George Cain

(c)Copyright 1999 by George Cain.

All rights reserved.

Table of Contents

Chapter One - Complex Numbers

1.1 Introduction

1.2 Geometry

1.3 Polar coordinates

Chapter Two - Complex Functions

2.1 Functions of a real variable

2.2 Functions of a complex variable

2.3 Derivatives

Chapter Three - Elementary Functions

3.1 Introduction

3.2 The exponential function

3.3 Trigonometric functions

3.4 Logarithms and complex exponents

Chapter Four - Integration

4.1 Introduction

4.2 Evaluating integrals

4.3 Antiderivatives

Chapter Five - Cauchy's Theorem

5.1 Homotopy

5.2 Cauchy's Theorem

Chapter Six - More Integration

6.1 Cauchy's Integral Formula

6.2 Functions defined by integrals

6.3 Liouville's Theorem

6.4 Maximum moduli

Chapter Seven - Harmonic Functions

7.1 The Laplace equation

7.2 Harmonic functions

7.3 Poisson's integral formula

Chapter Eight - Series

8.1 Sequences

8.2 Series

8.3 Power series

8.4 Integration of power series

8.5 Differentiation of power series

Chapter Nine - Taylor and Laurent Series

9.1 Taylor series

9.2 Laurent series

Chapter Ten - Poles, Residues, and All That

10.1 Residues

10.2 Poles and other singularities

Chapter Eleven - Argument Principle

11.1 Argument principle

11.2 Rouche's Theorem

----------------------------------------------------------------------------

George Cain

School of Mathematics

Georgia Institute of Technology

Atlanta, Georgia 0332-0160

cain@math.gatech.edu

Chapter One

Complex Numbers

1.1 Introduction. Let us hark back to the first grade when the only numbers you knew

were the ordinary everyday integers. You had no trouble solving problems in which you

were, for instance, asked to find a number x such that 3x 6. You were quick to answer

”2”. Then, in the second grade, Miss Holt asked you to find a number x such that 3x 8.

You were stumped”there was no such ”number”! You perhaps explained to Miss Holt that

32 6 and 33 9, and since 8 is between 6 and 9, you would somehow need a number

between 2 and 3, but there isn™t any such number. Thus were you introduced to ”fractions.”

These fractions, or rational numbers, were defined by Miss Holt to be ordered pairs of

integers”thus, for instance, 8, 3 is a rational number. Two rational numbers n, m and

p, q were defined to be equal whenever nq pm. (More precisely, in other words, a

rational number is an equivalence class of ordered pairs, etc.) Recall that the arithmetic of

these pairs was then introduced: the sum of n, m and p, q was defined by

n, m « p, q nq « pm, mq,

and the product by

n, mp, q np, mq.

Subtraction and division were defined, as usual, simply as the inverses of the two

operations.

In the second grade, you probably felt at first like you had thrown away the familiar

integers and were starting over. But no. You noticed that n, 1 « p, 1 n « p, 1 and

also n, 1p, 1 np, 1. Thus the set of all rational numbers whose second coordinate is

one behave just like the integers. If we simply abbreviate the rational number n, 1 by n,

there is absolutely no danger of confusion: 2 « 3 5 stands for 2, 1 « 3, 1 5, 1. The

equation 3x 8 that started this all may then be interpreted as shorthand for the equation

3, 1u, v 8, 1, and one easily verifies that x u, v 8, 3 is a solution. Now, if

someone runs at you in the night and hands you a note with 5 written on it, you do not

know whether this is simply the integer 5 or whether it is shorthand for the rational number

5, 1. What we see is that it really doesn™t matter. What we have ”really” done is

expanded the collection of integers to the collection of rational numbers. In other words,

we can think of the set of all rational numbers as including the integers“they are simply the

rationals with second coordinate 1.

One last observation about rational numbers. It is, as everyone must know, traditional to

1.1

n n

write the ordered pair n, m as . Thus n stands simply for the rational number , etc.

m 1

Now why have we spent this time on something everyone learned in the second grade?

Because this is almost a paradigm for what we do in constructing or defining the so-called

complex numbers. Watch.

Euclid showed us there is no rational solution to the equation x 2 2. We were thus led to

defining even more new numbers, the so-called real numbers, which, of course, include the

rationals. This is hard, and you likely did not see it done in elementary school, but we shall

assume you know all about it and move along to the equation x 2 1. Now we define

complex numbers. These are simply ordered pairs x, y of real numbers, just as the

rationals are ordered pairs of integers. Two complex numbers are equal only when there

are actually the same“that is x, y u, v precisely when x u and y v. We define the

sum and product of two complex numbers:

x, y « u, v x « u, y « v

and

x, yu, v xu yv, xv « yu

As always, subtraction and division are the inverses of these operations.

Now let™s consider the arithmetic of the complex numbers with second coordinate 0:

x, 0 « u, 0 x « u, 0,

and

x, 0u, 0 xu, 0.

Note that what happens is completely analogous to what happens with rationals with

second coordinate 1. We simply use x as an abbreviation for x, 0 and there is no danger of

confusion: x « u is short-hand for x, 0 « u, 0 x « u, 0 and xu is short-hand for

x, 0u, 0. We see that our new complex numbers include a copy of the real numbers, just

as the rational numbers include a copy of the integers.

Next, notice that xu, v u, vx x, 0u, v xu, xv. Now then, any complex number

z x, y may be written

1.2

z x, y x, 0 « 0, y

x « y0, 1

When we let 0, 1, then we have

z x, y x « y

Now, suppose z x, y x « y and w u, v u « v. Then we have

zw x « yu « v

xu « xv « yu « 2 yv

We need only see what 2 is: 2 0, 10, 1 1, 0, and we have agreed that we can

safely abbreviate 1, 0 as 1. Thus, 2 1, and so

zw xu yv « xv « yu

and we have reduced the fairly complicated definition of complex arithmetic simply to

ordinary real arithmetic together with the fact that 2 1.

z

Let™s take a look at division“the inverse of multiplication. Thus stands for that complex

w

number you must multiply w by in order to get z . An example:

z x « y x « y u v

u « v u v

w u « v

xu « yv « yu xv

u2 « v2

xu « yv yu xv

2 « 2

u « v2 u « v2

Note this is just fine except when u 2 « v 2 0; that is, when u v 0. We may thus divide

by any complex number except 0 0, 0.

One final note in all this. Almost everyone in the world except an electrical engineer uses

the letter i to denote the complex number we have called . We shall accordingly use i

rather than to stand for the number 0, 1.

Exercises

1.3

1. Find the following complex numbers in the form x « iy:

b) 1 i 3

a) 4 7i2 « 3i

b) 5«2i c) 1i

1«i

2. Find all complex z x, y such that

z2 « z « 1 0

3. Prove that if wz 0, then w 0 or z 0.

1.2. Geometry. We now have this collection of all ordered pairs of real numbers, and so

there is an uncontrollable urge to plot them on the usual coordinate axes. We see at once

then there is a one-to-one correspondence between the complex numbers and the points in

the plane. In the usual way, we can think of the sum of two complex numbers, the point in

the plane corresponding to z « w is the diagonal of the parallelogram having z and w as

sides:

We shall postpone until the next section the geometric interpretation of the product of two

complex numbers.

The modulus of a complex number z x « iy is defined to be the nonnegative real number

x 2 « y 2 , which is, of course, the length of the vector interpretation of z. This modulus is

traditionally denoted |z|, and is sometimes called the length of z. Note that

|x, 0| x 2 |x|, and so || is an excellent choice of notation for the modulus.

The conjugate z of a complex number z x « iy is defined by z x iy. Thus |z| 2 z z .

Geometrically, the conjugate of z is simply the reflection of z in the horizontal axis:

1.4

Observe that if z x « iy and w u « iv, then

z « w x « u iy « v

x iy « u iv

z « w.

In other words, the conjugate of the sum is the sum of the conjugates. It is also true that

zw z w. If z x « iy, then x is called the real part of z, and y is called the imaginary

part of z. These are usually denoted Re z and Im z, respectively. Observe then that

z « z 2 Re z and z z 2 Im z.

Now, for any two complex numbers z and w consider

|z « w| 2 z « wz « w z « w z « w

z z « w z « wz « ww

|z| 2 « 2 Rew z « |w| 2

‚ |z| 2 « 2|z||w| « |w| 2 |z| « |w| 2

In other words,

|z « w| ‚ |z| « |w|

the so-called triangle inequality. (This inequality is an obvious geometric fact“can you

guess why it is called the triangle inequality?)

Exercises

4. a)Prove that for any two complex numbers, zw z w.

z z

b)Prove that w w .

c)Prove that ||z| |w|| ‚ |z w|.

|z|

z

5. Prove that |zw| |z||w| and that | w | .

|w|

1.5

6. Sketch the set of points satisfying

a) |z 2 « 3i| 2 b)|z « 2i| ‚ 1

c) Re z « i 4 d) |z 1 « 2i| |z « 3 « i|

e)|z « 1| « |z 1| 4 f) |z « 1| |z 1| 4

1.3. Polar coordinates. Now let™s look at polar coordinates r, “ of complex numbers.

Then we may write z rcos “ « i sin “. In complex analysis, we do not allow r to be

negative; thus r is simply the modulus of z. The number “ is called an argument of z, and

there are, of course, many different possibilities for “. Thus a complex numbers has an

infinite number of arguments, any two of which differ by an integral multiple of 2. We

usually write “ arg z. The principal argument of z is the unique argument that lies on

the interval , .

Example. For 1 i, we have

2 cos 7 « i sin 7

1i

4 4

2 cos « i sin

4 4

2 cos 399 « i sin 399

4 4

, , and

7 399

is an argument of 1 i, but the

etc., etc., etc. Each of the numbers 4 4 4

principal argument is .

4

Suppose z rcos “ « i sin “ and w scos ™ « i sin ™. Then

zw rcos “ « i sin “scos ™ « i sin ™

rscos “ cos ™ sin “ sin ™ « isin “ cos ™ « sin ™ cos “

rscos“ « ™ « i sin“ « ™

We have the nice result that the product of two complex numbers is the complex number

whose modulus is the product of the moduli of the two factors and an argument is the sum

of arguments of the factors. A picture:

1.6

We now define expi“, or e i“ by

e i“ cos “ « i sin “

We shall see later as the drama of the term unfolds that this very suggestive notation is an

excellent choice. Now, we have in polar form

z re i“ ,

where r |z| and “ is any argument of z. Observe we have just shown that

e i“ e i™ e i“«™ .

It follows from this that e i“ e i“ 1. Thus

1 e i“

e i“

It is easy to see that

z re i“ r cos“ ™ « i sin“ ™

w s

se i™

Exercises

7. Write in polar form re i“ :

b) 1 « i

a) i

c) 2 d) 3i

e) 3 « 3i

8. Write in rectangular form”no decimal approximations, no trig functions:

a) 2e i3 b) e i100

c) 10e i/6 d) 2 e i5/4

9. a) Find a polar form of 1 « i1 « i 3 .

b) Use the result of a) to find cos 7 and sin 7

.

12 12

10. Find the rectangular form of 1 « i 100 .

1.7

11. Find all z such that z 3 1. (Again, rectangular form, no trig functions.)

12. Find all z such that z 4 16i. (Rectangular form, etc.)

1.8

Chapter Two

Complex Functions

2.1. Functions of a real variable. A function : I ‚ C from a set I of reals into the

complex numbers C is actually a familiar concept from elementary calculus. It is simply a

function from a subset of the reals into the plane, what we sometimes call a vector-valued

function. Assuming the function is nice, it provides a vector, or parametric, description

of a curve. Thus, the set of all t : t e it cos t « i sin t cos t, sin t, 0 ‚ t ‚ 2

is the circle of radius one, centered at the origin.

We also already know about the derivatives of such functions. If t xt « iyt, then

the derivative of is simply t x t « iy t, interpreted as a vector in the plane, it is

tangent to the curve described by at the point t.

Example. Let t t « it 2 , 1 ‚ t ‚ 1. One easily sees that this function describes that

part of the curve y x 2 between x 1 and x 1:

1

0

-1 -0.5 0.5 1

x

Another example. Suppose there is a body of mass M ”fixed” at the origin“perhaps the

sun“and there is a body of mass m which is free to move“perhaps a planet. Let the location

of this second body at time t be given by the complex-valued function zt. We assume the

only force on this mass is the gravitational force of the fixed body. This force f is thus

zt

f GMm

|zt| 2 |zt|

where G is the universal gravitational constant. Sir Isaac Newton tells us that

zt

mz t f GMm

|zt| 2 |zt|

2.1

Hence,

z GM z

|z| 3

Next, let™s write this in polar form, z re i“ :

d 2 re i“ k e i“

dt 2 r2

where we have written GM k. Now, let™s see what we have.

d re i“ r d e i“ « dr e i“

dt dt dt

Now,

d e i“ d cos “ « i sin “

dt dt

sin “ « i cos “ d“

dt

icos “ « i sin “ d“

dt

i d“ e i“ .

dt

(Additional evidence that our notation e i“ cos “ « i sin “ is reasonable.)

Thus,

d re i“ r d e i“ « dr e i“

dt dt dt

r i d“ e i“ « dr e i“

dt dt

dr « ir d“ e i“ .

dt dt

Now,

2.2

d“ « ir d 2 “ e i“ «

d 2 re i“ d 2 r « i dr

dt dt

dt 2 dt 2 dt 2

dr « ir d“ i d“ e i“

dt dt dt

d“ 2 « i r d 2 “ « 2 dr d“

d2r r e i“

dt dt dt

dt 2 dt 2

d2

re i“ rk2 e i“ becomes

Now, the equation dt 2

2 2

d 2 r r d“ “

« i r d 2 « 2 dr d“ k2 .

dt dt dt

dt 2 dt r

This gives us the two equations

2

d 2 r r d“ k2 ,

dt

dt 2 r

and,

2

“

r d 2 « 2 dr d“ 0.

dt dt

dt

Multiply by r and this second equation becomes

d r 2 d“ 0.

dt dt

This tells us that

r 2 d“

dt

is a constant. (This constant is called the angular momentum.) This result allows us to

get rid of d“ in the first of the two differential equations above:

dt

2

d2r r k2

dt 2 r2 r

or,

d2r 2 k .

dt 2 r3 r2

2.3

Although this now involves only the one unknown function r, as it stands it is tough to

solve. Let™s change variables and think of r as a function of “. Let™s also write things in

terms of the function s 1 . Then,

r

d d“ d d .

dt dt d“ r 2 d“

Hence,

dr dr ds ,

dt r 2 d“ d“

and so

d 2 r d ds s 2 d ds

dt d“ d“ d“

dt 2

2

2 s 2 d s ,

d“ 2

and our differential equation looks like

d 2 r 2 2 s 2 d 2 s 2 s 3 ks 2 ,

dt 2 r3 d“ 2

or,

d2s « s k .

d“ 2 2

This one is easy. From high school differential equations class, we remember that

s 1 A cos“ « « k2 ,

r

where A and are constants which depend on the initial conditions. At long last,

2 /k

r ,

1 « cos“ «

where we have set A 2 /k. The graph of this equation is, of course, a conic section of

eccentricity .

Exercises

2.4

1. a)What curve is described by the function t 3t « 4 « it 6, 0 ‚ t ‚ 1 ?

b)Suppose z and w are complex numbers. What is the curve described by

t 1 tw « tz, 0 ‚ t ‚ 1 ?

2. Find a function that describes that part of the curve y 4x 3 « 1 between x 0 and

x 10.

3. Find a function that describes the circle of radius 2 centered at z 3 2i .

4. Note that in the discussion of the motion of a body in a central gravitational force field,

it was assumed that the angular momentum is nonzero. Explain what happens in case

0.

2.2 Functions of a complex variable. The real excitement begins when we consider

function f : D ‚ C in which the domain D is a subset of the complex numbers. In some

sense, these too are familiar to us from elementary calculus”they are simply functions

from a subset of the plane into the plane:

fz fx, y ux, y « ivx, y ux, y, vx, y

Thus fz z 2 looks like fz z 2 x « iy 2 x 2 y 2 « 2xyi. In other words,

ux, y x 2 y 2 and vx, y 2xy. The complex perspective, as we shall see, generally

provides richer and more profitable insights into these functions.

The definition of the limit of a function f at a point z z 0 is essentially the same as that

which we learned in elementary calculus:

lim fz L

z‚z 0

means that given an 0, there is a so that |fz L| whenever 0 |z z 0 | . As

you could guess, we say that f is continuous at z 0 if it is true that lim fz fz 0 . If f is

z‚z 0

continuous at each point of its domain, we say simply that f is continuous.

Suppose both lim fz and lim gz exist. Then the following properties are easy to

z‚z 0 z‚z 0

establish:

2.5

lim fz ‚± gz lim fz ‚±lim gz

z‚z 0 z‚z 0 z‚z 0

lim fzgz lim fz lim gz

z‚z 0 z‚z 0 z‚z 0

and,

lim fz

fz

z‚z 0

lim

z‚z 0 gz lim gz

z‚z 0

provided, of course, that lim gz ‚® 0.

z‚z 0

It now follows at once from these properties that the sum, difference, product, and quotient

of two functions continuous at z 0 are also continuous at z 0 . (We must, as usual, except the

dreaded 0 in the denominator.)

It should not be too difficult to convince yourself that if z x, y, z 0 x 0 , y 0 , and

fz ux, y « ivx, y, then

lim fz ux, y « i

lim lim vx, y

z‚z 0 x,y‚x 0 ,y 0 x,y‚x 0 ,y 0

Thus f is continuous at z 0 x 0 , y 0 precisely when u and v are.

Our next step is the definition of the derivative of a complex function f. It is the obvious

thing. Suppose f is a function and z 0 is an interior point of the domain of f . The derivative

f z 0 of f is

fz fz 0

f z 0 lim z z0

z‚z 0

Example

Suppose fz z 2 . Then, letting z z z 0 , we have

2.6

fz fz 0 fz 0 « z fz 0

lim

lim z z0 z

z‚z 0 z‚0

z 0 « z 2 z 2

0

lim

z

z‚0

2z 0 z « z 2

lim

z

z‚0

lim 2z 0 « z

z‚0

2z 0

No surprise here“the function fz z 2 has a derivative at every z, and it™s simply 2z.

Another Example

Let fz zz. Then,

fz 0 « z fz 0 z 0 « zz 0 « z z 0 z 0

lim

lim

z z

z‚0 z‚0

lim z 0 z « z 0 z « zz

z

z‚0

z 0 « z « z 0 z

lim

z

z‚0

Suppose this limit exists, and choose z x, 0. Then,

z 0 « z « z 0 z z 0 « x « z 0 x

lim

lim

z x

z‚0 x‚0

z 0 « z0

Now, choose z 0, y. Then,

iy

z 0 « z « z 0 z lim z 0 iy z 0

lim

z iy

y‚0

z‚0

z 0 z0

Thus, we must have z 0 « z 0 z 0 z 0 , or z 0 0. In other words, there is no chance of

this limit™s existing, except possibly at z 0 0. So, this function does not have a derivative

at most places.

Now, take another look at the first of these two examples. It looks exactly like what you

2.7

did in Mrs. Turner™s 3 rd grade calculus class for plain old real-valued functions. Meditate

on this and you will be convinced that all the ”usual” results for real-valued functions also

hold for these new complex functions: the derivative of a constant is zero, the derivative of

the sum of two functions is the sum of the derivatives, the ”product” and ”quotient” rules

for derivatives are valid, the chain rule for the composition of functions holds, etc., etc. For

proofs, you need only go back to your elementary calculus book and change x™s to z™s.

A bit of jargon is in order. If f has a derivative at z 0 , we say that f is differentiable at z 0 . If

f is differentiable at every point of a neighborhood of z 0 , we say that f is analytic at z 0 . (A

set S is a neighborhood of z 0 if there is a disk D z : |z z 0 | r, r 0 so that D S.

) If f is analytic at every point of some set S, we say that f is analytic on S. A function that

is analytic on the set of all complex numbers is said to be an entire function.

Exercises

5. Suppose fz 3xy « ix y 2 . Find lim fz, or explain carefully why it does not

z‚3«2i

exist.

6. Prove that if f has a derivative at z, then f is continuous at z.

7. Find all points at which the valued function f defined by fz z has a derivative.

8. Find all points at which the valued function f defined by

fz 2 « iz 3 iz 2 « 4z 1 « 7i

has a derivative.

9. Is the function f given by

z 2

,z ‚® 0

z

fz

,z 0

0

differentiable at z 0? Explain.

2.3. Derivatives. Suppose the function f given by fz ux, y « ivx, y has a derivative

at z z 0 x 0 , y 0 . We know this means there is a number f z 0 so that

fz 0 « z fz 0

f z 0 lim .

z

z‚0

2.8

Choose z x, 0 x. Then,

fz 0 « z fz 0

f z 0 lim

z

z‚0

ux 0 « x, y 0 « ivx 0 « x, y 0 ux 0 , y 0 ivx 0 , y 0

lim

x

x‚0

ux 0 « x, y 0 ux 0 , y 0 vx 0 « x, y 0 vx 0 , y 0

lim «i

x x

x‚0

u x 0 , y 0 « i v x 0 , y 0

x x

Next, choose z 0, y iy. Then,

fz 0 « z fz 0

f z 0 lim

z

z‚0

ux 0 , y 0 « y « ivx 0 , y 0 « y ux 0 , y 0 ivx 0 , y 0

lim

iy

y‚0

vx 0 , y 0 « y vx 0 , y 0 ux 0 , y 0 « y ux 0 , y 0

lim i

y y

y‚0

v x 0 , y 0 i u x 0 , y 0

y y

We have two different expressions for the derivative f z 0 , and so

u x 0 , y 0 « i v x 0 , y 0 v x 0 , y 0 i u x 0 , y 0

x x y y

or,

u x 0 , y 0 v x 0 , y 0 ,

x y

u x 0 , y 0 v x 0 , y 0

y x

These equations are called the Cauchy-Riemann Equations.

We have shown that if f has a derivative at a point z 0 , then its real and imaginary parts

satisfy these equations. Even more exciting is the fact that if the real and imaginary parts of

f satisfy these equations and if in addition, they have continuous first partial derivatives,

then the function f has a derivative. Specifically, suppose ux, y and vx, y have partial

derivatives in a neighborhood of z 0 x 0 , y 0 , suppose these derivatives are continuous at

z 0 , and suppose

2.9

u x 0 , y 0 v x 0 , y 0 ,

x y

u x 0 , y 0 v x 0 , y 0 .

y x

We shall see that f is differentiable at z 0 .

fz 0 « z fz 0

z

ux 0 « x, y 0 « y ux 0 , y 0 « ivx 0 « x, y 0 « y vx 0 , y 0

.

x « iy

Observe that

ux 0 « x, y 0 « y ux 0 , y 0 ux 0 « x, y 0 « y ux 0 , y 0 « y «

ux 0 , y 0 « y ux 0 , y 0 .

Thus,

ux 0 « x, y 0 « y ux 0 , y 0 « y x u ™, y 0 « y,

x

and,

u ™, y « y u x , y « ,

0 00 1

x x

where,

lim 1 0.

z‚0

Thus,

ux 0 « x, y 0 « y ux 0 , y 0 « y x u x 0 , y 0 « 1 .

x

Proceeding similarly, we get

2.10

fz 0 « z fz 0

z

ux 0 « x, y 0 « y ux 0 , y 0 « ivx 0 « x, y 0 « y vx 0 , y 0

x « iy

u

x 0 , y 0 « 1 « i v x 0 , y 0 « i 2 « y u

x 0 , y 0 « 3 « i v x 0 , y 0 « i 4

x x x y y

,.

x « iy

where i ‚ 0 as z ‚ 0. Now, unleash the Cauchy-Riemann equations on this quotient and

obtain,

fz 0 « z fz 0

z

x u « i v « iy u « i v stuff

x x x x

«

x « iy x « iy

u « i v « stuff .

x « iy

x x

Here,

stuff x 1 « i 2 « y 3 « i 4 .

It™s easy to show that

lim stuff 0,

z

z‚0

and so,

fz 0 « z fz 0

u « i v .

lim

z x x

z‚0

In particular we have, as promised, shown that f is differentiable at z 0 .

Example

Let™s find all points at which the function f given by fz x 3 « i1 y 3 is differentiable.

Here we have u x 3 and v 1 y 3 . The Cauchy-Riemann equations thus look like

3x 2 31 y 2 , and

0 0.

2.11

The partial derivatives of u and v are nice and continuous everywhere, so f will be

differentiable everywhere the C-R equations are satisfied. That is, everywhere

x 2 1 y 2 ; that is, where

x 1 y, or x 1 « y.

This is simply the set of all points on the cross formed by the two straight lines

4

3

2

1

0

-3 -2 -1 1 2 3

x

-1

-2

Exercises

10. At what points is the function f given by fz x 3 « i1 y 3 analytic? Explain.

11. Do the real and imaginary parts of the function f in Exercise 9 satisfy the

Cauchy-Riemann equations at z 0? What do you make of your answer?

12. Find all points at which fz 2y ix is differentiable.

13. Suppose f is analytic on a connected open set D, and f z 0 for all zD. Prove that f

is constant.

14. Find all points at which

y

x

fz i 2

x2 « y2 x « y2

is differentiable. At what points is f analytic? Explain.

15. Suppose f is analytic on the set D, and suppose Re f is constant on D. Is f necessarily

2.12

constant on D? Explain.

16. Suppose f is analytic on the set D, and suppose |fz| is constant on D. Is f necessarily

constant on D? Explain.

2.13

Chapter Three

Elementary Functions

3.1. Introduction. Complex functions are, of course, quite easy to come by”they are

simply ordered pairs of real-valued functions of two variables. We have, however, already

seen enough to realize that it is those complex functions that are differentiable that are the

most interesting. It was important in our invention of the complex numbers that these new

numbers in some sense included the old real numbers”in other words, we extended the

reals. We shall find it most useful and profitable to do a similar thing with many of the

familiar real functions. That is, we seek complex functions such that when restricted to the

reals are familiar real functions. As we have seen, the extension of polynomials and

rational functions to complex functions is easy; we simply change x™s to z™s. Thus, for

instance, the function f defined by

2

fz z « z « 1

z«1

has a derivative at each point of its domain, and for z x « 0i, becomes a familiar real

rational function

2

fx x « x « 1 .

x«1

What happens with the trigonometric functions, exponentials, logarithms, etc., is not so

obvious. Let us begin.

3.2. The exponential function. Let the so-called exponential function exp be defined by

expz e x cos y « i sin y,

where, as usual, z x « iy. From the Cauchy-Riemann equations, we see at once that this

function has a derivative every where”it is an entire function. Moreover,

d expz expz.

dz

Note next that if z x « iy and w u « iv, then

3.1

expz « w e x«u cosy « v « i siny « v

e x e y cos y cos v sin y sin v « isin y cos v « cos y sin v

e x e y cos y « i sin ycos v « i sin v

expz expw.

We thus use the quite reasonable notation e z expz and observe that we have extended

the real exponential e x to the complex numbers.

Example

Recall from elementary circuit analysis that the relation between the voltage drop V and the

current flow I through a resistor is V RI, where R is the resistance. For an inductor, the

relation is V L dI , where L is the inductance; and for a capacitor, C dV I, where C is

dt dt

the capacitance. (The variable t is, of course, time.) Note that if V is sinusoidal with a

frequency §, then so also is I. Suppose then that V A sin§t « . We can write this as

V ImAe i e i§t ImBe i§t , where B is complex. We know the current I will have this

same form: I ImCe i§t . The relations between the voltage and the current are linear, and

so we can consider complex voltages and currents and use the fact that

e i§t cos §t « i sin §t. We thus assume a more or less fictional complex voltage V , the

imaginary part of which is the actual voltage, and then the actual current will be the

imaginary part of the resulting complex current.

What makes this a good idea is the fact that differentiation with respect to time t becomes

d

simply multiplication by i§: dt Ae i§t i§Ae i§t . If I be i§t , the above relations between

I

current and voltage become V i§LI for an inductor, and i§VC I, or V i§C for a

capacitor. Calculus is thereby turned into algebra. To illustrate, suppose we have a simple

RLC circuit with a voltage source V a sin §t. We let E ae iwt .

Then the fact that the voltage drop around a closed circuit must be zero (one of Kirchoff™s

celebrated laws) looks like

3.2

I « RI ae i§t , or

i§LI «

i§C

i§Lb « b « Rb a

i§C

Thus,

a

b .

1

R « i §L §C

In polar form,

a e i ,

b

2

1

R 2 « §L §C

where

1

§L §C

tan . (R ‚® 0)

R

Hence,

a

I Imbe i§t Im e i§t«

2

1

R 2 « §L §C

a

sin§t «

2

1

R 2 « §L §C

This result is well-known to all, but I hope you are convinced that this algebraic approach

afforded us by the use of complex numbers is far easier than solving the differential

equation. You should note that this method yields the steady state solution”the transient

solution is not necessarily sinusoidal.

Exercises

1. Show that expz « 2i expz.

expz

expz w.

2. Show that expw

3. Show that |expz| e x , and argexpz y « 2k for any argexpz and some

3.3

integer k.

4. Find all z such that expz 1, or explain why there are none.

5. Find all z such that expz 1 « i, or explain why there are none.

6. For what complex numbers w does the equation expz w have solutions? Explain.

7. Find the indicated mesh currents in the network:

3.3 Trigonometric functions. Define the functions cosine and sine as follows:

iz

iz

cos z e « e ,

2

iz

iz

sin z e e

2i

where we are using e z expz.

First, let™s verify that these are honest-to-goodness extensions of the familiar real functions,

cosine and sine“otherwise we have chosen very bad names for these complex functions.

So, suppose z x « 0i x. Then,

e ix cos x « i sin x, and

e ix cos x i sin x.

Thus,

3.4

ix

ix

cos x e « e ,

2

ix

ix

sin x e e ,

2i

and everything is just fine.

Next, observe that the sine and cosine functions are entire“they are simply linear

combinations of the entire functions e iz and e iz . Moreover, we see that

d sin z cos z, and d sin z,

dz dz

just as we would hope.

It may not have been clear to you back in elementary calculus what the so-called

hyperbolic sine and cosine functions had to do with the ordinary sine and cosine functions.

Now perhaps it will be evident. Recall that for real t,

t t

t t

sinh t e e , and cosh t e « e .

2 2

Thus,

iit t

iit t

sinit e e i e e i sinh t.

2

2i

Similarly,

cosit cosh t.

How nice!

Most of the identities you learned in the 3 rd grade for the real sine and cosine functions are

also valid in the general complex case. Let™s look at some.

sin 2 z « cos 2 z 1 e iz e iz 2 « e iz « e iz 2

4

1 e 2iz « 2e iz e iz e 2iz « e 2iz « 2e iz e iz « e 2iz

4

1 2 « 2 1

4

3.5

It is also relative straight-forward and easy to show that:

sinz ‚± w sin z cos w ‚± cos z sin w, and

cosz ‚± w cos z cos w sin z sin w

Other familiar ones follow from these in the usual elementary school trigonometry fashion.

Let™s find the real and imaginary parts of these functions:

sin z sinx « iy sin x cosiy « cos x siniy

sin x cosh y « i cos x sinh y.

In the same way, we get cos z cos x cosh y i sin x sinh y.

Exercises

8. Show that for all z,

a)sinz « 2 sin z; b)cosz « 2 cos z; c)sin z « cos z.

2

9. Show that |sin z| 2 sin 2 x « sinh 2 y and |cos z| 2 cos 2 x « sinh 2 y.

10. Find all z such that sin z 0.

11. Find all z such that cos z 2, or explain why there are none.

3.4. Logarithms and complex exponents. In the case of real functions, the logarithm

function was simply the inverse of the exponential function. Life is more complicated in

the complex case”as we have seen, the complex exponential function is not invertible.

There are many solutions to the equation e z w.

If z ‚® 0, we define log z by

log z ln|z| « i arg z.

There are thus many log z™s; one for each argument of z. The difference between any two of

these is thus an integral multiple of 2i. First, for any value of log z we have

3.6

e log z e ln |z|«i arg z e ln |z| e i arg z z.

This is familiar. But next there is a slight complication:

loge z ln e x « i arg e z x « y « 2k i

z « 2ki,

where k is an integer. We also have

logzw ln|z||w| « i argzw

ln |z| « i arg z « ln |w| « i arg w « 2ki

log z « log w « 2ki

for some integer k.

There is defined a function, called the principal logarithm, or principal branch of the

logarithm, function, given by

Log z ln|z| « iArg z,

where Arg z is the principal argument of z. Observe that for any log z, it is true that

log z Log z « 2ki for some integer k which depends on z. This new function is an

extension of the real logarithm function:

Log x ln x « iArg x ln x.

This function is analytic at a lot of places. First, note that it is not defined at z 0, and is

not continuous anywhere on the negative real axis (z x « 0i, where x 0.). So, let™s

suppose z 0 x 0 « iy 0 , where z 0 is not zero or on the negative real axis, and see about a

derivative of Log z :

Log z Log z 0 Log z Log z 0

lim

lim .

z z0 z‚z 0 e Log z e Log z 0

z‚z 0

Now if we let w Log z and w 0 Log z 0 , and notice that w ‚ w 0 as z ‚ z 0 , this becomes

3.7

Log z Log z 0

lim w ww 00

lim z z0 w

w‚w 0 e e

z‚z 0

1 0 z0 1

w

e

1

Thus, Log is differentiable at z 0 , and its derivative is .

z0

We are now ready to give meaning to z c , where c is a complex number. We do the obvious

and define

z c e c log z .

There are many values of log z, and so there can be many values of z c . As one might guess,

e cLog z is called the principal value of z c .

Note that we are faced with two different definitions of z c in case c is an integer. Let™s see

if we have anything to unlearn. Suppose c is simply an integer, c n. Then

z n e n log z e kLog z«2ki

e nLog z e 2kni e nLog z

There is thus just one value of z n , and it is exactly what it should be: e nLog z |z| n e in arg z . It

is easy to verify that in case c is a rational number, z c is also exactly what it should be.

Far more serious is the fact that we are faced with conflicting definitions of z c in case

z e. In the above discussion, we have assumed that e z stands for expz. Now we have a

definition for e z that implies that e z can have many values. For instance, if someone runs at

you in the night and hands you a note with e 1/2 written on it, how to you know whether this

means exp1/2 or the two values e and e ? Strictly speaking, you do not know. This

ambiguity could be avoided, of course, by always using the notation expz for e x e iy , but

almost everybody in the world uses e z with the understanding that this is expz, or

equivalently, the principal value of e z . This will be our practice.