AAU

2006

Complex Analysis Instructor Hunduma Legesse

M.T

W W W . A A U . E D U . E T

Hunduma Legesse(PhD)

Addis Ababa University

Mathematics Department

Hunduma Legesse

CHAPTER 1: COMPLEX NUMBERS

Why complex numbers?

Why do we need new numbers?

The hardest thing about working with complex numbers is, understanding why you might

want to. Before introducing complex numbers, let us backup and look at simpler

examples of the need to deal with new numbers.

If you are like most people, initially number meant whole number, 0, 1, 2, 3 whole

number makes sense. They provide a way to answer questions of the form How many

? You are learned about the operations of addition and subtraction, and you found that

while subtraction is a perfectly good operation, some subtraction problems, like 3-5,

dont have answers if we only work with whole numbers. Then you find that if you are

willing to work with integers, ,-3, -2, -1, 0, 1, 2, 3, ,then all subtraction problems do

have answers! Furthermore, by considering examples such as temperature scales, you see

that negative numbers often make sense.

Now in integers all subtraction problems have got answers. But if we want to deal with

division, some, in fact most, division problems do not have answers in integers. For

example 12, 32, 53 and the like are not integers. So we need new numbers! Now we

have rational numbers to furnish such problems.

There is more to this story. There are problems with square roots and other operations,

but we will not get into that here. The point is that you have had to expand your idea of

number on several occasions, and now we are going to do that again.

The problem that leads to complex numbers concerns solutions of equations like,

(1) 012 x (2) 012 x

Which equation has real roots? (Equation (1) or (2))

Can you explain?

To equation (1) -1 and 1 are the two real roots but equation (2) has no real root since

there is no real number whose square is negative. So if equation (2) is to be given

solutions then we must create a square root of -1.

In general for all quadratic equation of the form 02 cbxax to have a solution we

must need a number system in which acb 42 is defined for all numbers a, b and c. The

number system which we are going to define is called the complex number system.

The Complex Number System

Activity 1

1. Consider the set {(x, y)/ x and y are real numbers} in the coordinate plane.

a. What is the difference between (2, 3) and (3, 2)?

b. When do you say that (a, b) = (c, d)?

c. What is the sum (2, 3) + (5, 2)?

d. Can you generalize the sum for (a, b) + (c, d)?

2. Identify whether the following lies on the X-axis or the Y-axis.

(2, 0) (-3, 0)

(1/2, 0) (0.234, 0)

( 2 , 0) (0, 2)

(0, -3) (0, 0.123)

(0, y) (x, 0) for any real numbers x and y.

3. Suppose the product of two ordered pairs (a, b) and (c, d) is defined as follows:

(a, b).(c, d) = (ac-bd, ad + bc)

Try to find the following:

a. (2, 3).(5, 2)

b. (3, 0).(4, 0)

c. (0, 3).(0, 4)

d. (1, 0).(1, 0)

e. (0, 1).(0, 1)

f. ( )0,1x ( )0,2x

g. ( ),0)(,0 21 yy for any real numbers ,1x ,2x 21,, yy

Now we are in a position to define the system of complex numbers.

Definition:- The system of complex numbers denoted by C is the set of all ordered pairs

(x, y) of real numbers x and y which satisfies the following three conditions.

(i) (a, b) = (c, d) iff a = c and b = d

(ii) (a, b) + (c, d) = (a + b, c + d)

(iii) (a, b).(c, d) = (ac - bd, ad + bc)

Notation: For algebraic purposes we use the following notations.

- (x, 0) = x for any real number x.

- (0, 1) = i

- (0, y) = yi

Exercise

1. Use the above notation to show that i 2 = -1.

2. Any complex number (x, y) can be written us x + iy.

3. Simplify i3, i4, i8, i11

Now using the above notations we can redefine complex numbers as:

Definition: A complex number z is one of the form z = x + iy for some real numbers x

and y.

(i) the number x is called the real part of z and is denoted by Re(z).

(ii) the number y is called the imaginary part of z and is denoted by Im(z).

Example: a) In z = 2-5i, Re(z) = 2 and Im(z) = -5

b) In z = 6+4i, Re(z) = 6 and Im(z) = 4

c) In z = 0 +2i = 2i, Re(z) = 0 and Im(z) = 2

d) In z = 0 + 0i = 0, Re(z) = 0 and Im(z) = 0

e) In z = 4 + 0i = 4, Re(z) = 4 and Im(z) = 0

Now we need to define the four arithmetic operations and equality on complex numbers

using the new notation. If z = x + iy and w = a + bi then

Equality: z = w if and only if x = a and y = b.

Addition and Subtraction

To add or subtract two complex numbers, you add or subtract the real parts and the

imaginary parts.

(i) z + w = (x + a) + (y + b)i

(ii) z w = (x a) + (y b)i

Example: a) (3 5i) + (6 + 7i) = (3 + 6) + (-5 + 7)i = 9 +2i

b) (3 5i) - (6 + 7i) = (3 - 6) + (-5 - 7)i = -3 -12i

Note: These operations are the same as combining similar terms in expressions that have

a variable. For example, if we were to simplify the expression (3 5x) + (6 + 7x) by

combining similar terms, then the constant 3 and 6 would be combined, and the terms

(-5x) and (7x) would be combined to yield (9 + 2x).

Multiplication:

The formula for multiplying two complex numbers is

zw = (xa yb) + (xb ay)i.

You do not need to memorize the formula, because you can arrive at the same

result by treating the complex numbers like expressions with a variable,

multiply them as usual, then simplify. The only difference is that powers of i

do simplify, while powers of x do not.

Example: (2 + 3i)(4 + 7i) = 2x4 + 2x7i +4x3i + 3ix7i

= 8 + 14i + 12i -21

= (8 21) + (14 + 12)i

= -13 + 26i

Exercise: Perform the following operations.

1) (-3 + 4i) + (2 2i)

2) 3i (2 4i)

3) (2 7i)(3 + 4i)

4) (1 + i)(2 3i)

5) (2 i) i(1 2i)

1.3. Complex Conjugate and Modulus

Activity: Given a complex number z = x + iy and w = x iy, find

a) The product zw.

b) The sum z + w

c) The difference z w

d) If you represent z = x + iy as (x, y) and w = x iy as (x, -y) on the

number plane, then can you tell any relation between z and w?

We call w = x iy the conjugate (or complex conjugate) of the complex number z = x

+yi. Actually if you look both of them by putting on the number plane you find that one

is the reflection of the other. The conjugate of a complex number z is simply the

reflection of z through the X-axis. Conjugates are important because of the fact that a

complex number times its conjugate is real. That is, (x + yi)(x yi) = x2+ y2

DEFINITION: The complex conjugate or the conjugate of a complex number z = x+yi

is denoted by z is given by z x yi

Example 1:

1. If ,65 iz then iiz 65)6(5

2. If ,2

11 iz then iz

2

11

3. If ,044 iz then iz 4

4. If ,2iz then iz 2

Example 2:

Number z Conjugate of z Product

2 + 3i 2 -3i 13

3 5i 3 +5i 34

4i -4i 16

5 5 25

From the above activities, definition and examples we can draw the following

conclusions as a theorem:

THEOREM: For any complex numbers z and w, we have

i) zz

ii) )Re(2 zzz

iii) )Im(2 zizz

iv) wzwz

v) wzwz ..

vi) z z

w w

, if w 0

Note: (iv) and (v) of the above Theorem can be extended to any finite number of terms.

That is,

i) nn zzzzzz ....... 2121

ii) nn zzzzzzz ............. 32121

Division:

Division is the inverse process of multiplication; that is,

z

w u iff z = wu.provided w 0

If z = x + yi , w = a +bi and u = c + di, then one can solve

x + yi = (a +bi)(c + di) to get

c = 2 2

ax by

a b

and d=

2 2

ay bx

a b

Thus u = z

w

2 2

ax by

a b

+

2 2

ay bx

a b

i

Or we can use conjugates to simplify expressions of the form (x + yi) (a + bi) by

writing in the form of (x + yi)/ (a + bi) and multiplying both the numerator and

denominator by,

a bi which, is the conjugate of a + bi.

Example: if iz 32 and iw 5 than

ii

i

w

z

2222 )1(5

)1(2)3(5

)1(5

)1(35.2

5

32

i26

17

26

7

THEOREM: (C, +, ) is a field.

(Proof: exercise)

DEFINITION: The absolute value (or modulus) of a complex number ,yixz

denoted by z , is defined to be

22 yxz

Example:

1. If ,52 iz then 2952 22 z

2. If ,125 iz then 13169125 22 z

3. If ,iz then 112 z

4. If ,2z then 2/2/)2( 2 z

Note: If z1 = x1 +y1i and z2 = x2 +y2i then

2

1 2 1 2 1 2 1 2 1 2z z x x y y i x x y y

21 zz is the distance between 1z and 2z

THEOREM: For any two complex numbers z and w

(i) 2

. zzz

(ii) zz

(iii) zz Re

(iv) zz Im

(v) wzwz .

(vi) w

z

w

z if 0w

(vii) Triangle Inequality: wzwz

(viii) wzwz

Proof: Let yixz and viuw for some vuyx ,,,

(i) Since z x - yi

22222 zyxixyyxyxyixyixzz

22222 zxxoiyx

(ii) Since z x y i we have

22 2 2z x y x y z

(iii) Since 222 yxx , we have

(iv) Since 222 yxy , we have

zyxyyz 222Im

(v) wzzwz 22 bi (i)

zwzzwzwz .

22 wzwz w

wzwz

(vi)

2

2

2

w

z

w

z

ww

zz

w

z

w

z

w

z

w

z

w

zw

w

z

w

z Provided that 0w

zyxxxz 222Re

(vii) wzwzwzwzwz 2

wwzwwzzz

22wwzwzz

22 Re2 wwzz

222 2 wzwwzz

wzwz

(viii) 222 Re2 wwzzwzwzwz

222 2 wzwwzz

wzwz

Note: that the Triangle Inequality can be extended to any finite sum by induction to give

nxnz zzzzzz 12

Example: If 1z , then show that 1

abz

baz, where azb

Solution:

zbazbaz

zzab

baz

azb

baz

1as 1 zz

1

baz

baz

zbaz

baz

zbaz

baz

azb

naz

1.4. Polar Coordinates of Complex Numbers

A) Geometric Representation of Complex Numbers

The complex number yixz is uniquely determined by the ordered pair of real

numbers yx, . The same is true of the point yxp , in the plane with Cartesian

coordinates x and y. Hence it is possible to establish a 1-1 correspondence between

all complex numbers and all points of the given plane. We merely associate the

complex number z = x+iy with the point yxp , . The plane whose points represent

the complex numbers is called the complex plane or the z-plane. The real numbers

or points 0,xx are represented by points on the x-axis, hence the x-axis is called

the Real axis. And pure imaginary numbers or points yiy ,0 are represented by

points on the y-axis, and hence we call the y-axis the imaginary axis.

Instead of considering the point yxp , as the representation of yixz we may

equally well consider the directed segment or the vector extending from O to P as the

representation of the complex number yixz . In this case, any parallel segment of

the same length and direction is taken as corresponding to the same complex number

01

2

3

1 2 x

y

P (2,-3)z =2-3i

Real axis

Imaginary axis

z- Plane

p(-4,2)

z =4+2i

The representations of zz, , the sum and difference of complex numbers are as

follows:

z is the length of the vector

representing the complex number

z.

z is the reflection of z with respect

to the real axis.

The sum 21 zz is represented

by the vector sum of the vectors 1z and 2z

Real axis

z =z+yi

z =x-yi

z

Imaginary axis

Real axis

Imaginary axis

z1

z2 z1 +z2

The difference 21 zz is represented by a

vector from the point z1 to the point z2

Note that the product 21 zz of two complex numbers is a complex number. Therefore,

this product is neither the scalar product nor the vector product used in vector analysis.

Consequently, our complex numbers of z cannot be identified with the vectors of two

dimensional vector analyses.

Example:

a) If z1 and z2 are two given complex numbers as in the figure (a) below, construct

graphically 21 23 zz

b) If 321 ,, zzz and 4z are four given complex numbers as in the figure, construct

graphically 4321 zzzz

Solution:

Real axis

Imaginary axis

z1z2

z1 -z2-z2

Real axis

Imaginary axis

z1

z2

3z1

-2z2

3z1 -2z2

(a)

Real axis

Imaginary axis

z3z1 +z2 +z3 +z4

z4

z3

z4z2

z1

z2

(b)

B) POLAR FORM

L et ,r be the polar coordinates of the point representing the complex

number yixz , .0r

Then, cosrx

sinry

22 yxzr

x

y1tan

r

z =x +yi

Real axis

Imaginary axis

Hence we can write the complex number yixz as follows

sincos rrz sincos r

This form is called the polar form.

Notation: In short we can write, sincos rrz sincos r := rcis .

The angle is called the argument of z and we write as, zarg .

If zarg , then 2 , ,n n Z is also argument of z.. Arg(z) is called the principal

argument of z, is the value of argument of z in the interval , , that is,

zarg .

If zArg is the principal argument of z, then ZnnzArgz ,2arg describe all

possible values of zarg .

Example: Express the following complex number in polar form.

a) iz 322 b) iz 55

c) iz 3 d) 1z

Solution:

(a) 416124322 22 r

zArgx

y

33tan

2

32tantan 111

Therefore; 3

43

sin3

cos4322 cisiiz

is the polar form of z.

(b) 2550252555 22 r

zArgx

y

431tan

5

5tantan 111

Therefore; 4

325

cis is the polar form of z.

(c) Znnr

,

2

121sin,3930 122 and

Znn ,2

120cos 1 . In particular if n = 0 then

.2

In general, Zn

n

,

2

14

The principal argument: .2

zArg

Therefore, 2

3 cis is the polar form of z.

(d) Znnr ,121cos0sin,101 1122

The principal argument: .zArg

Therefore, cis is polar form of z.

DEFINITION: Let 111 cisrz and .222 cisrz Then

.,2212121 Zkkrrzz

Remark: If 111 cisrz and 222 cisrz then, 212121 cisrrzz

Proof: irrzz 22211121 sincossincos

irr 1221212121 sincossincossinsincoscos

irr 212121 sincos

2121 cisrr

The geometric representation of 21zz is best seen by use of the polar form of the product.

Given 111 cisrz and .222 cisrz Choose the number 1z on the real axis and form

the triangle .,1,0 1z Then with vector 2z as one side, construct triangle Pz ,2,0

similar to the first triangle, while keeping the orientation of the equal angles 1 the same.

By similarity of triangles

2121 rrzzOP , and 21arg P Hence 21 zzp

real axis

Imaginary axis

1

2

3

z2

P

z1

10

212

1

2

2

22

11

22

11

2

1

cis

r

r

cis

cis

cisr

cisr

cisr

cisr

z

z if 02 z

Hence the quotient of two complex numbers 1z and ,02 z in polar form is

212

1

2

1 cisr

r

z

z

The geometric representation of 2

1

z

z is as follows:

By similarity of triangles

,,,0,1,0 12 zQz

We have

OP

z

z

1

2

1

real axis

Imaginary axis

2

1

3

z1

z2

10

Q

1.5. Powers and Roots

THEOREM: Let nzzz ,, 21 be non-zero complex numbers such that ii rz and

iiz arg for .,,2,1 ni Then .21321 nn ciszzzz

Proof: We prove by Mathematical Induction. For 2121211,2 cisrrzn shown

above. Assume that it is true for ,kn that is, .12121 kkk cisrrrzzz

To show that it also holds true for ,1 kn

11121121 kkkkkk cisrcisrrrzzzz

11121 kkkk cisrrrr

121121 kkk cisrrrr

Hence it is true for all .2n Therefore,

2212121 ncisrrrzzz nnn

COROLLARY: (De Moivre's Formula)

If ,Nnandcisrz then

(i) ncisrz nn (ii) ncisr

zz nn

11 1

Examples: 1. Calculate

(a) 101 i (b) 163 i

1

21

z

z

OQ

2

1

2

1

r

r

z

zOQ and 21arg Q

Hence 1z

zQ

Solution: (a) since,

4

321

cisi , by De Moivre's Formula we have

2

1532

4

31021

1010 ciscisi

2

322

832 ciscis

ii 322

sin2

cos32

(b) Since,6

23 cisi , we have

3

82

623 1616

16 ciscisi

3

223

222 1616

ciscis

i

3

2sin

3

2cos216

ii 3122

3

2

12 1516

Example: 2. Use De Moivre's Formula to show that

(a) 23 sincos3cos3cos

(b) 32 sinsincos33sin

(c) 2224 sinsincos6cos4cos

(d) 33 sincos4sincos44sin

DEFINITION: A complex number z is called the nth root of a complex number w for

n if wz n and we write .n wz

Let iz sincos be the nth root of rcisw

Then cisrnciswz nn

rn and .,2 Zkkn

n r and .,2

Zkn

k

Hence Zkn

k

n ,

2 but these values of yield the same value of z for any two

integers that differ by a multiple of n . Therefore, there are n-distinct roots when ,0z

namely

1,,1,0,2

sin2

cos

nki

n

k

n

krz nk

Thus, the n roots are

ncisrz n

0

ncisrz n

21

.

.

.

n

ncisrz nn

121

Geometrically the length of each of the n vectors nz1

is the number .n r The argument

of one of those vectors is n

and the other arguments are obtained by adding multiples

of .2

nto

n

n2

n

2

n

zn-1

z0

z1z2

n r

Example: Find the fourth roots of .1 i

Solution: Since ,44

sin4

cos21 nandii

3,2,1,0,4

2

16sin

4

2

16cos21 8

14

ki

kki

Therefore; the fourth roots of i1 are

16sin

16cos2 8

1

0

z

16

9sin

16

9cos2 8

1

1

iz

16

17sin

16

17cos2 8

1

2

iz

16

25sin

16

25cos2 8

1

3

iz

DEFINITION: A complex number z is called the nth root of unity if .1nz

0sin0cos1 izz nn

1,,1,0,2

sin2

cos nkn

ki

n

kzk

are nth roots of unity.

If

ni

nw

2sin

2cos for 1k then by De Moivre's formula the roots of unity are

given by 12 ,,,,1 nwww . Since ,1z the nth root of unity are vertices of the n-sided

regular polygon inscribed in the unit circle.

Example 1: Find the third roots of unity

Solution: The cube roots of unity are

10sin0cos0 iz

iiz2

3

2

1

3

2sin

3

2cos1

iiz2

3

2

1

3

4sin

3

4cos2

z0

z1

z2

Real zxis

-1

1

1

-1

Imaginary axis

Example 2: If sincos2 irwz then

1,0,2

2sin

2

2cos

k

ki

krz

1,0,2

sin2

cos

kkikr

2sin

2cos

ir

But 2

cos1

2cos

and ,

2

cos1

2sin

where the sign infront of the radical

depends on the quadrant in which 2

lies.

Example 3: Solve .012 zforiziz

Solution: By the Quadratic Formula

2

43

12

1142

2,1

iiiiiz

Since 6.0cos,5

4

5

3543

iiw and ,80sin and hence lies in the 4th

quadrant so that2

lies in the second quadrant.

;5

2

25

31

2

cos1

2cos

and

5

1

25

31

2

cos1

2sin

Hence

2sin

2cos543

ii

ii

2

5

1

5

25

Take only ,2 i then

1

2

2

2

212,1

iiz

iiz and i

iiz

1

2

22

Therefore; the solution set is i 1,1

Note: If is any nth root of a complex number 0z , then ,,,,,,,12 nk zwzwzwzwz where

w is the nth root of unity when k =1, that is, ,2

sin2

cosn

in

w

are the nth roots of 0z .

Multiplying z by kw corresponds to increasing the argument of z by .2

n

k

Now let us define nm

z for ,sincos1,, mirzIfnmcdandnm then

mimrz mm sincos

and nmz 1 has n distinct roots

1,....,1,0,2sin2cos1

nh

n

h

n

mi

n

h

n

mrz n

mnm

hnm

wn

mi

n

mr .sincos

1....,1,0...sincos1

nkw

ni

nrz

m

kmnm

n

These two set of numbers are the same if the sets hw and kmw coincides when both h

and k range the values .1,...,2,1,0 n

a) We first show kmw has n distinct values. If some of its values correspond to two

distinct values kofkandk 111 are the same then the two part sets mkw1

and mkw11

coincide and there is Np such that,

Pn

mk

n

m

n

mk

n

m

2

22 111

and thus .111 Pn

mkk

Since p and 111 kk must be divisible by n . Contradiction as nkk 111 .

Hence kmw has exactly n distinct values.

b) We next show that for each fixed value of k the number kmw is one of the n distinct

numbers hw , so that the two set of numbers are identical.

Let 1,...,2,1,00 nk be fixed. Then either .00 nmkornmk

(i) If nmk 0 , we choose 0mkh . Therefore, the values corresponding to 0k in (**) is

the same as the value 0mk (*).

(ii) If nmk 0 , then by Division Algorithm

rqnmk 0 where .1,....,2,1,0 nr So .rrqnmk www In this case we choose

rh .

Hence we conclude that the two set of numbers in (*) and (**) are the same.

Thus, the n-numbers in either set can be written as

k

n

mik

n

mrz n mn

m

2sin2cos

.1,...,2,1,0 nk

Similarly, nmm

nnm

zzz11

Example: Find all values of 2331 i

Solution: 3

2sin3

2cos231 ii

kiki

2

3

2

2

3sin2

3

2

2

3cos831 2

3

1,0k

,okif then

32

2

3sin

3

2

2

3cos81

iz

.228sincos8 i

If 1k , then

2

3

2

2

3sin2

3

2

2

3cos82 iz

4sin4cos8 i

228018 i

Regions in the Complex Plane

1. A collection of points in the z-plane is called set of points or a point set.

2. Let 0z and .0 The nbhd of 0z , denoted by 00 zzzzB

The circular disk centered at 0z and radius .

0z

Real axis

Imaginary axis

0zB 0z

Real axis

Imaginary axis

,0* zB

3. The deleted nbhd of 0z , denoted by ,0zB is the set of all z such that

.0 0 zz 00* 0, zzzzB

4. A point 0z is called a limit point or a cluster point or an accumulation point if

every deleted nbhd of 0z contains points of S different from 0z .

Note that a limit point of S may or may not be element of S .

Example: 0 is the limit point of .1

nn

S

i is a limit point of .1 zzS

i21 is a limit point of .3 zzS

5. If every limit point of a set S belongs to S , we say that the set is closed.

6. A limit point 0z of S is said to be an interior point of S if 0zBnbhda such

that .0 SzB

A limit point 0z of S such that every nbhd of 0z contains both points of S and

points which don't belong to S is called a boundary point.

0z

PwR

If every point of a set S is an interior point we say that S is an open set.

7. A set of points S is said to be bounded if and only if there is a positive number M

such that Mz for all .Sz Otherwise, S is called unbounded.

8. A set of points S is said to be connected if any two of its points can be joined by a

continuous curve all of whose points belongs to S .

0z is interior point of S .

.SPandSR Every nbhd of w contains

both points of S and points not on S . Hence w

is a boundary point

S1

S3

P

QR O

SX

S2

1S and 2S are connected but 3S is not.

9. An open connected set is called a domain or an open region.

10. If we add to any set S those limit points which are not already in S , we obtain

closed set called the closure of S , often denoted by S .

11. The closure of an open region is called a closed region.

12. If to an open region we add all, or some or none of its limit points we obtain a set

called REGION.

13. A region which is both closed and bounded is called compact set.

Example: describe the set of points such that

(a) 0Re zz

(b) 1ImRe zzz

(c) 32 izz

(d) 32 izz

(e) 121 izizz

Solution:

(a) .00Re iyxzifxz

Open

Unbounded

Connected

R.A

I.A

(b) yixzifyxzz 11ImRe

R.A

I.A

1-1

i

-i

closed

bounded

connected

compact

(c) 323232 22 yxiyixiz

R.A

I.A

closed

bounded

2 4 6

-2i

-4i

-2-4-6

2i

4i

6i

(d) iyxzifyxiz 31232 22

R.A

I.A

Neither open nor closed

connected

2 4 6

-2i

-4i

-2-4

2i

4i

6i

(e) 1211121 222 yxyxiziz

Open

bounded

1 2 3 4

-i

-2i

-3i

i

2i

3i

-1-2-3

Not connected

MaEd 461 Functions of Complex Variables

Worksheet I

1. Write each of the following in the form bia where a and b are real numbers.

a) i

i

i

123

13 3

b) iii 321

5

c) 3194120 34 iii

d) i

i

i

i

5

2

43

21

2. If iziz 23,2 21 and ,2

3

2

13 iz

then find

a) 32

2

3

1 43 zzz

b) 4

3z

c) 321 43 zzz

3. If 321 ,, zzz are complex numbers prove that

a) 133 .zzzzi

b) 321321 zzzzzz

c) 321321 .... zzzzzz

d) 3121321 ... zzzzzzz

4. Prove that

a) 2222 2 wzwzwz b) zzz 2ImRe

5. Express the following in polar form

a) i322

b) i2222

c) i26

d) i2

3

2

3

6. Evaluate using polar form

a) 1022 i b) 8273 i

c) 21232 i d) 5144 i 7. Solve

a) iz 13 d) 0322 zz

b) 84 z e) 07122 iziz

c) iz 83 f) 0224 izz

8. Find the square roots of

a) i125 b) i548

9. Find the sixth roots of unity and show the roots graphically

10. Express 6cos and 6sin interms of sin and cos .

11. Find the locus of points in the z-plane satisfying

a) 2 iz d) 323 zz

b) 2Im iz e) 822 zz

c) 02Re zi f) 4

arg4

z

12. Express each equation interms of conjugate coordinates

a) 52 yx b) 3622 yx

13. Prove that the equation of any circle or straight line in z-plane can be written in the

form 0. rzzzzx

Were x and r are real numbers, and is a complex number.

CHAPTER 2: ANALYTIC FUNCTIONS

Introduction: In the previous section chapter 1, we defined complex

numbers and also considered the algebra of complex numbers. In this

section, we shall introduce the concepts of Limit and Contiunity which

underlie Analysis, and the notion of Differentiability applicable to the

complex valued functions of complex variables. It will be seen that the

notion of limit and continuity are development of the notion of

neighbourhood of a point, and the notion of differentiability constitutes a

departure from that of continuity in as much as the differentiability of a

function of a complex variable is not just equivalent to differentiability of

the real and imaginary parts thereof as is the case with continuity. This fact

leads to Cauchy-Riemann Equations and their important consequences. Far

reaching properties of analytic functions can only be arrived at with the help

of the Cauchys Integral Formula which is considered in the next chapter,

(Chapter 4).

2.1. Functions of a Complex Variable

A symbol such as z which can stand for any one of the numbers of a set S of complex

numbers is called a complex variable.

Let S be a set of points in the Z-plane. If to each value of z in S there corresponds a

unique value of a complex variable w we say that w is a function of a complex variable

z on S and write )(zfw or )(zw or ,zsw etc. The set S is called the domain

of the function and the quantity zf is called the value of f at z or the image of z

under f .

Example: The following are functions of a complex variable z where .iyxz

a) 2zzw d) 22 yxixyw

b) z

zw

1 e)

iyxiy

yxw

12

22

c) ziz

zw 1

As in the real analysis, algebraic operations with functions of complex variables are

defined by corresponding operations on function values when functions are written in the

form f(z) and z then the sum, the product, the quotient and composite are, respectively,

z

zfzzfzzf

,., and zf

Example: if izzzf 12 and ,3 izz then

a) 142 zzzzfzf

b) izizzzzfzf 31. 2

iziiz z 123332

3

c) 0f g z f g z

iizizizf 1333 2

iizzigz 23162

iigz 632

By separating into real part and imaginary part any function zf can be expressed

interms of two real functions, that is in the form

yxiyxuzf ,,

Similarly, in polar coordinates with sincos irz

,, riruzf

Example:

1. If ,2zw then 22, yxyxu and yyx 2,

iyyxiyxxw 22222 In polar coordinates

2sin2cos2sin2cos 2222 ririrzw and thus

2sin,2cos, 22 rrandrru

2. If ,1 z

z

then

2222

22

212111 yxx

yi

yxx

yxx

iyx

iyx

iyx

iyxw

2222

22

21,

21,

yxx

yyxand

yxx

yxxyxu

Graphical Representation of w = f(z)

By definition for each value of the independent variable iyxz ( in the domain of f )

the function yields a unique value ivuw of the dependent variable. Each of these

variables has two dimensions and their combined configuration is a four dimensional

entity that is rather difficult to graph. Because of this difficulty the graphical

representation of a complex function I effected by employing tow copies of the complex

plane labeled z plane and w plane. If zfw is a function for every yixz of its

domain in the z plane we calculate the corresponding ivuw and we locate if in the

w plane. Repetition of the same process of some set S in the domain of f which

yield "the image of S under f " in the w plane. The correspondence of points in z

plane into the w plane by is called mapping or transformation

Example:

1. Consider, 2 zw . This mapping is a translation by 2 units to the right.

Use two planes one for w and the other for z and hence we get each point in the z-plane

is translated by 2 units to the write. (Show on the plane: exercise)

2. If ,)( zzf then, f is a reflection of z with respect to the real axis. Thus the image of

every point in z-plane is the reflection of the point with respect to the real axis.

(Show on the plane: exercise)

2.2. Limits

At this level everybody knows how to find the limits and prove some

properties of limits of the case of real-valued functions of single variable and

real-valued functions of several variables. In this section we extend the

notion of limits to complex variable.

DEFINITION: Let f be a complex function with domain D. We say that the complex

number. 0w written

0

0lim

zz

wzf

iff for all .* 00 wBzfzBB

Restatement

0

0 0lim

zz

wzf

such that for any Dz

and ,0 0 zz we have 0wzf

Graphically

R.A

I.A

z0

R.A

I.A

w0

f(z)

Example 1: Show that iz

iz

2

2lim

Solution: Let 0 be given. We need to show that there exists 0 such that

.220 iziZ

Choose . Then

.2220 iiziz

iz

iz

2

2lim

Example 2: Show that .21

lim2

iiz

z

Solution: Let 0 . We observe that

.22212

izifiziiziiz

izizi

iz

z

Choose . Then

i

iz

ziziz 3

10

2

Example 3: Show that

0

0ImRe

lim

z

z

zz

Solution: Let 0 be given. We know that

zzandzz ImRe

01Im

1Re

zifz

zand

z

z

So zzz

z

z

z

zzImIm.

ReIm.Re

Choose . Then

.0Im.Re

00 z

zzzz

0

0Im.Re

lim

z

z

zz

Example 4 . Let show that 0

022lim

zz

zz

Solution: Let 0 . Since ,00022 zzzzzz assume that 10 zz . Then

000000 2122 zzzzzzzzz

0000022 21 zzzzzzzzz

0

0 21 zzz

.

Choose

021,1min

z . Then

.0 0220 zzzz

Exercise:

1. Show that iz

iyix

2

632lim

2. Show that

iz

iyix

1

2323lim 2

Theorem: (Uniqueness of Limit)

If a function f has a limit at Cz 0 then it is unique, that is, if

0

0lim

zz

wzf

and

0

1lim

zz

wzf

then .10 ww

Proof: Since

0

0lim

zz

wzf

and

0

1lim

zz

wzf

, for 0 there exist 1 and 2 such

that

,2

0 010 wzfzz and

.2

0 120 wzfzz

Choose 21 ,min . Then if 00 zzz then

20

wzf and .22

wzf

1010 wzfzfwww

10 wzfwzf

22

Sinceis arbitrary we have .0 0110 wwww Hence the limit of a function is

unique.

Remark: The definition of limit does not specify the direction form which z must

approach 0z . It requires that if a limit is to exist its value must be independent of the

direction of approach. This fact is useful in proving that a limit doesn't exist.

Example 5: Show that

0

lim

z

z

zdoes not exist.

Solution: Let yixz . Then iyxz

iyx

xy

yx

yx

yix

yix

z

z2222

22 2

If 0z along the real axis, 0y ,

00

1limlim2

2

xz

x

x

z

z

If 0z along the imaginary axis, 0y ,

00

.1limlim2

2

yz

y

y

z

z

Hence

0

lim

z

z

z does not exist

Theorem: Suppose that

(a) yxivyxuzf ,, has domain D

(b) 000 iyxz is in D or on boundary of D

Then

000000000

,,

,lim,limlim

yxyxyxyxzz

vyxvanduyxuiffivuzf

Proof:

Assume that

0

00 .lim

zz

ivuzf

Let 0 be given. Then 0 such that

.0 000 ivuzfzz

But since 000000 ivuzfvvandivuzfuu

We have 00 ,, vyxvanduyxu

Wherever .0 202202 yyxx

0000

00

,,,,

,lim,lim

yxyxyxyx

vyxvanduyxu

Assume that

000000

,,,,

,lim,lim

yxyxyxyx

vyxvanduyxu

Let 0 be given. Then 0, 21 such that

,2

,0 012

0

2

0 uyxuyyxx and

,2

,0 022

0

2

0 vyxvyyxx

Take 21 ,min . Then 22 20 vvuu

Wherever .0 202

0 yyxx

00 zz

0000 vviuuivuzf

2200

vvuu

0

00 .lim

zz

ivuzf

Example: 6. Evaluate

iz

ixyyxyx

1

52lim 232

Solution: Since 115,,2, 00232 yxandxyyxyxvyxyxu

1,1,1,1,

615,lim2,

yxyx

yxvandyxu

11

6252lim 232

z

iixyyxyx

Theorem: If

00

21 limlim

zzzz

wzandwzf

then

(a)

0

21lim

zz

wwzf

. . . . Sum Rule

(b)

0

21lim

zz

wwzf

. . . . Difference Rule

Proof: Let 0 be given. Then 0, 21 such that

,2

0 110 wzfzz

,2

0 220 wzfzz

Choose .,min 21

2

0 20 wzfzz

21 wwzzf

21 wzwzf

22

0

21lim

zz

wwzzf

The proof of (b) is similar to (a)

Theorem: If

00

21 limlim

zzzz

wzandwzf

then

0

21lim

zz

wwzzf

. . . . Product Rule

Proof: Assume that yxityxszyxivyxuzf ,,,,,

and 000 iyxz . Then 002001 itswandivuw

Where

000000000

,,,,,,

,lim,,lim,,lim

yxyxyxyxyxyx

yxssyxvvyxuu

and to

00 ,,,lim

yxyx

yxt

Since ,vtusivtuszzf by product rule for limit of two variable functions

we have

000000000000

,,

,lim,lim

yxyxyxyx

vstuyxvsutandtvsuyxvtus

0

00000000lim

zz

svtuitvsuzzf

210000 . wwtisivu

Lemma: If

0

2 ,0lim

zz

wz

then 0 such that

.2

02

0

wzzz

Proof: Since

0

2 0,lim

zz

wz

such that

.2

02

20

wwzzz

Writing ,22 zzww we have

02

22 02

zzifzw

zzww

.2

02

0

wzzz

Theorem: If

0

,0lim

zz

wz

then

0

11lim

zz

wz

Proof: Since

0

1 0,0lim

zz

wz

such that

02

10

2

10 forwwzzz

by the lemma above 02 such that

2

0 20w

zzz

Taking ,,min 21 we obtain

ww

w

wz

wz

wzzz

2

21110

2

0

wz

11lim

Theorem: (Quotient Rule). If

00

21 0limlim

zzzz

wzandwzf

Then

0

2

1lim

zz

w

w

z

zf

Proof:

0000

1lim.lim

1.limlim

zzzzzzzz

zzf

zzf

z

zf

2

1

2

1

1.

w

w

ww

Remark: By induction it follows that

1.

0

2121 ,.......lim

zz

andwwwzfzfzf nn

2.

0

2121 ,.........lim

zz

wwwzfzfzf nn

Provided that

0

lim

zz

wzf ii

Exercise: Evaluate the following limits.

(a)

iz

izzi

2

105lim 24

(b)

iz

iziziz

31

3.53lim 22

(c)

21432

lim

zi

izz

Limit of a Polynomial Function

If ,0,... 1 nbn

n aazazazP is a polynomial of degree n, then

0

00lim

zz

zzPzP

Limit of Rational Function

If

0

00 lim,0

zz

zRzRthenzQandzQ

zPzR

2.3. CONTINUITY

Definition: Let zf be a function on .D Then zf is said to be continuous at *

0 Dz (interior of D ) iff 00 such that

.00 zfzfzz

Restatement: f is said to be continuous at *0 Dz iff the following conditions must be

satisfied.

(i) 0zf is defined ;

(ii)

0

lim

zz

zf

exists; and

(iii)

0

0lim

zz

zfzf

Example: 1. Let,

iif

izifz

iz

zf

21

12 show that f is continuous at iz 0

Solution: (i) 2

1if It is defined at 0z

(ii)

iziziz

i

iizz

izzf

22

11lim

1limlim

2

(iii)

iz

izf

2lim

f is continuous at .0 iz

Definition: 1. A function f is said to be continuous on a set S if it is continuous at

every point S .

2. f is said to be continuous if ti is continuous on its domain

Theorem: Suppose that

(i) ,,, yxviyxuzf

(ii) zf is defined at every point of D ,

(iii) biaz 0 is an interior point of D .

Then f is continuous at

bayxbauyxuiffz

,,

,,lim0

and

bayx

bavyxv

,,

.,,lim

Example:

1. 432 3 xxyiyxzf is continuous because the polynomials 32 yx and 43 xxy are continuous

2. yxeixyz 2

cos is continuous because xycos and yxe 2 are continuous

every where.

Theorem: If zf and z are continuous at 0z , then

1. zzf is continuous at .0z

2. zzf . is continuous at .0z

3. zzf

is continuous at 0z provided .00 z

4. zf is continuous at 0z provided zf is continuous at 0z

Proof: The proof of this theorem is identical with the proof of the corresponding

theorem for real functions. It is left as exercise.

Example:

1. Every polynomial functions is continuous on .

2. Every rational function is continuous on its domain.

3. yiyezf x sincos is continuous on the entire complex plane.

4. xhyixhyxeeixeezfyyyy

sinsincoscossin2

cos2

is continuous

on the entire plane.

5. ,ln irzf where ,cisrz is continuous on the entire plane except .0z

Exercise:

Show that, the function ,,sincoscos.sin iyxzwhereyxiyhxzf is continuous

on the entire plane.

MaEd 461 Worksheet II

1. Decompose each of the given functions first in the form ),(),( yxivyxu and then in

the form ,, rivru

a) zzzf 32 c)

z

izizf Im

b) z

zzf

1

1 d) 3zzf

2. Describe the image of the rectangle with vertices 0,2,1 i and 12 i under the

transformation

a) zw Re c) izw

c) zw Im d) zizw 3. Using the definition of limit show that

a)

iz

iyi

3

8225lim 2

b) iz

ixyyiyx

2

822lim 22

c) iz

izz

122lim 2

d)

0

0lim2

z

z

x

4. Evaluate the following limits.

a)

iz

izzi

2

103lim 24

d)

0

3lim

2

23

z

z

zz

b)

2

1

1

432lim

2

z

zi

izz

e)

iz

zz

z

4

1lim

4

1

c)

iz

z

z

1

1lim

6

2

f)

iz

iziz

11

lim

5. Show that the following limits does not exist.

a)

0

2lim

2

2

22

z

x

yi

yx

xy

b)

0

lim24

2

z

iyx

yx

2.4. DERIVATIVES

DEFINITION: Let zfw be a complex function with domain D and 0z is in the

interior of . If zzz 0 be any other point in D , then f is said to be differentiable at

0z if

00

00

0

0

zzzz

z

zzzflimor

zz

zfzflim

exists. If the limit exists then we

denote it by ,01 zf that is,

00

00

0

00

1 limlim

zzzz

z

zfzzf

zz

zfzfzf

Notation: 0001 zz

dz

dforzz

dz

dforzf

Example 1. Let ,kzf where k is a complex constant. Find 01 zf for 0z .

Solution: Let 0z . Then

000

00

0 00limlimlim

zzzzzz

zz

kk

zz

zfzf

zzf 01

Example 2. Let 2zz . Show that .2 0001 zzz

Solution:

00

0

2

02

0

00

1 limlim

zzzz

zz

zz

zz

zzz

00

00

0

00 2limlim

zzzz

zzzzz

zzxz

Example: Given ,nzzf where n is a non-negative integer.

Show that 1001

nznzf for all .0 z

Solution:

0

0

00

1 lim

zz

zz

zzzf

nn

0

0

1

0

2

0

2

03

021

0 .....lim

zz

zz

zzzzzzzzzznnnnn

0

1

1

0lim

zz

zzn

k

kkn

0

10

1

1

0 1limlim

zz

nzzzn

k

n

k

kkn

1

0

nzn

Example: If ,1z

zf then show that .01 020

01 z

zzf

Solution: Let 0z and 00 z . Then

00

0

0

0

00

1 lim

11

lim

zzzz

zzzz

zz

zz

zzzf

0

2

0000

111lim

zz

zzzzz

Example: If ,2zzf show that 001 f .

Solution:

000

22

1 .0lim.

lim0

0lim0

zzzzzz

zz

zz

z

zf

Example: If ,iz

izzf

find .11 if

Solution: iz

iz

i

iz

iz

if

1

11

21

lim11

iz

iziz

iziiz

1

1

21lim

iziz

iziz

iiz

iziz

iiiz

11

1

222lim

1

22lim

iziz

iiziziz

izi

11

22

lim1

12lim

Example: If ,zzf then show that f is not different on the entire complex plane.

Solution: Consider the limit for 0z

0

limlim

0

00

0

0

zzz

z

zfzzf

zz

zfzf

00

limlim 00

zz

z

z

z

zzz

Since

0

lim

z

z

zdoes not exist,

0

lim

z

z

z does not exist. Therefore,

f is not differentiable at .0z

Example: Show that 2zzf is not differentiable at .00 z

Solution:

000

00

0

00

0

0 limlimlim

zzzzzz

z

zfzzf

zz

zzzz

z

zfzf

00

limlim 000000

zz

zzz

zz

z

zzzzzz

Along the real axis, 0000lim zzzzz

zz

Along the imaginary axis, ,0x

0

lim 0000

z

zzzzz

zz

and .000000 zzzzz Hence f is not differentiable at

.00 z

Theorem: If f is differentiable at 0z then f is continuous at 0z .

Proof: Suppose

0

0

0

00lim

zz

zzzz

zfzfzfzf

00

0

0

0 lim.lim

zzzz

zzzz

zfzf

00.01 zf

0

0lim

zz

zfzf

Therefore f is continuous at 0z

Remark: The converse of the above Theorem is not necessarily true. For instance

2zf is continuous on but the derivative of f exists at 0z only.

Note: The definition of the derivative zf 1 is identical in form to that of the derivative

of real valued function of one variable. But the significance difference is that the limit

involved in zf 1 is a two dimensional one. Many results from calculus of real variable

don't carry over to the calculus of complex variable. For example, the function 2xxf

has the derivative xx2 but the derivative of 2xxf exists only at 0z .

Theorem: If f and are differentiable at 0z are differentiable at 0z and k is a

complex constant, then .,,, fffkf and

f are differentiable at 0z , and their

derivatives are given by:

1. .01

0

1zfkzkf

2. 01

01

0

1zzfzf

3. 001

0 zzfzf

4. 01

0001

0

1... zzfzzfzf

5.

200

1000

1

0

1.

z

zzfzzfz

f

Provided .00 z

Theorem: (Chain Rule). If f is differentiable at 0z and if is differentiable at 0z ,

then zf is differentiable at 0z and

.. 01

01

0

1zzfzf

Exercise: Find the derivates of the following.

(a) nz

zfn

,1

is a non-negative integer

(b) iz

izizzf 6

23 23

(c) izz

ziz 21 2

2

3

(d) iz

zizzh

2

3

(e) 5

22

6

iz

z

izh

Complex Differentials: The concept of differential of a complex functions is identical

with that of the differential of real function. Let zfw be differentiable at the point z

And let zfzzfw so that

0

lim 1

z

zfz

w

zzzfw .1 where 00 zas zff 1 is called the differential w

and is denoted by zffdw 1 . In particular if ,zw we have dzdw and

zzdz .1 . Hence, dzffdw 1 .

Example: If ,21 32ziw then

.21124.213 22221 dfzizidzzizidzzfdw

2.5. CAUCHY-RIMANN EQUATIONS

Suppose that a function f has a derivative at 0z

Let iyxz 000

yxviyxuzf ,,

ibazf 01

00 zfzzff

0000 , yxuyyxxuu

0000 ,, yxvyyxxvv

Then

00

limlim

zz

ibayix

viu

z

f

00

Imlim,Relim

yy

byix

viua

yix

viu

In particular, when ,0y that is when ,xz we have

000

,2

2,,limlimRelim 00

0000

xxx

yxx

u

x

yxuyxxu

x

u

x

viua

000

,,,

limlimImlim 000000

xxx

yxx

v

x

yxvyxxv

x

v

x

viub

1...........,, 0000 byxx

vandayx

x

u

Similarly, when ,0x that is, when ,yiz we have

000

,,,

limlimRelim 000000

yyy

yxy

v

x

yxvyyxv

y

v

yi

viua

000

,,,

limlimImlim 000000

yyy

yxy

u

x

yxuyyxu

y

u

yi

viub

2...........,, 0000 byxy

uandayx

y

v

Hence, ),(),(),(),()( 000000000' yx

y

uiyx

y

vyx

x

viyx

x

uzf

If f is differentiable at 0z then

0000 ,, yxy

vyx

x

x

Cauchy Riemann Equations

0000 ,, yxy

uyx

x

v

Thus we have proved the following theorem.

Theorem 1: Let yxivyxuzf ,, . If f is differentiable at 000 yixz then the

partial derivatives y

vand

x

v

y

u

x

u

,, exist at 00 , yx and satisfy the Cauchy-Riemann

Equations. The derivative is given by

0000000001 ,,,, yx

y

uiyx

y

vyx

x

viyx

x

uzf

Example: Consider xyiyxzzf 2222 . We have seen that f is differentiable at every point Cz . Hence Cauchy Riemann Equations must be satisfied every where

xyyxvandyxyxu 2,, 22

xy

vandy

x

vy

y

ux

x

u22,2,2

,22,2y

vyy

x

vand

y

vx

x

u

and thus

ziyxyixx

vi

x

uzf 2)(2221

Remark: Theorem1 is a necessary condition for the existence of ,1 zf that is, the

Cauchy- Reiemann Equations are satisfied at 00 yx does not imply differentiability of

f at 0z .

For example

0,0

0,22

zif

zifyx

xyzf

is not differentiable at 0z but the Cauchy-

Riemann Equations are satisfied at (0,0).

Sufficient Conditions:

Conditions on vandu that ensure the existence of 01 zf are given in the following

theorem.

Theorem 2: Let 000,, yixzandyxviyxuzf .If

y

vand

y

v

x

v

x

uvu

,,,, are all continuous at 00 , yx and if the Cauchy-Riemann

Equations are satisfied at 00 , yx , then f is differentiable at 0z and the derivative is

given by

00000001 ,,, yx

y

uiyx

y

v

x

viyx

x

uzf

Proof: Since y

uand

x

u

are continuous, we have 0000 , yxuyyxuu

00000000 ,,,, yxuyyxuyyxuyyxxu

yy

ux

x

u

21

,21 yxyy

ux

x

u

where

0

00, 21

y

xas

Since the Cauchy Riemann Equations are satisfied at 00 yx , we can replace

x

uby

y

vand

x

vby

y

u

to obtain the equation in the form

viuf

yxyixx

viyix

x

u

21

Where 0000 422311 yxasiandi

z

y

z

x

x

xi

x

u

z

f

21

Since ,zyandzx we have

11

z

yand

z

x

So that

00

lim0lim 21

zz

z

y

z

x

. Therefore,

z

fz

lim0

x

vi

x

u

Hence f is differentiable at 0z and

.,,,, 0000000001 yx

y

uiyx

y

vyx

x

viyx

x

uzf

Example2. Let yeiyezf x sincos 2 . Find .1 zf

Solution: The functions ,sin,cos,cossin yex

vye

x

uye

y

vandyeu xxxx

yey

vandye

y

u xx cossin

all continuous at all points. It is easily seen

that the partial derivatives satisfy the Cauchy Riemann Equations every where. So f is

differentiable every where and

y

ui

y

v

x

vi

x

uzf

1

zfyeiye xx sincos

Example: Show that zzzf is not differentiable at any points, Cz .

Solution: yiyixyixyixzzzf 2

2,0,0,0,2,,0,

y

v

x

v

y

u

x

uyyxvyxu

(i) y

v

x

v

y

u

x

uvu

,,,,, are all continuous at every point.

(ii) y

v

x

u

20

The Cauchy Riemann equations are not satisfied at any point yx, . Hence

f is not differentiable at any point .iyxz

Example: Identify the points where 22 yixzf is differential.

Solution: 2, xyxu and 2, yyxv

xy

v

x

v

y

ux

x

u2,0,0,2

(i) y

v

x

v

y

u

x

uvu

,,,,, are all continuous

(ii) yxyxy

v

x

u

22 and

y

u

x

v

00

Hence the Cauchy Riemann Equations are satisfied iff .yx

f is differentiable at points xixz and at such points

.20211 xixx

vi

x

uixxfzf

Remark: In polar form the Cauchy Riemann equations can be written

v

rr

u 1 and

u

rr

v 1

Proof: We know that cosrx and

xyxrry

1tan,,sin 122 By Chain

Rule,

2222

.yx

yu

yx

x

r

u

x

u

x

r

r

u

x

u

r

u

r

u

sin.cos.

2222

.yx

xu

yx

y

r

u

y

u

y

r

r

u

y

u

r

u

y

u

cos.sin.

Similarly,

sin.1

cos.

v

rr

v

x

v

cos.1

sin.

v

rr

v

y

v

If follows from the Cauchy Riemann equation that

cos1

sinsin1

cos

v

rr

vu

rr

u

y

u

x

u

10sin1cos1

r

vu

r

v

rr

u

and r

u

r

uv

rr

v

y

u

x

v

cossinsin

1cos

20sin1cos1

v

rr

uu

rr

v

Multiplying (1) by cos and (2) by sin and adding yields

v

rr

uv

rr

u 10

1

Multiplying (1) by sin and (2) by cos , and adding yields

u

rr

vu

rr

v 10

1

Theorem 3: Let ,ru and ,rv have a single real-valued is some nbhd of 00r and

let

v

r

vu

r

uvu ,,,,, be continuous at 00r , and satisfy the Cauchy Riemann

equations in polar coordinate at 00r where .00 r Then 01 zf exists where

000000 sincos cisrrz and

r

vi

r

uzf 000

1 sincos

Proof: x

vi

x

uzf

0

1

sincossincos

r

v

r

vi

r

v

x

u

sincos ir

vi

r

u

Example 4: ,20,0,2

21

rcisrzzf using Theorem 3 show that zf 1

exists everywhere except along the real axis and at the origin and zf

zf2

11

Solution: 2

cos,

rru 2

sin,

rrv

2

cos2

1

rr

u

2sin

2

1

rr

v

2

sin2

1

r

u

2cos

2

1

r

v

(i)

v

r

vu

r

uvu ,,,,, are continuous if 0r

(ii) r

u

rr

v

r

2cos

2

1

2cos

2

11

r

v

rr

v

r

2sin

2

1

2sin

2

1.

2

11

Therefore, the Cauchy Riemann equations are satisfied if .0r

r

vi

r

uizf sincos1

2sin

2

1

2cos

2

1sincos

ri

ri

2cossin

2sincos

2sinsin

2coscos

2

1

i

r

2sin

2cos

2

1

i

r

2sin

2cos

2

1

2sin

2cos

2

1 i

ri

r

zfcisr

2

1

22

1

2.6. ANALYTIC FUNCTIONS

DEFINITION: A complex function zf is said to be analytic at Cz 0 iff f is

differentiable at every point of some nbhd of 0z . A function f is said to be analytic on

a set D if it analytic at every point of D . A function f is said to be an entire function

iff it is analytic on the entire complex plane.

Remark: if f analytic at 0z , then f is differentiable at 0z but not viceversa.

Example 1 The function 2zzf is differentiable at 0z but not analytic at 0z .

Note: .Re MonogeniccHolomorphigularAnalytic

Definition: if a function is analytic at some point in every nbhd of a point 0z except at

0z itself, then 0z is called a singular point, or a singularity, of the function.

Example 2: If z

zf1

, then 012

1

zz

zf . Thus f is analytic at every point

except the point 0z , where it is not continuous, so that 01f cannot exist.

The point If z

zf1

, then 012

1

zz

zf . Thus f is analytic at every

point except the point 0z is a singular point

Example3. Since the function 2zzf is nowhere analytic, it has no singular points at

all points in C.

Example 4: .,,0/13

iizz

zzf

i ,0 and i are singularities of .f

Theorem: Suppose that f and are analytic on S .

1. f is analytic on S .

2. .f is analytic on S .

3.

f is analytic S , provided .0 Sz

4. zf is analytic S , if f is analytic at every z for all z in S

Theorem: Given .,, yxivyxuzf Suppose that

(i) the functions ,,,,,,y

v

x

v

y

u

x

uvu

are continuous in some Nnbhd of

00 , yx and

(ii) the Cauchy Riemann equation, y

v

x

u

and

y

u

x

v

are satisfied at every

point of .N

Then f is analytic at 000 yixz

Theorem: Suppose that the function ivuzf is analytic at 0z then y

v

x

u

and

y

u

x

v

at every point of some neighborhood of the point 0z .

2.7. HARMONIC FUNCTIONS

Analytic functions possess the following property which will be established in the

nextchapter (Chapter 4).

f is analytic at 0z

1f is analytic at 0z

11f is analytic at 0z

.

.

nf is analytic at 0z

.

.

Let ...,,,,,, 4111111 ffffyxivyxzf are also analytic at 0z . Since

differentiability implies continuity it follows that ...,.,, 111111 fff are continuous at 0z .

We know that the derivatives of complex functions are given interms of the partial

derivatives of their component functions. Then since ...,.,, 111111 fff are continuous at

0z , it follows that the partial derivatives of all orders are continuous at 00 , yx . In

particular, the second order mixed partial derivatives are continues and hence equal, i.e

yx

v

xy

vand

yx

u

xy

u

2222

But f is analytic at 0z ,

y

u

x

vand

y

v

x

u

Which upon differentiation yield

(i) yx

v

x

u

2

2

2

(iii) 2

22

y

v

xy

u

(ii) xy

v

y

u

2

2

2

(iv) 2

22

x

v

yx

u

Adding (i) and (ii), we obtain

. . . .(*)

Adding (iii) and (iv), we obtain

. . . .(**)

Either one of the two equations in (*) and (**) is called LAPLACE'S Equation.

Definition: any functions yxu , satisfies Laplace's equations throughout some nbhd

Of 0z is said to be Harmonic at 0z , provided that it has continuous second order partials.

If yxviyxuzf ,, is analytic at 0z , then we have shown that u and v are

harmonic. Such pairs of Harmonic functions are called conjugate Harmonic functions.

Example 1: Let 22, yxyxu

(a) Show that u is Harmonic

(b) Determine the conjugate Harmonic

Solution: (a) 22, yxyxu

2,2,2,22

2

2

2

y

uy

y

u

x

ux

x

u

0222

2

2

2

y

u

x

u

u is a harmonic function

02

2

2

2

y

u

x

u

02

2

2

2

y

v

x

v

(b) To find the conjugate Harmonic, let

yxviyxuzf ,, be analytic

The Cauchy Riemann equations hold.

(i) xxyyxvxy

v

y

v

x

u

2,2 where x is an arbitrary

functions of x .

(ii) yyxyy

u

x

v222 1

01 x

,cx c a concstant

Hence cxyyxv 2, is the conjugate Harmonic, that is,

cxyiyxzf 222 is analytic function.

Example 2: Let xyeyxv x cos,

(a) Show that v is Harmonic

(b) Find u such that ivuzf is analytic.

Solution: a)

yey

v x cos2

2

0coscos2

2

2

2

yeye

y

v

x

v xx

v is a Harmonic function

b) Let ivuzf be analytic.

yyeuyex

u

y

v

x

u xx

sinsin

yyeyey

u

x

v xx 1cos1cos2

2

2

2

,11 Cyyy where c is a complex constant.

,sin, cyyeyxu x and thus

xyeicyyezf xx cossin

Note: The Laplace's Equations furnishes a necessary condition for a function to be the

real (imaginary part of analytic function).

Example 3: Show that 22 ,, yxyxu cannot be the real part of an analytic function

Solution: Since ,04222

2

2

2

y

u

x

uwe have no analytic function such tat u the

real part of it.

Remark: The real and imaginary prts of an analytic function of a complex variable

when expressed in polary form satisfy the equations

011

2

2

22

2

u

rr

u

rr

u

and 011

2

2

22

2

v

rr

v

rr

v Laplace's Equations in Polar Form

Proof: From Cauchy Riemann equations in polar form we have

2

2

r

vr

r

vv

and

2

22 1

u

rr

v

To eliminate v differentiate (*) with respect to r and (**) with respect to

2

22

r

vr

r

v

r

v

and

2

22 1

vr

rr

v

Assuming the second partial derivatives are continuous we haver

v

r

v

22

. Hence

2

2

2

2 1

u

rr

ur

r

u

011

2

2

22

2

u

rr

u

rr

u

To eliminate ,u differentiate (*) with respect to and (**) with respect to r we obtain:

r

u

r

u

rr

vand

r

ur

v 2

22

22

2

2 11

22

2

2

2

2

2

22

2

2

2 2

2

22

2

2

2

r

ur

u

r

vrand

r

ur

v

r

ur

r

vr

2

Since ,22

r

u

r

u we obtain

r

vr

r

vr

v

2

22

2

2

011

2

2

22

2

v

rr

v

rr

v

Example 4: In the domain ,20,0 r show that, the function, rru ln, is

Harmonic and find its conjugate Harmonic.

Solution:

2

2

22

2

,1

,0,1

u

rr

uu

rr

u

011

.111

222

2

2

2

rrrr

u

r

u

rr

u

and hence ,ru is Harmonic

function. To find its conjugate Harmonic, let viuzf be analytic.

rvr

rr

ur

v

1

1.

crr

ru

rr

v

0.

11 1

,, crv where c is a complex constant .Therefore,

cirxf ln is analytic.

CHAPTER 3: ELEMENTARY FUNCTIONS This chapter includes: elementary functions such as exponential functions, logarithmic functions, trigonometric functions and hyperbolic functions of a complex variable. The primarily objective of the chapter is to introduce elementary functions of complex variables and to study its analyticity.

Now we begin the chapter by recalling that the power series representation of xe where x is a real variable.

xe x 1 !2

2x...

!3

3

x

=

0 !n

n

n

x

....!5!3

sin53

xx

xx =

0

12

)!12(

)1(

n

nn

n

x

....!4!2

1cos42

xx

x =

0

2

)!2(

)1(

n

nn

n

x

Activity: in the above expressions replacing x by iy and using properties of the imaginary unit i a) find eiy. b) show that eiy = cosy + isiny The formula eiy = cosy + isiny is called Eulers Formula. Exponential Function

In this section we want to extend the notion of real exponential functions to complex exponential functions. Let z = x + iy be a complex number then we try to extend the exponential rules for real to complex and put,

ez = eiyxiyx ee = ex( cosy + isiny)

Definition:- Given a complex number z = x + iy we define the exponential function ez by the equation ez = ex(cosy + isiny).

If z is given in polar form as z = r( cos sini ) then using Eulers formula z can be written as,

z = rie and its conjugate z is given by,

z = ireirir )sin(cos)sin(cos .

Exercises

1. Given complex numbers z = )sin(cos ir and w = )sin(cos i show that

a) the product zw = )( ier

b) the quotient z w =

)(

ie

r

(w 0)

2. For any given complex number z and w show that

a) wzwz eee

b) wzwz eee

c) nznz ee )( for any integer n.

d) xz ee and arg(z) = y + 2k for some integer k when z = x + iy.

3. Show that the exponential function of a complex variable is periodic with period 2i .

4. Show that )(zz ee

5. Show that the exponential function is an entire function. 6. Find z such that ez = i.

Trigonometric Functions From Eulers formula, we have eiy = cosy + isiny (1) and e-iy = cosy isiny (2) Adding equation (1) and (2) and simplifying we get

2

cosiyiy ee

y

(3)

Subtracting equation (2) from equation (1) and simplifying we get

i

eey

iyiy

2sin

(4)

Based on equations (3) and (4) we have the following definition. Definition:- Given a complex variable z, the cosine of z denoted by zcos and the sine of z are defined as

2

cosiziz ee

z

i

eez

iziz

2sin

The other trigonometric functions are defined in terms of the sine and cosine functions.

z

zz

cos

sintan

z

zz

sin

coscot

zz

cos

1sec

zz

sin

1csc

Exercises 1. For any complex numbers z and w show that a) sin(-z) = -sinz b) cos(-z) = cosz c) sin2z + cos2z =1 d) cos(z+w) = coszcosw sinzsinw e) cos(z-w) = coszcosw +sinzsinw f) sin(z+w) = coszsinw +sinzcosw g) sin(z-w) = coszsinw sinzcosw

h) zsin = zsin

i) zcos = zcos 2. The function sinz is an entire function. 3. The function cosz is an entire function.

Hyperbolic Functions We know that the sine hyperbolic and cosine hyperbolic functions of real variable x denoted by sinhx and coshx respectively are defined in terms of exponential function ex as

2sinh

xx eex

and 2cosh

xx eex

In a similar way we define hyperbolic functions of complex variable z as

2sinh

zz eez

and 2cosh

zz eez

The other hyperbolic functions are defined in terms of sinhz and coshz as

z

zz

cosh

sinhtanh

z

zz

sinh

coshcoth

zhz

cosh

1sec

zhz

sinh

1csc

Exercises 1. For any complex numbers z and w show that a) sinh(-z) = -sinhz b) cosh(-z) = coshz c) cos2hz - sin2hz =1 d) cosh(z+w) = coshzcoshw +sinhzsinhw e) cosh(z-w) = coshzcoshw -sinhzsinhw f) sinh(z+w) = coshzsinhw +sinhzcoshw g) sinh(z-w) = coshzsinhw sinhzcoshw

h) zsinh = zsinh

i) zcosh = zcosh j) sin(iz) = isinhz k) cosh(iz) = cosz

l) cos(iz) = coshz 3. Given z = x + iy for any real numbers x and y show that

a) sinhz = sinhxcosy + icoshxsiny b) coshz = coshxcosy + isinhxsiny c) sinz = sinxcoshy + icosxsinhy d) cosz = cosxcoshy - isinxsinhy

e) yxz222 sinsinhsinh

f) yxz222 cossinhcosh

4. The function sinhz is an entire function. 5. The function coshz is an entire function. 6. Find z such that a) ez = i b) sinz = 0 c) cosz = 0 d) tanz = 0 e) sinz = sinw f) sinhz = 0 g) coshz = 0

Logarithmic Functions Consider w = ez , to each value of z corresponds a value of w. We shall now consider the inverse process and examine the existence of z such that ez has a given value w. Activities 1. Is there any complex number z such that ez = 0? 2. If there is a complex number z such that ez = w, then is z unique?

Let us try to solve for z, by putting ez = w where z = x + iy

weiyx

wyiyex )sin(cos

wex and y = arg(w)

Thus if 0w that is if w 0 , then x = wlog is a single value of x and stands for the

real logarithm of a the positive real number w . But y = arg(w) is infinite valued.

Hence z = x + iy = wlog + iarg(w), where arg(w) is infinite valued. We are now ready to define logarithmic functions of a complex variable z.

Notation:- Instead of log z we write the natural logarithm zln and use zlog for the

complex variable z. (Mostly instead of z we write r)

Definition:- Given a complex variable z = zie we define the logarithm of z by the

equation

zlog = )log(iez = iz ln ----------- (1)

This definition is the natural one in the sense that it is written by formally using properties of real logarithms.

Corresponding to the particular argument of z such that - < , we may write

z = )2( niez for any integer n.

Thus equation (1) can be written as

zlog = )2(ln niz where n is an integer. This shows that the function logz is multiple-valued with infinitely many values.

Now if n = 0 and - < we call the principal value of logz denoted by plogz and is given by,

plogz = iz ln .

Example:- Find the logarithms of the following complex numbers.

1. z = 1+i 2. z = i 3. z = 2 4. z = -1 5. z = -ei

Solutions:- logz = ln )2( niz where - <

1. 211122 iz and 4

)1

1(tan 1

Hence logz = ln 2)2

4(

ni =

)24

(2ln2

1

ni

and thus plogz = 42ln

2

1 i

which is obtained when n = 0

2. 1110022 iz and 2

Hence logz = ln1)2

2(

ni =

)22

(

ni

and thus plogz = 2

i

which is obtained when n = 0

3. 24020222 iz and 0

Hence logz = ln2 )20( ni = )2(2ln ni and thus plogz = 2ln which is obtained when n = 0

4. 110)1(0122 iz and

Hence logz = ln1 )2( ni = )2( ni and thus plogz = i which is obtained when n = 0

5. eeeeiz 222 )(00 and 2

Hence logz = lne)2

2(

ni

= )2

2(1

ni

and thus plogz = 21

i

which is obtained when n = 0

Branches of zlog

Consider the principal logarithm plogz at z = xo + 0i on the negative real axis (xo < 0).

At that point xoz and . Since ii as xoz from the second quadrant, and ii as xoz from the third quadrant, plogz is not continuous on the negative real axis. Thus, its derivative does not exist at each point on the negative real axis.

The single valued-function Plogz = iz ln , is defined at all points iezz except at the origin z = 0, and points on the negative real axis.

Plogz = iz ln , has continuous components

u = lnr and v = where rz > 0, throughout its domain of definition. Moreover, the derivatives

rr

u 1

, 0

u

and 0

r

v

, 1

v

are all continuous function of the point z in that domain, and satisfy Cauchy-Riemann conditions in polar coordinates. Thus plogz is differentiable and

))(sin(cos)log(

r

vi

r

uizp

dz

d

= )0

1( ir

e i

= ire

1

= z

1

Therefore, zzp

dz

d 1)log(

, ))arg(,0( zz Activity

1. Using the complex plane identify the region in which plogz is analytic and regions in which it is not analytic.

2. Consider irz lnlog where r = z > 0 and

2

44

a) Is irz lnlog single-valued or not?

b) Find )(log z

dz

d

c) Is logz analytic in the domain r = z > 0 and

2

44

?

3. Replace 4

by o and consider irz lnlog where r = z > 0 and 2 oo

d) Is irz lnlog single-valued or not?

e) Find )(log z

dz

d

f) Is logz analytic in the domain r = z > 0 and 2 oo ? The principal values of logz represented by plogz is said to be the principal branch of

logz. Each point of the negative real axis and the origin are singular points of

plogz. The ray is the branch cut for the plogz. The ray o is the branch cut for

irz lnlog where r = z > 0 and 2 oo . The singular point z = 0 is common to all branch cuts for logz is called a branch point. Exercise: Show that

1. zez log

2. If w = ez and z = x+iy, then logw = z + k2 for some integer k. 3. For complex numbers z and w in the domain of definition of log a) log(zw) = logz + logw.

b) wz

w

zloglog)log(

c) zmzm log)log(

d) .log

1)log(

1

zn

z n

e) z

n

mz n

m

log)log(

f) nzm

n

m

ez)(log

Complex Exponents

If c is a complex constant, cz and cz are defined as follows:

cz = zce log and cz =

cz

1

for z 0

Since logz is multiple-valued, cz is also multiple-valued. The same idea can be used for the principal of power function.

The principal value of cz is zce log where plogz is the principal logarithm of logz.

Hence cz = zce log is single-valued for

irez where r = 0z and .

Example:-

k

iki

iii eeei 42)4(

2log22

for some integer k.

The principal value of ii 2 is

e . Exercise

1. Find the domain in which cz is a single-valued

2. Find the domain in which cz is analytic.

3. In the domain where cz is analytic show that

1)( cc czzdz

d

4. Use exercise number (3) to find the derivatives of the following:

a) f(z) = 32

z

b) f(z) = iz 2

c) f(z) = iz 43

d) f(z) = zz 5. If c is a nonzero complex constant, then the exponential function with base c is

given by the equation,

czz ec log .

a) Show that the function f(z) = zc is an entire function.

b) Find)( zi

dz

d

.

c) Find))1(( zi

dz

d

.

Inverse Of Trigonometric And Hyperbolic Functions A complex function g(z) is said to be the inverse of a function f(z), if f(g(z)) = z = g(f(z)).

We denote the inverse of f(z) by ).(1 zf

Inverses of Trigonometric and Hyperbolic functions can be described in terms of logarithms.

Suppose w = )(sin1 z sinw = z

i

eez

iwiw

2

iwiw eeiz 2

iwiw eize 22

0122 iwiw izee

Solving and simplifying, we get

)1log()(sin 21 ziziz ,

which is a multiple-valued function.

Similarly, )1log()(cos 21 zziz

)log(2

)(tan 1zi

ziiz

Exercise: 1. Find an expression for the following:

a) )(sinh 1 z

b) )(cosh 1 z

c) )(tanh 1 z

2. Find the derivatives for the following:

a) )(sin 1 z

b) )(cos 1 z

c) )(tan 1 z

d) )(sinh 1 z

e) )(cosh 1 z

f) )(tanh 1 z

CHAPTER 4: THE INTEGRAL

Introduction: This chapter will deal with the notion of integrability and integral of a complex function along an oriented curve in the complex plane. We shall see that a function is integrable along a curve if the curve is piecewise smooth and the function is continuous along the curve. Some elementary properties of integration will also be obtained. The following section of the chapter deals with the notion of oriented and piecewise smooth curves. In order to introduce the line integral of f(z) in a simple way, first it is better to define the definite integral of a complex-valued function f of a real variable t. We write,

f(t) = u(t) + iv(t) for bta where u and v are real-valued piecewise continuous, functions of t on a bounded interval (a, b). Here u is the real part of f and v is the imaginary part of f.

Example: f(t) = ite is a complex-valued function defined on every interval. The real part

of f is cost and the imaginary part of is sint. In particular for 20 t , f(t) = ite is the

unit circle on the complex plane.

4.1. Complex valued Functions of a Real Variable

DEFINITION: A function f of the real variable t is said to be complex valued on an

interval, b