Essay Complex Analysis

HO CHI MINH CITY UNIVERSITY OF SCIENCE

FACULTY OF MATHEMATICS AND COMPUTER SCIENCE

________________________________________________________

________________________________________________________

LECTURER: PROF. DANG DUC TRONG

VO A. KHOA–NGUYEN T. HOAI–DANG P. NHAT

TRAN T. HUNG–LE T. NHAN

FUNCTIONS OF

ONE COMPLEX VARIABLE

FUNCTIONS OF ONE COMPLEX VARIABLEVO ANH KHOA - NGUYEN THANH HOAI - DANG PHUOC NHAT - TRAN

THE HUNG - LE THANH NHAN

February 15, 2012

Vo Anh Khoa - Nguyen Thanh Hoai - Dang Phuoc Nhat

Tran The Hung - Le Thanh Nhan

Vietnam National University

Ho Chi Minh City (HCMC) University of Science

Faculty of Mathematics and Computer Science

227 Nguyen Van Cu Street, District 5, Ho Chi Minh City

Vietnam

2

PREFACE

This book is intended as our English essay for students who study in university. In specific,the essay consists of some exercises which are chosen and solved by us; and these exercises istaken from A. David Wunsch, Complex Variables with Applications, 3rd ed. To read this essay,the readers need some knowledges of Functions of One Complex Variable (such as second - yearstudents or higher), but if the reader is a student in high school, still he or she can reads someexercises which is shown in chapter I. Then, we submit an essay to our tutor who is Prof. DangDuc Trong, that is also our lecturer.

Remark. We will present the following chapters :

1. Chapter 1 : Complex Number.

2. Chapter 2 : The Complex Function and Its Derivative.

3. Chapter 3 : The Basic Transcendental Functions.

4. Chapter 4 : Integration in the Complex Plane.

5. Chapter 5 : Infinite Series Involving a Complex Variable.

6. Chapter 6 : Residues and Their Use in Integration.

Acknowledgement. We thank the collaborators for all their helps. These include :

Name University

Mai Thanh Nhat Truong HCMC University of ScienceTran Hong Tai HCMC University of Science

Vu Tran Minh Khuong HCMC University of ScienceNgo Thanh Ha HCMC University of Science

The reader can find and download our “Functions of One Complex Variable” at :

http://anhkhoavo1210.wordpress.com/

And if you have any questions of feedback, please feel free to contact us at :

[email protected]

Note. Although this essay is free on sharing, we do not accept an arbitrary casethat relates to the copyright of it.

3

Bibliography

[1] A. David Wunsch (University of Massachusetts Lowell), Complex Variables with Applica-tions, 3rd ed, 2005.

[2] Dau The Cap (Ho Chi Minh City University of Pedagogy), Ham Bien Phuc & Phep TinhToan Tu, 2006.

[3] Terence Tao (University of California Los Angeles), Lecture Note of Complex Analysis.

[4] Dang Duc Trong - Dinh Ngoc Thanh - Pham Hoang Quan (Ho Chi Minh City Universityof Science), Giao trinh Giai Tich 2, 2008.

4

Contents

PREFACE 3

1 COMPLEX NUMBERS 71.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 MORE PROPERTIES OF COMPLEX NUMBERS . . . . . . . . . . . . . . . . . 131.3 COMPLEX NUMBERS AND THE ARGAND PLANE . . . . . . . . . . . . . . 181.4 INTEGER AND FRACTIONAL POWERS OF A COMPLEX NUMBER . . . . 271.5 POINTS, SETS, LOCI, AND REGIONS IN THE COMPLEX PLANE . . . . . . 39

2 THE COMPLEX FUNCTION AND ITS DERIVATIVE 412.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.2 LIMITS AND CONTINUITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.3 THE COMPLEX DERIVATIVE . . . . . . . . . . . . . . . . . . . . . . . . . . . 572.4 THE DERIVATIVE AND ANALYTICITY . . . . . . . . . . . . . . . . . . . . . 602.5 HARMONIC FUNCTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3 THE BASIC TRANSCENDENTAL FUNCTIONS 743.1 THE EXPONENTIAL FUNCTION . . . . . . . . . . . . . . . . . . . . . . . . . 743.2 TRIGONOMETRIC FUNCTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . 813.3 HYPERBOLIC FUNCTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 883.4 THE LOGARITHMIC FUNCTION . . . . . . . . . . . . . . . . . . . . . . . . . 913.5 ANALYTICITY OF THE LOGARITHMIC FUNCTION . . . . . . . . . . . . . 963.6 COMPLEX EXPONENTIALS . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1003.7 INVERSE TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS . . . . . . . 105

4 INTEGRATION IN THE COMPLEX PLANE 1094.1 INTRODUCTION TO LINE INTEGRATION . . . . . . . . . . . . . . . . . . . 1094.2 COMPLEX LINE INTEGRATION . . . . . . . . . . . . . . . . . . . . . . . . . . 1114.3 CONTOUR INTEGRATION AND GREEN’S THEOREM . . . . . . . . . . . . 1164.4 PATH INDEPENDENCE, INDEFINITE INTEGRALS, FUNDAMENTAL THE-

OREM OF CALCULUS IN THE COMPLEX PLANE . . . . . . . . . . . . . . . 1204.5 THE CAUCHY INTEGRAL FORMULA AND ITS EXTENSION . . . . . . . . 1244.6 SOME APPLICATIONS OF THE CAUCHY INTEGRAL FORMULA . . . . . . 127

5 INFINITE SERIES INVOLVING A COMPLEX VARIABLE 1315.1 INTRODUCTION AND REVIEW OF REAL SERIES . . . . . . . . . . . . . . . 1315.2 COMPLEX SEQUENCES AND CONVERGENCE OF COMPLEX SERIES . . 1355.3 UNIFORM CONVERGENCE OF SERIES . . . . . . . . . . . . . . . . . . . . . 140

5

Contents

5.4 POWER SERIES AND TAYLOR SERIES . . . . . . . . . . . . . . . . . . . . . 1455.5 TECHNIQUES FOR OBTAINING TAYLOR SERIES EXPANSIONS . . . . . . 1555.6 LAURENT SERIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

6 RESIDUES AND THEIR USE IN INTEGRATION 1696.1 INTRODUCTION AND DEFINITION OF THE RESIDUES . . . . . . . . . . . 1696.2 ISOLATED SINGULARITIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1736.3 FINDING THE RESIDUE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1776.4 EVALUATION OF REAL INTEGRALS WITH RESIDUE CALCULUS I . . . . 1806.5 EVALUATION OF INTEGRALS II . . . . . . . . . . . . . . . . . . . . . . . . . 1836.6 EVALUATION OF INTEGRALS III . . . . . . . . . . . . . . . . . . . . . . . . . 186

6

1 COMPLEX NUMBERS

1.1 INTRODUCTION

In this section, we show the exercise 10, 11, 20, 21, 22, 23, 29, 30, 32; and we note that E,P mean the exercise, the page which are numbered respectively. We aslo call that <,= arerespectively the symbols of real part and imaginary part of the complex number.

Exercise 1.1. (E. 10 P. 7)

An infinite decimal such as e = 2.718281... is an irrational number since there is no repetitivepattern in the successive digits. However, an infinite decimal such a 23.232323... is a rationalnumber. Because the digits do repeat in a cyclical manner, we can write this number as theration of two integers, as the following steps will show.First we rewrite the number as 23 (1.010101...) = 23

(1 + 10−2 + 10−4 + 10−6 + ...

).

1. Recall from your knowledge of infinite geometric series that1

1− r= 1 + r + r2 + ..., where

r is a real number such that −1 < r < 1.

Sum the series[1 + 10−2 + 10−4 + 10−6 + ...

].

2. Use the result of part (a) to show that 23.232323... equals2300

99. Verify this wit a divison

on a pocket calculator.

3. Using the same technique, express 376.376376... as a ration of integers.

Solution.

1. For the infinite geometric series, we can choose r = 10−2. Hence, we evaluate sum of thefollowing series.

1 + 10−2 + 10−4 + 10−6 + ... =1

1− 10−2

=100

99

2. In a similar fashion, we have

23.232323... = 23 (1.010101...)

= 23(1 + 10−2 + 10−4 + 10−6 + ...

)=

2300

99

7

1 COMPLEX NUMBERS

3. By choosing r = 10−3, we have

[1 + 10−3 + 10−6 + ...

]=

1

1− 10−3

=1000

999

Thus, we apply the above expression for expressing 376.376376...

376.376376... = 376 (1.001001...)

= 376(1 + 10−3 + 10−6 + ...

)= 376.

1000

999

=376000

999

Exercise 1.2. (E. 11 P. 7)

Recall from your knowledge of infinite geometric series that1

1− r= 1 + r + r2 + ..., where r is

a real number such that −1 < r < 1.

1. Using the method of this, express 3.04040404 . . . as the ratio of integers.

2. Using the above method show that .9999... is identical to 1. If you have any doubt aboutthis, try to find a number between .9999... and 1.

Solution.

1. We observe that

3.04040404... = 3 (1.01010101...) + 0.01010101...

= 3(1 + 10−2 + 10−4 + 10−6 + ...

)+ 10−2 + 10−4 + 10−6 + ...

We choose r = 10−2 for infinite geometric series, so the above equality is equivalent to

3.1

1− 10−2+

1

1− 10−2− 1 =

301

99

2. Using a similar fashion, we consider

8

1 COMPLEX NUMBERS

0.9999... = 9 (0.1111...)

= 9(10−1 + 10−2 + 10−3 + 10−4 + ...

)= 9

(1

1− 10−1− 1

)= 1

Completing the proof by using r = 10−1.

Exercise 1.3. (E. 20 P. 8)In the following exercise, perform the operations and express the result in the form a+ ib, wherea and b are real.

(x+ iy) (u− iv) (x− iy) (u+ iv)

where x, y, u, v are real.

Solution.We use the commutative rule to simplify this expression.

(x+ iy)(u− iv)(x− iy)(u+ iv) = (x+ iy)(x− iy)(u− iv)(u+ iv)

=(x2 − ixy + iyx− i2y2

) (u2 − iuv + ivu− i2v2

)= (x2 + y2)(u2 + v2)

Exercise 1.4. (E. 21 P. 8)Review the following binomial theorem.

(a+ b)n =

n∑k=0

an−kbkn!

(n− k)!k!

where n ≥ 0, a and b are complex numbers.

1. Use this theorem to find a sum to represent (1 + iy)n, where n is a positive integer.

2. Use your above results to find the real and imaginary parts of (1 + i2)5.

Solution.

1. For n is a positive integer, we can extend the expression (1 + iy)n.

(1 + iy)n =

n∑k=0

(iy)kn!

k! (n− k)!

2. By applying the above results, we have

9

1 COMPLEX NUMBERS

(1 + 2i)5 =n∑k=0

(2i)k5!

k!(5− k)!

= 5!

(1 +

2i

4!+

4i2

2!3!+

8i3

3!2!+

16i4

4!+

32i5

5!

)= 1 + 10i− 40− 80i+ 80 + 32i

= 41− 38i

So that, real part is 41 and imaginary part is −38.

Exercise 1.5. (E. 22, E. 23 P. 9)Let z1 = x1 + iy1 and z2 = x2 + iy2 be complex numbers, where the subscripted x and y arereal. Show that

1. < (z1z2) = < (z1)< (z2)−= (z1)= (z2)

2. = (z1z2) = < (z1)= (z2) + = (z1)< (z2)

Solution.

1. Let z1 = x1 + iy1 and z2 = x2 + iy2 be complex numbers with < (z1) = x1,< (z2) =

x2,= (z1) = y1 and = (z2) = y2 respectively. We have

z1z2 = x1x2 − y1y2 + i(x1y2 + x2y1)

Hence,

<(z1z2) = x1x2 − y1y2

= <(z1)<(z2)−=(z1)=(z2)

The exercise is completely shown.

2. Let z1 = x1+iy1 and z2 = x2+iy2 be complex number again. We have < (z1) = x1,= (z1) =

y1,< (z2) = x2,= (z2) = y2. Then we review z1z2 = x1x2 − y1y2 + i(x1y2 + x2y1). We willobtain

= (z1z2) = x1y2 + x2y1

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

, as desired.

10

1 COMPLEX NUMBERS

Exercise 1.6. (E. 29, E. 30, E. 32 P. 9)For the following equations, x and y are real numbers. Solve for x and y. Begin by equatingthe real parts on each side of the equation, and then the imaginary parts, thus obtaining tworeal equations. Obtain all possible solutions.

1. ex2+y2 + i2y = e−2xy + i

2. Log(x+ y) + iy = 1 + ixy

3. cosx+ i sinx = cosh(y − 1) + ixy

Solution.

1. By equating the real part of the left side to the real part on the right, and similarly forimaginary parts, we get a set of equationsex

2+y2 = e−2xy

2y = 1

This is equivalent to (x+

1

2

)2

= 0

y =1

2

So we have (x, y) =

(−1

2;1

2

)as a unique root.

2. According to (1), we obtain a set of equationsLog (x+ y) = 1

y = xy

In the second equation, we can derive x = 1 or y = 0.

By using first equation, we have y = e−1 for the case x = 1, and x = e for the case y = 0.

Hence, we investigate (x, y) = {(1; e− 1) ; (e; 0)} as the needed roots.

3. As a similar method, we have cosx = cosh (y − 1)

sinx = xy

We remember a formula

coshu =eu + e−u

2

By using AM-GM inequality, we obtain coshu ≥ 1 for all u. And we also know thatcos v ≤ 1 for all v. Hence

cosx ≤ 1 ≤ cosh (y − 1)

Thus that set of equations is equivalent to

11

1 COMPLEX NUMBERScosx = 1

cosh (y − 1) = 1

sinx = 1

Then, we find a root which is (0; 1).

12

1 COMPLEX NUMBERS

1.2 MORE PROPERTIES OF COMPLEX NUMBERS

We have the exercise 3, 6, 12, 14, 18, 19, 21, 22, 24 in this section.Exercise 1.7. (E. 3, E. 6 P. 12)

We showed in this section material that z1 + z2 = z1 +z2, where z1 = x1 + iy1and z2 = x2 + iy2.Follow a similar argument, to show the following.

1.(

1

z1

)=

1

z1

2. < (z1z2) = < (z1z2)

Solution.

1. Where z1 = x1 + iy1is a complex number, we consider the following expression.

1

z1=

1

x1 + iy1=x1 − iy1

x21 + y2

1

We derive the left side which is equivalent to

(1

z1

)=

(x1 − iy1

x21 + y2

1

)=

x1 − iy1

x21 + y2

1

=x1 + iy1

x21 + y2

1

=1

z1

Hence, we prove (1) completely.

2. In Exercise 1.5 (E. 22 P. 9), we already have <(z1z2) = x1x2− y1y2. Then, we also derivez1z2 = x1x2 − y1y2 + i (x1y2 + x2y1) by using z1 = x1 − iy1 and z2 = x2 − iy2 wherex1, x2, y1, y2 are real numbers.

At this point, we can understand why the proof is completed as desired.

Exercise 1.8. (E. 12, E. 14 P. 12)Compute the numerical values of the following expressions. Give the answers in the form a+ ib

where a and b are real.

1. 2i+3− 4i

1 + 2i

2.(

4− 4i

2 + 2i

)7

+

(4 + 4i

2− 2i

)7

Solution.

13

1 COMPLEX NUMBERS

1. For evaluating this expression, we can simplify the fraction3− 4i

1 + 2i. We have

3− 4i

1 + 2i=

(3− 4i) (1− 2i)

(1 + 2i) (1− 2i)

=3− 6i− 4i+ 8i2

1− 4i2

=−5− 10i

5

At this point, it is easy to compute this expression. For concretness let us write

2i+3− 4i

1 + 2i= −1

2. For calculating this expression, we can break it into smaller parts. In specific, we get

•(

4− 4i

2 + 2i

)7

=

(0− 16i

8

)7

= −27i

•(

4 + 4i

2− 2i

)7

=

(0 + 16i

8

)7

= 27i

So the sum of expression equals 0.

Exercise 1.9. (E. 18, E. 19, E. 21 P. 13)Let z1, z2 and z3 be three arbitrary complex numbers. Which of the following equations aretrue in general?

1. i (z1 + z2 + z3) = i (z1 + z2 + z3)

2. < (z1z2z3) = < (z1z2z3)

3. < (z1z2z3) = = (iz1z2z3)

Solution.

1. The equation i(z1 + z2 + z3) = i(z1 + z2 + z3) is not true in general.

For concreteness let us choose z1 = z2 = 0;z3 = i, we havei(z1 + z2 + z3) = −1

i(z1 + z2 + z3) = 1

We can see i(z1 + z2 + z3) = −i(z1 + z2 + z3) is true for any z1, z2, z3. If the readers haveany doubt about this, try to find a distinct equation.

2. This equation is true in general. According to Exercise 1.7 (E. 6 P. 12), we have <(k1k2

)=

<(k1k2

), with k1, k2 are two arbitrary complex numbers.

14

1 COMPLEX NUMBERS

Hence,

< (z1z2z3) = < [z1 (z2z3)]

= < [z1 (z2z3)]

= <[z1(z2z3)

]= < [z1 (z2z3)]

= < (z1z2z3)

3. In general, we assert this which is true. In details, we know that < (z1z2z3) = < (z1z2z3)

by applying the above exercise. Then, we put z1z2z3 = z4 because they are also a complexnumber. So this equation is equivalent to

<z4 = = (iz4)

At this point, we can use properties of complex numbers. For concretness let we write

<z4 =

z4 + z4

2

= (iz4) =iz4 − iz4

2i

We note that iz4 = −z4, hence the event <z4 = = (iz4) that is really true.

Exercise 1.10. (E. 22 P. 13)Consider this problem : (32 +52)(22 +72) = (p2 +q2). Our unknowns p and q are assumed to benonnegative integers. This problem has two sets of solution : p = 29, q = 31 and p = 41, q = 11,as reader can verify. We derive here a general solution to problems of the following type : weare given k, l,m, n which are nonnegative integers. We seek two sets of nonnegative integers pand q such that (p2 +q2) = (k2 + l2)(m2 +n2). It is not obvious that there are integers solutions,but complex will prove that there are.

1. Note that by factoring we have (p+ iq)(p− iq) = (k+ il)(k− il)(m+ in)(m− in). Explainwhy if (p + iq) = (k + il)(m + in) is satisfied, then (p − iq) = (k − il)(m − in) is also.With this hint, show that we can take p = |km− nl| and q = lm+ kn as solutions to ourproblem.

2. Rearrange the equation for (p + iq)(p − iq) given in (1) to show that we can also takep = km+ nl and q = |lm− kn| as a solution to our problem.

3. Using the results of parts (1) and (2), verify the values for p and q given at the start ofthe problem. In addition, solve the following problem for two sets of values for p and q :(p2 + q2) = (122)(53).

Solution.

1. If (p+iq) = (k+il)(m+in) is satisfied, then we have (p+iq)(p−iq) = (k+il)(m+in)(p−iq).

15

1 COMPLEX NUMBERS

For the case (k + il)(m + in) = 0, this means p + iq = 0 or p = q = 0. But we alreadyhave p, q which are nonnegative numberss. So this case does not happen.

Hence, if we want (p + iq)(p − iq) = (k + il)(k − il)(m + in)(m − in) to occur, then(p− iq) = (k − il)(m− in).

Take p = |km− nl| and q = lm+ kn, we can see that p2 + q2 =(k2 + l2

) (m2 + n2

). We

thus complete this proof.

2. At this time, we look at (p+ iq) (p− iq) = (k + il) (m− in) (m+ in) (k − il). So if(p+ iq) = (k + il) (m− in) is satisfied, then (p− iq) = (m+ in) (k − il).

We can prove that if (p+ iq) = (k + il) (m− in) is occur, we need take p = km + nl

and q = |lm− kn|. Hence, we verify this values of p, q, we also see that p2 + q2 =(k2 + l2

) (m2 + n2

).

We prove the exercise successfully.

3. Now we need to check with the solution : p2 + q2 = (122) (53) to find out whether theseabove results are true. We first divide 122 into sum of 112 and 12, and break 53 into sumof 72 and 22. So we have p2 + q2 =

(112 + 12

) (72 + 22

).

Then we use the formula in part (1) with k = 11, l = 1,m = 7 and n = 2, they are alreadynonnegative integers so let take p = |km− ln| = 75 and q = kn + lm = 29. For formulain part (2), we have p = km+ nl = 79 and q = |lm− kn| = |−15| = 15.

The readers can verify two p and q of each others.

Exercise 1.11. (E. 24 P. 13)

In Haminton’s formulation two complex numbers (a, b) and (c, d) are said to be equal if and onlyif a = c and b = d. These are necessary and sufficient conditions. In dealing with fractions weare to making a comparable statement in asserting their equality. Consider

p

qand

r

s, where the

numerators and denominators are complex numbers. Find the necessary and sufficient conditonsfor these fractions to be equal.

Solution.With a1, a2, b1, b2 are real numbers, we can build four complex numbers which is formed

p = a1 + ib1

q = a2 + ib2

r = c1 + id1

s = c2 + id2

But we must note that if we want to have the fractionsp

qand

r

s, then we need both denom-

inators are defined as the initial conditions. It means a22 + b22 6= 0 and c2

2 + d22 6= 0.

We consider two fractions and try to simplify these if we can.

16

1 COMPLEX NUMBERS

p

q=

a1 + ib1a2 + ib2

=(a1 + ib1) (a2 − ib2)

(a2 + ib2) (a2 − ib2)

=(a1a2 + b1b2) + i (a2b1 − a1b2)

a22 + b22

In a similar fashion, we also reduce the following fraction.

r

s=

c1 + id1

c2 + id2

=(c1 + id1) (c2 − id2)

(c2 + id2) (c2 − id2)

=(c1c2 + d1d2) + i (c2d1 − c1d2)

c22 + d2

2

At this point, we can apply Hamilton’s formulation for these fractions. In details,p

qand

r

sare said to be equal if and only if

a1a2 + b1b2a2

2 + b22=c1c2 + d1d2

c22 + d2

2a2b1 − a1b2a2

2 + b22=c2d1 − c1d2

c22 + d2

2

with the above initial conditions.

17

1 COMPLEX NUMBERS

1.3 COMPLEX NUMBERS AND THE ARGAND PLANE

We present the exercise 7, 8, 10, 11, 12, 13, 17, 35, 38, 39, 41, 43, 44, 45. And in this section,we have a convention cisθ which is equivalent to cos θ + i sin θ.

Exercise 1.12. (E. 7, E. 8 P. 26)Find the modulus of each of the following complex expressions :

1.(1 + i)5

(2 + 3i)5

2.(1− i)n

(2 + 2i)n, n > 0 is an integer.

Solution.

1. We have ∣∣∣∣∣ (1 + i)5

(2 + 3i)5

∣∣∣∣∣ =|1 + i|5

|2 + 3i|5

We evaluate the modulus of numerator and denominator.

• |1 + i|5 =(√

2)5

• |2 + 3i|5 =(√

13)5

So the consistent result is4

169

√2

13.

2. We have to compute the denominator and the numerator respectively after having thedifferent form of expression.

∣∣∣∣ (1− i)n

(2 + 2i)n

∣∣∣∣ =|1− i|n

|2 + 2i|n

Analyzing |2 + 2i|n = 8n2 and |1− i|n = 2

n2 , we have the following fraction

|1− i|n

|2 + 2i|n=

2n2

8n2

=1

2n

, n > 0 is an integer.

So this is the result of the exercise.

Exercise 1.13. (E. 10 P. 26)Find two complex numbers that are conjugates of each other. The magnitude of their sum is 1and the sum of their magnitudes is 2.

Solution.We derive that z1and z2 are two complex numbers that are conjugates of each other, with

z1 = a+ ib, z2 = a− bi, where a, b are real numbers.

18

1 COMPLEX NUMBERS

Therefore, we get a set of equations.|z1 + z2| = 1

|z1|+ |z2| = 2

Because of |z1 + z2| =√

(a+ a)2 + (b− b)2 and |z1| + |z2| =√a2 + b2 +

√a2 + b2, a set of

equations is equivalent to √

2a2 = 1√a2 + b2 = 1

Thus, we have 4 pairs complex numbers.

(a, b) =

{(1√2

;1√2

),

(− 1√

2;− 1√

2

),

(− 1√

2;

1√2

),

(1√2

;− 1√2

)}So we can choose

z1 =1√2

+ i1√2

z2 =1√2− i 1√

2

as the answer.

Exercise 1.14. (E. 11 P. 26)Find two complex numbers that are conjugates to each other. The magnitude of their sum is a

and the sum of their magnitude is1

2, where 0 < a < 1. Answer in terms of a.

Solution.We set z1 = x1 +iy1, z2 = x1−iy1 where x1, y1 are real numbers, because z1, z2 are conjugates

to each other as the exercise says.And we have these complex numbers satisfied by two initial conditions, namely|z1 + z2| = a

|z1|+ |z2| =1

a

, where 0 < a < 1.

We also have |z1 + z2| = 2 |x1| and |z1|+ |z2| = 2√x2

1 + y21. Hence, the above set of equation

is equivalent to 2 |x1| = a

2√x2

1 + y21 =

1

a

At this point, we can solve this set of equations to derive the roots by using a < 1. And wesee that the exercise only suggests finding two complex numbers, so we can choose

x1 =a

2

y1 =1

2

√1

a2− a2

19

1 COMPLEX NUMBERS

as a result.Hence,

z1 =a

2+

1

2i

√1

a2− a2

z2 =a

2− 1

2i

√1

a2− a2

Exercise 1.15. (E. 12 P. 26)

Find two complex numbers whose product is 2 and whose difference is i.

Solution.In a similar way of Exercise 1.14 (E. 11 P. 26), we build z1 = x1 + iy1, z2 = x1 − iy1 where

x1, y1 are real numbers.The initial conditions is known as two informations of this exercise : z1z2 = 2 and z1− z2 = i

because we can take z1 − z2 > 0. In details, we also write these to become a set of equationswith x1, y1 variables. x2

1 + y21 = 2

2iy1 = i

At this point, we can solve this set of equations easily. Thus we have the following result.z1 =

√7

2+

1

2i

z2 =

√7

2− 1

2i

Exercise 1.16. (E. 13, E. 17 P. 26)The following vectors represent complex numbers. State these numbers in the form a+ ib.

1. The vector beginning at (−1,−3) and terminating at (1, 4).

2. The vector of length3

2beginning at the origin and ending, in the first quadrant, on the

circle (x− 1)2 + y2 = 1.

Solution.

1. We show the vector in Figure 1 which is called −→a .

20

1 COMPLEX NUMBERS

Figure 1.

Hence, we have −→a = (1; 4)− (−1;−3) = (2; 7). Then the consistent form is 2 + 7i.

2. We see that the vector of length is known as its magnitude, so we have√x2 + y2 =

3

2.

Then, we can create a set of equations.√x2 + y2 =

3

2

(x− 1)2 + y2 = 1

By solving this set and noticing the first quadrant of circle, we obtain the consistent form

is z =9

8+ i

3√

7

8.

We can see it in Figure 2.

Figure 2.

21

1 COMPLEX NUMBERS

Exercise 1.17. (E. 35 P. 27)Reduce the following expression to the form rcisθ, giving only the principal value of the angle.

A =(−1− i) cis

(π4

)(√

3 + i)2

Solution.We consider the numerator and denominator of expression A, and we can present these into

trigonometric forms. −1− i = −√

2cis(π

4

)(√

3 + i)2

= 2cis(π

6

)Thus, expression A is equivalent to

A =−√

2cis(π

4

)cis(π

4

)2cis

(π6

)=−√

2

2

cis(π

2

)cis(π

6

)=−√

2

2cis(π

3

)

Exercise 1.18. (E. 38 P. 27)

1. Let z1 and z2 be complex numbers. By replacing z2 with −z2 in the inequality |z1 + z2| ≤|z1|+ |z2|, show that

|z1 − z2| ≤ |z1|+ |z2|

Interpret this result with the aid of triangle.

2. What must be the relationship between z1 and z2 in order to have the equality hold inpart (1) ?

Solution.

1. Using inequality |z1 + z2| ≤ |z1| + |z2| and replacing z2 with −z2 in this, we have −z2

which is also a complex number. Hence,

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

The readers can verify |z2| = |−z2| easily, so that we obtain the completed proof.

This inequality can be explain explicitly in Figure 3.

22

1 COMPLEX NUMBERS

Figure 3.

2. The equality hold in part (1) if only if z1 = −z2, namelycos (arg z1) = − cos (arg z2)

sin (arg z1) = − sin (arg z2)

For the first equation, we can deduce that arg z1 = π − arg z2 + k2π or arg z1 = arg z2 −π + k2π, for all k is an integer. In a similar way, for the second equation, we derive thatarg z1 = − arg z2 + k2π or arg z1 = π + arg z2 + k2π, for all k is an integer.

Finally, we have arg z1 = π + arg z2 + k2π for all k is an integer.

Exercise 1.19. (E. 39 P. 27)Let z1 and z2 be complex numbers. Show that

|z1 + z2| ≥ |z1| − |z2| ≥ 0 if |z1| ≥ |z2||z1 + z2| ≥ |z2| − |z1| ≥ 0 if |z2| ≥ |z1|

Explain why both formulas can be reduced to the single expression

|z1 + z2| ≥ ||z1| − |z2||

Solution.In a similar way of Exercise 1.18 (E. 38 P. 27), we still use inequality |z1 + z2| ≤ |z1| + |z2|

and try to get some distinct patterns to prove the inequalities of this exercise. We look at thefollowing technique.

|z1| = |z1 + z2 + (−z2)|

At this point, we have |z1 + z2 + (−z2)| ≤ |z1 + z2|+ |−z2|. It is easy to see that |−z2| = |z2|.Thus we prove |z1 + z2| ≥ |z1| − |z2| for all z1, z2 are complex numbers, so if |z1| ≥ |z2|, we candeduce an equality |z1 + z2| ≥ |z1| − |z2| ≥ 0.

23

1 COMPLEX NUMBERS

Hence, we also show the inequality |z1 + z2| ≥ |z2| − |z1| for all z1, z2 are complex numbersby using the same technique.

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

So we can derive an equality |z1 + z2| ≥ |z2| − |z1| ≥ 0 if |z2| ≥ |z1|.Because an inequality |z1 + z2| ≥ ||z1| − |z2|| can divide into two cases. These cases are really

two inequalities which is shown in this exercise. Hence, the proof is completed.

Exercise 1.20. (E. 41 P. 28)

1. By considering the expression (p− q)2, where p and q are nonnegative real numbers, showthat p+ q ≤

√2√p2 + q2.

2. Use the preceding result to show that for any complex number z we have |<z| + |=z| ≤√2 |z|.

Solution.

1. We consider the expression (p− q)2, where p and q are nonnegative real numbers. Wehave this expression which interpolates an inequality

(p+ q)2 ≤ 2(p2 + q2

)Thus we can complete this proof because of square roots.

2. For any complex number z, we put z = x+ iy, where x and y are real numbers. The proofis equivalent to

|x|+ |y| ≤√

2 |x+ iy|

We see that |x+ iy| =√x2 + y2, so this proof becomes part (1) which is shown.

Exercise 1.21. (E. 43 P. 28)

1. By considering the product of 1 + ia and 1 + ib, and the argument of each factor, wherea and b are real numbers. Show that

arctan(a) + arctan(b) = arctan

(a+ b

1− ab

)2. Use the preceding formula to prove that

π = 4

[arctan

(1

2

)+ arctan

(1

3

)]3. Extend the technique used in (1) to find a formula for arctan(a) + arctan(b) + arctan(c).

Solution.

1. Let z1 be 1 + ia and z2 be 1 + ib, where a, b are real numbers. We get

z1z2 = 1− ab+ i(a+ b)

24

1 COMPLEX NUMBERS

By using these knowledges : arg z1 + arg z2 = arg (z1z2), arg z1 = arctan a, arg z2 =

arctan b and arg (z1z2) = arctan

(a+ b

1− ab

), we can derive

arctan a+ arctan b = arctan

(a+ b

1− ab

), as desired.

2. Applying the above result, we take a =1

2and b =

1

3. Hence,

arctan1

2+ arctan

1

3= arctan (1)

=π

4

So we prove (2) completely.

3. By the same the technique used in (1), we can find that

arctan a+ arctan b+ arctan c = arctana+ b+ c− abc1− ab− bc− ca

where a, b, c are real numbers.

If the readers have any doubt about the above result, try to prove it in a similar interpo-lation.

Exercise 1.22. (E. 44 P. 28)

1. Consider the inequality |z1 + z2|2 ≤ |z1|2 + |z2|2 + 2 |z1| |z2|. Prove this expression byalgebraic means (no triangles).

2. Observe that |z1|2 + |z2|2 + 2 |z1| |z2| = (|z1|+ |z2|) 2. Show that the inequality proved inpart (1) leads to the triangle inequality |z1 + z2| ≤ |z1|+ |z2|.

Solution.

1. By using the formula of modulus of one complex number, we have

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

= (z1 + z2) (z1 + z2)

= |z1|2 + |z2|2 + z1z2 + z1z2

Then we notice that z1z2 + z1z2 = 2< (z1z2) ≤ 2 |z1z2|. Hence, the above expression isless than or equal to |z1|2 + |z2|2 + 2 |z1z2|. Next, we also notice that |z1z2| is equivalentto |z1| |z2|.

At this point, we have the successful proof.

25

1 COMPLEX NUMBERS

2. According to part (1), we get

|z1 + z2|2 ≤ |z1|2 + |z2|2 + 2 |z1| |z2|

, for all z1, z2 are complex numbers.

By using the initial condition that is

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

, we derive that

|z1 + z2|2 ≤ (|z1|+ |z2|) 2

And this inequality is equivalent to

|z1 + z2| ≤ |z1|+ |z2|

because the modulus of one complex number is already nonnegative real number. Hence,the proof is completed.

Exercise 1.23. (E. 45 P. 28)Beginning with the product (z1 − z2) (z1 − z2), show that

|z1 − z2|2 = |z1|2 + |z2|2 − 2< (z1z2)

Solution.Beginning with the product (z1 − z2) (z1 − z2), for all z1 = x1 + iy1 and z2 = x2 + iy2, where

x1, x2, y1, y2 are real numbers. We have

|z1 − z2|2 = (z1 − z2) (z1 − z2)

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

= x21 + x2

2 − 2x1x2 + y21 + y2

2 − 2y1y2

= |z1|2 + |z2|2 − 2 (x1x2 + y1y2)

At this point, we get z1z2 = x1x2 +y1y2 + i (−x1y2 + y1x2), so < (z1z2) is exactly x1x2 +y1y2.Therefore, we obtain what we need to prove.

|z1 − z2|2 = |z1|2 + |z2|2 − 2< (z1z2)

26

1 COMPLEX NUMBERS

1.4 INTEGER AND FRACTIONAL POWERS OF A COMPLEXNUMBER

We choose the exercise 7, 8, 9, 20, 27, 28, 29, 34, 35 for illustrational section.Exercise 1.24. (E. 7 P. 35)

1. Using DeMoivre’s theorem, the binomial formula, and an obvious trigonometric identity,show that for integer n,

cosnθ = <n∑k=0

(cosn−k θ

) (√1− cos2 θ

)kik

n!

(n− k)!k!

2. Show that the preceding expression can be written as

cosnθ =

n/2∑m=0

(cos θ)n−2m (1− cos2 θ)m

(−1)mn!

(n− 2m)! (2m)!, if n is even,

cosnθ =

(n−1)/2∑m=0

(cos θ)n−2m (1− cos2 θ)m

(−1)mn!

(n− 2m)! (2m)!, if n is odd.

Solution.

1. We can change the right side for using the binomial formula.

<n∑k=0

(cosn−k θ

)(√1− cos2 θ

)kik

n!

(n− k)!k!= <

n∑k=0

(cos θ)n−k (i |sin θ|)k n!

(n− k)!k!

= < (cos θ + i |sin θ|)n

For the case sin θ ≥ 0, the above expression is equivalent to

< (cos θ + i sin θ)n = < (cosnθ + i sinnθ) = cosnθ

For the case sin θ < 0, we obtain

< (cos θ − i sin θ)n = < [cos (−θ) + i sin (−θ)]n

= cos (−nθ)

= cosnθ

The proof is completed by using DeMoivre’s theorem.

2. If n is even, we put n = 2p, where p is integer. Hence that first proof is equivalent to

27

1 COMPLEX NUMBERS

cos 2pθ =

p∑m=0

(cos θ)2p−2m (1− cos2 θ)m

(−1)m(2p)!

(2p− 2m)! (2m)!

=

p∑m=0

(cos θ)2p−2m(√

1− cos2 θ)2m

(i)2m (2p)!

(2p− 2m)! (2m)!

= <2p∑k=0

(cos2p−k θ

)(√1− cos2 θ

)kik

(2p)!

(2p− k)!k!

This is proved in part (1), so the first proof is completely true.

In a similar fashion, if n is odd, we put n = 2q + 1, where q is integer. Thus the secondproof is equivalent to

cos (2q + 1) θ =

q∑m=0

(cos θ)2q+1−2m (1− cos2 θ)m

(−1)m(2q + 1)!

(2q + 1− 2m)! (2m)!

=

q∑m=0

(cos θ)2q+1−2m(√

1− cos2 θ)2m

(i)2m (2q + 1)!

(2q + 1− 2m)! (2m)!

= <2q+1∑k=0

(cos2q+1−k θ

)(√1− cos2 θ

)kik

(2q + 1)!

(2q + 1− k)!k!

This is shown in part (1), hence the second proof is exact.

Exercise 1.25. (E. 8 P. 36)Prove that (

1 + i tan θ

1− i tan θ

)n=

1 + i tannθ

1− i tannθ

where n is any integer.

Solution.For n is an arbitrary integer and an angle of θ that tan θ is defined. We have

(1 + i tan θ

1− i tan θ

)n=

(cos θ + i sin θ)n

(cos θ − i sin θ)n

=cosnθ + i sinnθ

cosnθ − i sinnθ

=1 + i tannθ

1− i tannθ

Completing the proof by using DeMoivre’s theorem.

28

1 COMPLEX NUMBERS

Exercise 1.26. (E. 9, E. 20 P.36)Express the following in the form a+ ib. Give all values and make a polar plot of the points orthe vectors that represent your results.

1. (9i)12

2. (1 + i)−54

Solution.

1. We put z = i, then we can derive the trigonometric form of z. This means z = cis(π

2

)easily. Applying the formula z

1m = m

√rcis

(θ

m+

2kπ

m

), k = 0,m− 1, with z = rcis (θ),

we have

(9i)12 = 3

[cos(π

2

)+ i sin

(π2

)] 12

= 3[cos(π

4+ kπ

)+ i sin

(π4

+ kπ)]

, k = 0, 1.

Expressed as decimals, these answers become approximately

2.121 +i2.121 , k = 0

−2.121 −i2.121 , k = 1

Vectors representing the roots are plotted in Figure 4. We can see that they are spaced πor 180◦ apart.

Figure 4.

29

1 COMPLEX NUMBERS

2. We put z = 1 + i and first compute r =√

2 and argument of z which we call θ, equalsπ

4by evaluating tan θ = 1.

Therefore, we have a following trigonometric form of z.

z =√

2(

cosπ

4+ i sin

π

4

)And then, our result is

z−54 =

(√2)−5

4

[cos

(−5π

16− 5kπ

2

)+ i sin

(−5π

16− 5kπ

2

)]=

(√2)−5

4

[cos

(−5π

16− kπ

2

)+ i sin

(−5π

16− kπ

2

)], k = 0, 1, 2, 3.


0.360 − i0.539 , k = 0

−0.539 − i0.360 , k = 1

−0.360 + i0.539 , k = 2

0.539 + i0.360 , k = 3

Vectors representing the roots are plotted in Figure 5. We can see that they are spacedπ

2or 90◦ apart.

Figure 5.

30

1 COMPLEX NUMBERS

Exercise 1.27. (E. 27 P. 36)

1. Show that zn+1 − 1 = (z − 1)(zn + zn−1 + ...+ z + 1

), where n ≥ 0 is an integer annd z

is any complex number.

The preceding implies that

zn+1 − 1

z − 1= zn + zn−1 + ...+ z+1 for z 6= 1

which the reader should recognize as the sum of a geometric series.

2. Use the preceding result to find and plot all solutions of z4 + z3 + z2 + z + 1 = 0.

Solution.

1. We prove it by mathematical induction :

For the case n = 0, the expression says

z − 1 = (z − 1) 1

This assertion is really true. Given the artificiality of these assumptions, we may behappier if the base case for n = 1 is also given.

For the case n = 1, the expression says

z2 − 1 = (z − 1) (z + 1)

It is easy to see that the above assertion is true. Thus we have the base cases for ourinduction.

For the induction step we assume that

zk+1 − 1 = (z − 1)(zk + zk−1 + ...+ z + 1

)and show that

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

)We consider the left side zk+2 − 1 which can be analyse the sum of (z − 1) zk+1 andzk+1 − 1. Hence,

(z − 1) zk+1 +(zk+1 − 1

)= (z − 1) zk+1 + (z − 1)

(zk + zk−1 + ...+ z + 1

)= (z − 1)

(zk+1 + zk + zk−1 + ...+ z + 1

)At this point, we can have completely the proof.

2. By using the result of part (1), we can solve this equation. Moreover, we might presentthese roots by plotting of trigonometric forms. This equation is equivalent to

z5 − 1

z − 1= 0

31

1 COMPLEX NUMBERS

Hence, we derive the roots because of z 6= 1.

z5 = cosπ

2+ i sin

π

2

Applying the DeMoivre’s theorem, we thus have

z = cos

(π

10+k2π

5

)+ i sin

(π

10+k2π

5

), k = 0, 1, 2, 3, 4


z0 = cosπ

10+ i sin

π

10= 0.951 + i0.309

z1 = cos

(π

10+

2π

5

)+ i sin

(π

10+

2π

5

)= i

z2 = cos

(π

10+

4π

5

)+ i sin

(π

10+

4π

5

)= −0.951 + i0.309

z3 = cos

(π

10+

6π

5

)+ i sin

(π

10+

6π

5

)= −0.587− i0.809

z4 = cos

(π

10+

8π

5

)+ i sin

(π

10+

8π

5

)= 0.587− i0.809

Then, according to the suggestion, we will plot all the above roots. The readers can seeit in Figure 6.

Figure 6.

32

1 COMPLEX NUMBERS

Exercise 1.28. (E. 28 P. 37)

1. If z = rcisθ, show that the sum of the values of z1n is given by

n−1∑k=0

n√r

(cisθ

n

)[cis(

2π

n

)]kwhere n ≥ 2 is an integer.

2. Show that the sum of the values of z1n is zero. Do this by rewriting the preceding problem

in Exercise 1.26 (E. 27 P. 36), but with n−1 used in place of n, and employing the formulaof part (1).

Solution.

1. If z = rcisθ, we use the formula of z1n , where n ≥ 2 is an integer.

z1n = n

√r

[cos

(θ

n+

2kπ

n

)+ i sin

(θ

n+

2kπ

n

)], k ∈ N, k = 0, n− 1.

For the case k = 0, this formula says

z1n = n

√r

(cos

θ

n+ i sin

θ

n

)= n√rcis

(θ

n

)

For the case k = 1, this formula says

z1n = n

√r

[cos

(θ

n+

2π

n

)+ i sin

(θ

n+

2π

n

)]= n√rcis

(θ

n

)cis(

2π

n

)

For the case k = 2, this formula also says

z1n = n

√r

[cos

(θ

n+

4π

n

)+ i sin

(θ

n+

4π

n

)]= n√rcis

(θ

n

)cis(

4π

n

)

In a similar fashion, for the case k = n− 1, we obtain

z1n = n

√rcis

(θ

n

)cis[

2 (n− 1)π

n

]

33

1 COMPLEX NUMBERS

Hence, the sum of the values of z1n , which is called S, is created by the following expression.

S = n√rcis

(θ

n

){1 + cis

(2π

n

)+ cis

(4π

n

)+ ...+ cis

[2 (n− 1)π

n

]}=

n−1∑k=0

n√r

(cisθ

n

)[cis(

2π

n

)]k

, as desired.

2. We get the following formula

zn − 1

z − 1=

n−1∑k=0

zk

, where n ≥ 0 is an integer and z is any complex number.

Thus we can choose z = cis(

2π

n

), where n ≥ 2 is an integer. The above expression

becomes

[cis(

2π

n

)]n− 1

cis(

2π

n

)− 1

=

n−1∑k=0

[cis(

2π

n

)]k

Because of[cis(

2π

n

)]n− 1 = cis2π − 1 = 0, we can completely prove that the sum of

the values of z1n is zero.

Exercise 1.29. (E. 29 P. 37)

1. Suppose a complex number is given in the form z = rcisθ. Recalling the indentities

sin

(θ

2

)= ±

√1

2−(

1

2

)cos θ and cos

(θ

2

)= ±

√1

2+

(1

2

)cos θ, show that

z12 = ±

√r

(√1 + cos θ

2+ i

√1− cos θ

2

)for 0 ≤ θ ≤ π.

2. Explain why the preceding formula is invalid for −π < θ < 0, and find the correspondingcorrect formula for this interval.

3. Use the formulas derived in (1) and (2) to find the square roots of 2cis(π

6

)and 2cis

(−π

6

),

respectively.

Solution.

1. Using the form z = rcisθ, we have

34

1 COMPLEX NUMBERS

z12 =√r

[cos

(θ

2+ kπ

)+ i sin

(θ

2+ kπ

)], k = 0, 1. For 0 ≤ θ ≤ π, we see that if we take k = 0, the above expression is equivalentto

z12 =

√r

[cos

(θ

2

)+ i sin

(θ

2

)]=√r

(√1 + cos θ

2+ i

√1− cos θ

2

)

And for the case k = 1, we get

z12 =

√r

[cos

(θ

2+ π

)+ i sin

(θ

2+ π

)]= −

√r

(√1 + cos θ

2+ i

√1− cos θ

2

)

Then we prove part (1) completely.

2. For −π < θ < 0 because of principal argument. We observe that cos

(θ

2

)and sin

(θ

2

)are negative numbers, we obtain

z12 = ±

√r

(√1 + cosα

2− i√

1− cosα

2

)

So this formula is not the above formula of part (1). Hence we find out the correspondingcorrect formula for this interval.

3. Applying the formula of part (1), the square roots of 2cis(π

6

)equals

±√

2

√√√√1 + cos

π

62

+ i

√√√√1− cosπ

62

= ±[(√

3 + 1)

+ i(√

3− 1)]

And using the formula of part (2), the square roots of 2cis(−π

6

)is±

[(√3− 1

)− i(√

3 + 1)].

35

1 COMPLEX NUMBERS

Exercise 1.30. (E. 34 P. 38)Use the formula for the sum of a geometric series in Exercise 27 and DeMoivre’s theorem toderive the following formulas for 0 <θ < 2π :

1 + cos θ + cos 2θ + · · ·+ cosnθ =

cos

(nθ

2

)sin

[(n+ 1) θ

2

]sin

(θ

2

)

sin θ + sin 2θ + sin 3θ + · · ·+ sinnθ =

sin

(nθ

2

)sin

[(n+ 1) θ

2

]sin

(θ

2

)Solution.By using DeMoivre’s theorem, we can have

n∑k=0

cos kθ + i

(n∑k=0

sin kθ

)= 1 + (cisθ) + (cis2θ)2 + · · ·+ (cisnθ)n

According to the formula zn+1−1 = (z − 1)(zn + zn−1 + · · ·+ z + 1

), where n is an nonneg-

ative integer and z is any complex number, we have

36

1 COMPLEX NUMBERS

1 + (cisθ) + (cis2θ)2 + · · ·+ (cisnθ)n =(cisθ)n+1 − 1

cisθ − 1

=cos [(n+ 1) θ] + i sin [(n+ 1) θ]− 1

cos θ + i sin θ − 1

=(cos θ − 1− i sin θ) {cos [(n+ 1) θ]− 1 + i sin [(n+ 1) θ]}

(cos θ − 1 + i sin θ) (cos θ − 1− i sin θ)

=(cos θ − 1) {cos [(n+ 1) θ]− 1}+ sin θ sin [(n+ 1) θ]

2 (1− cos θ)

+i(cos θ − 1) sin [(n+ 1) θ]− sin θ {cos [(n+ 1) θ]− 1}

2 (1− cos θ)

=cos [nθ]− cos θ − cos [(n+ 1) θ] + 1

4 sin2 θ

2

+isin (nθ)− sin [(n+ 1) θ] + sin θ

4 sin2 θ

2

=

2 sin

[(2n+ 1

2

)θ

]sin

θ

2+ 2 sin2 θ

2

4 sin2 θ

2

+i

2 sin θ2 cos

θ

2− 2 cos

[(2n+ 1

2

)θ

]sin

θ

2

4 sin2 θ

2

=

cos(n

2θ)

sin

[(n+ 1

2

)θ

]sin

θ

2

+ i

sinnθ

2sin

[(n+ 1

2

)θ

]sin

θ

2

Using the homogeneous method, we can deduce that

1 + cos θ + cos 2θ + · · ·+ cosnθ =

cos(n

2θ)

sin

[(n+ 1

2

)θ

]sin

θ

2

sin θ + sin 2θ + sin 3θ + · · ·+ sinnθ =

sinnθ

2sin

[(n+ 1

2

)θ

]sin

θ

2

, as desired.

37

1 COMPLEX NUMBERS

Exercise 1.31. (E. 35 P. 38)If n is an integer greater than or equal to 2, prove that

cos

(2π

n

)+ cos

(4π

n

)+ ...+ cos

[2 (n− 1)π

n

]= −1

and that

sin

(2π

n

)+ sin

(4π

n

)+ ...+ sin

[2 (n− 1)π

n

]= 0

Solution.For n is an integer greater than or equal to 2, we have this sum in Exercise 1.28 (E. 28 P. 37)

n−1∑k=0

n√r

(cisθ

n

)[cis(

2π

n

)]k=

n−1∑k=0

z1n

[cis(

2π

n

)]k

By choosing z = 1, the above formula becomesn−1∑k=0

[cis(

2π

n

)]k.

And then, using general DeMoivre’s theorem, we get

n−1∑k=0

[cis(

2π

n

)]k=

n−1∑k=0

cis(

2kπ

n

)= cos 0 + cos

(2π

n

)+ cos

(4π

n

)+ ...+ cos

[2 (n− 1)π

n

]+i

{sin 0 + sin

(2π

n

)+ sin

(4π

n

)+ ...+ sin

[2 (n− 1)π

n

]}Using again the result of Exercise 1.28 (E. 28 P. 37), this says that the above sum is zero,

namely cos 0 + cos

(2π

n

)+ cos

(4π

n

)+ ...+ cos

[2 (n− 1)π

n

]= 0

sin 0 + sin

(2π

n

)+ sin

(4π

n

)+ ...+ sin

[2 (n− 1)π

n

]= 0

We note that cos 0 = 1 and sin 0 = 0, we thus have the proof successfully.

38

1 COMPLEX NUMBERS

1.5 POINTS, SETS, LOCI, AND REGIONS IN THE COMPLEXPLANE

We call the exercise 12, 31 as the presentation.Exercise 1.32. (E. 12 P. 46)

Find the points on the circle |z − 1− i| = 1 that have the nearest and furthest linear distanceto the point z = −1 + i0. In addition, state what these two distances are.

Solution.We put z = x+ iy, where x, y are real numbers. Since |z − 1− i| = 1, we have

1 = |x+ iy − 1− i|

1 = |(x− 1) + i (y − 1)|

1 = (x− 1)2 + (y − 1)2

We then consider the above equation, we have center of there circle which is called I and hasthe coordinate I (1, 1). And the modulus of radius equals 1.

Put A(−1, 0), we obtain that the equation of line AB is y =1

2x+

1

2. Thus, by replacing this

into expression (x− 1)2 + (y − 1)2 = 1, we get that the left side is equivalent to

(x− 1)2 + (y − 1)2 = (x− 1)2 +

(1

2x+

1

2− 1

)2

= (x− 1)2 +1

4(x− 1)2

=5

4(x− 1)2

Then, we can deduce x = 1 +2√5

or x = 1 − 2√5. And we call K1

(1 +

2√5, 1 +

1√5

);

K2

(1− 2√

5, 1− 1√

5

)are the points which is recently found.

It is easy too see that

AK1 =

√(2 + 2

√5)2

+

(1 +

1√5

)2

> AK2 =

√(2− 2

√5)2

+

(1 +

1√5

)2

So we have the conclusions :K2 is the point on the circle |z − 1− i| = 1 that have the nearest linear distance to the point

z = −1 + 0i,K1 is the point on the circle |z − 1− i| = 1 that have the furthest linear distance to the point

z = −1 + 0i.

Exercise 1.33. (E. 31 P. 47)Is a boundary point of a set necessarily an accumulation point of that set?

39

1 COMPLEX NUMBERS

Solution.We display 2 definitions of a boundary point and an accumulation point.

• u is a boundary point of set D if it is a limit point of D and E\D.

• u is an accumulation of D if for all r > 0 : (B (u; r) \ {u}) ∩D 6= ∅.

In Z space, given A = [0; 1], we can see that 0 is a boundary point of this set but is not anaccumulation point.

40

2 THE COMPLEX FUNCTION AND ITSDERIVATIVE

2.1 INTRODUCTION

For the introduction, we have some exercises which are 2, 4, 6, 8, 10, 13, 15, 17, 19, 21, 24, 27.Exercise 2.1. (E. 2, E. 4 P. 53)

Suppose z = x + iy. Let f (z) =(z − i) (z − 2)

(z2 + 1) cosx. State where in the following domains this

functions fails to be defined.

1. |z| < 1.1

2.∣∣∣z − (1 + i)

π

2

∣∣∣ < π

2

Solution.

1. It is easy to observe that f (z) is defined when its denominator is not equal to 0. It meansz 6= ±i and z 6= π

2+ kπ + iy for k is an integer. So that z = ±i, z =

π

2+ kπ + iy make

this function fail to be defined basically. Then we see that z = ±i belong in inequality|z| < 1.1 but z =

π

2+ kπ + iy are not here for all k is an integer.

Hence, this function is not defined at two points z = ±i.

2. We simplify the modulus of z− (1 + i)π

2which equals

√(x− π

2

)2+(y − π

2

)2. Thus we

have the coordinate of origin of this circle that is(π

2;π

2

)and the radius of domain is aslo

π

2. We have both points z = ±i which are not in inequality

∣∣∣z − (1 + i)π

2

∣∣∣ < π

2. For the

only case k = 0, we have a point(π

2; y)

that belongs in this inequality. Then we candeduce y satisfying 0 < y < π.

Hence, f (z) is not defined on line x =π

2, 0 < y < π.

Exercise 2.2. (E. 6, E. 8 P. 53)For each of the following functions, find f (1 + 2i) in the form a+ib. If the function is undefinedat 1 + 2i, state this fact.

1.1

zz − 5

2.z

cosx+ i sin y

Solution.

41

2 THE COMPLEX FUNCTION AND ITS DERIVATIVE

1. With z = 1 + 2i, we have zz = |z|2 = 5. So that the fraction1

zz − 5is not defined. That

means the function is undefined at 1 + 2i.

Thus we cannot find f (1 + 2i).

2. In this function, we can get a condition of denominator that is cosx 6= 0 and sin y 6= 0.So the function is defined at 1 + 2i. Then f (1 + 2i) becomes

f (1 + 2i) =1 + 2i

cos 1 + i sin 2

=cos 1 + 2 sin 2 + i (2 cos 1− sin 2)

cos2 1 + sin2 2

Exercise 2.3. (E. 10, E. 13 P. 54)Write the following functions of z in the form u (x, y) + iv (x, y), where u (x, y) and v (x, y) areexplicit real functions of x and y.

1.1

z+ i

2. z̄3 + z̄

Solution.

1. We put z = x + iy, where x, y are real numbers. We can call f (z) for this function andfind u (x, y) and v (x, y) by simplifying it. For concreteness let us write

f (z) =1

z+ i

=1

x+ iy+ i

=x− iyx2 + y2

+ i

=x

x2 + y2+ i

(1 +

−yx2 + y2

)

We deduce the following result.

u (x, y) = xx2+y2

v (x, y) = 1 + −yx2+y2

2. In a similar pattern, we also reduce this function.

42


f (z) = z̄3 + z̄

= (x− iy)3 + x− iy

= x3 − 3x2iy − 3xy2 + iy3 + x− iy

= x3 − 3xy2 + x+ i(y3 − 3x2y − y

)Hence, we get

u (x, y) = x3 − 3xy2 + x

v (x, y) = y3 − 3x2y − y

Exercise 2.4. (E. 15, E. 17 P. 55)Rewrite the following functions in terms of z and if necessary z as well as constants. Thus xand y must not appear in your answer. Simplify your answer as much as possible.

1.1

x+

1

iy

2. x+x

x2 + y2+ iy +

iy

x2 + y2

Solution.

1. Using the following formula

x =

z + z

2

y =1

i

z − z2

we can reduce a function of part (1). Moreover, we have no need of existent condition.

1

x+

1

iy=

2

z + z+

2

z − z

=2z − 2z + 2z + 2z

z2 − z2

=4z

z2 − z2

2. In a similar way, we can apply the above method to simplify a function. But we note thata denominator is exactly |z|. Hence, we have a distinct way as a solution.

x+x

x2 + y2+ iy +

iy

x2 + y2= (x+ iy) +

x+ iy

x2 + y2

= z +z

|z|2

43


Furthermore, we remember a formula of magnitude of one complex number that is |z| =√zz. So we obtain that expression is equivalent to

z +1

z

This is a result which we find.

Exercise 2.5. (E. 19, E. 21 P. 55)For each of the following functions, tabulate the value of the function for these values of z :1, 1 + i, i, ,−1 + i,−1. Indicate graphically the correspondence between values of w and valuesof z.

1. w =i

z

2. w = z3

Solution.

1. In the following table and in Figure 1, we have investigated a few points in the case of

w = f (z) =i

z.

z w =i

zA = 1 A′ = i

B = 1 + i B′ =i+ 1

2C = i C ′ = 1

D = −1 + i D′ =1− i

2E = −1 E′ = −i

Then, we show the Figure 1 to clarify the exercise.

44


z-plane

w-plane

Figure 1.

2. In the following table and in Figure 1, we have interpolated a few points in the case ofw = f (z) = z3.

z w = z3

A = 1 A′ = 1

B = 1 + i B′ = −2 + 2i

C = i C ′ = −iD = −1 + i D′ = 2 + 2i

E = −1 E′ = −1

Next, we present the Figure 1 to clarify the exercise.

45


z-plane

w-plane

Figure 1.

Exercise 2.6. (E. 24 P. 55)

Let f (z) =1

z + i. Find the following.

f

(1

f (z)

)Solution.

46


By considering f (z), we must have the initial condition which makes f (z)’s existence. Butin this exercise, this can be omitted. So we have

f

(1

f (z)

)=

11

f (z)+ i

=f (z)

if (z) + 1

=

1

z + ii

z + i+ 1

=1

z + 2i

, as a result.

Exercise 2.7. (E. 27 P. 55)

Using MATLAB obtain three-dimensional plots of the function f (z) =1

z − 3

2iand allow z to

assume values over a grid in the region of the complex plane defined by −1 ≤ x ≤ 1,−1 ≤ y ≤ 1.

Solution.In this exercise, we present three kinds of three-dimensional plots. These are plots of modulus,

real part and imaginary part of f (z). Hence, we will respectively show these plots with thecorresponding scripts which are typed in MATLAB.

1. Plot of Modulus of f (z) :

>‌> x = -1 : 0.2 : 1;

>‌> y = -1 : 0.2 : 1;

>‌> [x , y] = meshgrid(x , y);

>‌> z = x + i*y;

>‌> w = 1./(z - 1.5*i);

>‌> surfc(x , y , abs(w))

47


Figure 1.

2. Plot of Real Part of f (z) :

>‌> x = -1 : 0.2 : 1;

>‌> y = -1 : 0.2 : 1;


>‌> z = x + i*y;

>‌> w = 1./(z - 1.5*i);

>‌> surfc(x , y , real(w))

48


Figure 1.

3. Plot of Imaginary Part of f (z) :

>‌> x = -1 : 0.2 : 1;

>‌> y = -1 : 0.2 : 1;


>‌> z = x + i*y;

>‌> w = 1./(z - 1.5*i);

>‌> surfc(x , y , imag(w))

49


Figure 1.

50


2.2 LIMITS AND CONTINUITY

This is an important section, so we show the exercise 5, 10, 11, 12, 14, 15, 16, 17, 18 becausewe think that these are so necessary for solutions.

Exercise 2.8. (E. 5 P. 62)Assuming the continuity of the functions f (z) = z and f (z) = c, where c is any constant. Provethe continuity of the following function in the domain indicated. Take z = x+ iy.

z4 +1 + i

z2 + 3z + 2

, all z 6= −1,−2.

Solution.In this exercise, we can break the function into smaller parts. We then determine two functions

which is defined for all z 6= −1,−2.h (z) = z4

g (z) =1 + i

z2 + 3z + 2

We must notice that for any z 6= −1,−2, we can write down a denominator of g (z) which is(z + 1) (z + 2). Of course, this denominator is continuous function. By applying respectivelytheorem of products and quotient of continuous functions, we gain h (z) that is continuousfunction. Thus applying theorem of sums of continuous functions, we can derive that h (z)+g (z)

is continuous function.Completing the proof by using theorem of continuous functions.

Exercise 2.9. (E. 10 P. 62)Prove that the following function is continuous at z = i.

f (z) =

z − i

z2 − 3iz − 2, z 6= i

i , z = i

Solution.Because f (i) is defined, we must determine lim

z→i

z − iz2 − 3iz − 2

. For all z 6= i, we have a

denominator of f (z) which is equivalent to (z − i) (z − 2i). Moreover, we can state that

limz→i

z − iz2 − 3iz − 2

= limz→i

z − i(z − i) (z − 2i)

= limz→i

1

z − 2i

From this we might conclude that f (z) is a continuous function because of limz→i

f (z) = f (i) =

i.Hence, we have the completed proof.

51


Exercise 2.10. (E. 11 P. 62)

1. Consider the function f (z) =z2 − 5z + 6

z2 − 4defined for z 6= ±2. How should this function

be defined at z = 2 so that f (z) is continuous at z = 2 ?

2. Consider the function f (z) =z4 + 10z2 + 9

z2 − 4iz − 3defined for z 6= 3i and z 6= i. How should

this function be defined at z = 3i and z = i so that f (z) is continuous everywhere ?

Solution.

1. Consider this function f (z), we note that we can simplify it because of factoring numeratorand denominator.

f (z) =z2 − 5z + 6

z2 − 4

=(z − 2) (z − 3)

(z − 2) (z + 2)

=z − 3

z + 2

This function does not belong to one of first conditions. This is z 6= 2 because theexpression z − 2 is cancelled. Hence this function is really defined at z = 2. Moreover,the following conditions are both satisfied easily : f (2) is defined; lim

z→2f (z) exists and

limz→2

f (z) = f (2). According to definition of continuity of the complex function, we assertf (z) is continuous at z = 2.

2. In a similar way, it is easy to reduce this function, which is not defined at z 6= 3i andz 6= i, to define for these points.

f (z) =z4 + 10z2 + 9

z2 − 4iz − 3

=

(z2 + 9

) (z2 + 1

)(z − 3i) (z − i)

=(z − 3i) (z + 3i) (z − i) (z + i)

(z − 3i) (z − i)= (z + 3i) (z + i)

At this point, we can observe that f (z) is continuous everywhere. In details, if we takean arbitrary complex number z0, we obtain explicitly that both conditions of continuityof the comlex function are satisfied.

52


Exercise 2.11. (E. 12 P. 62)In this problem we prove rigorously, using the definition of the limit at infinity, that

limz→∞

z

1 + z= 1

1. Explain why, given ε > 0, we must find a function r (ε) such that∣∣∣∣ 1

z + 1

∣∣∣∣ < ε for all

|z| > r.

2. Using one of the triangle inequalities, show that the preceding inequality is satisfied if we

take r > 1 +1

ε.

Solution.

1. We present the following definition of the limit at infinity :

If for every real number ε > 0 there exists a real number r (ε) > 0 such that∣∣∣∣ 1

z + 1− 1

∣∣∣∣ < ε

for all |z| > r (ε).

If we have limz→∞

z

1 + z= 1 firstly, we can say the above definition. But if we need prove

limz→∞

z

1 + z= 1 as the mention of this exercise, we must alter the phrase “for every” into

“given” and the phrase “there exists” into “find”, to see some works need to be make theproof. For concretness let us write down

Given real number ε > 0, find a real number r (ε) > 0 such that∣∣∣∣ 1

z + 1

∣∣∣∣ < ε

for all |z| > r (ε).

2. If we take r (ε) > 1+1

ε, we obtain |z| > 1+

1

ε. We need verify that this r (ε) is used in this

proof, which is either true or false. It means that the inequality∣∣∣∣ 1

z + 1

∣∣∣∣ < ε is satisfied.

In explicit, we have the triangle inequality |z + 1| ≥ |z| − 1; so if we want to have the

inequality∣∣∣∣ 1

z + 1

∣∣∣∣ < ε, then we must have the inequality1

|z| − 1< ε or |z| > 1 +

1

ε. This

fact is true because this inequality obtained.

Hence, we prove successfully that the preceding inequality is satisfied if we take r (ε) >

1 +1

ε.

Exercise 2.12. (E. 14 P. 63)

1. Knowing that f (z) = z2 is everywhere continuous. Explain why the real function xy iseverywhere continuous.

2. Explain why the function g (x, y) = xy + i (x+ y) is everywhere continuous.

Solution.

53


1. The form of z must be x+iy, where x, y are real numbers. Thus we have z2 = x2−y2+i2xy.Because f (z) = z2 is everywhere continuous and 2xy is an imaginary of this function, thereal function v (x, y) = 2xy is obviously everywhere continuous. Hence it is explicit to endthe explaination.

2. The continuity of complex function belongs to the continuity of both parts of that complexfunction. We have that xy is really a continuous function. We observe that function x+y

is also a continuous function because it does not belong to an arbitrary condition. Thusthe function g (x, y) must be continuous everywhere.

Exercise 2.13. (E. 15 P. 63)Show by finding an example, that the sum of two functions, neither of which possesses a limitat a point z0, can have a limit at this point.

Solution.Firstly, we take z0 = 1 for the following example. Then, we will choose two functions which

have no a limit at z0 = 1, witth assuming existent condition of denominator. These areg (z) =

z

z2 + z − 2

h (z) =−1

z2 + z − 2

The reader can see easily these limits at z0 = 1 that do not exist by factoring these denomi-nators. But the function f (z) = g (z)+h (z) can have a limit exactly at z0 = 1. For concretnesslet us write

limz→1

z − 1

(z − 1) (z + 2)= lim

z→1

1

z + 2=

1

3

Finally, we show the exercise by finding an example.

Exercise 2.14. (E. 16 P. 63)Show by finding an example, that the product of two functions, neither of which possesses alimit at a point z0, can have a limit at this point.

Solution.In a similar fashion, for the exercise 2.13 (E. 15 P. 63), we show that the product of two

functions, neither of which possesses a limit at a point z0, can have a limit at that point byfinding an example. This fact is not only true for the sum of two functions, but also the productof two functions.At this time, we take z0 = 0, and choose g (z) =

|z|z

;h (z) =|z|z

with z 6= 0 as the initialcondition. Although |z| is really a real number and is not relate to limit of the consideringfunctions, its effect is important for our proof. In details, we consider

limz→0

g (z)h (z) = limz→0

|z|2

zz= 1

54


We see that the limit of this product exists and is always equal to 1. But it is easy to showthat g (z) and h (z) have no limit at z0 = 0.Then, we prove this exercise completely.

Exercise 2.15. (E. 17 P. 63)Show that, in general, if g (z) has a limit as z tends to z0 but h (z) does not have such a limit,then f (z) = g (z) + h (z) does not have a limit as z tends to z0 either.

Solution.By showing mathematically, we must present the definition of limit and unlimitedness at a

point z0. We need analyse the assumption and conclusion in this needful proof. And we considerthree definitions of three clauses respectively.

• “g (z) has a limit as z tends to z0”

Let g0 be a complex constant.

If for every real number ε > 0 there exists a real number δ (ε) > 0 such that |g (z)− g0| < ε,for all z satisfying 0 < |z − z0| < δ (ε).

• For the clause “h (z) does not have such a limit”, we must negate the definition of limit.

Let h0 be a complex constant.

There exists a real number ε1 > 0, for every real number δ1 (z, ε1) > 0, have one z whichis satisfied 0 < |z − z0| < δ1 (z, ε1), such that |h (z)− h0| ≥ ε1.

• We also negate the definition of limit for the conclusion “f (z) = g (z) + h (z) does nothave a limit as z tends to z0”.

Let f0 = g0 + h0 be a complex constant.

We need find a real number ε2 > 0, for every given real number δ2 (z, ε2) > 0, have one zwhich is satisfied 0 < |z − z0| < δ2 (z, ε2), such that |f (z)− f0| ≥ ε2.

We look at the inequality |f (z)− f0| ≥ ε2 and do not forget that f (z) = g (z) + h (z) andf0 = g0 +h0. We deduce that |h (z)− h0 − [g0 − g (z)]| ≥ ε2. If we want to have this inequality,we must have the inequality |h (z)− h0|−|g (z)− g0| ≥ ε2 because of |h (z)− h0 − [g0 − g (z)]| ≥|h (z)− h0| − |g (z)− g0|. According to the first and the second clauses, we have |h (z)− h0| −|g (z)− g0| > ε1 − ε. Hence, we can take ε =

ε2

2, ε1 = 2ε2 and δ (ε) = δ1 (z, ε1) = δ2 (z, ε2) to

clarify the proof.

55


Exercise 2.16. (E. 18 P. 63)This problem deals with functions of a complex variable having a limit of infinity (or ∞). Wesay that lim

z→z0f (z) =∞ if, given p > 0, there exists a δ > 0 such that |f (z)| > ρ for all

0 < |z − z0| < δ. In other words, one can make the magnitude of f (z) exceed any preassignedpositive real number ρ if one remains anywhere within a deleted neighborhood of z0. The radiusof this neighborhood, δ, typically depends on ρ and shrinks as ρ increases.

1. Using this definition, show that limz→0

1

z=∞. What should we takes as δ?

2. Repeat the previous problem, but use the function1

(z − i)2 as z → i.

3. Consult a textbook on real calculus concerning the subject of infinite limits and explain

why one does not say that limx→0

1

x= ∞. What is the correct statement? Contrast this to

the result in (1).

4. The definition used above and in parts (1) and (2) cannot be used for functions whoselimits at infinity are infinite. Here we modify the definition as follows : We say thatlimz→∞

f (z) = ∞ if, given ρ > 0, there exists r > 0 such that |f (z)| > ρ for all r < |z|. Inother words, one can make the magnitude of f (z) exceed any preassigned positive realnumber ρ if one is at any point at least a distance r from the origin. Using this definition,show that lim

z→∞z2 =∞. How should we choose r?

Solution.

1. To show limz→0

1

z=∞, according to the definition, it says that :

Given p > 0, we must find a δp > 0 such that1

|z|> p, for all 0 < |z| < δp.

At this point, we observe that inequality 0 < |z| < δp can be conclude a distinct form1

|z|>

1

δp. This fact is the same as

1

|z|> p if we compare these. Hence, the δp, which we

need find, is obviously p.

Thus the proof is completed by choosing δp = p.

2. We still use the above definition for proving limz→i

1

(z − i)2 = ∞. We have the following

clause

Given p > 0, we must find a δp > 0 such that1

|z − i|2> p, for all 0 < |z − i| < δp.

If we unchange this clause to see the similarities, we cannot find out δp. We note that0 < |z − i| < δp can modify into 0 < |z − i|2 < δ2

p. Then if we put γp = δ2p > 0 and

|z − i|2 =∣∣∣(z − i)2

∣∣∣ = |w (z)|, we will see clearly that the clause is really the definition.For concretness let us write again

Given p > 0, we must find a γp > 0 such that1

|w (z)|> p, for all 0 < |w (z)| < γp.

At this point, we only prove this clause in a same way of part (1).

56


2.3 THE COMPLEX DERIVATIVE

In this section, we present exercise 3, 12, 14, 18, 19 for examples.Exercise 2.17. (E. 3 P. 70)Assume that f (z) = z is not differentiable. Obtain this conclusion by using the definition ofthe derivative of a function of a complex variable and show that this results in your having toevaluate lim

∆z→02 arg (∆z). Why does this limit not exist?

Solution.By using the definition of the derivative of a function of a complex variable, we take an

arbitrary point z0 = x0 + iy0, where x0, y0 are real numbers. We have f (z0) = x0 − iy0 andwith ∆z = ∆x+ i∆y, then f (z0 + ∆z) = (x0 + ∆x)− i (y0 + ∆y). Thus

f ′ (z0) = lim∆z→0

f (z0 + ∆z)− f (z0)

∆z

= lim∆x,∆y→0

(x0 + ∆x)− i (y0 + ∆y)−∆x+ i∆y

∆x+ i∆y

= lim∆x,∆y→0

∆x− i∆y∆x+ i∆y

We put g (∆x,∆y) =∆x− i∆y∆x+ i∆y

, we must prove this limit which does not exist. In explicit,

we can find two sequences which are also approach to 0, to prove that. For all n is an integer,

we take two sequences an =

(1

n; 0

)and bn =

(0;

1

n

). At infinity, they approach to (0; 0) but

g (an) = 1 differs from g (bn) = −1. Thus these mean that lim∆x,∆y→0

g (∆x,∆y) is does exist.

Hence, f (z) is not differentiable.

In a similar fashion, with ∆z = ∆x+ i∆y, we can see that arg (∆z) =∆y

∆x. Hence,

lim∆x,∆y→0

2 arg (∆z) = lim∆x,∆y→0

2∆y

∆x

Then we put h (∆x,∆y) = 2∆y

∆xand choose two sequences cn =

(1

n;

1

n

)and dn =

(1

n;− 1

n

)for all n is an integer. They also approach to (0; 0) at infinity but h (cn) = 2 differs fromh (dn) = −2. Thus this limit does not exist.

Exercise 2.18. (E. 12, E. 14 P. 70)For what values of the complex variable z do the following functions have derivatives?

1. ex + ie2y

2. (x− 1)2 + iy2 + z2

Solution.

1. It is easy to know that u (x, y) = ex and v (x, y) = e2y. So we consider

57


∂u

∂x= ex

∂u

∂y= 0

∂v

∂x= 0

∂v

∂y= 2e2y

as a test. And then we see that the Cauchy-Riemann equations is satisfied if only ifex = 2e2y since u (x, y) , v (x, y) are continuous in R.

Hence, if this function want to have derivate, the form of z must be log 2 + 2 log y + iy.

2. By using z = x+ iy, we can simplify this function as follows

(x− 1)2 + iy2 + z2 = (x− 1)2 + iy2 + (x+ iy)2

= (x− 1)2 + x2 − y2 + i(y2 + 2xy

)Then, we have u (x, y) = (x− 1)2 + x2 − y2 and v (x, y) = y2 + 2xy. At this point, wederive

∂u

∂x= 4x− 2

∂u

∂y= −2y

∂v

∂x= 2y

∂v

∂y= 2y + 2x

In a similar fashion, if this function want to have the derivate, we need 4x− 2 = 2y+ 2x.From this we can say that the form of z must be y + 1 + iy.

Exercise 2.19. (E. 18 P. 70)Let f (z) = u (x, y) + iv (x, y). Assume that the second derivative f ′′ (z) exists. Show that

f ′′ (z) =∂2u

∂x2+ i

∂2v

∂x2

and

f ′′ (z) = −∂2u

∂y2− i∂

2v

∂y2

Solution.We showed that

f ′ (z) =∂u

∂x+ i

∂v

∂x

and

58


f ′ (z) =∂v

∂y− i∂u

∂y

as the exercise. So in the first expression, we take the second derivate for x variable and fory variable in second expression. We thus assert that

f ′′ (z) =∂2u

∂x2+ i

∂2v

∂x2

and

f ′′ (z) = −∂2u

∂y2− i∂

2v

∂y2

since the second derivative f ′′ (z) exists.Hence, the proof is completed.

Exercise 2.20. (E. 19 P. 70)Show that if f ′ (z0) exists, then f (z) must be continuous at z0.

Solution.The existence of f ′ (z0) means that f (z) has a derivate at point z0. The definition says that

this is defined by the following expression, provided the limit exists

f ′ (z0) = lim∆z→0

f (z0 + ∆z)− f (z0)

∆z

At this point, if we desire that f (z) is contiunous at z0, we need prove

limz→z0

f (z) = f (z0)

If we put z = z0 + ∆z in the above expression and turn the right side to the left side becauseof definement of lim

z→z0f (z0), it becomes

lim∆z→0

[f (z0 + ∆z)− f (z0)] = 0

, as the consistent proof.Next, we consider this consistent proof which can be break into products of two limit. For

concretness let us write

lim∆z→0

[f (z0 + ∆z)− f (z0)] = lim∆z→0

f (z0 + ∆z)− f (z0)

∆z∆z

= lim∆z→0

f (z0 + ∆z)− f (z0)

∆zlim

∆z→0∆z

,∆z approachs to 0 but equals 0, that means ∆z 6= 0. And it is easy to see that lim∆z→0

∆z = 0,

we can derive the consistent proof because of the existence of lim∆z→0

f (z0 + ∆z)− f (z0)

∆z.

Therefore, we can end this proof successfully.

59


2.4 THE DERIVATIVE AND ANALYTICITY

We have exercise 2, 3, 9, 10, 11, 23 as the practice.Exercise 2.21. (E. 2 P. 77)

1. Find the derivative of f (z) =1

z+(x− 1)2 + ixy at any points where the derivative exists.

Give the numerial value.

2. Where is this function analytic ?

Solution.

1. We have f(z) =x− iyx2 + y2

+ (x− 1)2 + ixy (z 6= 0, it means that x, y don’t equal to zero at

the same time)

f(x, y) = u(x, y) + iv(x, y) with u(x, y) =x

x2 + y2+ (x− 1)2 and v(x, y) = xy − y

x2 + y2

We use Cauchy-Riemann equations to find this f(z)’s derivative

∂u

∂x= 2(x− 1) +

x2 + y2 − 2x2

(x2 + y2)2= 2(x− 1) +

y2 − x2

(x2 + y2)2

∂v

∂y= x− x2 + y2 − 2y2

(x2 + y2)2= x− x2 − y2

(x2 + y2)2

Let∂u

∂x(x0, y0) =

∂v

∂y(x0, y0) and we have 2(x0 − 1) +

y20 − x2

0

(x20 + y2

0)2= x0 −

x20 − y2

0

(x20 + y2

0)2

Thus x0 = 1

The second equations are

∂u∂y = −2xy

(x2+y2)2

− ∂v∂x = −y − 2xy

(x2+y2)2

By equality of these equations, we get y0 = 0. So f (z) is derivative at z = 1 andf ′ (1) = −1.

2. Because f (z) is only derivative at one point, f (z) is nowhere analytic.

Exercise 2.22. (E. 3 P. 77)

1. Where is the function f (z) = z3 + z2 + 1 analytic ?

2. Find an expression for f ′ (z) and give the derivative at 1 + i.

Solution.

1. Let z be x+ iy for all x, y in R, we have

f (z) = u (x, y) + iv (x, y)

60


with u (x, y) = x3 − 3xy2 + x2 − y2 and v (x, y) = −y3 + 3x2y + 2xy

Then we examine that f is statisfies Cauchy-Riemann equations. Indeed, we see that

∂u

∂x= 3x2 − 3y2 + 2x = ∂v

∂y

∂u

∂y= −6xy − 2y = −∂v

∂x

So this function is analytic everywhere.

2. We can get f ′ (z) = 3z2 + 2z easily because of similarity of real number. And then

f ′ (1 + i) = 3 (1 + i)2 + 2 (1 + i)

= 8i+ 2

Exercise 2.23. (E. 9, E. 10 P. 77)

Use L’Hopital’s Rule to establish these limits :

1.(z − i) +

(z2 + 1

)z2 − 3iz − 2

as z → i.

2.(z3 + i

)(z2 + 1) z

as z → i.

Solution.

1. We set g(z) = z − i+(z2 + 1

)and h(z) = z2 − 3iz − 2

We can easily see that g(i) = h(i) = 0 as z = i and g(z), h(z) are differentiable at z = i

(the reader can verify that). Moreover h′(i) = −i 6= 0 so we can use L’Hopital’s rule tosolve this limit

We have g′(z) = 1 + 2z and h′(z) = 2z − 3i

Thus limz→i

(z − i) + (z2 + 1)

z2 − 3iz − 2= lim

z→i

1 + 2z

2z − 3i= −1 + 2i

i= i− 2

2. Like that exercise above, we set g(z) = z3 + i and h(z) =(z2 + 1

)z. Of course g(i) =

h(i) = 0

g(z) and h(z) are differentiable and h′(i) = −2 6= 0, therefore we can use L’Hopital’s rule

limz→i

z3 + 1

(z2 + 1)z= lim

z→i

3z2

3z2 + 1=−3

−2=

3

2

Exercise 2.24. (E. 11 P. 77)If g (z) has a derivative at z0 and h (z) does not have a derivative at z0, explain why g (z)+h (z)

cannot have a derivative at z0.

Solution.

61


We can put g (z) = u (x, y) + iv (x, y) and h (z) = m (x, y) + in (x, y) to show the assumptionexplicitly. With z0 = (x0, y0), because of derivation of g (z), we have Cauchy-Riemann equationsat point z0.

∂u

∂x(x0, y0) =

∂v

∂y(x0, y0)

∂v

∂x(x0, y0) = −∂u

∂y(x0, y0)

And for the function h (z), it does not have a derivative at z0; this means one of Cauchy-Riemann equations is not satisfied at least. We thus choose the following expression for illus-tration.

∂m

∂x(x0, y0) 6= ∂n

∂y(x0, y0)

Then, we see that

∂m

∂x(x0, y0) +

∂u

∂x(x0, y0) 6= ∂n

∂y(x0, y0) +

∂v

∂y(x0, y0)

We note that if we take g (z) + h (z) = u (x, y) +m (x, y) + i [v (x, y) + n (x, y)], it is easily todeduce the sum of these functions cannot have a derivative at z0 by the above fact.

62


Exercise 2.25. (E. 23 P. 79)Polar form of the Cauchy-Riemann equations.

1. Suppose, for the analytic function f (z) = u (x, y) + iv (x, y), that we express x

and y in terms of the polar variables r and θ, where x = r cos θ and y = r sin θ(r =

√x2 + y2, θ = arctan

y

x

). Then f (z) = u (r, θ) + iv (r, θ). We want to rewrite the

Cauchy-Riemann equations entirely in the polar variables. From the chain rule for partialdifferentiation, we have

∂u

∂x=

(∂u

∂r

)θ

(∂r

∂x

)y

+

(∂u

∂θ

)r

(∂θ

∂x

)y

Give the corresponding expressions for∂u

∂y,∂v

∂x,∂v

∂y.

2. Show that (∂r

∂x

)y

= cos θ and(∂θ

∂x

)y

=− sin θ

r

and find corresponding expressions for(∂r

∂y

)x

and(∂θ

∂y

)x

. Use these four expressions

in the equations for∂u

∂x,∂u

∂y,∂v

∂xand

∂v

∂yfound in part (1). Show that u and v satisfy the

equations

∂h

∂x=∂h

∂rcos θ − 1

r

∂h

∂θsin θ and

∂h

∂y=∂h

∂rsin θ +

1

r

∂h

∂θcos θ

where h can equal u or v.

3. Rewrite the Cauchy-Riemann equations using the two equations from part (2) of thisexercise. Multiply the first Cauchy-Riemann equation by cos θ, multiply the second bysin θ, and add to show that

∂u

∂r=

1

r

∂v

∂θ.

Now multiply the first Cauchy-Riemann equation by − sin θ, the second by cos θ, and addto show that

∂v

∂r=−1

r

∂u

∂θ.

These are the polar form of the Cauchy-Riemann equations. If the first partial derivativesof u and v are continuous at some point whose polar coordinates are r, θ (r 6= 0), then thepolar form of the Cauchy-Riemann equations provide a necessary and sufficient conditionfor the existence of the derivative at this point.

4. Use equation f ′ (z0) =

(∂u

∂x+ i

∂v

∂x

)x0,y0

and the Cauchy-Riemann equations in polar

form to show that if the derivative of f (r, θ) exists, it can be found from

f ′ (z) =

[∂u

∂r+ i

∂v

∂r

][cos θ − i sin θ] or from f ′ (z) =

[∂u

∂θ+ i

∂v

∂θ

](−ir

)[cos θ − i sin θ]

Solution.

63


1. The exercise has already given us

∂u

∂x=

(∂u

∂r

)θ

(∂r

∂x

)y

+

(∂u

∂θ

)r

(∂θ

∂x

)y

so∂u

∂y,∂v

∂x,∂v

∂yare just the same expression as above, using Cauchy-Riemann equations

we have

∂u

∂y=

(∂u

∂r

)θ

(∂r

∂y

)x

+

(∂u

∂θ

)r

(∂θ

∂y

)x

∂v

∂x=

(∂v

∂r

)θ

(∂r

∂x

)y

+

(∂v

∂θ

)r

(∂θ

∂x

)y

∂v

∂y=

(∂v

∂r

)θ

(∂r

∂y

)x

+

(∂u

∂θ

)r

(∂θ

∂y

)x

2. We get r =√x2 + y2 and tan θ =

y

xso that

(∂r

∂x

)y

=x√

x2 + y2= cos θ

To solve(∂θ

∂x

)y

, first we have to compute (tan θ)′ =θ′

cos2 θ. Then we obtain

(∂θ

∂x

)y

=−yx2

cos2 θ

=−yx2

1

1 + y2

x2

=− sin θ

r

Let h be u, we replace∂r

∂x,∂θ

∂xwith these above and see that

∂h

∂x=∂h

∂rcos θ − 1

r

∂h

∂θsin θ

Hence, we can do the same for the others.

3. Using the above part, we have

∂u

∂x=

∂u

∂rcos θ − 1

r

∂u

∂θsin θ

∂v

∂y=

∂v

∂rsin θ +

1

r

∂v

∂θcos θ

We multiply them to cos θ and let them equal as Cauchy-Riemann first equation. It means

64


∂u

∂xcos θ =

∂v

∂ycos θ

Express this and then

∂u

∂rcos2 θ − 1

r

∂u

∂θsin θ cos θ =

∂v

∂rsin θ cos θ +

1

r

∂v

∂θcos2 θ

We do the same with Cauchy-Riemann second equation but we multiply them to sin θ

instead of cos θ and express them too. We obtain

∂v

∂rcos θ sin θ − 1

r

∂v

∂θsin2 θ = −∂v

∂rsin2 θ − 1

r

∂u

∂θcos θ sin θ

The next is adddition of these two expression to each other and simplification of them.Thus

∂u

∂rcos2 θ − 1

r

∂v

∂θsin2 θ =

1

r

∂v

∂θcos2 θ −−∂v

∂rsin2 θ

Because of sin2 θ + cos2 θ = 1, we deduce that

∂u

∂r=

1

r

∂v

∂θ

Hence we prove this part successfully, the reader can do the others.

4. We need to prove

f ′ (z) =

[∂u

∂r+ i

∂v

∂r

](cos θ − i sin θ)

And we have

f ′ (z) =∂u

∂x+ i

∂v

∂x

We use part (b) to express∂u

∂x,∂v

∂x. Hence,

∂u

∂x=∂u

∂rcos θ − 1

r

∂u

∂θsin θ (1)

∂v

∂x=∂v

∂rcos θ − 1

r

∂v

∂θsin θ (2)

Then, we replace (1) and (2) into f ’(z).

∂u

∂rcos θ−1

r

∂u

∂θsin θ+i

(∂v

∂rcos θ − 1

r

∂v

∂θsin θ

)= cos θ

(∂u

∂r+ i

∂v

∂r

)−1

rsin θ

(∂u

∂θ+ i

∂v

∂θ

)

We are now using part (c) to simplify above equation

65


cos θ

(∂u

∂r+ i

∂v

∂r

)− 1

rsin θ

(−r∂v

∂r+ ri

∂u

∂r

)Note that i2 = −1, so we can deduce that

cos θ

(∂u

∂r+ i

∂v

∂r

)− 1

rsin θ

(−r∂v

∂r+ ri

∂u

∂r

)= cos θ

(∂u

∂r+ i

∂v

∂r

)− i sin θ

(i∂v

∂r+∂u

∂r

)=

[∂u

∂r+ i

∂v

∂r

](cos θ − i sin θ)

We prove the exercise for the first feedback, and in the second, the reader can showsimilarly.

66


2.5 HARMONIC FUNCTIONS

The reader can see some exercises we will show such as 1, 2, 3, 6, 7, 8, 9, 11, 12, 13, 18, 20.Exercise 2.26. (E. 1 P. 85)Where in the complex plane will the function φ(x, y) = x2−y4 satisfy Laplace’s equation? Whyisn’t this a harmonic function?

Solution.Take (x0, y0) is a point of function, we will find this point such that φ (x0, y0) satisfies Laplace’s

equation.

∂2φ

∂x2(x0, y0) +

∂2φ

∂y2(x0, y0) = 0

This is equivalent to x0 ∈ R

y0 =±1√

6

Because the domain of function which satisfies Laplace’s equation is a subset of R2. Thus itis not a harmonic function.

Exercise 2.27. (E. 2 P. 86)Where in the complex plane will the function φ(x, y) = sin(xy) satisfy Laplace’s equation? Whyisn’t this a harmonic function?

Solution.We have φ (x, y) = sin (xy), let take the first derivative and second derivative of φ (x, y)

• ∂φ

∂x= y cos (xy) and then

∂2φ

∂x2= −y2 sin (xy)

• ∂φ

∂y= x cos (xy)so that

∂2φ

∂y2= −x2 sin (xy)

Hence we obtain

∂2φ

∂x2+∂2φ

∂y2= −

(x2 + y2

)sin (xy)

Dealing with the equation(x2 + y2

)sin (xy) = 0. we have the result is sin (xy) = 0 or x2+y2 =

0. Then we can derive the roots belowy0 =kπ

x0, x0 6= 0

y0 = 0, x0 = 0

Therefore, φis not harmonic on its domain but harmonic on (0, 0) and(x0,

kπ

x0

)when x0 6= 0.

67


Exercise 2.28. (E. 3 P. 86)Consider the function φ(x, y) = eky sin(mx). Assuming this function is harmonic throughoutthe complex plane, what must be he relationship between the real constants k and m? Assumethat m 6= 0.

Solution.This is a harmonic function then we can display Laplace’s equation there.

∂2φ

∂x2+∂2φ

∂y2= 0

It is equivalent to

−m2eky sin(mx) + k2eky sin(mx) = 0

So we can derive that k2 = m2.

Exercise 2.29. (E. 6 P. 86)Putting z = x+ iy. Show that <

(z3)is harmonic in any domain.

Solution.Factoring out z3, we obtain

z3 = x3 − 3xy2 + i(3x2y − y3

)Thus <

(z3)

= x3 − 3xy2. We will examine this function whether or not satisfying Laplace’sequation.

∂2

∂x2

(x3 − 3xy2

)+

∂2

∂y2

(x3 − 3xy2

)= 6x− 6x = 0

Hence, it is harmonic in any domain.

Exercise 2.30. (E. 7 P. 86)Find two values of k such that cosx

[ey + eky

]is harmonic.

Solution.We put u(x, y) = cosx

[ey + eky

]is harmonic function. Then we get

∂2u

∂x2+∂2u

∂y2= 0

It means that

−(ey + eky

)cosx+

(ey + k2eky

)cosx = 0

After solving the above equation, we have k = ±1.

Exercise 2.31. (E. 8 P. 86)If g(x)

[e2y − e−2y

]is harmonic, g(0) = 0, g′(0) = 1, find g(x).

68


Solution.Like the above exercise, we put u(x, y) = g(x)

[e2y − e−2y

]is harmonic function and get

Laplace’s equation.

∂2u

∂x2+∂2u

∂y2= 0

This is equivalent to

[e2y − e−2y

]g′′(x) + 4g(x)

[e2y − e−2y

]= 0

Then we need to solve the following equation

g′′(x) + 4g(x) = 0

We can have general root is g(x) = C1 cos 2x+ C2 sin 2x.

Exercise 2.32. (E. 9 P. 86)

1. Consider φ(x, y) = x3y−y3x+y2−x2+x. Show that this can be the part or the imaginarypart of an analytic function.

2. Assuming the preceding is the real part of an analytic function, find the imaginary part.

3. Assuming that φ(x, y) is the imaginary part of an analytic function, find the real part.

4. If φ(x, y) + iv(x, y) is an analytic function and if u(x, y) + iφ(x, y) is also analytic whereφ(x, y) is an arbitrary harmonic function, prove that, neglecting constants, u(x, y) andv(x, y) must be negatives of each other.

Solution.

1. With φ (x, y) = x3y− y3x+ y2 − x2 + x, we prove this can be real part or imaginary partof an analytic function by proving this satisfy Laplace’s equation.

• ∂φ

∂x= 3x2y − y3 − 2x+ 1 so

∂2φ

∂x2= 6xy − 2

• ∂φ

∂y= x3 − 3y2x+ 2y so

∂2φ

∂y2= −6xy + 2

Let∂2φ

∂x2+∂2φ

∂y2= 6xy − 2− 6xy + 2 = 0

Thus φ (x, y)satisfy Laplace’s equation. Hence it is harmonic, therefore φ (x, y)can be realpart or imaginary part of an analytic function.

2. Assuming φ (x, y) is the real part of an analytic function, we need to find the imaginarypart v (x, y) of this function.

By applying Cauchy-Riemann equations

• ∂φ

∂x=∂v

∂yso

∂v

∂y= 3x2y − y3 − 2x+ 1

• v =3

2x2y2 − y4

4− 2xy + y + C ′ (x)

69


The second equation is −x3 + 3xy2 − 2y .

Hence C ′ (x) = −x3, it means that C (x) = −x4

4+ C.

Thus v (x, y) =3

2x2y2 − y4

4− 2xy + y +−x

4

4+ C

3. As the same of part (b), let φ (x, y) be the imaginary part of an anlytic function and wefind u (x, y) is the real part of this function

We found u (x, y) = −v (x, y) which we found in part (b)

4. We have f = u+ iφ analytic so we deduce that

∂u

∂x=∂φ

∂y∂u

∂y= −∂φ

∂x

As the same g = φ+ iv analytic so we derive that

∂φ

∂x=∂v

∂y∂φ

∂y= −∂v

∂x

By transforming these, we have ∂u

∂x= −∂v

∂x∂u

∂y= −∂v

∂y

By neglecting constants, we see that u (x, y) = −v (x, y) so we have what we need to prove.

Exercise 2.33. (E. 11 P. 86)Find the harmonic conjugate of ex cos y + ey cosx+ xy.

Solution.Putting u the harmonic conjugate of v(x, y) = ex cos y + ey cosx + xy, we need to apply

Cauchy-Riemann equations : ∂u∂x = ∂v

∂y

∂u∂y = − ∂v

∂x

We have∂u

∂x= ex cos y−ey sinx+y. Then we can deduce that v = ex sin y−ey sinx+

y2

2+C(x).

Hence, we obtain that

∂v

∂x= ex sin y − ey cosx+ C ′(x)

But we can see that

∂u

∂y= −ex sin y + ex cosx+ x

Thus C ′(x) = x. It means that C(x) =x2

2+ C. Then the harmonic conjugate is ex sin y −

ey sinx+y2

2+x2

2+ C.

70


Exercise 2.34. (E. 13 P. 86)Show, if u(x, y) and v(x, y) are harmonic functions, that u+v must be a harmonic function butthat uv need not be a harmonic function.

Solution.We get u, v are harmonic functions. That means

∂2u

∂x2+∂2u

∂y2= 0

∂2v

∂x2+∂2v

∂y2= 0

If u+ v is harmonic, then need to show

∂2(u+ v)

∂x2+∂2(u+ v)

∂y2= 0


∂2u

∂x2+∂2v

∂x2+∂2u

∂y2+∂2v

∂y2= 0

This can be derived from Cauchy-Riemann equations above.But if (uv) is harmonic, then we need to prove

∂2(uv)

∂x2+∂2(uv)

∂y2= 0

We can see that

∂2(uv)

∂x2= v

∂2u

∂x2+ u

∂2v

∂x2+ 2

∂v

∂x

∂u

∂x∂2(uv)

∂y2= v

∂2u

∂y2+ u

∂2v

∂y2+ 2

∂v

∂y

∂u

∂y

Then by using Cauchy-Riemann equations above, we get only that

∂2(uv)

∂x2+∂2(uv)

∂y2= 2

(∂v

∂x

∂u

∂x+∂v

∂y

∂u

∂y

)It is not really equal to 0. Hence, we complete the proof.

Exercise 2.35. (E. 18 P. 86)Show that f(z) = ex cos y + iex sin y = u+ iv is entire.

Solution.Examining Cauchy-Riemann equations, we obtain

∂u

∂x− ∂v

∂y= ex cos y − ex cos y = 0

∂u

∂y+∂v

∂x= −ex sin y + ex sin y = 0

Because it is analytic throughout finite z-plane, it is entire function.

71


Exercise 2.36. (E. 20 P. 87)

1. Let x = r cos θ and y = r sin θ, where r and θ are the usual polar coordinate variables.Let f(z) = u(r, θ) + iv(r, θ) be a function that is analytic in some domain that does notinclude z = 0. An assumed continuity of second partial derivatives to show that in thisdomain u and v satisfy the differential equation

∂2φ

∂r2+

1

r2

∂2φ

∂θ2+

1

r

∂φ

∂r= 0

This is Laplace’s equation in the polar variables r and θ.

2. Show that u(r, θ) = r2 cos 2θ is a harmonic function.

3. Find v(r, θ), the harmonic conjugate of u(r, θ), and show that it too satisfies Laplace’sequation everywhere.

Solution.

1. By using the polar coordinate variables, we have the form of the Cauchy-Riemann equa-tions is now

∂u

∂r=

1

r

∂v

∂θ∂v

∂r= −1

r

∂u

∂θ

In the system of equations, we deduce that

∂2u

∂r2=−1

r2

∂v

∂θ+

1

r

∂2v

∂θ∂r∂2u

∂θ2= −r ∂

2v

∂r∂θ

Hence,

∂2u

∂r2+

1

r2

∂2u

∂θ2+

1

r

∂u

∂r=−1

r2

∂v

∂θ+

1

r

∂2v

∂θ∂r− 1

r

∂2v

∂r∂θ+

1

r2

∂v

∂θ= 0

It is the same for case of v.

2. We see that u has continuity of second partial derivatives in the given domain. Thus wecan easily show that it is harmonic function by applying equation above.

3. Using the Cauchy-Riemann equations above, we have

∂v

∂θ= r

∂u

∂r= 2r2 cos 2θ

Then we deduce that v = r2 sin 2θ + C(r). Hence

72


∂v

∂r= 2r sin 2θ + C ′(r)

But we note that−1

r

∂u

∂θ= 2r sin 2θ, then C ′(r) = 0. It means that C(r) = C. So that

v = r2 sin 2θ + C.

Thus it too satisfies Laplace’s equation everywhere.

73

3 THE BASIC TRANSCENDENTALFUNCTIONS

3.1 THE EXPONENTIAL FUNCTION

The reader can see some exercises below such as 2, 3, 4, 5, 6, 13, 14, 15, 16, 19, 20, 22, 23, 25,26.

Exercise 3.1. (E. 2, E. 3, E. 4, E. 5, E. 6 P. 106)Express each of the following in the form a + ib where a and b are real numbers. If the resultis multivalued, be sure to state all the values.

1. e1/2+2i

2. e1/2−2i

3. e−i

4. e1/2+2ie−1/2−2i

5. e(−i)7

Solution.It is easy for the reader or even the beginners. Indeed, we only need to get the brief solutions.

1. e12

+2i = e12 (cos 2 + i sin 2) = e

12 cos 2 + ie

12 sin 2

2. e12−2i = e

12 (cos 2− i sin 2)

3. e−i = cos (−1) + i sin (−1)

4. e12

+2ie−12−2i = e

12

+2i− 12−2i = e0 = 1

5. e(−i)7 = e(−i)6(−i) = ei = e0 (cos 1− i sin 1)

Exercise 3.2. (E. 13 P. 106)

Find all solution of ez = e by equating corresponding parts (reals and imaginaries) on bith sidesof the equation.

Solution.We have

74

3 THE BASIC TRANSCENDENTAL FUNCTIONS

ex+iy = ex cos y + iex sin y

e1+i0 = e+ i0

So ex cos y + iex sin y = e+ i0 and we can deduce thatex cos y = e

ex sin y = 0

Then we take x = 1 and notice that cosx = 1 and sinx = 0 which mean tan y = 0. Thusy = kπwhere k ∈ Z

Exercise 3.3. (E. 14, E. 15, E. 16 P. 106)

Recalling that an analytic functionn of an analytic function is analytic, state the domain ofanalyticity of each the following functions. Find the real and imaginary parts u (x; y) and v (x; y)

of the function, show that these satisfy the Cauchy-Riemann equations and find f ′ (x) in termof z.

1. f (z) = eiz

2. e1/z

3. eez

Solution.

1. Let z = x+ iy. We have

f (z) = eiz = ei(x+iy) = eix−y = e−y (cosx+ i sinx) = e−y cosx+ ie−y sinx

Thus

u (x; y) = e−y cosx

v (x; y) = e−y sin y

Check the Cauchy-Riemann equations

∂u

∂x= −e−y sinx =

∂v

∂y

∂u

∂y= −e−y cos y = −∂v

∂x

Thus f (z) analytic everywhere

75


f ′ (z) = ieiz

2. We have

e1z = e

x−iyx2+y2 = e

xx2+y2

(cos

y

x2 + y2+ i sin

y

x2 + y2

)u (x; y) = e

xx2+y2 cos y

x2+y2

v (x; y) = ex

x2+y2 sin yx2+y2

We compute some following derivative and consider the Cauchy-Riemann equations

∂u

∂x=−x2 + y2

(x2 + y2)2 .ex

x2+y2 . cos2xy

(x2 + y2)+

2xy

(x2 + y2)2 .ex

x2+y2 . siny

x2 + y2=∂v

∂y∂u

∂y=

−2xy

(x2 + y2)2 .ex

x2+y2 . cos2xy

(x2 + y2)+−x2 + y2

(x2 + y2)2 .ex

x2+y2 . siny

x2 + y2=−∂v∂x

Thus f (z) = e1z is an analytic function, so we can compute f ′ (z) is term of z

f ′ (z) = − 1

z2e

1z

3. In a similar way, we obtain

eez

= eex(cos y+i sin y) = ee

x cos y [cos (ex sin y) + i sin (ex sin y)]

Then we set

u (x, y) = eex cos y cos (ex sin y)

v (x, y) = eex cos y sin (ex sin y)

We will use Cauchy-Riemann equations to examine its analytic domain

∂u

∂x= exee

x cos y cos (y) cos (ex sin y)− eex cos y sin (ex sin y) ex sin y

∂u

∂y= −exeex cos y sin (y) cos (ex sin y)− eex cos y sin (ex sin y) ex cos y

∂v

∂x= exee

x cos y cos (y) sin (ex sin y) + eex cos y cos (ex sin y) ex sin y

∂v

∂y= −exeex cos y sin (y) sin (ex sin y) + ee

x cos y cos (ex sin y) ex cos y

From the following equations we can see that f (z) satisfy Cauchy-Riemann equations onentire Z plane. So f (z) is an entire function.

Thus we have f ′ (z) = ezeez

76


Exercise 3.4. (E. 19 P. 106)

Consider the indentity e(z1+z2) = ez1ez2 , which we proved somewhat tediously in this section.

Here is an elegant the proof which relies on our knowing thatdez

dz= ez and e0 = 1.

1. Taking a as a constant, show thatd (ezea−z)

dz= 0 by using the usual formula for the

derivative of a product, as well as the derivative of ez, and the chain rule. Note that youcannot combine the exponents, as this has not been justified.

2. Since ezea−z has just been shown to bae a constant, which we will call k, evaluate k interms of a by using the fact that e0 = 1.

3. Using ezea−z = k as well as k found above, and z = z1, a = z1 + z2, show that ez1+z2 =

ez1 .ez2 .

Solution.

1. We have

d (ezea−z)

dz= ez.ea−z +

(ea−z

)′.ez

= ez.ea−z + ez(−ea−z

)= 0

2. Because of ez.ea−z = k where k is a constant, we obtainezea

ez= k then k = ea.

3. We see the equations below

ea = k

z = z1

a = z1 + z2

And notice that k = ezea−z is equivalent to ez1 .ez1+z2−z1 = ez1+z2 , hence ez1ez2 = ez1+z2 .

Exercise 3.5. (E. 22, E. 23 P. 106)For the following closed bounded regions, R, where does the given |f (z)| achieve its maximumand minimum values, and what are these values?

1. R is |z − 1− i| ≤ 2 and f (z) = ez

2. R is |z| ≤ 1 and f (z) = e(z2)

Solution.

1. We have

77


|f (z)| = |ez| = ex

And,

|z − 1− i| ≤ 2

is the circle (inside and on the bounded as well) centered 1 + i with radius 2.

So |f (z)|has maximum and minimum when |ez| = ex has too that mean x is maximumand minimum.

From the region R we can see that x achieve maximum when x = 3, y = 1 and minimumwhen x = −1, y = 1. So that

max |f (z)| = e3

min |f (z)| = e−1

2. We have

e(z2) = e(x+iy)2 = ex

2−y2e2xyi = ex2−y2 (cos 2xy + i sin 2xy)∣∣∣e(z2)∣∣∣ =

∣∣∣ex2−y2 (cos 2xy + i sin 2xy) = ex2−y2

∣∣∣And f (z) has maximum value of ex

2−y2when x2 has maximum and y2 has minimum whichmean x = ±1, y = 0. Thus max |f (z)| = e.

Similarly, f (z) has minimum value of ex2−y2 when x2 has minimum and y2 has maximum

which mean x = 0, y = ±1. Hence, min |f (z)| = e−1.

78


Exercise 3.6. (E. 25 P. 107)

The absolute magnitude of the expression:

P (ψ) = 1 + eiψ + e2iψ + . . .+ ei(N−1)ψ =

N−1∑n=0

einψ

is of interest in many problems involving radiation from N identical physical elements (e.g.,antennas, loudspeakers). Here ψ is real quantity that depends on the separation of the elementand the position of an observer of the radiation. |P | can tell us the strength of the radiationobserved.

1. Using the formula for the sum of finite geometric series (see Exercise 27, section 1.4), showthat

|P (ψ)| =

∣∣∣∣∣∣∣∣sin

(Nψ

2

)sin

(ψ

2

)∣∣∣∣∣∣∣∣

2. Find limψ→0 |P (ψ)|

Solution.

1. Use the formular P (ψ) = 1 + eiψ + e2iψ + . . .+ ei(N−1)ψ =

(eiψ)N − 1

eiψ − 1We obtain

|P (ψ)| =

∣∣∣∣∣(eiψ)N − 1

eiψ − 1

∣∣∣∣∣=

∣∣∣(eiψ)N − 1∣∣∣

|eiψ − 1|

=|cos (Nψ)− 1 + i sin (Nψ)||cos (ψ)− 1 + i sin (ψ)|

=

√2− 2 cos (Nψ)√

2− 2 cosψ

=

√4 sin2 Nψ

2√4 sin2 ψ

2

=

∣∣∣∣∣sin Nψ2

sin ψ2

∣∣∣∣∣2. We get eiM0 = e0 = 1 for all M ∈ N. Hence,

79


lim→ψ|P (ψ)| = lim

ψ→0

∣∣∣1 + eiψ + ei2ψ + . . .+ ei(N−1)ψ∣∣∣

=∣∣∣1 + ei.0 + ei.2.0 + . . .+ ei.(N−1).0

∣∣∣= |1 + 1 + 1 + . . .+ 1| = N

Exercise 3.7. (E. 26 P. 107)

Let z = reiθ, where r and θ are the usual polar variables.

Show that <[

1 + z

1− z

]=

1− r2

1 + r2 − 2r cos θ. Why must this function satisfy Eq. (2.5-14) through-

out any domain not containing z = 1.

Solution.We have

1 + z

1− z=

1 + reiθ

1− reiθ=

1 + r (cos θ + i sin θ)

1− r (cos θ + i sin θ)

=1 + r cos θ + ir sin θ

1− r cos θ − ir sin θ

=(1 + r cos θ + ir cos θ) (1− r cos θ + ir sin θ)

(1− r cos θ)2 + r2 sin2 θ

=1− r2 + i2r sin θ

1 + r2 − 2r cos θ

Hence, we easily see that

<[

1 + z

1− z

]=

1− r2

1 + r2 − 2r cos θ

Because we can easily prove that1 + z

1− zis analytic on the Z plane except z = 1 so when it is

analytic, its real and imaginary part must be harmonic so <(

1 + z

1− z

)must satisfy equation.

80


3.2 TRIGONOMETRIC FUNCTIONS

The next are exercises 5, 9, 10, 12, 13, 14, 15, 16, 17, 25, 26, 27, 28, 29, 30.Exercise 3.8. (E. 5, E. 9, E. 10 P. 112)Using Eqs. (3.2-9) and (3.2-10), find the numerical values of the following in the form a + ib,where a and b are real numbers. If there is more than one numerical value, state all of them.

1. sin(i1/2)

2. tan(i arg

(1 +√

3i))

3. arg (tan i)

Solution.This exercise is basic for the beginner. We have 3 brief solutions there.

1. sin(i12

)= sin

(√2

2+ i

√2

2

)= sin

(√2

2

)cosh

(√2

2

)+ i cos

(√2

2

)sinh

(√2

2

)

2. tan(i arg

(1 + i

√3))

= tan(i(π

3+ k2π

))=

sin(i(π

3+ k2π

))cos(i(π

3+ k2π

)) =i sinh

(π3

+ k2π)

cosh(π

3+ k2π

)

3. arg (tan i) = arg

(sin i

cos i

)= arg

(sin 0 cosh 1 + i cos 0 sinh 1

cos 0 cosh 1− i sin 0 sinh 1

)= arg

i(e2−1)

2(e2 + 1

)2

= arg

[i(e2 − 1

)e2 + 1

]=

π

2+ k2π

Exercise 3.9. (E. 12 P. 112)

Prove the identity sin2 z + cos2 z = 1 by the following two methods

1. Use the definitions of sine and cosine contained in Eq.(3.2-5) and (3.2-6).

2. Use cos2 z + sin2 z=(cos z + i sin z) (cos z − i sin z) as well as Euler’s identity generalizedto complex z.

Solution.

1. From the equation 3.2-5, we can get

sin2 z =

(eiz − e−iz

2i

)2

=e2iz + e−2iz − 2

−4

And the equation 3.2-6 is that

cos2 z =

(eiz + e−iz

2

)2

=e2iz + e−2iz + 2

4

We take a sum of sin2 z and cos2 z is

81


cos2 z + sin2 z =2 + 2

4= 1

2. It is easy to see that

cos2 z + sin2 z = (cos z + i sin z) (cos z − i sin z)

= eiz.e−iz = e0 = 1.

Exercise 3.10. (E. 13, E. 14, E. 15 P. 112)Using the definitions of the sine and cosine, Eqs. (3.2-5) and (3.2-6), prove the following.

1.d

dzsin z = cos z and

d

dzcos z = − sin z

2. cos2 z =1

2+

1

2cos 2z

3. sin (z + 2π) = sin z and cos (z + 2π) = cos z

Solution.

1. We will solve them directly.

d

dz(sin z) =

d

dz

(eiz − e−iz

2i

)1

2i

(d

dz

(eiz)− d

dz

(e−iz

))1

2i

(ieiz − (−i) e−iz

)1

2

(eiz + e−iz

)= cos z

d

dz(cos z) =

d

dz

(eiz + e−iz

2

)=

1

2

(d

dz

(eiz)

+d

dz

(e−iz

))=

1

2i

(−eiz − (−1) e−iz

)= − 1

2i

(eiz − e−iz

)= sin z

2. The next is similar.

82


cos2 z =

(eiz + e−iz

2

)2

=ei2z + e−i2z + 2

4

=1

2+

1

2cos 2z

3. Furthermore, we obtain

sin (z + 2π) =ei(z+2π) − e−i(z+2π)

2i

=eizei2π − e−ize−i2π

2i

=eiz − e−iz

2i= sin z

cos (z + 2π) =ei(z+2π) + e−i(z+2π)

2

=eizei2π + e−ize−i2π

2

=eiz + e−iz

2= cos z

Exercise 3.11. (E. 16, E. 17 P. 112)

1. Show that the equation sin z = 0 has solutions in the complex z−plane only where z = nπ

and n = 0,±1,±2, . . . Thus like cos z; sin z has zeros only on the real axis.

2. Show that sin z − cos z = 0 has solution only for real values of z. What are the solution?

Solution.

1. From the equation 3.2-5, we have

sin z =eiz − e−iz

2i

Hence,

eiz − e−iz

2i= 0

Simplify this and we get e2iz = 1. Thus the solution of sin z = 0 is 2z = k2π where k ∈ Z,so that z = kπ.

83


We can see that this solution is real so sin z only has zeros on the real axis.

In a similar fashion, we can get the solution of cos z = 0 which are z =π

2+ mπ where

m ∈ Z. Then cos z also has zeros only on the real axis.

2. From the equations 3.2-5 and 3.2-6, we have cos z =eiz + e−iz

2and sin z =

eiz − e−iz

2i.

Take a subtraction, we see that

sin z − cos z =eiz − e−iz

2i− eiz + e−iz

2

=eiz (1− i)− e−iz (1 + i)

2i

Let sin z − cos z = 0 which meanseiz (1− i)− e−iz (1 + i)

2i= 0

Transform this and we get√

2

(ei

(z−π

4

)− e

i

(π4−z))

= 0

Hence the solutions of this equation are 2z−π2

= k2π where k ∈ Z. ThenWe get z =π

4+kπ

where k ∈ Z.

The reader can see that these solutions are real values so sin z − cos z = 0 has solutiononly for real values of z and its solutions are z =

π

4+ kπ where k ∈ Z

Exercise 3.12. (E. 25, E. 26, E. 27, E. 28 P. 112)

1. Show that |cos z| =√

sinh2 y + cos2 x

2. Show that |sin z| =√

sinh2 y + sin2 x

3. Show that |sin z|2 + |cos z|2 = sinh2 y + cosh2 y

4. Show that

tan z =sin 2x+ i sinh 2y

cos 2x+ cosh 2y

Solution.

84


|cos z| = |cosx cosh y − i sinx sinh y|

=

√[cosx cosh y]2 + [sinx sinh y]2

=√

cos2 x[1 + sinh2 y

]+ [1− cos2 x] sinh2 y

=

√cos2 x+ cos2 x sinh2 y + sinh2 y − cos2 x sinh2 y

=

√cos2 x+ sinh2 y

|sin z| =

√sin2 x. cosh2 y + cos2 x. sinh2 y

=√

sin2 x.(1 + sinh2 y

)+ cos2 x. sinh2 y

=

√sin2 x+ sinh2 y

|sin z|2 + |cos z|2 = sinh2 y + cos2 x+ sinh2 y + sin2 x

= 2 sinh2 y + 1

= 2 sinh2 y + cosh2 y − sinh2 y

= sinh2 y + cosh2 y

tan z =sin z

cos z

=sin (x+ iy)

cos (x+ iy)

=sinx cosh y + i cosx sinh y

cosx cosh y − i sinx sinh y

=(sinx cosh y + i cosx sinh y) (cosx cosh y + i sinx sinh y)

cos2 x cosh2 y + sin2 x sinh2 y

We have

cos2 x cosh2 y + sin2 x sinh2 y = cos2 x cos2 (iy)− sin2 x sin2 (iy)

=(1 + cos 2x) (1 + cos (2iy))− (1− cos 2x) (1− cos (2iy))

4

=cos 2x+ cos (2iy)

2

=cos 2x+ cosh (2iy)

2

Then

(sinx cosh y + i cosx sinh y) (cosx cosh y + i sinx sinh y) =1

2sin 2x

(cosh2 y − sinh2 y

)+ i sinh y. cosh y

(sin2 x+ cos2 x

)=

1

2sin 2x+ i sinh y. cosh y

=1

2sin 2x+ sin (iy) . cos (iy)

=1

2(sin 2x+ i sinh 2y)

85


So that

tan z =sin 2x+ i sinh 2y

cos 2x+ cosh 2y

Exercise 3.13. (E. 29, E. 30 P. 113)

1. Show that

cot z =sin 2x− i sinh 2y

cosh 2y − cos 2x

2. Since sin z = sinx cosh y+ i cosx sinh y and |sinh y| ≤ cosh y, show that |sinh y| ≤ |sin z| ≤|cosh y|

3. Derive a comparable double inequality for |cos z|.

Solution.

1. We have

cot z =cos z

sin z

=cos (x+ iy)

sin (x+ iy)

=cosx cosh y − i sinx sinh y

sinx cos y + i cosx sinh y

=(cosx cosh y − i sinx sinh y) (sinx cos y − i cosx sinh y)

sin2 x cos2 y + cos2 x sinh2 y

At this time, we do this is the same as exercise 28. Then we can complete the solution.

2. We get the inequality below.

|sin z| =√

sin2 x+ sinh2 y ≥√

sinh2 y ≥ |sinh y|

And

|sin z| =√

sin2 x cosh2 y + cos2 x sinh2 y ≤√

sin2 x cosh2 y + cos2 cosh2 y ≤ cosh2 y

Thus|sinh y| ≤ |sin z| ≤ |cosh y|

3. We obtain that

|cos z| =√

cos2 x+ sinh2 y ≥√

sinh2 y = |sinh y|

86


And

|cos z| =√

cos2 x+ cosh2 y − 1 =

√cosh2 y − sin2 x ≤

√cosh2 y = |cosh y|

So a comparable double inequality is |sinh y| ≤ |cos z| ≤ |cosh y|

87


3.3 HYPERBOLIC FUNCTIONS

We show the solutions of exercises 1, 2, 5, 10, 11, 12, 18, 19.Exercise 3.14. (E. 1, E. 2, E. 5 P. 114)

Use Equations. (3.3-1) and (3.3-2) to prove the following

1. sinh z = sinhx. cos y + i coshx. sin y

2. cosh z = coshx. cos y + i sinhx. sin y

3. sinh (iθ) = i sin θ and cosh (iθ) = cos θ. Thus the hyperbolic sine of a pure imaginarynumber is a pure imaginary number while the hyperbolic cosine of a pure imaginarynumber is a real number.

Solution.

1. We have

sinh z =ez − e−z

2

=ex+iy − e−x−iy

2=ex (cos y + i sin y)− e−x (cos y − i sin y)

2

= cos y.

(ex − e−x

2

)+ i sin y.

(ex + e−x

2

)= cos y. sinhx+ i sin y. coshx

2. We get

cosh z =ez + e−z

2

=ex+iy + e−x−iy

2

=ex (cos y + i sin y) + e−x (cos y − i sin y)

2

= cos y.

(ex + e−x

2

)+ i sin y.

(ex − e−x

2

)= cos y. coshx+ i sin y. sinhx

3. From exercises 1, 2, we subtitute z = iθ and notice that sinh 0 = 0 and cosh 0 = 1.

sinh iθ = cos θ. sinh 0 + i sin θ. cosh 0 = i sin θ

cosh iθ = cos θ. cosh 0 + i sin θ. sinh 0 = cos θ

Thus the hyperbolic sine of a pure imaginary number is a pure imaginary number whilethe hyperbolic cosine of a pure imaginary number is a real number.

88


Exercise 3.15. (E. 10, E. 11 P. 114)Find the numerical value of the following derivatives.

1.d

dzsinh (sin z) at z = i

2.d

dzsin (sinh z) at z = i

Solution.

1. We see that

d

dzsinh (sin z) = cosh (sin z) . cos z

Then with z = i and the aid of Exercise 5,

cosh (sin i) . cos i = cosh (i sinh 1) . cos i

= cosh (sinh 1) . cosh 1 = 2.44

2. In a similar way, we consider

d

dzsin (sinh z) = cos (sinh z) . cosh z

Then with z = i and the aid of Exercise 5 plus equation 3.2-12,

cos (sinh i) . cosh i = cos (i sin 1) .cos1

= cosh (sin 1) . cos 1 = 0.7431

Exercise 3.16. (E. 12 P. 114)

Consider the equation sinh (x+ iy) = 0. Use equation 3.3-6 to equate the real and imaginaryparts of sinh z to zero. Show that this pair of equations can be satisfied if and only if z = inπ

where n ∈ Z . Thus the zeros of all sinh z all lie along the imaginary axis in the z-plane.

Solution.In equation 3.3-6, we have sinh z = sinhx. cos y + i coshx. sin y

Then let sinhx. cos y + i coshx. sin y = 0, this means sinhx. cos y = 0 and coshx. sin y = 0

The solutions are x = 0 and y = nπ where n ∈ ZHence z = inπ where n ∈ ZThus the zeros of all sinh z all lie along the imaginary axis in the z-plane.

89


Exercise 3.17. (E. 18, E. 19 P. 115)Show that

1. |sinh z|2 = sinh2 x+ sin2 y

2. |cosh z|2 = sinh2 x+ cos2 y = cosh2 x− sin2 y

Solution.

1. By using equation 3.3-6, we obtain

|sinh z|2 = |sinhx. cos y + i coshx. sin y|2

= sinh2 x. cos2 y + cosh2 x. sin2 y

= sinh2 x. cos2 y +(1 + sinh2 x

)sin2 y

= sinh2 x.(cos2 y + sin2 y

)+ sin2 y

Hence we get |sinh z|2 = sinh2 x+ sin2 y

2. Use equation 3.3-7, we solve respectively.

|cosh z|2 = |coshx. cos y + i sinhx. sin y|2

= cosh2 x. cos2 y + sinh2 x. sin2 y

=(1 + sinh2 x

). cos2 y + sinh2 x. sin2 y

= cos2 y + sinh2 x.(cos2 y + sin2 y

)= sinh2 x+ cos2 y

But notice that

cosh2 x. cos2 y + sinh2 x. sin2 y = cosh2 x.(1− sin2 y

)+(cosh2 x− 1

). sin2 y

= cosh2 x− cosh2 x. sin2 y + cosh2 x. sin2 y − sin2 y

= cosh2 x− sin2 y

Thus we get

|cosh z|2 = sinh2 x+ cos2 y = cosh2 x− sin2 y

90


3.4 THE LOGARITHMIC FUNCTION

The reader can see some exercises as examples such as 8, 9, 10, 11, 13, 18, 19, 23, 24, 26.Exercise 3.18. (E. 8, E. 9, E. 10 P. 119)Find all values of the logarithm of each of the following numbers and state in each case theprincipal value. Put answers in the form a+ ib.

1. elog(i sinh 1)

2. eei

3. Log (Logi)

Solution.

1. We deduce that

log z = log elog(i sinh 1)

= log (i sinh 1)

= log [sinh 1 (0 + i)]

= log[sinh 1

(cos

π

2+ i sin

π

2

)]= Log (sinh 1) + i

(π2

+ k2π), k ∈ Z

Then we set θ =π

2+ k2π and get

Logz = Log (sinh 1) + iπ

2

2. We have

log(eei)

= log(ecos 1+i sin 1

)= log

[ecos 1 (cos (sin 1) + i sin (sin 1))

]= Log

(ecos 1

)+ i [arccos (cos (sin 1)) + k2π] , k ∈ Z

The principal value is Log(ecos 1

)+ i arccos (cos (sin 1)) when k = 0.

3. We get

logLog (Logi) = logLog(iπ

2

)= log

(Log

π

2+ i arcsin

π

2

)= Log

√Log2π

2+ arcsin2 π

2+ i (1 + k2π)

91


The principal value is Log√

Log2π

2+ arcsin2 π

2+ i1 when k = 0.

Exercise 3.19. (E. 11 P. 119)For what values of z is the equation Logz = Logz true?

Solution.We have equation 3.4-5

Logz = Log |z|+ arg z

Logz̄ = Log |z̄|+ arg z̄ = Log |z| − arg z

Logz̄ = Log |z̄| − arg z̄ = Log |z|+ arg z

The question is Logz = Logz̄, it means that

Log |z|+ arg z = Log |z̄| − arg z̄


arg z = − arg z̄

With z = x+ iy, x 6= 0, we see that

argy

x= − arg

(−yx

)Then we obtain the fact that

y

x=y

x.

Thus, the solution of equation for z ∈ R with <z 6= 0.

Exercise 3.20. (E. 13 P. 119)Give solutions to the following equations in Cartesian form.

(Logz)2 + Logz = −1

Solution.From the equation above, we see that

(Logz)2 + Logz + 1 = 0

We have ∆ = 3i2, so that Logz =

−1 + i√

3

2

Logz =−1− i

√3

2

92


Exercise 3.21. (E. 18, E. 19 P. 119)Use logarithm to find all solutions of the following equations.

1. (ez − 1)2 = ez

2. (ez − 1)3 = 1

Solution.

1. The equation given is equivalent to

(ez)2 − 3ez + 1 = 0

Then we can derive the roots below.

ez =3 +√

5

2

ez =3−√

5

2

From these roots, we take logarithm both sides to get

z = log3 +√

5

2+ il2π

z = log3−√

5

2+ ik2π

where k, l ∈ Z.

2. Because the case ez = 0 is discarded, the equation becomes

ez = 2

It is very easy to deal with by taking logarithm both sides, so we obtain

z = Log2 + ik2π, k ∈ Z

Exercise 3.22. (E. 23, E. 24 P. 119)Prove that if θ is real, then

1. <[log(1 + eiθ

)]= Log

∣∣∣2 cosπ

2

∣∣∣ if eiθ 6= −1

2. <[log(reiθ − 1

)]=

1

2Log

(1− 2r cos θ + r2

)if r ≥ 0, reiθ 6= 1

Solution.

1. We have

93


1 + eiθ = cos θ + i sin θ + cos 0 + i sin 0

= cos θ + cos 0 + i (sin θ + sin 0)

= 2 cosθ

2cos

θ

2+ i2 sin

θ

2cos

θ

2

= 2 cosθ

2

(cos

θ

2+ i sin

θ

2

)

So that, we conclude

log(

1 + eiθ)

= Log∣∣∣∣2 cos

θ

2

∣∣∣∣+ i

(θ

2+ k2π

)where k ∈ Z

Hence,

<[log(

1 + eiθ)]

= Log

∣∣∣∣2 cosθ

2

∣∣∣∣2. We have

eiθ = cos θ + i sin θ

Then taking logarithm both sides, we see that

log (r cos θ − 1 + ir sin θ) = Log√r2 cos2 θ + 1− 2r cos θ + r2 sin2 θ + i

(arcsin

r sin θ√r2 cos2 θ + 1− 2r cos θ + r2 sin2 θ

+ k2π

)

= Log√r2 + 1− 2r cos θ + i

(arcsin

r sin θ√r2 + 1− 2 cos θ

+ k2π

)=

1

2Log

(r2 + 1− 2r cos θ

)+ i

(arcsin

r sin θ√r2 + 1− 2 cos θ

+ k2π

)

Hence, we get the proof completely.

<[log(reiθ − 1

)]=

1

2Log

(1− 2r cos θ + r2

), r ≥ 0, reiθ 6= 1

Exercise 3.23. (E. 26 P.119)Consider the identity log zn = n log z, where n is an integer, which is valid for appropriatechoices of the logarithms on each side of the equation. Let z = 1 + i and n = 5.

1. Find values of log zn and log z that satisfy n log z = log zn

2. For the given z and n is nLogz = Logzn satisfied?

3. Suppose n = 2 and z is unchanged. Is nLogz = Logzn then satisfied?

94


Solution.

1. By letting the condition above, we have

log (1 + i)5 = 5 log (1 + i)

We see that

log (1 + i) = log√

2 + i(π

4+ k2π

)Then

log (1 + i)5 = 5 log√

2 + 5i(π

4+ k2π

), k ∈ Z

2. Take k = 0, we obtain

Log (1 + i)5 = Log√

25

+ 5i(π

4

)It means that

5Log (1 + i) = 5(

log√

2 + iπ

4

)= 5 log

√2 + i5

π

4

Thus, with n = 5, z = 1 + i then we get Log (1 + i)5 = 5Log (1 + i)

3. We still analyse like the above.

Log (1 + i)2 = log 2 + iπ

2

2Log (1 + i) = 2(

log√

2 + iπ

4

)= log 2 + i

π

2

Thus, with n = 2, z = 1 + i, we obtain Log (1 + i)2 = 2Log (1 + i)

95


3.5 ANALYTICITY OF THE LOGARITHMIC FUNCTION

We consider the exercises 2, 3, 4, 6, 7, 8, 10, 12.

Exercise 3.24. (E. 2 P. 126)

Suppose that: f (z) = log z = Logr + iθ, 0 ≤ θ < 2π

1. Find the largest domain of analyticity of this function.

2. Find the numerical value of f(−e2

)3. Explain why we cannot determine f

(e2)within the domain of analyticity.

Solution.

1. f (z) is discontinuous at origin z = 0, since log 0 not identified. θ discontinuous at thepoint on the positive real axis. Because the point on the positive real axis value θ = 0,while points located in quadrant-four while

limθ→0

θ = 2π

Cut branches located in the plane xy, so f (z) is a branch of logarithm analytic in thedomain D as follows

D = C \ {z : =z = 0,<z ≥ 0}

2. We consider the function f(−e2

)= log

(−e2

)and have −e2 = e2eiπ. So we take logarithm

both sides as follows

log(−e2

)= log

(e2eiπ

)= Log

(e2)

+ i (π + k2π) , k ∈ Z

Then, because of 0 ≤ θ = π + k2π < 2π, we choose k = 0, it means θ = π.

Hence,

log(e2)

= 2Loge+ iπ

3. We have

log(e2)

= Log(e2)

+ i (0 + k2π) , k ∈ Z

When we choose k = 0, at that point ,log(e2)is (2; 0) located on the positive real axis in

the plane xy, in which the function is not analytic. Therefore, we can not define f(e2)in

the analytic domain.

96


Exercise 3.25. (E. 3, E. 4, E. 6 P. 126)

Consider a branch of log z analytic in the domain created with the branch cut x = 0, y ≥ 0. Iffor this branch, log (−1) = −iπ, find the following:

1. log (1)

2. log (−ie)

3. log(−√

3 + i)

Solution.

1. Using the equation 3.4-2 along with the definition of branch cut and we achieve

log (1) = log(ei0)

= Log1 + i (0 + 2kπ) , k ∈ Z

We need to find θ, consider that

−3π

2< θ = 0 + k2π <

π

2

So θ = 0

Thus, we obtain

log 1 = Log1 + i0 = 0


arg (−ie) = −π +π

2= −π

2

Hence,

log (−ie) = log e+ i(−π

2

)= 1− iπ

2


log(−√

3 + i)

= log(

2ei5π6

)= i

(5π

6+ k2π

), k ∈ Z

We find θ from the inequality below.

−3π

2< θ =

5π

6+ k2π <

π

2

97


So θ = −7π

6.

Thus

log(−√

3 + i)

= Log2 + i

(−7π

6

)

Exercise 3.26. (E. 8, E. 10 P. 127)

Consider a branch of log z analytic in the domain created with the branch cut x = −y, x ≥ 0.If, for this branch , log (1) = −2πi, find the following.

1. log (i)

2. log (−ie)

Solution.

1. We see that x = −y and x > 0 is the branch cut so −9π

4< θ < −π

4Using this branch cut and we get the value of θ

arg z = −2π +π

2= −3π

2

log i = log |i| − i3π2

= i

(−3π

2− 2π

)

2. We consider that x = −y and x > 0 is the branch cut so −9π

4< θ < −π

4Using this branch cut and we get the value of θ

log (−ie)− log(e.ei(−

π2 ))

= Loge+ i(−π

2+ k2π

), k ∈ Z

We find k from θ

−9π

4< θ = −π

2+ k2π < −π

4, k ∈ Z

So we have θ = −π2, k = 0.

Thus

log (−ie) = 1− π

2i

98


Exercise 3.27. (E. 12 P. 127)

1. Show that −Logz = Log1

zis valid throughout the domain of analyticity of Logz.

2. Find a nonprincipal branch of log z such that − log z = log1

zis not satisfied somewhere

in your domain of analyticity of log z. Prove your result.

Solution.

1. We have log z = log |z|+ i arg where −π < arg z < π

Now, if arg z lies between –π and π, we can take arg1

z= − arg z

And we will get −π < arg1

z< π

Thus,

log1

z= log

∣∣∣∣1z∣∣∣∣+ i arg z

= − log |z| − i arg z

Or we can say that

− log z = log1

z

2. We obtain that log i =iπ

2

log1

i= log (−i) = i

3π

2

And note that

log i 6= log1

i

99


3.6 COMPLEX EXPONENTIALS

We go together to solve exercises 8, 9, 11, 13, 16, 18, 19, 20, 24, 25.Exercise 3.28. (E. 8, E. 9, E. 10 P. 132)Find all values of the following in the form a+ ib and state the principal value.

1. (Logi)π/2

2. (1 + i tan i)√

2

Solution.

1. We have

Logi = Log(

cosπ

2+ i sin

π

2

)= Log1 + i

π

2+ k2π

= iπ

2+ k2π

=π

2

(cos

π

2+ i sin

π

2

)+ k2π

Thus

log (Logi) = Logπ

2+ i(π

2+ 2kπ

)Moreover, we get

eπ2Logπ

2

[cos

(π2

4+ kπ2

)+ i

(π2

4+ kπ2

)]When k = 0, the principal value is

(Logi)π2 = e

π2Logπ

2+(π2

4

)

= eπ2Logπ

2

[cos

π2

4+ i sin

π2

4

]

2. We have

(1 + i tan 1)√

2 = e√

2. log(1+i tan 1)

We see the exponent here and make some changes

log (1 + i tan 1) = Log√

1 + tan2 1 + i arccos1√

1 + tan2 1

= Log

∣∣∣∣ 1

cos 1

∣∣∣∣+ i (1 + 2kπ)

100


Hence,

e√

2 log(1+i tan 1) = e√

2(Log| 1cos 1 |+i(1+k2π))

=

(1

cos 1

)√2

.e(√

2+2√

2kπ)i

And the principal value is

(1

cos 1

)√2

.e√

2i

Exercise 3.29. (E. 11 P. 132)Show that all possible values of zi are real if |z| = enπ, where n is any integer.

Solution.We have

zi = ei log z

= ei(log|z|+i(θ+k2π))

= ei log|z|−(θ+k2π)

With |z| = enπ, the equality above is equivalent to

ei log enπ .e−(θ+k2π) = einπ.e−(θ+k2π)

= cos (nπ) .e−(θk2π) ∈ R, ∀n ∈ N

Exercise 3.30. (E. 13 P. 132)Using Eq. (3.1-5) of Eq. (3.1-7) and the definition in Eq. (3.6-1), prove that for any complexvalues α, β and z, we have the following.The values of zαzβ are identical to the values of zα+β.

Solution.From the equation 3.6-1, we have

zα = eα log z

zβ = eβ log z

Then we multiply them side by side, we get

101


zαzβ = eα log zeβ log z

= eα log z+β log z

= e(α+β) log z

= zα+β

Exercise 3.31. (E. 16, E. 18 P. 132)Using the principal branch of the function, evaluate the following.

1. f ′ (i) if f (z) = z2+i

2. f ′ (−8i) if f (z) = z1/3+i

Solution.

1. We must get the derivative of this function, but we can change the form before.

f (z) = z2+i = e(2+i)Logz

f ′ (z) =(e(2+i)Logz

)′=

2 + i

ze(2+i)Logz

With z = i, we have

f ′ (i) =2 + i

ie(2+i)Logi

=2 + i

ie(2+i)iπ

2

=2 + i

ie−

π2

+iπ

2. In exercise above, we can use

f ′ (z) =(e(2+i)Logz

)′=

2 + i

ze(2+i)Logz

f ′ (−8i) =

(−1 + i

1

3

)e(

13

+i)Log(−8i)

8

Then we evaluate them respectively.

Log (−8i) = Log8− iπ2(

1

3+ i

)Log (−8i) = Log2 +

π

2+ i(Log8− π

6

)102


So that, we conclude

f ′ (−8i) =(−3 + i)

24eLog2+π

2 cis(Log8− π

6

)=

1

12(−3 + i) e

π2 cis

(Log8− π

6

)

Exercise 3.32. (E. 19, E. 20 P. 132)Let f (z) = zz, where the principal branch is used. Evaluate the following.

1. f ′ (z)

2. f ′ (i)

Solution.We need to find the derivative now.

f ′ (z) =d

dzzz

=d

dzez. log z

= (log z + 1) ez. log z

= (log z + 1) .zz

So take z = i here, we obtain

f ′ (i) = (log i+ 1) ei log i

=

(iπ

2+ 1

)eiiπ2

=

(iπ

2+ 1

)e−

π2

Exercise 3.33. (E. 21, E. 24, E. 25 P. 132)

1. Let f (z) = zsin z, where the principal branch is used. Find f ′ (i).

2. Let f (z) = 10z3 . This function is evaluated such that f ′ (z) is real when z = 1. Find

f ′ (1 + i). Where in the complex plane is f (z) analytic?

3. Let f (z) = 10ez . This function is evaluated such that |f (iπ/2)| = e−2π. Find f ′ (z) and

f ′ (iπ/2).

Solution.

1. We still use old method above to get the derivative.

103


f (z) = zsin z = esin zLogz

f ′ (z) =(esin zLogz

)′=

(cos zLogz +

sin z

z

)esin zLogz, z 6= 0

With z = i, we have

f ′ (i) =

(cos iLogi+

sin i

i

)esin iLogi

=(iπ

2cos i− i sin i

)eiπ2

sin i

2. Like the exercise above, we get

f ′ (z) =d

dz

(ez

3. log 10)

= 3z2 log 10.ez3. log 10

= 3z2 log 10.f (z)

With z = 1 + i, then we deduce that

f ′ (1 + i) = 3 log 10. (1 + i)2 .10(1+i)3

= 3 log 10.2i.102i−2

And notice that f (z)analytic when z 6= 0.

3. In a similar fashion, we change the form of the function

f (z) = 10(ez) = eez log 10

So that the derivative is

f ′ (z) =[10(ez)

]′= e(Log10+i2π)ez (Log10 + i2π) ez

With z =iπ

2, the result we need is

f ′(iπ

2

)= 0.00464− i0.0116

104


3.7 INVERSE TRIGONOMETRIC AND HYPERBOLICFUNCTIONS

In this section, we have exercise 2, 4, 7, 8, 16, 17.Exercise 3.34. (E. 2 P. 137)

1. Show that if we differentiate a branch of arccos z, we obtain

d

dzarccos z =

−1

(1− z2)1/2

2. Obtain

d

dzarcsin z =

d

dz

(−i log

[zi+

(1− z2

)1/2])=

1

(1− z2)1/2

by noting that

z2 = sin2w = 1− cos2w = 1−(dz

dw

)2

can be solved for dw/dz.

3. Obtain

d

dzsinh−1 z =

1

(1 + z2)1/2

directly from sinh−1 z = log(z +

(z2 + 1

)1/2); obtain it also by a procedure similar topart (2) of this exercise.

Solution.

1. We need to find the inverse function of cosine from the equation z = cosw. We have

z = cosw =eiw + e−iw

2

And then we put y = eiw, the equation above becomes y2 − 2zy + 1 = 0, so the root isy = z + i

√1− z2. It means that

w = −i log(z + i

√1− z2

)Hence,

arccos z = −i log(z + i

√1− z2

)Then, the derivative is

105


d

dzarccos z =

d

dz

(−i log

(z + i

√1− z2

))=

−1

(1− z2)1/2

2. In a similar fashion, from the given equality, we deduce that

(dz

dw

)2

= z2 − 1

Next, we inverse both sides to see the proof needed.

3. We have

z2 = sinh2w = −1 + cosh2w


z2 = −1 +

(dz

dw

)2

Thus we see that

dw

dz=

1

(z2 + 1)1/2

Exercise 3.35. (E. 4, E. 7, E. 8 P. 137)Find all solutions to the following equations.

1. cosw = 3

2. sinhw = i√

2

3. cosh2w = −1

Solution.

1. Because of the exercise above, we obtain

w = arccos 3 = i log(

3− 2√

2)

+ k2π

2. We have

sinh−1 z = log(z +

√z2 + 1

)So that from this, we obtain

106


w = sinh−1(i√

2)

= log

(i√

2 +

√(i√

2)2

+ 1

)= log

(i√

2 + i)

Hence,

w = log(√

2 + 1)

+ i(π

2+ k2π

)3. We see the formular below.

coshw =1

2

(ew + e−w

)Then we square both sides and get the equation, it means that

1

4

(e2w + e−2w + 2

)= −1

We easily solve this equation and obtain the root is e−2w = 1. Hence,

w = ikπ

Exercise 3.36. (E. 16 P. 137)

Show that tanh−1(eiθ)

=1

2log

(i cot

(θ

2

)).

Solution.We start with the left side.

tanh−1(eiθ)

=1

2log

(1 + eiθ

1− eiθ

)=

1

2log

(e−iθ2 + e

iθ2

e−iθ2 − e

iθ2

)

Moreover, we have

cot

(θ

2

)= i

e−iθ2 + e

iθ2

eiθ2 − e

−iθ2

or we can say that

e−iθ2 + e

iθ2

e−iθ2 − e

iθ2

= i cot

(θ

2

)

107


So that we can complete the proof here.

Exercise 3.37. (E. 17 P. 137)Find a formula similar to the one above for tan−1

(eiθ).

Solution.We have a formula below.

tan−1(eiθ)

=i

2log

(i+ eiθ

i− eiθ

)Also, we can observe that

i+ eiθ

i− eiθ=

(i+ eiθ

)2−1− e2iθ

= −(−1 + e2iθ

1 + e2iθ+

2ieiθ

1 + e2iθ

)= −

(eiθ − e−iθ

eiθ + e−iθ+

2i

eiθ + e−iθ

)= −isin θ + 1

cos θ

So that we obtain the proof by replacing the equality above into a formula of tanh−1(eiθ).

108

4 INTEGRATION IN THE COMPLEX PLANE

4.1 INTRODUCTION TO LINE INTEGRATION

We can evaluate exercises 2, 3, 6, 7 to get some examples.Exercise 4.1. (E. 2, E. 3 P. 160)Let C be that portion of the curve y = x2 lying between (0, 0) and (1, 1). Let F (x, y) = x+y+1.Evaluate these integrals along C.

1.ˆ 1,1

0,0F (x, y) dx

2.ˆ 1,1

0,0F (x, y) dy

Solution.

1. Along the path of integration, y changes with x. The equation of the contour of integrationy = x2 can be used to express y as a function of x in the preceding integrand. Thus

ˆ 1,1

0,0F (x, y) dx =

ˆ 1,1

0,0

(x+ x2 + 1

)dx

=x2

2+x3

3+ x

∣∣∣∣x=1

x=0

=11

6

2. To change this to a conventional integral, we may regard x as a function of y along C.Since y = x2 and x ≥ 0 on the path of integration, we obtain x =

√y. Hence

ˆ 1,1

0,0F (x, y) dy =

ˆ 1,1

0,0(√y + y + 1) dy =

13

6

Exercise 4.2. (E. 6 P. 160)Let C be that portion of the curve x2 + y2 = 1 lying in the first quadrant. Let F (x, y) = x2y.Evaluate the integral along C.

ˆ 1,0

0,1F (x, y) ds

Solution.Because of portion of the curve x2 +y2 = 1 lying in the first quadrant, we will use parametric

representation to solve this problem. Indeed, we put

109


x = r cos θ

y = r sin θ, r ≤ 1, θ ∈

[0;π

2

]Then it is easy to get the following integral

ˆ π2

0

ˆ 1

0r3 cos2 θ sin θrdrdθ =

ˆ π2

0cos2 θ sin θdθ

ˆ 1

0r4dr

=

(−cos3 θ

3

∣∣∣∣π2

0

)(r5

5

∣∣∣∣10

)=

1

15

Exercise 4.3. (E. 7 P. 160)

Show thatˆ 0,1

0,−1ydx =

−π2.The integration is along that portion of the circle x2 + y2 = 1 lying

in the half plane x ≥ 0. Be sure to consider signs in taking square roots.

Solution.Because the integration is along that portion of the circle x2 + y2 = 1 lying in the half plane

x ≥ 0 so we put x = cos t

y = sin t

(−π2≤ t ≤ π

2

)Therefore we have dx = − sin tdt.Thus we consider the integration is

ˆ 0,1

0,−1ydx =

ˆ π2

−π2

sin t (− sin t) dt

=1

2

ˆ π2

−π2

(cos 2t− 1) dt

= −π2

110


4.2 COMPLEX LINE INTEGRATION

We will write down some exercises there such as 4, 5, 6, 7, 8, 10, 13, 14, 17.Exercise 4.4. (E. 4, E. 5, E. 6 P. 170)

Evaluateˆ 1

iz̄dz along the contour C, where C is

1. the straight line segment lying along x+ y = 1.

2. the parabola y = (1− x)2.

3. the portion of the circle x2 + y2 = 1 in the first quadrant.

Solution.

1. Now we put

x = t

y = 1− twhere 0 ≤ t ≤ 1.

And we notice that dz = (1− i) dt. Then by replacing all above, we get

ˆ 1

iz̄dz =

ˆ 1

0(t+ (t− 1) i) (1− i) dt

=

ˆ 1

0(2t− 1− i) dt

= −i

2. Like a method of exercise above, by setting up

x = t

y = (1− t)2with 0 ≤ t ≤ 1.

We obtain dz = [1− 2 (1− t) i] dt

Hence, the result is

ˆ 1

iz̄dz =

ˆ 1

0

[t− (t− 1)2 i

][1− 2 (1− t) i] dt

=

ˆ 1

0

[t+ 2 (t− 1)3 +

(t2 − 1

)i]dt

=−2

3i

3. Here the change of variables is give by

x = cos t

y = sin t

(0 ≤ t ≤ π

2

)

So we get z = cos t+ i sin t. This deduce that dz = (− sin t+ i cos t) dt.

Thus the result is

111


I =

ˆ π2

0(cos t− i sin t) (− sin t+ i cos t) dt

= i

ˆ π2

0dt

= iπ

2

Exercise 4.5. (E. 7 P. 171)

Evaluateêzdz

1. from z = 0 to z = 1 along the line y = 0;

2. from z = 1 to z = 1 + i along the line x = 1;

3. from z = 1 + i to z = 0 along the line y = x. Verify that the sum of your three answers iszero.

Solution.

1. From z = 0 to z = 1 along the line y = 0, we take x = t and y = 0, so z (t) = t anddz = dt.

This intergral becomes

ˆ 1,0

0,0ezdz =

ˆ 1

0etdt

= e− 1

2. From z = 1 to z = 1 + i along the line x = 1, this time we set x = 1 and y = t soz = x+ iy = 1 + it. Therefore, we get dz = idt.

Now the result is

ˆ 1,0

1,1ezdz =

ˆ 1

0ie1+itdt

= ie

ˆ 1

0eitdt

= ei+1 − e

3. From z = 1 + i to z = 0 along the line y = x, by taking x = y = t and z = t (1 + i), weobtain dz = (1 + i) dt.

The intergral becomes

112


ˆ 0,0

1,1ezdz =

ˆ 0

1et(1+i) (1 + i) dt

= 1− e1+i

Note that when we sum three answers, the result is e− 1 + ei+1 − e+ 1− e1+i = 0.

Exercise 4.6. (E. 8 P. 171)The function z(t) = eit = cos t+ i sin t can provide a useful parametric representation of circulararcs. If t ranges from 0 to 2π we have a representation of the whole unit circle, while if t goesfrom α to β we generate an arc extending from eiα to eiβ on the unit circle. Use this parametrictechnique to perform the following intergrations.

1.ˆ −1

1

1

zdz along |z| = 1, upper half plane.

2.ˆ i

1z̄4dz along |z| = 1, first quadrant.

Solution.By letting z (t) = cos t + i sin t above, we deduce that dz = (− sin t+ i cos t) dt. So we will

apply this result to solve 2 problems.

1. Because of upper half plane, it is easy to obtain that

ˆ 1

−1

1

zdz =

ˆ π

0(cos t− i sin t) (− sin t+ i cos t) dt

=

ˆ π

0idt

= πi

2. And for the first quadrant, it follows that

ˆ i

1z̄4dz =

ˆ π2

0(cos t− i sin t)4 (− sin t+ i cos t) dt

=

ˆ π2

0(cos 4t− i sin 4t) (− sin t+ i cos t) dt

=

ˆ π2

0(sin 3t+ i cos 3t) dt

=1

3+ i

1

3

113


Exercise 4.7. (E. 13 P. 171)

1. Find a parametric representation of the shorter of the two arcs lying along lying along(x− 1)2 + (y − 1)2 = 1 that connects z = 1 with z = i.

Hint: See discussion preceding Exercises 8-10 above, where parametization of a circlediscussed.

2. Findˆ i

1zdz along the arc of (a), using the parametrization you have found.

Solution.

1. The parametization of a circle (x− 1)2 + (y − 1)2 = 1 that connects z = 1 with z = i is

x = 1 + cos t

y = 1 + sin t

(−π ≤ t ≤ −π

2

)

2. From the above, we have z = 1 + cos t+ i (1 + sin t)

And then the following is

dz = (− sin t+ i cos t) dt

Thus we replace all to get the result.

ˆ i

1zdz =

ˆ −π2

−π[1 + cos t− i (1 + sin t)] (− sin t+ i cos t) dt

= i(

2− π

2

)

Exercise 4.8. (E. 14 P. 171)

Consider I =

ˆ 2+i

0+i0ez

2dz taken along the line x = 2y. Without actually doing the integration,

show that |I| ≤√

5e3.

Solution.

We need to find M that is∣∣∣ez2∣∣∣ ≤ M , so M = max

(ex

2−y2)when x : 0 → 2 and y : 0 → 1.

Thus we have M = e3.Next, we must find L. Because this path in this exercise is a line so the length of it is

L =√

(2− 0)2 + (1− 0)2 =√

5.Finally, we have the result |I| ≤ML = e3

√5

Hence, we complete the proof.

114


Exercise 4.9. (E. 17 P. 171)

1. Let g (t) be a complex function of the real variable t. Expressˆ b

ag (t) dt as the limit of a

sum. Show that for b > a we have

∣∣∣∣ˆ b

ag (t) dt

∣∣∣∣ ≤ ˆ b

a|g (t)| dt

2. Use the equation above to prove that

∣∣∣∣ˆ 1

0

√teitdt

∣∣∣∣ ≤ 2

3

Solution.

1. We express the integral that

ˆ b

ag (t) dt = lim

n→∞

n∑k=1

[u (x (tk) , y (tk)) + iv (x (tk) , y (tk))] [∆x (tk) + i∆y (tk)]

Then we have the inequality

∣∣∣∣ˆ b

ag (t) dt

∣∣∣∣ ≤ limn→∞

n∑k=1

|u (x (tk) , y (tk)) + iv (x (tk) , y (tk))| [∆x (tk) + i∆y (tk)]

≤ˆ b

a|g (t)| dt

2. By using the equation above, we can complete the proof easily. Indeed, we consider

∣∣∣∣ˆ 1

0

√teitdt

∣∣∣∣ ≤ ˆ 1

0

∣∣∣√teit∣∣∣ dt=

ˆ 1

0

√tdt =

2

3

115


4.3 CONTOUR INTEGRATION AND GREEN’S THEOREM

In this section, the examples is exercises 1, 2, 12, 13, 16, 18.Exercise 4.10. (E. 1 P. 180)

1. Let C be an arbitrary simple closed contour. Use Green’s theorem to find a simple inter-

pretation of the line integral1

2

‰C

(−ydx+ xdy) .

2. Consider‰

C

[cos ydx+ sinxdy] performed around the square with corners at

(0, 0) , (1, 0) , (0, 1) , (1, 1). Evaluate this integral by doing an equivalent integral over thearea enclosed by the square.

3. Suppose you know the area enclosed by a simple closed contour C. Show with the aid of

Green’s theorem that you can easily evaluate‰

C

z̄dz around C.

Solution.

1. Applying Green’s theorem, we have

1

2

‰C

(−ydx+ xdy) =1

2

¨R

(1 + 1) dxdy

= |R|

2.

‰

C

[cos ydx+ sinxdy] =

¨R

(∂ (sinx)

∂x− ∂ (cos y)

∂y

)dxdy

=

ˆ 1

0

ˆ 1

0(cosx+ sin y) dxdy

= sin 1− cos 1 + 1

3. Put z = x+ iy, so z̄ = x− iy and applying Green’s theorem for‰

C

z̄dz, we get

116


‰

C

z̄dz =

‰

C

(x− iy) (dx+ idy)

=

‰

C

(x− iy) dx+ (y + ix) dy

=

¨R

[∂ (y + ix)

∂x− ∂ (x− iy)

∂y

]dxdy

= 2i

¨Rdxdy

= 2i |R|

Exercise 4.11. (E. 2 P. 180)To which of the following integral is the Cauchy-Goursat theorem directly applicable?

‰

|z|=1

sin z

z + 2idz

Solution.By setting up z = x+ iy, this integral becomes

‰|x+iy|=1

sin (x+ iy)

x+ iy + 2i(dx+ idy)

Let f (x, y) =sin (x+ iy)

x+ iy + 2i, we note that the function f (x, y) do not define at the point

(0,−2), but (0,−2) /∈ D with D ={

(x, y) |x2 + y2 ≤ 1}

Therefore, f is analytic and we can deduce that the integral is the Cauchy-Goursat theoremdirectly applicable.

Exercise 4.12. (E. 12, E. 13 P. 180)

Prove the following result by means of Cauchy-Gorsat theorem. Begin with‰ezdz per-formed

around |z| = 1. Use the parametric representation z = eiθ, 0 ≤ θ ≤ 2π. Separate your equationinto real and imaginary parts.

1.ˆ 2π

0ecos θ [cos (sin θ + θ)] dθ = 0

2.ˆ 2π

0ecos θ [sin (sin θ + θ)] dθ = 0

Solution.By using the parametric representation z = eiθ, we have dz = ieiθdθ. Hence,

117


‰ezdz =

ˆ 2π

0eeiθieiθdθ

= i

ˆ 2π

0ecos θ+i sin θ(cos θ + i sin θ)dθ

=

ˆ 2π

0ecos θ [cos (sin θ) + i sin (sin θ)] (− sin θ + i cos θ)dθ

= −ˆ 2π

0ecos θ [sin (sin θ + θ)] dθ + i

ˆ 2π

0ecos θ [cos (sin θ + θ)] dθ

According to Cauchy-Gorsat theorem‰ezdz = 0 performed around |z| = 1. Therefore,

ˆ 2π

0ecos θ [sin (sin θ + θ)] dθ = 0

ˆ 2π

0ecos θ [cos (sin θ + θ)] dθ = 0

Exercise 4.13. (E. 16 P. 181)Show that for real a, where |a| > 1, we have

ˆ 2π

0

1− a cos θ

1− 2a cos θ + a2dθ = 0

Solution.We consider the integral

‰1

z − adz, where the integral is taken around the unit circle. Now

we use the parametric representation like exercise 12, 13 above. Indeed, if we put z = eiθ, wewill have dz = ieiθdθ. Hence,

‰1

z − adz =

ˆ 2π

0

ieiθdθ

eiθ − a

=

ˆ 2π

0

i (cos θ + i sin θ) (cos θ − a− i sin θ)

(cos θ − a)2 + sin2 θdθ

=

ˆ 2π

0

a sin θ

1− 2a cos θ + a2dθ + i

ˆ 2π

0

1− a cos θ

1− 2a cos θ + a2dθ

According to Cauchy-Gorsat theorem‰

1

z − az = 0 performed around |z| = 1. Therefore, we

get the proof completely.

118


Exercise 4.14. (E. 18 P. 181)Evaluate the following integral. The contour is the square centered at the origin with cornersat ± (2± 2i) (C1) . The result contained in the previous problem as well as the principle ofdeformation of contours will be useful.

‰C1

dz

z − i

Solution.We let f (z) =

1

z − iand consider (C2) : |z| = 3

2. Since a function f (z) is a analytic not only

on C1 and C2 but also all points lying between C1and C2 so

‰C1

f (z) dz =

‰C2

f (z) dz

According to the following equation, with z0 an arbitrary complex constant,

‰|z−z0|=r

(z − z0)n dz =

0, n 6= −1

2πi, n = −1

where the contour of integration is a circle centered at z0.We have

‰C2

f (z) dz = 2πi

Thus this means that

‰C1

dz

z − i= 2πi

119


4.4 PATH INDEPENDENCE, INDEFINITE INTEGRALS,FUNDAMENTAL THEOREM OF CALCULUS IN THECOMPLEX PLANE

In this section, we have some exercises to deal with, these are exercise 2, 3, 5, 7, 8.Exercise 4.15. (E. 2, E. 3, E. 5, E. 7 P. 190)Use theorem (Integration of Functions that are the Derivatives of Analtyic Functions) to evaluatethe following integrals along the curve y =

√x.

1.

4+2iˆ

0

eizdz

2.

4+2iˆ

0

(1 + z2

)dz

3.

4+2iˆ

0

ez sinh zdz

4.

4+2iˆ

1+i

z

z2 − 1dz

Solution.

1. Withd(eiz)

dz= ieiz, we can compute this integral as follows.

4+2iˆ

0

eizdz = −i4+2iˆ

0

ieizdz

= −i. eiz∣∣4+2i

0

= −i(e−2+4i − 1

)= i− ie−2+4i

2. Withd

(z +

z3

3

)dz

= 1 + z2, we can evaluate this integral.

120


4+2iˆ

0

(1 + z2

)dz =

(z +

z3

3

)∣∣∣∣4+2i

0

= 4 + 2i+(4 + 2i)3

3

= 4 + 2i+16 + 88i

3

=28 + 94i

3

3. Withd(e2z − 2z

)dz

= 2e2z − 2, like two exercises above, we have

4+2iˆ

0

ez sinh zdz =

4+2iˆ

0

ezez − e−z

2dz

=

4+2iˆ

0

e2z − 1

2dz

=1

4

4+2iˆ

0

(2e2z − 2

)dz

=1

4

(e2z − 2z

)∣∣4+2i

0

=1

4

(e8+4i − 9− 4i

)

4. Withd(Log

∣∣z2 − 1∣∣)

dz=

d1

z2 − 1, we obtain

4+2iˆ

1+i

z

z2 − 1dz =

1

2

4+2iˆ

1+i

2z

z2 − 1dz

=1

2

4+2iˆ

1+i

d(z2 − 1

)z2 − 1

=1

2

(Log

∣∣z2 − 1∣∣)∣∣4+2i

1+i

=1

2

[Log

∣∣∣(4 + 2i)2 − 1∣∣∣− Log ∣∣∣(1 + i)2 − 1

∣∣∣]=

1

2(Log |11 + 16i| − Log |2i− 1|)

=1

2Log

∣∣∣∣11 + 16i

−1 + 2i

∣∣∣∣=

1

2Log

∣∣∣∣21− 38i

5

∣∣∣∣121


Exercise 4.16. (E. 8 P. 190)

1. What, if anything, is incorrect about the following two integrations? The integrals areboth along the line y = x.

1+iˆ

0+0i

zdz =z2

2

∣∣∣∣1+i

0+0i

=(1 + i)2

2= i (∗)

1+iˆ

0+0i

z̄dz =z̄2

2

∣∣∣∣1+i

0+0i

=(1− i)2

2= −i (∗∗)

2. What is the correct numerical value of each of the above integrals?

Solution.

The integrals (*),(**) are incorrect since with z = x + iy and seeing Fig.(a), Fig(b) above,we have

122


ˆ

C

zdz =

ˆ

I

zdz +

ˆ

II

zdz

=

1ˆ

0

xdx+

1ˆ

0

(1 + yi) dy

=

1ˆ

0

xdx+

1ˆ

0

dy + i

1ˆ

0

ydy

=1

2+ 1 +

i

2

=3

2+i

2

ˆ

C

z̄dz =

ˆ

I

z̄dz +

ˆ

II

z̄dz

=

1ˆ

0

xdx+

1ˆ

0

(1− yi) dy

=

1ˆ

0

xdx+

1ˆ

0

dy − i1ˆ

0

ydy

=1

2+ 1− i

2

=3

2− i

2

123


4.5 THE CAUCHY INTEGRAL FORMULA AND ITS EXTENSION

We go to see together some exercises such as 2, 6, 17, 19.Exercise 4.17. (E. 2, E. 6 P. 199)Evaluate the following integrals using the Cauchy integral formula, its extension, or the Cauchy-Goursat theorem where appropriate.

1.‰

sin z

z − 2dz around |z| = 3.

2.‰

eiz

z2 + z + 1dz around

∣∣∣∣z +1

2− 2i

∣∣∣∣ = 2.

Solution.

1. We can see that the term z − 2 does become 0 at the point z = 2 inside C is the circle|z| = 3. We therefore apply Cauchy integral formula with f (z) = sin z. Thus

‰sin z

z − 2dz = 2πif (2) = 2πi sin 2

2. We need to solve the equation z2 + z + 1 = 0, then we get the roots are z =−1± i

√3

2.

And the next is that we need to definite the point makes the denominator becomes 0 such

that inside C is the circle∣∣∣∣z +

1

2− 2i

∣∣∣∣ = 2.

We consider that

∣∣∣∣∣−1 + i√

3

2+

1

2− 2i

∣∣∣∣∣ =

∣∣∣∣∣i(√

3

2− 2

)∣∣∣∣∣ < 2

So this is the point we find and we can verify the others is outside C.

Then we apply Cauchy integral formula with f (z) =eiz

z + 1+i√

32

. Thus

‰eiz

z2 + z + 1dz = 2πif

(−1 + i

√3

2

)=

2π√

3ei2

+√32

Exercise 4.18. (E. 17 P. 200)If a is a real number and |a| < 1 show that

ˆ 2π

0

1− a cos θ

1− 2a cos θ + a2dθ = 2π

Solution.We consider

‰dz

z − aaround |z| = 1 and because of |a| < 1, the point which makes the

denominator becomes 0 is inside C is the circle |z| = 1. Hence, we apply Cauchy integral

124


formula with f (z) = 1 for all z inside C, we have

‰dz

z − a= 2πi

And in the section 4.3, by putting parametric representation, we have proved that

‰1

z − adz =

ˆ 2π

0

ieiθdθ

eiθ − a

=

ˆ 2π

0

i (cos θ + i sin θ) (cos θ − a− i sin θ)

(cos θ − a)2 + sin2 θdθ

=

ˆ 2π

0

a sin θ

1− 2a cos θ + a2dθ + i

ˆ 2π

0

1− a cos θ

1− 2a cos θ + a2dθ

At this time, we can deduce that

ˆ 2π

0

1− a cos θ

1− 2a cos θ + a2dθ = 2π

Exercise 4.19. (E. 19 P. 201)Let f (z) be analytic on and inside a simple closed contour C. Let z1 and z2 lie inside C (seefigure below). Show that

1

2πi

‰C

f (z)

(z − z1) (z − z2)dz =

f (z1)

z1 − z2+

f (z2)

z2 − z1

Solution.We note that

f (z)

(z − z1) (z − z2)=

f (z)

z2 − z1

(1

z − z2− 1

z − z1

)Then we have two integrals. These are

‰C

f (z)

(z − z1) (z − z2)dz =

1

z2 − z1

[‰C2

f (z)

z − z2dz −

‰C1

f (z)

z − z1dz

]125


We see in the figure above and consider the first integral, at the point z2 which makes thedenominator becomes 0 is inside C2 so we can apply Cauchy integral formula there. And it issimilar for the case of z1. Thus we obtain

‰C2

f (z)

z − z2dz −

‰C1

f (z)

z − z1dz = 2πi [f (z2)− f (z1)]

Hence, this is really the proof we want to solve.

1

2πi

‰C

f (z)

(z − z1) (z − z2)dz =

1

z2 − z1[f (z2)− f (z1)]

=f (z1)

z1 − z2+

f (z2)

z2 − z1

126


4.6 SOME APPLICATIONS OF THE CAUCHY INTEGRALFORMULA

In this section, we get the exercises 1, 3, 5, 8, 9, 10, 13, 15 to show.Exercise 4.20. (E. 1, E. 3, E. 5 P. 211)Use Gauss’ mean value theorem in its various versions and integrations around appropriatecircles to prove the following:

1.1

2π

2πˆ

0

eeiθdθ = 1

2.1

2π

π̂

−π

cos2(π

6+ aeiθ

)dθ =

3

4, where a > 0

3.2πˆ

0

Log[a2 + 1 + 2a cos (nθ)

]dθ = 4πLog a where a > 1, n integer

Solution.

1. Let f (z) = ez

Since function f is analytic on C, applying Gauss’ mean value theorem we have

1

2π

2πˆ

0

eeiθdθ =

1

2π

2πˆ

0

f(

0 + eiθ)dθ where r = 1

= f (0)

= 1

2. Let f (z) = cos2 (z), like the exericise above :

1

2π

π̂

−π

cos2(π

6+ aeiθ

)dθ =

1

2π

π̂

−π

f(π

6+ aeiθ

)dθ

= f(π

6

)= cos2

(π6

)=

3

4

3. We have Log[a2 + 1 + 2a cos (nθ)

]= Log

∣∣a+ einθ∣∣2 completely.

Let f (z) = 2Log |z|, since function f is analytic on D = {z : |z| 6= 0}, applying Gauss’mean value theorem with z =

∣∣a+ einθ∣∣ ∈ D127


2πˆ

0

Log[a2 + 1 + 2a cos (nθ)

]dθ =

2πˆ

0

Log∣∣∣a+ einθ

∣∣∣2 dθ=

2πˆ

0

f(∣∣∣a+ einθ

∣∣∣) dθ= 2π2Log a

= 4πLog a

Exercise 4.21. (E. 8 P. 212)Let f (z)be a nonconstant function that is continuous and nonzero throughout a closed boundedregion R. Let f (z) be analytic at every interior point of R. Show that the minimum value of|f (z)| in R must occur on the boundary of R.

Solution.We consider the function g (z) =

1

f (z). So we can get the comment is f (z) has a minimum

if and only if g (z) has a maximum.Since f (z)is a nonconstant function that is continuous and nonzero throughout a closed

bounded region R and f (z) is analytic at every interior point of R so g(z)f (z)is a nonconstantfunction that is continuous and nonzero throughout a closed bounded region R and analytic atevery interior point of R.Therefore, apllying Maximum Modulus Theorem we have the maximum value of |g (z)| in R

must occur on the boundary of R.Thus, that the minimum value of |f (z)| in R must occur on the boundary of R.

Exercise 4.22. (E. 9, E. 10 P. 212)For the following closed regions R and functions f (z), find the values of z in R where |f (z)|archieves its maximum and minimum values. If your answers do not lie on the boundary of R,give an explanation. Give the values of |f (z)|at its maximum and minimum in R.

1. f (z) = z,R is |z − 1− i| ≤ 1.

2. f (z) = z2, R is |z − 1− i| ≤ 2.

Solution.

1. Because f (z) = z is analytic in R, module of f (z), |f (z)|, archieves maximum on|z − 1− i| = 1. We put z = 1 + i+ eiθ, 0 ≤ θ ≤ 2π, then we obtain

f(z) = 1 + i+ eiθ

= 1 + i+ cos θ + i sin θ

= 1 + cos θ + (1 + sin θ)i

128


Moreover, module of this function is

|f(z)| =√

(1 + cos θ)2 + (1 + sin θ)2

=√

3 + 2 cos θ + 2 sin θ

=

√3 + 2

√2 cos

(−π

4+ θ)

Next, we consider the function g(θ) = 3 + 2√

2 cos(−π

4+ θ), 0 ≤ θ ≤ 2π, we get

cos(−π

4+ θ)≤ 1 easily. Then we can deduce that g(θ) ≤ 3 + 2

√2. At this time,

we conclude |f(z)|max = 3 + 2√

2 at the point is θ =π

4and the function is z = 1 + i+ ei

π4 .

Assuming that f(z) = 0, we have the contradiction is | − 1 − i| ≤ 1 and get f(z) 6= 0.Thus |f (z)| has a minimum on |z − 1− i| = 1. Like the maximum, we obtain |f(z)|min =√

3− 2√

2 at the point θ =5π

4and the function is z = 1 + i+ ei

5π4 .

2. We have f (z) = 0 if and only if z = 0 and this point is in |z − i− 1| ≤ 2 by considering.So |f (z)|min = 0 at z = 0. Then since f (z) is analytic in R, |f(z)| has a minimum on|z − i− 1| = 2. We will set up

z = i+ 1 + 2eiθ, θ ∈ [−π, π]

= i+ 1 + 2 cos θ + 2i sin θ

Evaluating the module of f (z) follows

|z|2 = (1 + 2 cos θ)2 + (1 + 2 sin θ)2

= 6 + 4√

2 cos(−π

4+ θ)

≤ 6 + 4√

2

Then |f(z)|max = 6+4√

2 at the point θ =π

4and the function is z = (1+

√2)+ i(1+

√2).

Exercise 4.23. (E. 13 P. 212)Let u (x, y) be real, nonconstant and continuous in a closed bounded region R . Let u (x, y) baharmonic in the interior of R . Prove that the maximum value of u (x, y)in the region occurs onthe boundary. This is known as the maximum priciple.

Solution.Consider F (x, y) = u (x, y) + iv (x, y), where v is the harmonic conjugate of u.Let f (z) = eF (z), we have f (z) is an entire function and is never 0 in R ,and f (z) continuous

in closed bounded region R ( because u (x, y)continuous in a closed bounded region R) so theminimum value of |f (z)| in R must occur on the boundary of R.

129


With |f (z)| =∣∣eF (z)

∣∣ =∣∣eu(x,y)+iv(x,y)

∣∣ = eu(x,y), we say that |f (z)| has a maximum if andonly if u (x, y) has a maximum too.Thus, the maximum value of u (x, y)in the region occurs on the boundary.

Exercise 4.24. (E. 15 P. 212)Consider the closed bounded region R given by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. Now u = x2 − y2 isharmonic in R. Find the maximum and minimum values of u in R and state where they areachieved.

Solution.In fact, the function u (x, y) = x2 − y2 is harmonic then have maximum and minimum on

boundary of R. And we can see thatx2 − y2 ≤ x2 ≤ 1

x2 − y2 ≥ −y2 ≥ −1

It means that −1 ≤ u(x, y) ≤ 1 for all x, y in R. We go on to note that on boundary of R,the unique point (1, 0) makes u = 1 and the same is (0, 1) which makes u = −1.Then we obtain the result above.

130

5 INFINITE SERIES INVOLVING ACOMPLEX VARIABLE

5.1 INTRODUCTION AND REVIEW OF REAL SERIES

We will solve exercises 1, 3, 4, 10, 11, 16, 17, 18, 20, 21.Exercise 5.1. (E. 3, E. 3, E. 4 P. 231)Use Eq (5.1-4) to obtain the first nonzero terms in the Taylor series expansions of the followingreal functions. Give a formula for the general or nth term.

1.1

1− xexpanded about x = 0

2.1

1− xexpanded about x = −1

3.√x expanded about x = 1

Solution.

1. We have

c0 =f (0)

0!= 1

c1 =f ′ (0)

1!= 1

c2 =f ′′ (0)

2!= 1

c3 =f ′′′ (0)

3!= 1

We can deduce that cn = 1 by using induction.

2. Apply Taylor series expansions, we get

f (z) =

∞∑n=0

cn (x− x0)n =

∞∑n=0

cn (x+ 1)n

Then, we list some coefficients as follow

131

5 INFINITE SERIES INVOLVING A COMPLEX VARIABLE

c0 = f (−1) =1

2

c1 = f ′ (−1) =1

22

c2 =f ′′ (−1)

2!=

1

23

c3 =f ′′′ (−1)

3!=

1

24

So that the general coefficient is cn =1

2n+1.

3. In a similar fashion, we observe that

f ′ (x) =1

2√x

f ′′ (x) =−1

4x√x

f ′′′ (x) =3

8x2√x

Then we have the respectively coefficients are c0 = 1, c1 =1

2, c2 =

−1

8, c3 =

1

16. So the

general is cn = (−1)n+1 1

2n, n 6= 0.

Exercise 5.2. (E. 10, E. 11 P. 231)Study the ratio test and the notion of absolute convergence in a book on elementary calculusand show that the following real series are absolutely convergent or are divergent in the intervalsspecified.

1.∞∑n=1

2nxn

nabs.conv. for |x| < 1

2, div. for |x| > 1

2

2.∞∑n=1

sinhn

en(x+ 1)n abs.conv. for −2 < x < 0, div. for x < −2 and x > 0

Solution.

1. We see the ratio test for this exercise as follows

∣∣∣∣un+1

un

∣∣∣∣ =2 |x|nn+ 1

Hence,

limn→∞

∣∣∣∣un+1

un

∣∣∣∣ = limn→∞

2 |x|nn+ 1

= 2 |x|

In order to this series are absolutely convergent, we need

132


limn→∞

∣∣∣∣un+1

un

∣∣∣∣ = 2 |x| < 1

It means that the proof we find. And the divergent is created simiilarly.

2. Also, we consider the ratio test below.

∣∣∣∣∣∣∣sinh (n+ 1)

en+1(x+ 1)n+1

sinhn

en(x+ 1)n

∣∣∣∣∣∣∣ =

(1− e−2π−2

)(x+ 1)

1− e−2π

Thus if we want to solve the absolutely convergent, we must have

limn→∞

∣∣∣∣∣(1− e−2π−2

)(x+ 1)

1− e−2π

∣∣∣∣∣ = |x+ 1| < 1

It means that −2 < x < 0 and vice versa for the interval of divergent.

Exercise 5.3. (E. 16, E. 17 P. 232)Recall the comparison test for convergence, from elementary calculus : If a series of positive

constants∞∑n=1

cn is known to converge, and if one is given a series∞∑n=1

un (x), where 0 ≤ un (x) ≤

cn, then the latter series converges. Assume that we know that the p-series∞∑n=1

1

npconverges,

where p > 1, to show the convergence of the following. Use the positive value of any quantitythat appears to be multivalued.

1.∞∑n=1

cos2 nx

n3/2for −∞ < x <∞

2.∞∑n=1

tanhnx

n1.1for −∞ < x <∞

3.∞∑n=1

1

n1+nxfor x > 0

Solution.

1. We have an inequality below.

∣∣∣∣cos2 nx

n3/2

∣∣∣∣ ≤ 1

n3/2

And see that the series∞∑n=1

1

n3/2converges, so the series given converges.

2. In a similar way, we get

133


∞∑n=1

tanhnx

n1.1≤∞∑n=1

1

n1.1

And the series on the right side converges, the series given thus converges.

3. Because of 1 + nx > 1, there exists ε > 0 such that 1 + nx ≥ 1 + ε > 1. Then

1

n1+nx≤ 1

n1+ε

And the series on the right side converges, the series given thus converges.

Exercise 5.4. (E. 20, E. 21 P. 232)

Recall the comparison test for divergence. If a series of positive constants∞∑n=1

cn is known to

diverge, and if one is given a series∞∑n=1

un (x) where 0 < cn ≤ un (x), then the latter series

diverges. Assuming that we know that the p-series diverges if p ≤ 1, prove the following seriesdiverge where indicated.

1.∞∑n=1

1 + cos2 nx√n

for −∞ < x <∞

2.∞∑n=1

cothnx

nfor |x| > 0

Solution.

1. We consider the inequality.

1√n≤ 1 + cos2 nx√

n

But the series on the left side diverges, hence the series given too.

2. Because of |x| > 0, we have |cothnx| ≥ 1. Then

1

n≤∣∣∣∣cothnx

n

∣∣∣∣But the series on the left side diverges, hence the series given too.

134


5.2 COMPLEX SEQUENCES AND CONVERGENCE OFCOMPLEX SERIES

We have exercises 4, 5, 6, 8, 9, 10, 11.Exercise 5.5. (E. 4, E. 5, E. 6 P. 241)Use the nth term test to prove that the following series are divergent in the indicated regions.

1.∞∑n=0

(n+ 1) (i+ 1)n (z + 1)n for |z + 1| ≥ 1√2

2.∞∑n=2

n (i− 1)n

(z − 2i)nfor |z − 2i| ≤

√2

3.∞∑n=1

(2n+ 2

n

)n(z + 1 + i)n for |z + 1 + i| ≥ 1

2

Solution.

1. We have

|un (z) | = | (n+ 1) (i+ 1)n (z + 1)n |

= |n+ 1|.|i+ 1|n.|z + 1|n

= (n+ 1)(√

2)n|z + 1|n

If |z + 1| = 1√2, then

(√2)n |z + 1|n → 1 when n→∞.

Thus the series is divergent. And for the case of |z + 1| > 1√2, we still have too.

2. We get

|un (z) | =

∣∣∣∣n (i− 1)n

(z − 2i)n

∣∣∣∣=|n (i− 1)|n

|(z − 2i)|n

=nn√

2n

|z − 2i|n

We consider a sequence√

2n

|z − 2i|nbecause of nn →∞. If |z−2i| =

√2, you have

√2n

|z − 2i|n=

1 when n→∞. So that the series is divergent.

3. We obtain that

135


|un (z) | =

∣∣∣∣(2n+ 2

n

)n(z + 1 + i)n

∣∣∣∣=

(2n+ 2

n

)n|z + 1 + i|n

= 2n(

1 +1

n

)n|z + 1 + i|n

Moreover, we remember that(

1 +1

n

)n→ e, n→∞

And if |z + 1 + i| = 1

2, then 2n|z + 1 + i|n = 1. Hence,

|un (z) | =(

1 +1

n

)n→ e 6= 0

The proof is completed.

Exercise 5.6. (E. 8, E. 9, E. 10 P. 241)Use the ratio test to prove the absolute convergence, in the indicated domains, of the followingseries. Where does the ratio test assert that each series diverges?

1.∞∑n=0

n!en2z for <z < 0

2.∞∑n=0

(2 + i)n

(z + i)n (n+ i)2 for |z + i| >√

5

3.∞∑n=0

1

n!

(nz

)nfor |z| > e

Solution.

1. We have

∣∣∣∣un+1

un

∣∣∣∣ =

∣∣∣∣∣(n+ 1)!e(n+1)2z

n!en2z

∣∣∣∣∣=

∣∣∣(n+ 1) e(2n+1)z∣∣∣

= (n+ 1) e(2n+1)x

For the case of x < 0, we see that

∣∣∣∣un+1

un

∣∣∣∣ = (n+ 1) e(2n+1)x → 0 < 1

136


So the series is convergent.

For the case of x ≥ 0, we consider that

∣∣∣∣un+1

un

∣∣∣∣ = (n+ 1) e(2n+1)x →∞ > 1

Hence, the series is divergent.

2. In a similar fashion, we take the ratio test here.

∣∣∣∣un+1

un

∣∣∣∣ =

√5(n2 + 1

)|z + i| (n2 + 2n+ 2)

And if we make n approachs to ∞, the limit of this will depend on

limn→∞

∣∣∣∣un+1

un

∣∣∣∣ =

√5

|z + i|

At this time, we have a notice.

For the case of |z + i| >√

5, the series is convergent.

For the case of |z + i| <√

5, the series is divergent.

For the case of |z + i| =√

5, we cannot define the convergence of this series.

3. We get

∣∣∣∣un+1

un

∣∣∣∣ =

(n+ 1

n

)n 1

|z|

Then

limn→∞

∣∣∣∣un+1

un

∣∣∣∣ =1

|z|limn→∞

(n+ 1

n

)n=

e

|z|

At this time, we still have 3 cases

If |z| > e, the series is convergent.

If |z| < e, the series is divergent,

If |z| = e, we cannot define the convergence of this series.

Exercise 5.7. (E. 11 P. 242)Make the substitution z = eiθ in Eq. (5.2-9), N = n − 1, assume θ is real, and separate Eq.(5.2-9) into its real and imaginary parts to show that

1. 1 + cos θ + cos 2θ + ...+ cos (Nθ) = cos [Nθ/2]sin [(N + 1) θ/2]

sin [θ/2]

2. sin θ + sin 2θ + ...+ sin (Nθ) = sin [Nθ/2]sin [(N + 1) θ/2]

sin [θ/2]

137


Solution.We have

1 + z + z2 + ...+ zn−1 = 1 + eiθ + e2iθ + ...+ eiNθ

= 1 + cos θ + i sin θ + cos 2θ + i sin 2θ + ...+ cos (Nθ) + i sin (Nθ)

= (1 + cos θ + cos 2θ + ...+ cos (Nθ)) + i (sin θ + sin 2θ + ...+ sin (Nθ))

So that

1− zn

1− z=

1− ei(N+1)θ

1− eiθ

=1− cos (N + 1) θ − i sin (N + 1) θ

1− cos θ − i sin θ

=

2 sin2

[(N + 1) θ

2

]− i2 sin

[(N + 1) θ

2

]cos

[(N + 1) θ

2

]2 sin

(θ

2

)[sin

(θ

2

)− i cos

(θ

2

)]

=

2 sin

[(N + 1) θ

2

](sin

[(N + 1) θ

2

]− i cos

[(N + 1) θ

2

])2 sin

(θ

2

)[sin

(θ

2

)− i cos

(θ

2

)]

=

sin

[(N + 1) θ

2

](sin

[(N + 1) θ

2

]− i cos

[(N + 1) θ

2

])[sin

(θ

2

)− i cos

(θ

2

)]sin

(θ

2

)

=

sin

[(N + 1) θ

2

](cos

[Nθ

2

]− i sin

[Nθ

2

])sin

(θ

2

)

Because of1− zn

1− z= 1 + z + z2 + ...+ zn−1, we deduce that

sin

[(N + 1) θ

2

](cos

[Nθ

2

]− i sin

[Nθ

2

])sin

(θ

2

) = (1 + cos θ + cos 2θ + ...+ cos (Nθ))+i (sin θ + sin 2θ + ...+ sin (Nθ))

Hence,

sin

[(N + 1) θ

2

]cos

[Nθ

2

]sin

(θ

2

) = 1 + cos θ + cos 2θ + ...+ cos (Nθ)

138


sin

[(N + 1) θ

2

]sin

[Nθ

2

]sin

(θ

2

) = (sin θ + sin 2θ + ...+ sin (Nθ))

139


5.3 UNIFORM CONVERGENCE OF SERIES

In this section, we face with some exercises such as 1, 2, 3, 4, 6, 8.Use the Weierstrass M test to establish the uniform convergence of the following series in theindicated regions. State the convergent series of constants that is employed.

1.∞∑j=1

(−z)j−1 for |z| ≤ 0.999

2.∞∑j=1

j

j + 1(−z)j for |z| ≤ r, where r < 1

3.∞∑j=1

zj

j!for |z| ≤ r, where r <∞

4.∞∑n=1

|n− i|n3

zn for |z| ≤ r, where r < 1

Solution.

1. Apply Leibnitz’s theorem for alternating series, we have (0.999)j−1 is a sequence reducedand approachs to 0. Hence

∞∑j=1

(−1)j−1 (0.999)j−1

is converging. Then, we put Mj = (0.999)j−1 and obtain that

|uj | ≤Mj

So that this series is uniformly converging.

2. We consider the series∞∑j=0

(−z)j and easily prove that it converges because of the con-

verging series∞∑j=0

rj with r < 1. And then, according to Weierstrass theorem, we obtain

the series given is uniformly converging.

3. Because of |z| ≤ r, we see that

|z|j

j!≤ rj

j!

And∞∑j=0

rj

j!= er is converging. So that the series given is uniformly converging.

4. We observe the inequality below.

140


∣∣∣∣ |n− i|n3zn∣∣∣∣ =

√n2 + 1

n3|z|n

≤√n2 + 1

n3rn

≤√n2 + 1

n3

≤√

2

n2

And the series∞∑n=1

√2

n2is converging. Hence, the given series is converging uniformly.

Exercise 5.8. (E. 6 P. 247)

In this problem, we prove the Weierstrass M test. We are given a series∞∑j=1

uj (z) whose sum

is S (z) when z lies in a region R. We have also at our disposal a convergent series of constants∞∑j=1

Mj such that throughout R we have |uj (z)| ≤Mj and wish to show this guarantees uniform

convergence of the original series as well as absolute convergence.

1. Using the comparison test from real calculus, explain why the series∞∑j=1

uj (z) must be

absolutely convergent. Recall that absolute convergence guarantees ordinary convergence.

2. Using the definition of convergence explain why

|S (z)− Sn (z)| =

∣∣∣∣∣∣ limk→∞

k∑j=1

uj (z)−n∑j=1

uj (z)

∣∣∣∣∣∣=

∣∣∣∣∣∣ limk→∞

k∑j=n+1

uj (z)

∣∣∣∣∣∣if we take k > n.

3. Explain why

∣∣∣∣∣∣ limk→∞

k∑j=n+1

uj (z)

∣∣∣∣∣∣ ≤ limk→∞

k∑j=n+1

Mj

4. Prove that given ε > 0 there must exist a number N such that limk→∞

k∑j=n+1

Mj < ε for

n > N .

Solution.

141


1. With the converging series∞∑j=1

Mj , we have for all z in R, Mj > 0

|uj (z)| ≤Mj

In contrast, we see that

∞∑j=1

|uj (z)| = |u1 (z)|+ ...+ |un (z)|+ ...

≤ M1 + ...+Mn + ...

=∞∑j=1

Mj

Hence, the series∞∑j=1

|uj (z)| is converging and then∞∑j=1

uj (z) is absolutely converging.

2. We get

S (z) =∞∑j=1

uj (z) = limk→∞

k∑j=1

uj (z)

Sn (z) =

n∑j=1

uj (z)

So that we take a subtraction below.

S (z)− Sn (z) =∞∑

j=n+1

uj (z)

= limk→∞

k∑j=n+1

uj (z) , k > n

3. We obtain the inequality below.

∣∣∣∣∣∣k∑

j=n+1

uj (z)

∣∣∣∣∣∣ ≤∞∑

j=n+1

|uj (z)|

=∞∑

j=n+1

Mj

Thus we deduce that

142


∣∣∣∣∣∣ limk→∞

k∑j=n+1

uj (z)

∣∣∣∣∣∣ ≤ limk→∞

k∑j=n+1

Mj

4. With M =

∞∑j=1

Mj , we create a sequence Mn =

n∑j=1

Mj and obtain obviously that

M −Mn =∞∑j=1

Mj −n∑j=1

Mj

=∞∑

j=n+1

Mj

At this time, Mn approachs to M if for all ε > 0, there exists a positive integer N (ε) suchthat

|M −Mn| < ε,∀n > N


∣∣∣∣∣∣∞∑

j=n+1

Mj

∣∣∣∣∣∣ < ε,∀n > N

Or we can say that

limk→∞

k∑j=n+1

Mj < ε

Exercise 5.9. (E. 8 P. 248)

In this exercise, we show that the series∞∑j=1

zj−1 converges uniformly to 1/ (1− z) in the disc

|z| ≤ r, where r < 1. The proof requires that we obtain a value for N satisfying the definitionof uniform convergence.

1. Explain why Log (1/ |z|) ≥ Log (1/r) in the disc.

2. Prove 1/ |1− z| ≤ 1/ (1− r) and Log [1/ (ε |1− z|)] ≤ Log [1/ (ε (1− r))] for |z| ≤ r. Takeε > 0.

3. Assume that 0 < ε < 1/2. Show that in the disc |z| ≤ r we have

Log1

ε |1− z|

Log1

|z|

≤Log

1

ε (1− r)

Log1

r

143


Solution.

1. It is easy to see that1

|z|≥ 1

r, thus we get the inequality need to explain.

2. In a similar fashion, we can observe carefully the inequality

|1− z| ≥ 1− z ≥ 1− r

So that the next is

1

ε |1− z|≤ 1

ε (1− r)

Hence, the proof is shown completely.

3. We have had 2 inequalities above as follows

Log1

ε |1− z|≤ Log

1

ε (1− r)1

Log

(1

|z|

) ≤ 1

Log

(1

r

)

Because of ε < 1/2, we obtain the proof.

144


5.4 POWER SERIES AND TAYLOR SERIES

We get excercises 27, 28, 29, 30, 31, 32, 33.Exercise 5.10. (E. 27 P. 261)Use the equation below and triangle inequality to prove that in the disc |z| ≤ 1 we have|ez − 1| ≤ (e− 1) |z|.

ez =∞∑n=0

1

n!zn

Solution.By using the given equation, we have

|ez − 1| =

∣∣∣∣∣∞∑n=1

1

n!zn

∣∣∣∣∣Because of |z| ≤ 1, we can easily deduce that |zn| ≤ |z|for all n.Hence, apply that inequality, we get

|ez − 1| =

∣∣∣∣∣∞∑n=1

1

n!zn

∣∣∣∣∣≤

∣∣∣∣∣∞∑n=1

1

n!

∣∣∣∣∣ |z|= (e− 1) |z|

So this is the inequality we need to find.

Exercise 5.11. (E. 28 P. 261)

1. Let zN − zN0 =N∑n=1

cn (z − z0)n valid for all z. N is a positive integer. Show that cn =

N !zN−n0 / [n! (N − n)!].

2. Replace z in the above with z + z0 and show that

(z + z0)N =

N∑n=0

N !zN−n0

n! (N − n)!zn

This is the familiar binomial expansion.

Solution.

1. We have zN − zN0 =N∑n=1

cn (z − z0)ndefine for all z where N is a positive integer

From Talor series formular, we need to evaluate the first and the second coefficients.

c1 =f (1) (z0)

1!= NzN−1

0

145


c2 =f (2) (z0)

2!=N (N − 1) zN−2

0

2!

So we can easily see that cn =N !

n! (N − n)!zN−n0 by using mathematical induction.

2. By replacing z in the above with z+z0, the function f (z) becomes f (z) = (z + z0)N−zN0 .

We will do the same as above. First we have

(z + z0)N − zN0 =N∑n=1

cnzn


(z + z0)N =

N∑n=1

cnzn + zN0

We already have cn above. Then

(z + z0)N =N∑n=1

N !zN−n0

n! (N − n)!zn + zN0

At this time, we notice that

c0 =N !zN−0

0

0! (N − 0)!z0 = zN0

Hence we have (z + z0)N =N∑n=0

N !zN−n0

n! (N − n)!zn

The proof is found completely.

146


Exercise 5.12. (E. 29 P. 262)

1. Let f (z) = Log (z), where the principal branch of the logarithm is used . Thus f (z) isdefined by means of the branch cut y = 0, x ≤ 0. Show that

f (z) =∞∑n=0

cn (z + 1− i)n

where

c0 = Log√

2 + i3π

4

cn =(−1)n+1 e−i(3π/4)n

n(√

2)n , n 6= 0

2. What is the radius of the largest circle centered at −1 + i within which the series of part(1) converges to f (z)?

3. Use the ratio test to establish that the series of part (1) converges inside |z − (−1 + i)| =√2 and diverges outside this circle. Compare this circle with the one in part (2).

Solution.

1. We have f (z) = Log (z) is the principal branch of the logarithm is used, and take thecircle |z − (−1 + i)| <

√2.

First c0 = f (z0) = Log (z0) = Log(√

2)

+ i3π

4because of θ =

3π

4by deriving from

z0 = −1 + i and |z − (−1 + i)| <√

2 with the principal value.

With n 6= 0, we still use Taylor series formular for general coefficient, and compute thefirst, the second respectively.

c1 =f (1) (z0)

1!=e−i

3π

4√

2

c2 =f (2) (z0)

2!= − e

−2i3π

4

2(√

2)2

We have f ′ (z) =1

zso fn (z) = (−1)n+1 (n− 1)!z−n

Thus

cn =f (n) (z0)

n!= (−1)n+1 (n− 1)!z−n0

n!= (−1)n+1 e

−ni3π

4

n(√

2)n

147


Hence we have what we need to prove and that is f (z) =∞∑n=0

cn (z + 1− i)n with c0 =

Log(√

2)

+ i3π

4and cn =

f (n) (z0)

n!= (−1)n+1 e

−ni3π

4

n(√

2)n , n 6= 0.

2. Because the branch cut of the principal Logz is y = 0 and x ≥ 0 so the largest radius ofcircle center at −1 + i must be from the origin to −1 + i and that is d = |−1 + i| =

√2.

3. We will examine

limn→∞

∣∣∣∣∣cn+1 (z + 1− i)n+1 (z + 1− i)n+1

cn (z + 1− i)n (z + 1− i)n

∣∣∣∣∣ = limn→∞

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

(−1)n+2 e−i(n+1)

3π

4

(n+ 1)(√

2)n+1

(−1)n+1 e−in

3π

4

n(√

2)n

(z + 1− i)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣= lim

n→∞

∣∣∣∣∣∣∣∣(−1)ne

−i3π

4

(n+ 1)√

2(z + 1− i)

∣∣∣∣∣∣∣∣= lim

n→∞

1√2

∣∣∣∣ n

n+ 1(z + 1− i)

∣∣∣∣limn→∞

n

(n+ 1)√

2|(z + 1− i)| = lim

n→∞

|(z + 1− i)|√2

=|(z + 1− i)|√

2

Now with the ratio test, we set|(z + 1− i)|√

2< 1 or |(z + 1− i)| <

√2

Our series converges absolutely if z lies inside a circle of radius√

2 centered at the point−1 + i and diverges outside the circle. We need to observe on |(z + 1− i)| =

√2.

We have

∣∣∣∣∣∣∣∣(−1)n+1 e

−in3π

4

n(√

2)n (z + 1− i)n

∣∣∣∣∣∣∣∣ =e−in

3π

4

n

148


Exercise 5.13. (E. 30 P. 262)

1. Let f (z) be analytic in a domain containing z = 0. Assume that f (z) is an even functionof z in this domain, i.e., f (z) = f (−z). Show that in the Maclaurin expansion f (z) =∞∑n=0

cnzn the coefficients of odd order, c1, c3, c5..., must be zero.

2. Show that for an odd function, f (z) = −f (−z), analytic in the same kind of domain, thecoefficients of even order, c0, c2, c4..., are zero.

3. What coefficients vanish in the Maclaurin expansions of z sin z, z2 tan z, andcosh z/

(1 + z2

)?

Solution.

1. f (z) is analytic in a domain containing z = 0 and it is an even function in this domain.

We have Maclaurin expansion f (z) =

∞∑n=0

cnzn =

∑∞n=0 cn (−z)n

So c0z0 + c1z

1 + c2z2 + ... = c0 (−z)0 + c1 (−z)1 + c2 (−z)2 + ...

With even order we see (z)n = (−z)n. Then we simplify the even order and get c1z1 +

c3z3 + c5z

5 + ... = 0 because n is an odd positive integer and zn 6= zn+1for all n. Inorder to get this sum equal to zero then all the odd coefficients must be zero. That meansc1 = c3 = ... = c2n+1 = 0

2. We do the same as part (1)

∞∑n=0

cnzn = −

∞∑n=0

cn (−z)n

So we deduce that

c0z0 + c1z

1 + c2z2 + ... = −c0 (−z)0 − c1 (−z)1 − c2 (−z)2 − ...

With even order we see (z)n = (−z)n. Then this time it is the even order that remain inthis sum

c0z0 + c2z

4 + c4z4 + ... = 0

To get the odd function, we must have even coefficients equal to zero. That meansc0 = c2 = ... = c2n = 0

3. From part (1) and (2), we will now examine what coefficients is zero

z sin (z) = −z sin (−z) because of sin (−z) = − sin (z) so z sin (z)is an even function, thusits odd coefficients are zero.

149


z2 tan (z) is an odd function since tan (z) =sin (z)

cos (z)and sin (−z) = − sin (z) , cos (z) =

cos (−z) so its even coefficients are zero.cosh (z)

1 + z2=

ez + e−z

2 (1 + z2)=

e−z + ez

2 (1 + z2)so it is an even function then its odd coefficients are

zero.

Exercise 5.14. (E. 31 P. 262)Suppose a function f(z) is analytic in a domain which contains the origin, and we expand it in a

Maclaurin series f(z) =

∞∑n=0

cnzn. If the circle |z| = r lies within the domain and if on the circle

we have the bound |f (z)| ≤ K, then we can place a bound on |cn|, namely |cn| ≤ K/rn, forn ∈ N. This is called Cauchy’s inequality and it is sometimes useful in telling us if we have madean error in computing the coefficients in a Maclaurin series expansion; i.e., if the inequality isnot satisfied, there is a mistake. The inequality is easily generalized to Taylor series.

1. Derive the Cauchy inequality by using the following equation

cn =f (n) (0)

n!=

1

2πi

‰C′

f (z)

zn+1dz, n = 1, 2, 3, ...

, taking the contour C ′ as |z| = r, and using the ML inequality.

2. Consider the expansion ez =∞∑n=0

cnzn. We want to obtain a bound on |cn| without actually

obtaining the coefficients. Through suitable choices of r show that for all n ≥ 0 we have|cn| ≤ e and |cn| ≤ e2/2n. Do the known coefficients satisfy this inequality?

3. Generalize Cauchy’s inequality so that it can be applied to a Taylor series expansion f(z) =∞∑n=0

cn (z − z0)n. Use your result to explain why in the expansion ez =

∞∑n=0

cn (z − 3)n we

can say, without finding any of the coefficients, that |cn| ≤ e4.

Solution.

1. We have

|cn| =∣∣∣∣ 1

2πi

‰C′

f (z)

zn+1dz

∣∣∣∣ ≤ 1

2πML

by using ML inequality where M is the magnitude off (z)

zn+1and L is the length of the cicle

and that is L = 2πr.

First∣∣∣∣f (z)

zn+1

∣∣∣∣ ≤ K

|zn+1|=

K

rn+1= M

Then ML inequality is that

|cn| =1

2π

∣∣∣∣‰C′

f (z)

zn+1dz

∣∣∣∣ ≤ 1

2πML =

1

2π

(K

rn+1

)2πr =

K

rn

150


Hence we obtain Cauchy’s inequality |cn| ≤K

rn, n = 1, 2, 3, ...

2. We observe that |cn| ≤ e and get f (z) = ez =∞∑n=0

cnzn and |cn| ≤

∣∣∣∣ ezzn+1

∣∣∣∣ r =er

rnby

choosing r = 1. Like above but with r = 2 then we get |cn| ≤e2

2n

The known coefficients satisfy this inequality because we using Maclaurin expansion forz = 0 so it can satisfy.

3. We generalize the Cauchy inequality in Taylor expansion. This time is f (z) =

∞∑n=0

cn (z − z0)n.

And

cn =f (n) (z0)

n!=

1

2πi

‰C′

f (z)

(z − z0)n+1dz, n = 1, 2, 3, ...

where the circle is |z − z0| = r

If we use the same technique as part (1), then we will get the Cauchy’s inequality forTaylor series expansion.

|cn| ≤K

rn

where |f (z)| ≤ K and |z − z0| = r.

151


Exercise 5.15. (E. 32 P. 263)

1. Refer to the proof of theorem of Taylor series and use the equation

f (z1) =1

2πi

‰C′

f (z)

z − z1dz =

1

2πi

‰C′

f (z)

z(

1− z1

z

)dzto show that

f (z1) =1

2πi

‰C′f (z)

1

z+z1

z2+z2

1

z3+ ...+

zn−11

zn+

(zn1zn

)z − z1

dz

2. Use the expression for f (z1) given in part (1) to show, after integration, that

f (z1) = f (0) + f ′ (0) z1 +f ′′ (0)

2z2

1 + ...+f (n−1) (0)

(n− 1)!zn−1

1 +Rn

where

Rn =zn12πi

‰C′

f(z)

zn (z − z1)dz

3. We can place an upper bound on the remainder in equation above. Assume |f (z)| ≤ m

everywhere on |z| = b (the contour C ′). Use the ML inequality to show that

|Rn| ≤∣∣∣z1

b

∣∣∣n mb

b− |z1|

4. Suppose we wish to determine the approximate value of cosh i by the finite series inEq 5.4-21. Taking the contour C ′ as |z| = 2, show that the error made cannot exceed(cosh 2) /210.

Solution.

1. Now we recall equation that is1

1− z= 1 + z + z2 + ...+ zn−1 +

zn

1− zSo for substitution z by

z1

zthen we get

1

1− z1

z

= 1 +z1

z+z2

1

z2+ ...+

zn−11

zn−1+

(z1

z

)n1− z1

z

Therefore,

152


f (z1) =1

2πi

‰C′

f (z)

z(

1− z1

z

)dz=

1

2πi

‰C′

f (z)

z

1 +z1

z+z2

1

z2+ ...+

zn−11

zn−1+

(z1

z

)n1− z1

z

dz

=1

2πi

‰C′f (z)

1

z+z1

z2+z2

1

z3+ ...+

zn−11

zn+

(zn1zn

)z − z1

dz

2. From part (1), we have had

f (z1) =1

2πi

‰C′f (z)

1

z+z1

z2+z2

1

z3+ ...+

zn−11

zn+

(zn1zn

)z − z1

dz

=1

2πi

‰C′f (z)

1

zdz +

1

2πi

‰C′f (z)

z1

z2dz + ...+

1

2πi

‰C′f (z)

zn−11

zndz +

1

2πi

‰C′f (z)

(zn1zn

)z − z1

dz

From equation 5.4-17 we construct cn =f (n) (0)

n!=

1

2πi

�C′f (z)

zn+1dz, we get

f (z1) = f0 (0) +f (1) (0)

1!z1 + ...+

f (n−1) (0)

(n− 1)!zn−1

1 +zn12πi

‰C′

f (z)

zn (z − z1)dz

3. We set Rn =zn12πi

�C′

f (z)

zn (z − z1)dz is the remainder and assume |f (z)| ≤ m everywhere

on |z| = b (the contour C ′).

Let take |Rn| =∣∣∣∣ zn12πi

�C′

f (z)

zn (z − z1)dz

∣∣∣∣ ≤ML (ML inequality)

First we examine L = 2πb is the length of C ′. And the second is

∣∣∣∣ zn12πi.

f (z)

zn (z − z1)

∣∣∣∣ ≤ 1

2π

∣∣∣∣ zn1zn (z − z1)

∣∣∣∣ |f (z)| ≤ 1

2π

∣∣∣∣zn1bn∣∣∣∣ m

b− |z1|= M

Hence,

ML =1

2π

∣∣∣∣zn1bn∣∣∣∣ m

b− |z1|2πb =

∣∣∣∣zn1bn∣∣∣∣ mb

b− |z1|

Thus we have |Rn| ≤∣∣∣∣zn1bn∣∣∣∣ mb

b− |z1|.

4. We already have cosh (z) =ez + e−z

2so |cosh (z)| =

∣∣∣∣ez + e−z

2

∣∣∣∣ ≤ ex + e−x

2= cosh (x).

Then |cosh (z)| ≤ cosh (x).

153


Using part (3) with |z| = 2 = b and f (z) = cosh (z). Therefore, we get |f (z)| ≤ cosh (2) =

m

|Rn| ≤∣∣∣∣zn1bn∣∣∣∣ mb

b− |z1|=

∣∣∣∣ i10

210

∣∣∣∣ 2 cosh (2)

2− |i|=

cosh (2)

29

with n = 10

Hence the remainder iscosh (2)

29and the error can not exceed

cosh (2)

210

Exercise 5.16. (E. 33 P. 263)Let a function f (z) be entire and have property that f (n) (z) = f (z) , n = 1, 2, ... Show by usinga Taylor series expansion that this function must have this “addition property” : f (z1 + z2) =

f (z1) f (z2).

Solution.

We get the Maclaurin expansion of f (z2) =

∞∑n=0

cnzn2 where cn =

f (0)

n!for all n ∈ N (in

equation 5.4-18 and 5.4-19). Now we use Taylor expansion (equation 5.4-10)

f (z) =

∞∑n=0

cn (z − z0)n, |z − z0| ≤ a

Let z = z1 + z2 and z0 = z1 then

f (z1 + z2) =∞∑n=0

bn (z1 + z2 − z1)n =∞∑n=0

bn (z2)n =∑

cn (z2)n

So the Taylor series becomes the Maclaurin. Hence, we want to get the Taylor so it must be

bn =f (z1) f (0)

(n!).

We replace with the one into

f (z1 + z2) =∞∑n=0

bn (z2)n

= f (z1)

∞∑n=0

f (0)

n!(z2)n

= f (z1) f (z2)

154


5.5 TECHNIQUES FOR OBTAINING TAYLOR SERIESEXPANSIONS

In this section, the exercises will be shown are 1, 2, 3, 4, 5, 7, 8, 11, 12, 15, 16, 19, 23.Exercise 5.17. (E. 1, E. 2, E. 3, E. 4 P. 274)

The following exercises involve our generating a new Taylor series through a change of variablesin the geometric series Eq.(5.2-8) or some other familiar expansion. Here a is any constant.Explain how the following are derived

1.1

1 + az= 1− az + a2z2 − a3z3 + ...

(|z| <

∣∣∣∣1a∣∣∣∣)

2.1

1 + z2= 1− z2 + z4 − z6 + ... (|z| < 1)

3.1

1 + a+ z= 1− (z + a) + (z + a)2 − (z + a)3 + ... (|z − a| < 1)

4.

a) e−z2 = 1− z2 +z4

2− z6

6+ ... all z

b) Use the preceding result to find the 10th derivative of e−z2 at z = 0.

Solution.

1. Because of |z| <∣∣∣∣1a∣∣∣∣, we get

1

1 + az=

1

1− (−az)

=∞∑n=0

(−az)n

2. We use the formular

1

1 + w= 1− w + w2 − w3 + ..., |w| < 1

And set w = z2. we can infer

1

1 + z2= 1− z2 + z4 − z6 + ..., |z| < 1

3. In a similar fashion, we also use the formular

1

1− w= 1 + w + w2 + w3 + ..., w < 1

We can deduce that

1

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

155


for |−a− z| < 1 or |z + a| < 1

4.

a) The formular is ew = 1 + w +w2

2!+ ... for all w

Then with w = −z2, we replace into this formular and get

e−z2

= 1 +−z2 +z4

2!− z6

3!+z8

4!− ...

b) Apply Taylor series expansion, we see the general coefficient is cn =f (n) (z0)

n!So with z0 = 0, we have

f (n) (0) = cnn!

Hence, the coefficient of z10 is − 1

5!in the series of (a)

Thus f (10) (0) = −10!

5!= 30240

Exercise 5.18. (E. 5, E. 7, E. 8 P. 274)

1. Differentiate the series of equation 5.5-2 to show that

1

z3= 1− 3.2

2(z − 1) +

4.3

2(z − 1)2 − 5.4

2(z − 1)3 + ..., |z − 1| < 1

2. Use the series in equation 5.2-8 and successive differentiation to show that, for N ≥ 0

1

(1− z)N=∞∑n=0

cnzn, cn =

(N − 1 + n)!

n! (N − 1)!, |z| < 1

3. Integrate the series of Exercise 2 along a contour connecting the origin to an arbitrarypoint z, where |z| < 1 to show that

tan−1 z =∞∑n=0

(−1)nz2n+1

2n+ 1, |z| < 1

Solution.

1. We can divide it as follows

156


1

z3=

1

z2.z, |z − 1| < 1

=

[ ∞∑n=0

(n+ 1) (1− z)n][ ∞∑

n=0

(1− z)n]

∞∑n=0

(n+ 2) (n+ 1)

2(1− z)n

This series express as the right side in the proof.

2. In equation 5.2-8, we have

∞∑j=1

zj−1 = 1 + z + z2 + ... = S (z) =1

1− z, |z| < 1

Successive differentiation is that

1

(1− z)N= z0 +

N !

(N − 1)!z +

(N + 1)!

(N − 1)!2!z2 +

(N + 2)!

(N − 1)!3!z3 + ...

=

∞∑n=0

(N − 1 + n)!

n! (N − 1)!zn

where |z| < 1.

3. We observe the formular below

1

1 + w2= 1− w2 + w4 − w6 + ..., |w| < 1

And with |z| < 1, we have

ˆ 0

z

1

1 + w2dw =

ˆ 0

z

(1− w2 + w4 − w6 + ...

)dw

= z − z2

3+z5

5− z7

7+ ...

Then

ˆ 0

z

1

1 + w2dw = tan−1 z

Thus tan−1 z =

∞∑n=0

(−1)nz2n+1

2n+ 1with |z| < 1.

157


Exercise 5.19. (E. 11 P. 276)

By considering the first and second derivatives of the geometric series in Equation 5.2–8, showthat

∞∑n=1

n2zn =

(z + z2

)(1− z)3

for |z| < 1

Solution.In equation 5.2-8, we have had

∞∑j=1

zj−1 = 1 + z + z2 + z3 + z4 + z5 + ... =1

1− z, |z| < 1

Successive differentiation the first time, we obtain

−1

(1− z)2 = 1 + 2z + 3z2 + 4z3 + 5z4 + 6z5 + ...

Successive differentiation the second time, we get

2

(1− z)3 = 2 + 6z + 12z2 + 10z3 + ...

We compute the following series

z

(1− z)2 = z + 2z2 + 3z3 + 4z4 + ... =∞∑n=1

nzn

2z

(1− z)3 = 2z + 6z2 + 12z3 + 20z4 + ... =

∞∑n=1

n (n+ 1) zn

So2z

(1− z)3 −z

(1− z)2 =2z − z (1− z)

(1− z)3 =z2 + z

(1− z)3 =∞∑n=1

n (n+ 1) zn −∞∑n=1

nzn =∞∑n=1

n2zn

Obtain the following Taylor expansions. Give a general formula for the nth coefficient, andstate the circle within which your expansion is valid.

Exercise 5.20. (E. 12 P. 276)Use series multiplication to find a formula for cn in these Maclaurin expansion. In what circleis each series valid?

cosh z

1− z=∞∑n=0

cnzn

Solution.We have that for all z then

cosh z = 1 +z2

2!+z4

4!+z6

6!+ ...

158


And with |z| < 1

1

1− z= 1 + z + z2 + z3 + ...

Then we multiply them and get

cosh z

1− z=

(1 +

z2

2!+z4

4!+z6

6!+ ...

)(1 + z + z2 + z3 + ...

)=

∞∑n=0

cnzn

= 1 + z +

(1 +

1

2!

)z2 +

(1 +

1

2!

)z3 +

(1 +

1

2!+

1

4!

)z2 + ...

Hence, the general coefficient is

cn =

1 +

1

2!+

1

4!+ ...+

1

n!, neven

1 +1

2!+

1

4!+ ...+

1

(n− 1)!, nodd

Exercise 5.21. (E. 15, E. 16, E. 19, E. 23 P. 276)Obtain the following Taylor expansions. Give a general formula for the nth coefficient, and statethe circle within which your expansion is valid.

1.z

(z − 1) (z + 2)expanded about z = 0

2.z + 1

(z − 1)2 (z + 2)expanded about z = 2

3.1

z3expanded about z = i

4.z3 + 2z2 + z − 1

z2 − 4expanded about z = 1

Solution.

1. We have, from Rule I,

f (z) =z

(z − 1) (z + 2)

=A

(z − 1)+

B

(z + 2)

=z (A+B) + 2A−B

(z − 1) (z + 2)

We obtain a system of equations

159


A+B = 1

2A−B = 0

Then we get the root is

A =

1

3

B =2

3

Thus f (z) be rewritten

f (z) =1

3

1

z − 1+

2

3

1

z + 2

Then, we see 2 familiar series below.

1

3

1

z − 1= −1

3

(1 + z + z2 + z3 + ...

)= −1

3

∞∑n=0

zn

2

3

1

z + 2=

2

3

1

1 +z

2

=2

3

∞∑n=0

(−1)n 2−nzn

We can infer

f (z) = −1

3

∞∑n=0

zn +2

3

∞∑n=0

(−1)n 2−nzn

=∞∑n=0

(−1

3+

2

3(−1)n 2−n

)zn

2. In a similar way, we get

2

z + 2=

2

z − 1 + 3=

2

3

1 +z − 1

3

=2

3

[1− z − 1

3+

(z − 1

3

)2

− ...

]

− 1

z + 1= − 1

z − 1 + 2=

−1

2

1 +z − 1

2

= −1

2

[1− z − 1

2+

(z − 1

2

)2

− ...

]

Add the two previous series

z

(z + 1) (z + 2)=

∞∑n=0

cn (z − 1)n

where cn = −1

2

(−1

2

)n+

2

3

(−1

3

)n160


3. We have

f (z) =z + 1

(z − 1)2 (z + 2)=

A

(z − 1)+

B

(z − 1)2 +C

(z + 2)

=A (z − 1) (z + 2) +B (z + 2) + C (z − 1)2

(z − 1)2 (z + 2)

=z2 (A+ C) + z (A+B − 2C)− 2A+ 2B + C

(z − 1)2 (z + 2)

Let the numerator be equal to z + 1 and we get A =1

9, B =

2

3, C = −1

9Hence,

f (z) =1

9.

1

(z − 1)+

2

3.

1

(z − 1)2 −1

9.

1

(z + 2)

Then

1

9.

1

(z − 1)=

1

9.

1

(1 + z − 2)=

1

9.∞∑n=0

(−1)n (z − 2)n, |z − 2| < 1

2

3.

1

(z − 1)2 =2

3.

1

(1 + z − 2)2 =2

3.

∞∑n=0

(−1)n (n+ 1) (z − 2)n, |z − 1| < 1

−1

9.

1

(z + 2)= −1

9.

1

(4 + z − 2)= − 1

36

1

1 +z − 2

4

= − 1

36.∞∑n=0

(−1)n 4−n (z − 2)n,

∣∣∣∣z − 1

4

∣∣∣∣ <1

We can infer

f (z) =∞∑n=0

cnzn

where cn =1

9. (−1)n +

2

3. (−1)n (n+ 1)− 1

36. (−1)n 4−n

4. We see that

f (z) =z3 + 2z2 + z − 1

z2 − 4= (z − 1) + 3 +

5z + 7

z2 − 4

And analyse that

5z + 7

z2 − 4=

A

(z + 2)+

B

(z − 2)

=z (A+B)− 2A+ 2B

z2 − 4

161


Let the numerator equal to 5z + 7 and we get A =3

4, B =

17

4Then

5z + 7

z2 − 4=

3

4.

1

(z + 2)+

17

4.

1

(z − 2)

3

4.

1

(z + 2)=

1

4.

1(1 +

z − 1

3

) =

∞∑n=0

1

4. (−1)n .

1

3n. (z − 1)n , |z − 1|<3

17

4.

1

(z − 2)= −17

4.

1

(1− (z − 1))=∞∑n=0

−17

4. (z − 1)n , |z − 1|<1

We can infer

f (z) =

∞∑n=0

cn (z − 1)n

where cn =1

4. (−1)n .

1

3n− 17

4

162


5.6 LAURENT SERIES

We consider some excercises 1, 2, 3, 4, 5, 6, 7, 9, 11.Exercise 5.22. (E. 1, E. 2, E. 3, E. 4, E. 5 P. 293)Obtain the following Laurent expansions. State the first four nonzero terms. State explicitlythe nth term in the series, and state the largest possible annular domain in which your series isa valid representation of the function.

1.sinh z

z3expanded in powers of z

2.cos (1/z)

z3expanded in powers of z

3. sin

(1 +

1

z − 1

)expanded in powers of z − 1

4. Log[1 +

1

z − 1

]expanded in powers of z − 1

5.(z +

1

z

)7

expanded in powers of z

Solution.

1. We have

sinh z = z +z3

3!+z5

5!+z7

7!+ ...

So that

f(z) = z−2 +1

3!+z2

5!+z4

7!+ ...

=

∞∑n=−1

z2n

(2n+ 3)!, |z| > 0

2. We get

cos

(1

z

)= 1− z−2

2!+z−4

4!− z−6

6!+ ...

Hence,

163


f(z) = z−3 − 1

2!z−5 +

z−7

4!+z−9

6!+ ...

=−3∑

n=−∞

z−(2n+1)

(2n− 1)!

=

−3∑n=−∞

zm

(−m− 3), |z| > 0

3. We use the formula of sin (a+ b) for this function.

sin

(1 +

1

z − 1

)= sin(1) cos

(1

z − 1

)− cos (1) sin

(1

z − 1

)cos

(1

z − 1

)=

∞∑n=1

(1

z − 1

)2n

× 1

(2n)!(−1)m

sin

(1

z − 1

)=

∞∑m=0

(1

z − 1

)2m+1

× 1

(2m+ 1)!(−1)m

Thus

f(z) =

n∑i=0

ci (z − 1)i

with the general coefficient

ci = (−1)i/2 sin(1)(−i)! , ieven

ci = (−1)(i+1)/2 cos(1)(−i)! , iodd

4. We can see that

Log(

1 +1

z − 1

)= Log (z) + Log

(1

z − 1

)Log (z) =

∞∑n=0

(−1)2n+1

nzn

Log(

1

z − 1

)= −

ˆ z

0

1

1− wdw

= −ˆ z

0

∞∑n=0

wndw

= −∞∑n=0

ˆ z

0wndw

= −∞∑m=1

zm

m

164


So the result is

f(z) =

∞∑n=0

(−1)2n+1

nzn−∞∑m=1

zm

m

=∞∑−∞

ckzk

with the general coefficient

ck = (−1)−2k+1

−k , k ≤ 0

ck = −1m , k ≥ 1;

5. We solve this excercise easily.

f(z) =

(z +

1

z

)7

=7∑i=0

Ci7z7−2i

= z7 + 7z5 + 21z3 + 35z + ...

=

7∑i=0

7!

(7− i)!i!; |z| > 0

Exercise 5.23. (E. 6, E. 7 P. 293)

Obtain the indicated Laurent expansions of1

z + i. State the nth terms of the series.

1. An expansion valid for |z| > 1

2. An expansion valid for |z − i| > 2

Solution.

f(z) =1

i+ z; |z| > 1

=1

z

∞∑n=0

(i

z

)n(−1)n

=∞∑n=0

in (−1)n z−n−1

=−1∑

m=−∞

(1

i

)m+1

(−1)m+1 zm

165


f(z) =1

i+ z; |z − i| > 2

1

z − i

∞∑n=0

(2i

z − i

)n(−1)n =

∞∑n=0

2nin (−1)n (z − i)−n−1

=−1∑

m−∞

(1

2i

)m+1

(−1)m+1 (z − i)m

Exercise 5.24. (E. 9 P. 293)Expand the following functions in a Laurent series valid in a domain whose outer radius isinfinite. State the center and inner radius of the domain. Give the nth term of the series.

1

z + 2expanded in powers of z − i

Solution.

f(z) =1

z + 2

=1

z − i+ i+ 2

=1

(z − i)(

1 + i+2z−i

)=

∞∑n=0

((i+ 2)n (−1)n

(z − i)n+1

); |z − i| >

√5

=−1∑

m=−∞(z − i)m (i+ 2)−m−1 (−1)m+1 ; |z − i| > 1

Exercise 5.25. (E. 11 P. 293)

1. Consider f (z) =1

z (z − 1) (z + 3). This function is expanded in three different Laurent

series involving powers of z. State the three domains in which Laurent series are available.

2. Find each series and give an explicit formula for the nth term.

Solution.

1. We will expand on 3 domains as follow

M1 = {z : 0 < |z| < 1}

M2 = {z : 1 < |z| < 3}

M3 = {z : |z| > 3}

2. Consider the domain M1, we have

166


f(z) =1

[z (z − 1) (z + 3)]

=1

4z

(1

z − 1− 1

z + 3

)=

1

4z

(−∞∑n=0

zn − 1

3

∞∑n=0

(z3

)n(−1)n

)

= −∞∑n=0

1

4zn−1 −

∞∑n=0

zn−1

(1

4× 1

3n+1

)(−1)n

=

∞∑n=0

1

4zn−1

(−1− 1

3n+1(−1)n

)

=∞∑

m=−1

1

4zn−1

(−1− 1

3m+2(−1)m+1

)

The next domain M2, we obtain

f(z) =1

[z (z − 1) (z + 3)]

=1

4z

(1

z − 1− 1

z + 3

)=

1

4z

(1

z

∞∑n=0

(1

z

)n− 1

3

∞∑m=0

(−1)m(z

3

)m)

=

−2∑k=−∞

zk(

1

4

)−∞∑

t=−1

(−1)t+1 zi(

1

4× 1

3t+2

)

=

∞∑−∞

ctzt

with the general coefficient is

ct = 14 t ≤ −2

ct = (−1)t(

14 ×

13t+2

), t ≥ −1

And the last is M3

167


f(z) =1

[z (z − 1) (z + 3)]

=1

4z

(1

z − 1− 1

z + 3

)=

1

4z

(1

z

∞∑n=0

(1

z

)n− 1

z

∞∑m=0

(−1)m(

3

z

)m)

=∞∑n=0

1

4z−n−2 −

∞∑m=0

1

4(−1)m 3mz−m−2

=

∞∑n=0

1

4z−n−2

(1 + (−1)n+1 3n

)=

−2∑n=−∞

1

4zk(

1 + (−1)k+1 3−k−2)

168

6 RESIDUES AND THEIR USE ININTEGRATION

6.1 INTRODUCTION AND DEFINITION OF THE RESIDUES

The first is shown by the exercises such as 1, 2, 3, 4, 5, 6, 7, 10.Exercise 6.1. (E. 1, E. 2, E. 3 P. 341)Using the method of residues, evaluate the integral

‰C

1

(z − 1)2 +i

z − 1+ 2 (z − 1) +

3

z − 4dz

, where the contour C is given below.

1. |z − 1| = 2

2. |z − 5| = 2

3. The rectangle with corners at ± (5± i)

Solution.

1. We can see that the function f (z) = 2 (z − 1) +3

z − 4is integrability on and even in the

contour C, soCf (z) dz = 0. Then we need to compute

‰C

1

(z − 1)2 +i

z − 1+ f (z) dz = 2πi

(Res

[1

(z − 1)2 , 1

]+Res

[i

z − 1, 1

])= 2πi (0 + i) = −2π

2. Using a similar comment, we have‰C

1

(z − 1)2 +i

z − 1+ 2 (z − 1) dz = 0. Hence,

‰C

1

(z − 1)2 +i

z − 1+ 2 (z − 1) +

3

z − 4dz = 2πiRes

[3

z − 4, 4

]= 6πi

3. In this exercise, in the given rectangle, we obtainC

2 (z − 1) dz = 0. Thus

‰C

1

(z − 1)2 +i

z − 1+2 (z − 1)+

3

z − 4dz = 2πi

(Res

[1

(z − 1)2 , 1

]+Res

[i

z − 1, 1

]+Res

[3

z − 4, 4

])

169

6 RESIDUES AND THEIR USE IN INTEGRATION

According to exercise 1 and 2, we get the result is

‰C

1

(z − 1)2 +i

z − 1+ 2 (z − 1) +

3

z − 4dz = 2πi (3 + i)

Exercise 6.2. (E. 4, E. 5, E. 6, E. 7 P. 341)Evaluate the following integrals by using the method of residues. In Problems 6-8 use Laurentexpansions valid in deleted neiborghoods of the singular points to get the residue.

1.‰ ∞∑

n=−∞e−n

2(n− 1) (z − 1)n dz around |z| = 2

2.‰ ∞∑

n=−5

1

(z + i)n (n+ 6)!dz around |z − i| = 3

3.‰ ∞∑

n=−∞cosh (1/z) dz around the square with corners at ± (1± i)

4.‰z sin

(1

z − 1

)dz around |z| = 2

Solution.

1. We consider the function f (z) =∞∑

n=−∞e−n

2(n− 1) (z − 1)n on the contour |z| = 2 and

see that f is integrability for all z 6= 1. So we have a singular points z0 = 1 where n < 0.

Applying the method of residues, we get

‰Cf (z) dz = Res [f (z) , 1]

= −2e−1

2. In this exercise, we note that

‰C

−4∑n=−∞

1

(z + i)n (n+ 6)!= 0

So that the first integral becomes

‰C

∞∑n=−5

1

(z + i)n (n+ 6)!=

‰C

∞∑n=−∞

1

(z + i)n (n+ 6)!

In a similar fashion, we compute the result is

‰C

∞∑n=−5

1

(z + i)n (n+ 6)!=

2πi

7

170


3. We have z0 = 0 is a singular point of the function f (z) = cosh (1/z) in the square withthe corners ± (1± i). Then we need to extense the function by Laurent series.

cosh (1/z) = 1 +1

2!z2+

1

4!z4+ ...

Hence,

‰C

∞∑n=−∞

cosh (1/z) dz = Res [cosh (1/z) , 0] = 0

4. Like the exercise above, we see that z0 = 1 is a singular point of the function f (z) =

z sin

(1

z − 1

)in the contour |z| = 2. Next, we must expand the function by Laurent

series.

z sin

(1

z − 1

)= (z − 1) sin

(1

z − 1

)+ sin

(1

z − 1

)= (z − 1)

[(1

z − 1

)− 1

3!

(1

z − 1

)3

+ . . .

]+

[(1

z − 1

)− 1

3!

(1

z − 1

)3

+ . . .

]

= 1 +

(1

z − 1

)− 1

3!

(1

z − 1

)2

− 1

3!

(1

z − 1

)3

+ . . .

Thus the result is

‰Cz sin

(1

z − 1

)dz = 2πiRes

[z sin

(1

z − 1

), 1

]= 2πi

Exercise 6.3. (E. 10 P. 341)Use residue calculus to show that if n ≥ 1 is an integer, then

‰C

(z +

1

z

)ndz =

2πin!(

n− 1

2

)!

(n+ 1

2

)!

, nodd

0, neven

where C is any simple closed contour encircling the origin.

Solution.We consider the function as follows

f (z) =

(z +

1

z

)n=

n∑k=0

Cknzn−kz−k

=

n∑k=0

Cknzn−2k

171


There are 2 cases below.

• n = 2m. We note that

‰|z−z0|=r

(z − z0)l dz =

0, l 6= −1

2πi, l = −1

Because of n− 2k = 2m− 2k 6= −1 for all m, k, we obtain‰cf (z) dz = 0.

• n = 2m+ 1. We see that

‰cf (z) dz =

‰c

2m+1∑k=0

Ck2m+1z2m+1−2kdz

=2m+1∑k=0

Ck2m+1

‰cz2m+1−2kdz

Then, by solving the equation 2m+ 1− 2k = −1, we have k = m+ 1. Hence,

‰cf (z) dz = 2πiCm+1

2m+1

=2πi (2m+ 1)

(m+ 1)! (2m+ 1−m− 1)!

=2πin!(

n− 1

2

)!

(n+ 1

2

)!

So we complete the proof.

172


6.2 ISOLATED SINGULARITIES

We use the exercises 1, 2, 3, 8, 9, 10, 17, 18, 19, 23 for our examples.Exercise 6.4. (E. 1, E. 2, E. 5 P. 350)

Show by means of a Laurent series expansion∞∑

n=−∞cn (z − z0)n that the following functions

have essential singularities at the points stated. State the residue and give c−2, c−1, c0, c1.

1. sinh (1/z) at z = 0

2. (z − 1)3 cosh (1/ (z − 1)) at z = 1

3. e1/(z−i)ez−i at z = i

Solution.By using Laurent series for the given function, we can easily compute the coefficients are

needed. Moreover, the first negative coefficient is really the residue. Indeed, the reader can seeas follows

1. We have

sinh

(1

z

)=

1

z+

1

3!z3+

1

5!z5+ . . . , z 6= 0

Then c−2 = 0, c−1 = 1, c0 = 0, c1 = 0.

2. We get

(z − 1)3 cosh

(1

z − 1

)= (z − 1)3

[1 +

1

2! (z − 1)2 +1

4! (z − 1)4 + . . .

]= (z − 1)3 +

1

2!(z − 1) +

1

4! (z − 1)+ . . .

Then c−2 = 0, c−1 =1

4!, c0 =

1

2, c1 = 0.

3. We obtain

e1z−i ez−i =

[1 +

1

1! (z − i)+

1

2! (z − i)2 + . . .

] [1 + (z − i) +

1

2!(z − i)2 + . . .

]=

∞∑n=0

∞∑k=0

1

n!

1

k!(z − i)n−k

Then

173


c−2 =∞∑n=2

1

n! (n− 2)!

c−1 =

∞∑n=1

1

n! (n− 1)!

c0 =∞∑n=0

(1

(n!)2

)

c1 =∞∑n=0

1

n! (n+ 1)!

Exercise 6.5. (E. 8, E. 9, E. 10 P. 350)Use series expansions or L’Hopital’s rule to show that the following functions possess removablesingularities at the indicated singular points. You must show that lim

z→z0f (z) exists and is finite

at the singular point. Also, state how f (z0) should be defined at each point in order to removethe singularity. Use principal branches where there is any ambiguity.

1.ez − 1

zat z = 0

2.ez − eLogz

at z = 1

3.sinh z

z2 + π2at the two singular points

Solution.In three exercises, we use L’Hopital rule to show the functions possess removable singularities

at the indicated singular points.

1. We have

limz→0

ez − 1

z= lim

z→0ez = 1

Hence,

f (z) =

ez − 1

z, z 6= 0

1 , z = 0

2. Next, we get

limz→1

ez − eLogz

= limz→1

ez

1

z

= e

Thus

174


f (z) =

ez − elog z

, z 6= 1

e , z = 1

3. We are carefully because we obtain 2 roots in the equation z2 + π2 = 0. Then, we have

limz→πi

sinh z

z2 + π2= lim

z→πi

1

(1 + z2)1/2

2z=

1

2πi (1 + πi)1/2

limz→−πi

sinh z

z2 + π2= lim

z→πi

1

(1 + z2)1/2

2z=

1

2πi (1− πi)1/2

So the result is

f (z) =

sinh z

z2 + π2, z 6= ±πi

1

2πi (1 + πi)1/2, z = πi

1

2πi (1− πi)1/2, z = −πi

Exercise 6.6. (E. 17, E. 18, E. 19, E. 23 P. 351)State the location of all the poles for each of the following functions and give the order of eachpole. Use the principal branch of the given functions if there is any ambiguity.

1.1

z2 + 2z + 1

2.1

z2 + z + 1

3.1

z3 − 1

4.1

10z − ez

Solution.In four exercises, we need to deal with the equation of denominator for each function.

1. The root in there is z = −1, so we try assuming that z = −1 is a pole of order two.

limz→−1

(z + 1)2 1

z2 + 2z + 1= 0

Hence, our assumption is true.

2. In a similar fashion, the roots are z = −1

2+ i

√3

2, z = −1

2− i√

3

2. Then this means that

two roots are really two simple poles.

175


3. We see easily that the roots in this function are z = 1, z = −1

2+ i

√3

2, z = −1

2− i√

3

2.

We deduce similarly that three roots are simple poles.

4. We carefully solve the equation 10z = ez and obtain the root is z =i2kπ

log 10− 1. So that

this is a simple pole.

176


6.3 FINDING THE RESIDUE

Exercise 6.7. (E. 3, E. 4, E. 5 P. 357)For each of the following functions state the location and order of each pole and find thecorresponding residue. Use the principal branch of any multivalued function given below.

1.cos z

z2 + z + 1

2.1

z− ez

z (z + 1)+

1

(z − 1)4

3.1

z1/2 (z2 − 9)2

Solution.

1. We factor the denominator and obtain

f (z) =cos z(

z − −1− i√

3

2

)(z − −1 + i

√3

2

)

We can easily examine that both z =−1± i

√3

2are single poles of f (z).

So we will compute the residue of each pole

Res

[f (z) ,

−1− i√

3

2

]= lim

z→−1−i√3

2

(z − −1− i

√3

2

)f (z)

=

cos

(−1− i

√3

2

)−i√

3

Res

[f (z) ,

−1 + i√

3

2

]= lim

z→−1+i√3

2

(z − −1 + i

√3

2

)f (z)

=

cos

(−1 + i

√3

2

)i√

3

2. Take f (z) =1

z− ez

z (z + 1)+

1

(z − 1)4

Just like Exercise 3 we can see that f (z) has two poles at z = −1 with order 1 and z = 1

with order 4 as the reader can verify themselves.

Because limz→0

zf (z) = 0, z = 0 is not a pole of f (z)

So we compute easily

177


Res [f (z) ,−1] = limz→−1

(z + 1) f (z) = e−1

Res [f (z) , 1] = limz→1

d3

2dz3

[(z − 1)4 f (z)

]= 0

3. Because we use principle branch of z1

2 , f (z) have no pole at z = 0 and z = −3 but it hasa pole z = 3 with order 2.

Res [f (z) , 3] = limz→3

d

dz

((z − 3)2 f (z)

)

= limz→3

− 1

2√z

(z + 3)− 2√z

z (z + 3)3 = − 1

72√

3

Exercise 6.8. (E. 16, E. 18, E. 20 P. 358)Find the residue of the following functions at the indicated point.

1.z + 1

zsin (1/z) at 0

2.1

(z + i)5 at −i

3.z12

(z − 1)10 at 1

Solution.

1. We see that the function f (z) =z + 1

zsin (1/z) has a pole of order 2 at z = 0. We need

to prove that

limz→0

z2f (z) = 0

Indeed, we have

∣∣∣∣z2 z + 1

zsin (1/z)

∣∣∣∣ ≤ |z (z + 1)|

We know that limz→0

z (z + 1) = 0, so we obtain the proof.

2. We see that the function has a pole of order 6 at z = −i. Indeed, we have

limz→−i

(z + i)6 1

(z + i)5 = limz→−i

(z + i) = 0

3. In a similar fashion, the function has a pole of order 11 at z = 1. We see this as follows

178


limz→1

(z − 1)11 z12

(z − 1)10 = limz→1

z12 (z − 1) = 0

Exercise 6.9. (E. 27, E. 31 P. 358)

Use residues to evaluate the following integrals. Use the principal branch of multivalued func-tions.

1.‰

dz

sin zaround |z − 6| = 4

2.‰

e1/z

z2 − 1dz around |z − 1| = 3/2

Solution.

179


6.4 EVALUATION OF REAL INTEGRALS WITH RESIDUECALCULUS I

The reader can see exercises 3, 4, 7, 8 for examples of residue calculus I.Exercise 6.10. (E. 3, E. 4 P. 364)Using residue calculus, establish the following identities.

1.ˆ π

−π

dθ

a+ b cos θ=

2π√a2 − b2

for a > b ≥ 0.

2.ˆ 3π

2

−π2

cos θ

a+ b cos θdθ =

2π

b

[1− a√

a2 − b2

]for a > b > 0.

Solution.

1. We put t = π + θ and see that

ˆ π

−π

dθ

a+ b cos θ=

ˆ 2π

0

dθ

a− b cos θ

The next is the denominator is definited because of a > b ≥ 0. Then we apply the methodof evaluation of real integrals with residue calculus 1. We get z = eiθ, so that

dθ = dziz

cos θ = z+z−1

2

Then the integral becomes

ˆ 2π

0

dθ

a− b cos θ=

‰|z|=1

2dz

2iaz − ibz2 − ib

We need to face with the equation ibz2 − 2iaz + ib = 0 and have 2 roots easily.

z =a±√a2 − b2b

But the only point, z =a−√a2 − b2b

, is in the unit circle, the reader can verify themselves.

Now we only compute the residue at this point.

ˆ 2π

0

2dθ

2iaz − ibz2 − ib= 4πiRes

(1

2iaz − ibz2 − ib;a−√a2 − b2b

)=

2π√a2 − b2

Note that, we only get the result in the case a > b > 0, but do not worry, it is very easyto get the proof for the cases a > b = 0.

180


2. Like the exercise above, we put t = θ +π

2and see that

ˆ 3π2

−π2

cos θ

a+ b cos θdθ =

ˆ 2π

0

sin θ

a+ b sin θdθ

The next is the denominator is definited because of a > b > 0. Then we apply the methodof evaluation of real integrals with residue calculus 1. We get z = eiθ, so that

dθ = dziz

sin θ = z−z−1

2i


ˆ 2π

0

sin θ

a+ b sin θdθ =

‰|z|=1

(z2 − 1

)dz

iz (bz2 + 2iaz − b)

We need to face with the equation bz2 + 2iaz − b = 0 and have 2 roots easily.

z =−ia± i

√a2 − b2

b

But the only point, z =−ia+ i

√a2 − b2

b, is in the unit circle, the reader can verify them-

selves.

Now we only compute the residue at this point and the point z = 0.

ˆ 2π

0

sin θ

a+ b sin θdθ = 2πi

[Res

(z2 − 1

bz2 + 2iaz − b;−ia+ i

√a2 − b2

b

)+Res

(z2 − 1

bz2 + 2iaz − b; 0

)]

=2π

b

[1− a√

a2 − b2

]

Exercise 6.11. (E. 7, E. 8, E. 9 P. 364)Using residue calculus, establish the following identities.

1.ˆ 2π

0

dθ

(a+ b sin θ)2 =2πa(√a2 − b2

)3 for a > b ≥ 0.

2.ˆ 2π

0

dθ

a+ sin2 θ=

2π√a (a+ 1)

for a > 0.

Solution.

1. Like the exercises above, the denominator is definited because of a > b ≥ 0. Then weapply the method of evaluation of real integrals with residue calculus 1. We get z = eiθ,so that

181


dθ = dziz

sin θ = z−z−1

2i


ˆ 2π

0

dθ

(a+ b sin θ)2 =

‰|z|=1

−4zdz

i (bz2 + 2iaz − b)2

Using the root of exercise 4, we still choose z =−ia+ i

√a2 − b2

bis in the unit circle but

it is a second-order pole. Then we have

ˆ 2π

0

dθ

(a+ b sin θ)2 = 2πiRes

[−4z

i (bz2 + 2iaz − b)2 ; z1 =−ia+ i

√a2 − b2

b

]

= 2π limz→z1

d

dz

[(z − z1)2 −4z

(bz2 + 2iaz − b)2

]=

2π(√a2 − b2

)3

2. We will reduce the degree of sin2 θ by using formula sin2 θ =1− cos 2θ

2and the denom-

inator is definited because of a > 0. Then we apply the method of evaluation of realintegrals with residue calculus 1. We get z = eiθ, so that

dθ = dziz

cos 2θ = z2+z−2

2


ˆ 2π

0

2dθ

2a+ 1− cos 2θ=

‰|z|=1

−4zdz

i [z4 − z2 (4a+ 2) + 1]

We need to solve the equation z4 − z2 (4a+ 1) + 1 = 0 then we have 4 roots.

z = ±√

2a+ 1±√a (4a+ 1)

But there are only 2 points in the unit circle, these are z = ±√

2a+ 1−√a (4a+ 1).

Thus if we call these are z1; z2 respectively, we will have

ˆ 2π

0

2dθ

2a+ 1− cos 2θ= −8πi

[Res

(z

i [z4 − z2 (4a+ 2) + 1]; z1

)+Res

(z

i [z4 − z2 (4a+ 2) + 1]; z2

)]=

2π√a (a+ 1)

182


6.5 EVALUATION OF INTEGRALS II

We think 4 exercises below are the basic things to help the reader understanding in this section.Exercise 6.12. (E. 19, E. 20, E. 21 P. 371)Evaluate the following integrals by means of residue calculus.

1.ˆ ∞−∞

dx

x2 + x+ 1

2.ˆ ∞−∞

dx

(x2 + x+ 1) (x2 + 1)

3.ˆ ∞−∞

x4dx

x6 + 1

Solution.In 3 exercises, the denominator is nonzero for all real numbers and the degree of it differs

from the numerator by 2 or more. Thus at all poles in u.h.p, we have the result respectively.

1. This follows that

ˆ ∞−∞

dx

x2 + x+ 1= 2πi

∑Res

1

z2 + z + 1

Then we solve the equation z2 + z + 1 = 0 and get

z =−1± i

√3

2= ei

2π3 ; e−i

2π3

Hence, this function has a simple pole in the u.h.p at ei2π3 . We will evaluate the residue

at this point.

Res

(1

z2 + z + 1; ei

2π3

)=

1

(z2 + z + 1)

∣∣∣∣z=ei

2π3

=1

2ei2π3 + 1

=1

i√

3

So that the result is2π√

3.


ˆ ∞−∞

dx

(x2 + x+ 1) (x2 + 1)= 2πi

∑Res

1

(z2 + z + 1) (z2 + 1)

Then by solving the equations z2 + z + 1 = 0 and z2 + 1 = 0, we get

z =−1± i

√3

2= ei

2π3 ; e−i

2π3

183


and

z = ±i

Hence, this function has 2 simple poles in the u.h.p at ei2π3 and i . We will evaluate the

residues at these point.

Res

[1

(z2 + z + 1) (z2 + 1); ei

2π3

]=

1

(z2 + z + 1)′ (z2 + 1)

∣∣∣∣z=ei

2π3

=1(

2ei2π3 + 1

)(ei

4π3 + 1

)=

2

i√

3(1− i

√3)

Res

[1

(z2 + z + 1) (z2 + 1); i

]=

1

(z2 + z + 1) (z2 + 1)′

∣∣∣∣z=i

=−1

2

So that the result isπ√3.


ˆ ∞−∞

x4dx

x6 + 1= 2πi

∑Res

z4

(z4 − z2 + 1) (z2 + 1)

Then by solving the equations z4 − z2 + 1 = 0 and z2 + 1 = 0, we get

z =−√

3± i2

= ei5π6 ; e−i

5π6

and

z =

√3± i2

= eiπ6 ; e−i

π6

and

z = ±i

Hence, this function has 2 simple poles in the u.h.p at eiπ6 , ei

5π6 and i . We will evaluate

the residues at these point.

184


Res

[z4

(z4 − z2 + 1) (z2 + 1); ei

π6

]= lim

z→eiπ6

z4(z − e−i

π6

)(z − ei

5π6

)(z − ei

−5π6

)(z2 + 1)

=−1 + i

√3

3i(√

3 + i)2

Res

[z4

(z4 − z2 + 1) (z2 + 1); ei

5π6

]= lim

z→ei5π6

z4(z − e−i

π6

)(z − ei

π6

)(z − ei

−5π6

)(z2 + 1)

=−1− i

√3

3i(√

3− i)2

Res

[z4

(z4 − z2 + 1) (z2 + 1); i

]=

z4

(z4 − z2 + 1) (z2 + 1)′

∣∣∣∣z=i

=1

6i

So that the result is2π

3.

Exercise 6.13. (E. 28 P. 372)Show that for a, b, c real and b2 < 4ac, the following hold

ˆ ∞−∞

dx

ax2 + bx+ c=

2π√4ac− b2

Solution.Because of b2 < 4ac, the denominator on the right hand is definited and the equation ax2 +

bx+ c = 0 has 2 complex roots. This means that

x =−b± i

√4ac− b2

2a

Assume that a > 0, we choose the point−b+ i

√4ac− b2

2a, we will evaluate the residue at this

point.

ˆ ∞−∞

dx

ax2 + bx+ c= 2πiRes

(1

az2 + bz + c;−b+ i

√4ac− b2

2a

)

= 2πi

(1

2az + b

∣∣∣∣z=−b+i

√4ac−b2

2a

)

=2π√

4ac− b2

And if we consider the case a < 0, by choosing the point−b− i

√4ac− b2

2aand applying

method above, we still have the result is the right hand.

185


6.6 EVALUATION OF INTEGRALS III

Exercises 1, 3, 5, 7, 13, 24 are expressed there to solve easily.

Exercise 6.14. (E. 1, E. 3, E. 5, E. 7, E. 13 P. 382)Evaluate the following integrals by residue calculus. Use Cauchy principal values and takeadvantage of even and odd symmetries where appropriate

1.

∞̂

−∞

cos (2x)

x2 + 9dx

2.

∞̂

−∞

xeix

(x− 1)2 + 9dx

3.

∞̂

−∞

x sin (2x)

x2 + x+ 1dx

4.

∞̂

−∞

(x3 + x2

)cos(√

2x)

x4 + 1dx

5.

∞̂

0

cos (x)

(x2 + 4)2dx

Solution.We consider arc C below.

Now we know completely how to evaluate the residue at each point in some sections before.Then we only need to solve some exercises by using equations which is gotten in this section.

1.

186


∞̂

−∞

cos (2x)

x2 + 9dx = <

∞̂

−∞

ei2x

x2 + 9dx

= <

[2πi∑

Res

(ei2z

z2 + 9

)]= <

[2πi.Res

(ei2z

z2 + 9, z1 = 3i

)]= <

[2πi.

ei2z

2z

∣∣∣∣z=3i

]= <

[2πi.

e−6

6i

]=

π

3e−6

2.

∞̂

−∞

xeix

(x− 1)2 + 9dx = 2πi

∑Res

[zeiz

(z − 1)2 + 9

]

= 2πi.Res

[zeiz

(z − 1)2 + 9, z1 = 1 + 3i

]= 2πi.

zeiz

2 (z − 1)

∣∣∣∣z=1+3i

= 2πi.(1 + 3i) e−3+i

6i

=π

3. (1 + 3i) e−3+i

3.

187


∞̂

−∞

x sin (2x)

x2 + x+ 1dx = =

[2πi∑

Res

(ze2zi

z2 + z + 1

)]

= =

[2πi.Res

(ze2zi

z2 + z + 1, z1 =

−1 +√

3i

2

)]

= =

[2πi.

ze2zi

2z + 1

∣∣∣∣z=(−1+

√3i)/2

]

= =

[π

(−1 +

√3i)e−√

3−i√

3

]

= =

[π.e−

√3

√3

(−1 +

√3i)

(cos 1− i sin 1)

]

= =

[π.e−

√3

√3

(− cos 1 + i sin 1 +

√3i cos 1 +

√3 sin 1

)]

=π.e−

√3

√3

.(

sin 1 +√

3 cos 1)

=2π.e−

√3

√3

sin

(2π

3− 1

)

4.

∞̂

−∞

(x3 + x2

)cos(√

2x)

x4 + 1dx = <

[2πi

∑Res

((z3 + z2

)ei√

2z

z4 + 1

)]

= <

[2πi.Res

((z3 + z2

)ei√

2z

z4 + 1, e

iπ4

)+ 2πi.Res

((z3 + z2

)ei√

2z

z4 + 1, e

i3π4

)]

= <

[2πi.

(z3 + z2

)ei√

2z

4z3

∣∣∣∣∣z=e

iπ4

+ 2πi.

(z3 + z2

)ei√

2z

4z3

∣∣∣∣∣z=e

i3π4

]

= <

2πi.

ei3π4 + e

iπ

2

e−1+i

4e

i3π

4

+ 2πi.

ei9π4 + e

i3π

2

e−1−i

4e

i9π

4

=

π√2e−1 (cos 1− sin 1)

5. Sincecos (x)

(x2 + 4)2 =cos (−x)(

(−x)2 + 4)2 , we can derive that

188


∞̂

0

cos (x)

(x2 + 4)2dx =1

2<[2πi

∑Res

(eiz

(z2 + 4)2

)]

= <[πiRes

(eiz

(z2 + 4)2 , z1 = 2i

)]

= <

πi limz−→2i

d

[(z − 2i)2 eiz

(z2 + 4)2

]dz

= <

πi limz−→2i

d

[eiz

(z + 2i)2

]dz

= <

[πi limz−→2i

ieiz (z + 2i)2 − 2 (z + 2i) eiz

(z + 2i)4

]

= <

[πiie−2 (2i+ 2i)2 − 2 (2i+ 2i) e−2

(2i+ 2i)4

]

= <[πe−2 16 + 8

256

]=

3πe−2

32

189


Exercise 6.15. (E. 24 P. 384)To establish the well-known result

∞̂

0

sinx

xdx =

π

2

we proceed as follows:

1. Show that

f (z) =eiz − 1

z

has a removable singularity at z = 0. How should f (0)be defined to remove the singularity?

2. Using the contour of Fig 6.5-1, prove that

R̂

−R

eix − 1

xdx+

ˆC1

eiz − 1

z= 0

and also

R̂

−R

cosx− 1

xdx+ i

R̂

−R

sinx

xdx =

ˆ

C1

1

zdz −

ˆ

C1

eiz

zdz

3. Evaluate the first integral on the above right by using the polar representation of C1:z = Reiθ, 0 ≤ θ ≤ π. Pass to the limit R −→ ∞ and explain why the second integral onthe right goes to zero. Thus prove that

∞̂

−∞

cosx− 1

xdx = 0

∞̂

−∞

sinx

xdx = 0

and finally that

∞̂

0

sinx

xdx =

π

2

190


Solution.

1. We have

f (z) =eiz − 1

z

=∞∑n=0

in+1

(n+ 1)!zn

Thus, f (z) has a removable singularity at z = 0.

And we define that

f (0) = limz−→0

f (z)

= limz−→0

ieiz

1= i

2. By using a contour of integration containing a semicircular arc in the u.h.p, we get

R̂

−R

eix − 1

xdx+

ˆC1

eiz − 1

z= 2πi

∑Res

(eiz − 1

z

)

= 2πiRes

(eiz − 1

z, z1 = 0

)= 2πi. lim

z−→0

(z.eiz − 1

z

)= 2πi. lim

z−→0

(eiz − 1

)= 0

Thus,

R̂

−R

cosx− 1

xdx+

R̂

−R

sinx

xdx =

ˆC1

1

zdz −

ˆC1

eiz

zdz

We symbolize the equation above is (*).

3. We will evaluate I =Ć1

1

zdz before by using the polar representation of C1 : z = Reiθ, 0 ≤

θ ≤ π. Then we obtain dz = iReiθdθ.

Thus

191


I =

π̂

0

1

ReiθiReiθdθ

=

π̂

0

idθ = iπ

We continue to symbolize this equation is (**). The next is evaluation of J =

ˆC1

eiz

zdz.

We get a comment that since limz−→∞

∣∣∣∣1z∣∣∣∣ = 0, lim

R−→∞

Ć1

eiz

zdz = 0 (***)

From (*),(**) and (***), we have

∞̂

−∞

sinx

xdx = π

Andsinx

xis even function, so that

∞̂

0

sinx

xdx =

π

2

Hence, we complete the proof.

192

MATHEMATICAL BOOKS

“To myself I am only a child playing on the beach,

while vast oceans of truth lie undiscovered before me.” – I. Newton.

“I am sure that no subject loses more than mathematics

by any attempt to dissociate it from its history.” – J.W.L. Glaisher.

Documents

Essay Complex Analysis