y)

ǫ dθ = 2h(ξ) tan−1 . y

(17.19)

y and we put the results (17.16), (17.17) and (17.19) into the expression

lim

Z

β

y→0 α

1 h(u)y du = lim 2h(ξ) tan−1 3 . 2 2 y→0 (u − x) + y y4

Chapter 17. Different Forms of Analytic Functions 3

Since tan−1 (y − 4 ) →

π 2

218

and ξ → x as y → 0, we conclude that lim

Z

β

y→0 α

h(u)y du = πh(x), (u − x)2 + y 2

completing the proof of the problem.



Problem 17.13 Bak and Newman Chapter 17 Exercise 13.

Proof. According to direct integration, it is clear that Z 1 dt log(1 − z) f (z) = =− . z 0 1 − zt

(17.20)

Recall from [4, p. 115] that log z is analytic in C \ (−∞, 0], so log(1 − z) and then f is analytic in D = C \ [1, ∞). Suppose that γ is a simple closed curve encircling the point z = 1. Since the function F (z) = 1 − z is analytic inside and on γ and F (z) 6= 0 on γ, Corollary 10.9 (The Argument Principle) implies that Z F ′ (z) 1 1 ∆Arg (1 − z) = dz = 1 2π 2πi γ F (z) which means that ∆Arg (1 − z) = 2πi as z traverses along γ. Hence it follows from the representation (17.20) that f has a “jump” of 2πi x as z crosses from the upper half-plane to the lower half-plane through any point x > 1. This completes the analysis of the problem.  Problem 17.14 Bak and Newman Chapter 17 Exercise 14.∗

Proof. (a) By the definition, |φ(n)| = φ(n) ≤ n. Since |nz | = nRe z , we obtain ∞ ∞ ∞ ∞ X X 1 φ(n) X φ(n) X n ≤ = . z = z Re z Re n |n | n n z−1 n=1 n=1 n=1 n=1

By [27, Theorems 6.6, 6.10, pp. 76, 77], the series ∞ X φ(n) n=1

nz

converges absolutely for Re z − 1 > 1 or equivalently, Re z > 2. (b) By [4, Example, pp. 254, 255], we have ζ(z)

∞ X φ(n)

n=1

nz

where cn =

=

∞ ∞ ∞ X 1 X φ(n) X cn · = , nz nz nz

n=1

n=1

n=1

X n X = φ(n) = n. φ d d|n

d|n

219 Consequently, we note that ζ(z)

∞ X φ(n)

n=1

nz

=

∞ X

n=1

1 = ζ(z − 1) nz−1

for Re z > 2. Hence we end the proof of the problem.



Chapter 17. Different Forms of Analytic Functions

220

CHAPTER

18

Analytic Continuation; The Gamma and Zeta Functions

Problem 18.1 Bak and Newman Chapter 18 Exercise 1.

Proof. Let D1 = C \ {(x, 0) | x ≤ 0} and D2 = C \ {(0, y) | y ≤ 0}. Define g1 : D1 → C by g1 (z) = log |z| + iArg z,

(18.1)

where Arg z ∈ (−π, π). Similarly, we define g2 : D2 → C by g2 (z) = log |z| + iArg z,

(18.2)

where Arg z ∈ (− π2 , 3π 2 ). Obviously, we have eg1 (z) = eg2 (z) = z, so g1 and g2 are analytic branches of log z by Definition 8.7. If we let D be the first quadrant and define f : D → C by f (z) = log |z| + iArg z, where Arg z ∈ (0, π2 ), then each of g1 and g2 is an analytic continuation of f . Suppose that ζ is a point lying in the third quadrant. On the one hand, we know from the definition (18.1) that  π Im g1 (ζ) = Arg ζ ∈ − π, − . 2 On the other hand, the definition (18.2) gives

 3π  . Im g2 (ζ) = Arg ζ ∈ π, 2

Therefore, we establish that g1 (ζ) 6= g2 (ζ), completing the proof of the problem. Problem 18.2 Bak and Newman Chapter 18 Exercise 2.∗

221



Chapter 18. Analytic Continuation; The Gamma and Zeta Functions

222

Proof. (a) Assume that there was a point |z0 | = 1 such that the series

∞ X

an z0n converges. Without

n=0

loss of generality, we may assume that z0 = 1. Otherwise, if z0 = eiθ0 for some θ0 ∈ [0, 2π], then we have ∞ ∞ X X (an einθ0 ) · 1n . an z0n = n=0

n=0

Thus it must be true that an → 0 as n → ∞. Given ǫ > 0. Then there exists an N ∈ N such that |an | < ǫ for all n ≥ N . Therefore, for every z ∈ D(0; 1), we see that |f (z)| ≤

∞ X

n=0

|an | · |z|n =

N −1 X n=0

|an | · |z|n +

∞ X

n=N

|an | · |z|n ≤

N −1 X n=0

|an | +

ǫ 1 − |z|

which gives, for 0 < x < 1, lim (1 − x)f (x) ≤ ǫ.

x→1− x∈R

Since ǫ is arbitrary, we actually have (1 − x)f (x) → 0 as x → 1− and x ∈ R. However, this contradicts the hypothesis that f has a pole at z = 1. Hence the power series diverges at every point on C(0; 1). (b) Suppose that the radius of convergence of f (z) =

∞ X

an z n

(18.3)

n=0

is R > 0 and f has a pole at z0 = Reiθ0 . Then the function g(z) = f (Rz) =

∞ X

(an Rn )z n

(18.4)

n=0

has radius of convergence 1 and g has a pole at z = eiθ0 . By the argument in part (a), we may assume that the pole of g is at z = 1. Hence part (a) implies that the power series (18.4) diverges at every point on C(0; 1) which is equivalent to saying that the power series (18.3) diverges at all points on C(0; R).  This completes the proof of the problem. Problem 18.3 Bak and Newman Chapter 18 Exercise 3.

Proof. Define f (z) =

∞ X

(−1)n an z n and its radius of convergence to be R < ∞. Then we have

n=0

f (−z) =

∞ X

an z n

n=0

with R as its radius of convergence. Now the function f (−z) satisfies the hypotheses of Theorem 18.3, so it has a singularity at z = R. Hence we conclude that f (z) has a singularity at z = −R which ends the proof of the problem. 

223 Problem 18.4 Bak and Newman Chapter 18 Exercise 4.

Proof. (a) By the application of Problem 18.5 in advance, we have Z ∞ 2 1 1 √ e−nt t− 3 dt. = 3 1 n Γ( 3 ) 0 If |z| < 1, then we derive

Z ∞ ∞ h ∞ i X X 1 zn n −nt − 23 √ = dt z e t 3 n Γ( 31 ) 0 n=1 n=1 Z ∞hX ∞ i 1 −t n − 32 = dt (ze ) t Γ( 13 ) 0 n=1 Z ∞ −2 t 3 · ze−t 1 = dt Γ( 13 ) 0 1 − ze−t Z ∞ −2 t 3 z dt = Γ( 13 ) 0 et − z Z N −2 t 3 z lim = dt. 1 t Γ( 3 ) ǫ→0 ǫ e − z

(18.5) (18.6)

N →∞

− 32

Let f (t) = et and g(t) = t . Clearly, f and g are continuous real-valued functions on [ǫ, N ]. Furthermore, f ′ (t) = et > 0 is also continuous on [ǫ, N ]. By Proposition 17.10, the integral (18.6) is analytic in C \ [eǫ , eN ]. Since ǫ → 0 and N → ∞, the integral (18.5) is actually analytic in C \ [1, ∞) as required. (b) Now we apply the second well-known formula on [4, p. 262], we have Z ∞ 1 e−nt sin t dt = n2 + 1 0 which show that if |z| < 1, then

∞ hZ ∞ i X zn −t n (ze ) sin t dt = n2 + 1 n=0 0 n=0 Z ∞hX ∞ i (ze−t )n sin t dt = ∞ X

0

=

Z

∞

n=0 t e sin t

dt et − z Z N t e sin t dt. = lim ǫ→0 et − z ǫ

(18.7)

0

(18.8)

N →∞

et

et sin t,

Again, if we let f (t) = and g(t) = then both are continuous real-valued functions ′ on [ǫ, N ] and f > 0 is also continuous on [ǫ, N ]. Hence we derive from Proposition 17.10 that the integral (18.8) is analytic outside [eǫ , eN ]. Finally, by letting ǫ → 0 and N → ∞, the integral (18.7) is analytic outside [1, ∞).  We have ended the proof of the problem.

Chapter 18. Analytic Continuation; The Gamma and Zeta Functions

224

Problem 18.5 Bak and Newman Chapter 18 Exercise 5.

Proof. Suppose that x = nt. Thus it follows from the definition [4, Eqn. (1), p. 265] that Z ∞ Z ∞ Z ∞ p−1 Γ(p) 1 dx −x x −nt p−1 e · p−1 · e t dt = e−x xp−1 dx = p = p n n n 0 n 0 0 for p > 0. This completes the proof of the problem.



Problem 18.6 Bak and Newman Chapter 18 Exercise 6.

Proof. By the substitution x = t2 , we deduce from the definition [4, Eqn. (1), p. 265] and then √ the fact Γ( 21 ) = π (see [4, p. 268]) that √ Z ∞ Z 1 ∞ −x − 1 1 1 π −t2 2 = e dt = e x dx = Γ 2 2 2 2 0 0 as required, completing the proof of the problem.



Problem 18.7 Bak and Newman Chapter 18 Exercise 7.

Proof. Define Γn (z) =

Z

0

n

 t n tz−1 1 − dt, n

where Re z > 0 and n ∈ N. Then we have Z ∞ Z nh  t n i z−1 −t e − 1− e−t tz−1 dt. t dt + Γ(z) − Γn (z) = n n 0

(18.9)

It is easy to see that the second integral in the expression (18.9) tends to 0 as n → ∞. Let t ≤ n, i.e., nt ≤ 1. By the power series expansion of e−x , we see that t

e− n = 1 −

t2 t3 t + − + ··· n 2!n2 3!n3

so that on the one hand,   t t t2 t t2 t3 t2  1 e− n − 1 − = − + · · · ≤ − + · · · = . n 2!n2 3!n3 n2 2! 3!n 2n2

We derive from the given identity with a = exp(− nt ) and b = 1 −

t n

that

h (n − 1)t i t   t et2 t n t  t2 −n −t · ≤ n exp − . − 1 − − 1 − e ≤ ne−t · e n · 2 ≤ e−t · e n n n 2n 2n

Consequently, it implies that Z n Z n Z nh   t n e t n i z−1 z−1 −t −t |t | · e − 1 − t dt = e − 1− e−t tRe z+1 dt. dt ≤ n n 2n 0 0 0

225 By the definition, we know that Z Z n e−t tRe z+1 dt = lim n→∞ 0

so that

∞

0

Z lim

n→∞

0

nh

e−t tRe z+1 dt = Γ(Re z + 2) < ∞

 t n i z−1 e−t − 1 − t dt = 0 n

and then it yields from the expression (18.9) immediately that Γ(z) = lim Γn (z), n→∞

ending the proof of the problem.



Problem 18.8 Bak and Newman Chapter 18 Exercise 8.∗

Proof. Suppose that z = x > 0. If we take the logarithm on both sides of the product formula Γ

−1

γx

(x) = xe

∞  Y x −x e k, 1+ k

k=1

then we arrive at − log Γ(x) = log x + γx + and

thusa −

∞ h X k=1

 x xi − log 1 + k k

∞  1 d Xh x xi Γ′ (x) − . = +γ+ log 1 + Γ(x) x dx k k

(18.10)

k=1

Now termwise differentiation is permitted for the series (18.10) provided that the differentiated series is uniformly convergent. To see this, we consider the series ∞ ∞ ∞  X X dh x x xi X  1 1 log 1 + − = =− − . dx k k x+k k k(x + k) k=1

k=1

k=1

Let x ∈ [ 12 , 1]. Then we have k(x + k) ≥ k2 which gives ∞ X k=1

∞

X 1 x ≤ . k(x + k) k2 k=1

By the Comparison Test, the differentiated series is uniformly convergent on [ 12 , 1]. Hence it follows from [28, Theorem 10.8, p. 4] that we can take term by term differentiation to get −

∞

X x 1 Γ′ (x) = +γ− . Γ(x) x k(x + k)

(18.11)

k=1

Put x = 1 into the expression (18.11) and use the fact Γ(1) = 1, we conclude that −Γ′ (1) = 1 + γ − 1 = γ which means that Γ′ (1) = −γ, completing the proof of the problem. a

The logarithm derivative ψ(z) = (log Γ(z))′ is called the digamma function.



Chapter 18. Analytic Continuation; The Gamma and Zeta Functions

226

Problem 18.9 Bak and Newman Chapter 18 Exercise 9.

Proof. Recall from the factb  1 1 1 1 − z ζ(z) = 1 + z + z + · · · 2 3 5

that

 f (z) = 1 −  = 1−  = 1+

2 ζ(z) 2z 1 1 ζ(z) − z ζ(z) z 2 2  1 1 1 1 1 + + · · · − z − z − z − ··· z z 3 5 2 4 6 1 1 1 = 1 − z + z − z + ··· . 2 3 4 By Theorem 18.9, f is analytic in C \ {1} and furthermore   2 2 lim (z − 1)f (z) = lim 1 − z (z − 1)ζ(z) = lim 1 − z lim (z − 1)ζ(z) = 0 · 1 = 0, z→1 z→1 z→1 2 2 z→1 so f has a removable singularity at z = 1. Therefore, f is entire which completes the proof of  the problem. Problem 18.10 Bak and Newman Chapter 18 Exercise 10.

Proof. Let {pk } be the set of all primes. Since ζ has a singularity at z = 1, we have ζ(1+ǫ) → ∞ as ǫ → 0+ so that the identity [4, Eqn. (5), p. 269] ensures that ∞  Y 1 (18.12) 1− pk k=1

diverges to 0. Clearly, we observe that

∞ ∞ X X 1 1 ≤ < ∞. k2 p2k k=1

∞ X

k=1

1 was convergent. Then Problem 17.4 guarantees that the product (18.12) pk k=1 is convergent which is a contradiction. Hence the series

Assume that

−

∞ X k=1

−

1 pk

must diverge, completing the proof of the problem. Remark 18.1 For another proof of Problem 18.10, please refer to [26, Problem 8.10, p. 180]. b

Read [4, p. 268].



CHAPTER

19

Applications to Other Areas of Mathematics

Problem 19.1 Bak and Newman Chapter 19 Exercise 1.

Proof. Let f (z) = z − tan z and N ∈ N. Consider the square SN with vertices N π(±1 ± i). Then f is meromorphic in SN , so Theorem 10.8 implies that Z Z 1 tan2 z 1 sec2 z − 1 dz = dz = Zf − Pf . (19.1) 2πi ∂SN tan z − z 2πi ∂SN tan z − z On the one hand, if we plot the graphs of y = x and y = tan x on [−N π, N π], then we see that the straight line y = x has one and only one intersection with the curve y = tan x in each of the following intervals [−(k + 1/2)π, −kπ] and [kπ, (k + 1/2)π], where k = 1, 2, . . . , N − 1, see Figure 19.1 for an example.

Figure 19.1: The intersections of y = x and y = tan x. 227

Chapter 19. Applications to Other Areas of Mathematics

228

Furthermore, the power series of tan z (see [1, p. 184]) indicates that f (0) = f ′ (0) = f ′′ (0) = 0 but f (3) (0) 6= 0, so f has a zero of order 3 at z = 0. In conclusion, if we denote Z′f to be the number of nonreal zeros of f , then we establish that Zf = Z′f + 2(N − 1) + 3 = Z′f + 2N + 1

(19.2)

sin z inside SN . On the other hand, since tan z = cos z , we have Pf = Zcos z inside SN . It is clear that 1 cos(n + 2 )π = 0 for all −N ≤ n ≤ N − 1, we know that

Pf = 2N

(19.3)

inside SN . Putting the values (19.2) and (19.3) back into the expression (19.1), we get Z 1 tan2 z dz = 1 + Z′f . 2πi ∂SN tan z − z

(19.4)

Now it remains to estimate the integral in the formula (19.4). On the vertical side Re z = N π and y = Im z ∈ [−N π, N π], we have | tan z| =

|e−y − ey | |eiz − e−iz | = < 1. |eiz + e−iz | |e−y + ey |

(19.5)

Similarly, the bound (19.5) holds on the other vertical side. Next, if we take N large enough, then along the horizontal side x = Re z ∈ [−N π, N π] and Im z = N π, we obtain | tan z| =

|e−N π eix − eN π e−ix | |e2N π − e2ix | 1 + e−2N π 10 = ≤ ≤ . −N π ix N π −ix 2N π 2ix −2N π |e e +e e | |e +e | 1−e 9

Consequently, we always have | tan z| ≤ 10 9 on ∂SN for large enough N . By the triangle inequal10 ity, | tan z − z| ≥ |z| − | tan z| ≥ N − 9 on ∂SN so that tan2 z 100 ≤ tan z − z 81N − 90

on ∂SN . Since the length of ∂SN is 8N , it follows from Theorem 4.10 (The M -L Formula) that 1 Z tan2 z 400N dz ≤ 0.

so that

on [−N π, N π]. Let F (x) =

ex +1 ex −x .

ez − 1 ex + 1 z ≤ x e −z e −x

Then we have F ′ (x) =

1 − xex . (ex − x)2

(19.12)

231 Thus F ′ (x0 ) = 0 if and only if x0 ex0 = 1. Clearly, x0 ∈ (0, 1), so we can establish from the First Derivative Test that F attains its absolute maximum at x = x0 and the maximum value is given by ex0 + 1 1 x0 (ex0 + 1) F (x0 ) = x0 = > 0. (19.13) = x e − x0 x0 (e 0 − x0 ) 1 − x0 Combining the inequality (19.12) and the result (19.13), we obtain the bound ez − 1 1 21 , we have px > inequality (19.23) reduces to

√

2 which implies that

px 2(px −1)

< 2. Therefore, the

X 1 2 2 nz < 2x ≤ √ < ∞. np p p

(19.24)

n≥2

As a result, we have

X 1 X 2 . nz < np p2x

p prime

(19.25)

p prime

n≥2

Suppose that K is a compact subset of Re z > 12 and F = {z ∈ C | Re z ≤ 12 }. Now F is closed in C and F ∩ K = ∅. By [22, Exercise 21, p. 101], there exists a δ > 0 such that d(K, F ) > 2δ. Furthermore, we have K ⊆ {z ∈ C | Re z ≥ 12 + δ}. Given a prime p ≥ 2, for every n ∈ N, consider the function 1 fn,p (z) = nz np which is clearly analytic in Re z > 12 . Now it follows from Theorem 1.9 (The Weierstrass M ∞ X fn,p (z) converges uniformly to a function Test) and the inequality (19.24) that the sequence n=2

fp (z) in Re z > 12 . Hence we conclude from Theorem 7.6 that fp (z) is analytic in Re z > 21 . Next, if Re z ≥ X

p prime

1 2

+ δ, then we get from the inequality (19.25) that

|fp (z)| ≤

∞ X 1 X X 2 X 1 2 < ≤ < 2 12 , as desired. This completes the proof of the problem.

Index

B Bolzano-Weierstrass Theorem, 71, 85, 177

H homotopy of path, 101

C Chain Rule, 38

I inversion through the unit circle, 22

D digamma function, 225 Dirichlet’s Test, 28

J Joukowsky transform, 189

E Enestr¨om and Kakeya Theorem, 16 Euler’s constant, 166

L Leibniz’s Rule, 93 lemniscate of Bernoulli, 10

F Fresnel integral, 149 Fubini Theorem, 93

T Topologist’s sine curve, 18, 103

237

Index

238

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