lecture 23 25

BITS Pilani Pilani Campus

MATH F112 (Mathematics-II)

Complex Analysis

BITS Pilani Pilani Campus

Lecture 23-24 Complex Functions

Dr Trilok Mathur, Assistant Professor, Department of Mathematics

BITS Pilani, Pilani Campus

• Function of a complex variable Let S be a set complex numbers. A function f defined

on S is a rule that assigns to each z in S a complex number w.

3

S

Complex numbers

f

S’

Complex numbers

z

The domain of definition of f The range of f

w


Suppose that w = u + iv is the value of a function f at z = x + iy, so that

Thus each of real number u and v depends on the real variables x and y, meaning that

Similarly if the polar coordinates r and θ, instead of x and y, are used, we get

4

( )u iv f x iy

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

( ) ( , ) ( , )f z u r iv r


• Properties of a real-valued function of a real variable are often exhibited by the graph of the function. But when w = f(z), where z and w are complex, no such convenient graphical representation is available because each of the numbers z and w is located in a plane rather than a line.

• We can display some information about the function by indicating pairs of corresponding points z = (x, y) and w = (u, v). To do this, it is usually easiest to draw the z and w planes separately.


x u

y v

z-plane w-plane

domain of definition

range

w = f(z)


• Example: If f (z) = z2, then

case #1: case #2:

7

2 2 2( ) ( ) 2f z x iy x y i xy

2 2( , ) ; ( , ) 2u x y x y v x y xy

z x iy

iz re 2 2 2 2 2( ) ( ) cos2 sin 2i if z re r e r ir

2 2( , ) cos2 ; ( , ) sin 2u r r v r r

When v = 0, f is a real-valued function.


• Example: A real-valued function is used to illustrate some

important concepts later in this chapter is • Polynomial function:

where n is zero or a positive integer and a0, a1, …an

are complex constants, an is not 0;

• Rational function: the quotients P(z)/Q(z) of polynomials

8

2 2 2( ) | | 0f z z x y i

20 1 2( ) ... n

nP z a a z a z a z

The domain of definition is the entire z-plane

The domain of definition is Q(z)≠0


9

Suppose the function f is defined in some neighborhood of z0, except possibly at z0 itself. Then f is said to possess a limit at z0, written

0)(lim0

wzfzz

δz-z-wzf

δ,

00 0)(

00

whenever

that such a given for if


10

meaning the point w = f (z) can be made arbitrarily chose to w0 if we choose the point z close enough to z0 but distinct from it.


,,

),,(),()(

000000 ivuwiyxz

yxivyxuzf

Let Thm 1

0)(),(

0)(),(

0

)(lim)(

)(lim)(

)(lim

0,0

0,0

0

vx,yvii

ux,yui

wzf

yxyx

yxyx

zz

Then


Then .

, Let

0

0

0

0

lim

)(lim

W F(z)

wzf

zz

zz

Thm 2

.)]()(lim 000

WwzFzfzz

[ (i)


.W)]()([ lim)( 000

wzFzfiizz

.0 Wif ,W)(

)( lim)( 0

0

0

0

w

zF

zfiii

zz

BITS Pilani, Pilani Campus 14

• Example Show that in the open disk | z | < 1, then

Proof:

( ) / 2f z iz

1lim ( )

2z

if z

| || 1| | 1|| ( ) | | |

2 2 2 2 2

i iz i i z zf z

0, 2 , . .s t

when 0 | 1| ( 2 )z

| 1|0 | ( ) |

2 2

z if z


15

• Example If then the limit does not exist.

( ,0)z x

( )z

f zz

0

lim ( )z

f z

0

0lim 1

0x

x i

x i

(0, )z y0

0lim 1

0y

iy

iy

≠


The point at infinity is denoted by

, and the complex plane together

with the point at infinity is called

the Extended complex Plane.


Riemann Sphere & Stereographic Projection

N: The North pole


18

• The -Neighborhood of Infinity

O x

y

R1

R2

When the radius R is large enough

i.e. for each small positive number

The region of |z| > R = is called the -Neighborhood of Infinity(∞)

0

1R

1


.0)(

1lim)(lim.1

00

zf

zfzzzz

00

0

1lim)(lim.2 w

zfwzf

zz

01

1lim)(lim.3

0

zf

zfzz


δz-zzf

ei

z-z zf

a

zfzz

0

0

00)(

1..

01

)(

0

,0)(lim)10

whenever

whenever

that such

eachfor


.0)(

1lim)(lim

00

zf

zfzzzz

Thus,


22

A function f is continuous at a point z0 if

meaning that 1. the function f has a limit at point z0 and 2. the limit is equal to the value of f (z0)

00lim ( ) ( )

z zf z f z

For a given positive number ε, there exists a positive number δ, s.t.

0| |z z When ⇒ 0| ( ) ( ) |f z f z

00 | | ?z z


23

The function f(z) is said to be

continuous in a region R if it is

continuous at all points of the

region R.


24

Theorem 1. A composition of two continuous functions is itself

continuous. Theorem 2. If f (z) = u(x, y) + iv(x, y), then f (z) is continuous iff Re(f (z)) = u(x, y) and Im(f (z)) = v(x, y)

are continuous functions.


25

.continuousallare

(c)

(b)

(a)

then ,continuous are and If 3.Theorem

0)(,)(

)(

)()(

)()(

)()(

zgzg

zf

zgzf

zgzf

zgzf


26

Theorem 4. If a function f (z) is continuous and nonzero at a point z0, then f (z) ≠ 0 throughout some neighborhood of that point.

Proof 0

0lim ( ) ( ) 0z z

f z f z

0| ( ) |0, 0, . .

2

f zs t

0| |z z When 0

0

| ( ) || ( ) ( ) |

2

f zf z f z

f(z0) f(z)

If f (z) = 0, then 00

| ( ) || ( ) |

2

f zf z

Contradiction!

0| ( ) |f z

Why?


27

Theorem 5. If a function f is continuous throughout a region R that is both closed and bounded, there exists a nonnegative real number M such that

where equality holds for at least one such z.

| ( ) |f z M for all points z in R


Example: Discuss the continuity of f (z) at z = 0 if

zz zf ii

z

zzf i

Re

Re

1)()(

1)()(


29

Derivatives: Let f (z) be a function defined

on a set S and S contains a nbd of z0. Then

derivative of f (z) at z0, written as

is defined by the equation:

),( 0zf

exists. RHS on limit the provided

)1(,)()(

lim)(0

00

0 zz

zfzfzf

zz


exists. at derivative its if

at abledifferenti be to said is function The

0

0)(

z

zzf

z

zfzzfzf

zz-z

z

)()(lim)(

,

00

00

0 to reduces (1) then If

30


ablityDifferenti Continuity

Continuityablity Differenti :Problem

0

00

0

)()(lim)(

)(

0 zz

zfzfzf

zzf

zz

at abledifferenti is Let :Proof

)()(lim 00

zfzfzz

Now

)(

)()(lim 0

0

0

0

zzzz

zfzfzz

31


00)('

)(lim)()(

lim

0

00

0

00

zf

zzzz

zfzfzzzz

)()(lim 00

zfzfzz

0)( zzf at continuousis


Continuity Differentiability

To show this consider the function

f (z) =z2 = x2 + y2 = u(x, y)+ i v(x, y)

everywhere continuous is hence

,everywhere continuous are and Since

)(

.0),(,),( 22

zf

vu

yxvyxyxu


0

00

0

2

0

2

0

0

0

)()(

,

zz

zzzz

zz

zz

zz

zf zf

zz

have we For

0

0000 zz

zzzzzzzz

0

000 )()(

zz

zzzzzz


zzzz

zzz

00 ,.

yix

yixzz

.0

x

y

C1

C2

20

10

,

,

Czz

Czz

paththealong

paththealong


unique.notis

then if Thus,

0

0

0

)()(lim

)00(

0 zz

zfzf

,,z

zz

else. whereno

and origin the at abledifferenti is

then When

)(

.0)()(

lim

)00(

00

0

0

0

zf

zzz

zfzf

,,z

zz


)()( zfdz

d zf 1.

,0.2 cdz

d

,1.3 zdz

d


,.4 1 nn nzzdz

d

zfdz

dczcf

dz

d .5


)()()()(.6 zgzfzgzfdz

d

)()()()()()(.7 zgzfzgzfzgzfdz

d


0)(

,))((

)()()()(

)(

)(.8

2

zg

zg

zfzgzfzg

zg

zf

dz

d

if


Let F(z) = g(f (z)), and assume that f (z) is

differentiable at z0 & g is differentiable at

f (z0), then F(z) is differentiable at z0 and

)())(()( 000 zfzfgzF


dz

dw

dw

dW

dz

dWzFW

w gW zfw

rule Chainby hence

and Let Ex.

),(

)()(

. pointany at

exist not does thatshow (a)8Q

z

zfzzf )(,)(:

0

0

0

0

0

0

0

)()(

,

zz

zz

zz

zz

zz

zfzf

zz

then Let :Solution


yix

yix

z

z

z

zfzzf

)()( 00

any whereexist not does)(

1

1

)()(

0

lim

2

1

00

zf

Calong

Calong

z

zfzzf

z

x

y

C1

C2


exist. NOT does thatShow

by defined function a be Let Q.9

)0(

0,0

0,

)(

2

f

z

zz

z

zf

f


000

)(

)()()(

iyxz

zf

yx,ivyx,uzf

point a at exists that and

that Suppose

equations-CR thesatisfy they and

at exist must and

sderivative partial order-first the Then

)( 00 y,xv,v,uu yxyx


).,(),()('

)(

00000

0

yxivyxuzf

zf

xx

as writtenbe can

Also,

Proof

)1(......)()(

lim)(

)(

00

00

0

z

zfzzfzf

zzf

z

at abledifferenti is Since


),(),()(

,

00000

000

yxviyxuzf

yixz

iyxziyxz

that Note

),(

),()(

00

000

yyxxiv

yyxxuzzf


yix

yxvyyxxvi

yix

yxuyyxxu

yx

zf

),(),(

),(),(

)0,0(),(

lim

)(

0000

0000

0

givesEq. )1(


x

y

C1

C2

2

0000

1

0000

0

),,(),(

),()(

)(

C

yxvyxui

C

yxviyxu

zf

yy

xx

along

along


???WHY

),()(

),,(,

000

00

yxatviuzfand

yxatvuvu

xx

xyyx


:thatsuch

point the of oodneighbourh

some in throughout defined function

any be Let

000

)()()(

iyxz

x, yi v x,y u zf

,,,,)( 0zvvuui yxyx of nbd. that in exist

),(,,,)( 00 yxvvuuii yxyx at continuous are


).,(,

)(

00 yxvuvu

iii

xyyx atequations,-CR the

satisfy sderivative partial order first the

.)( 0zzf at exists Then


satisfy

they and( at exist sderivative

partial the Then point givenany at

abledifferenti be Let

),,,

.

)()(

00

000

θ,rvvuu

erz

θr,vir,θ uf(z)

rr

iθ

ur

vvr

u rr

1,

1

oorrri viuezf

,0 )(

and


find out the points where the function is differentiable. Also find

,z zf 2)(

function the For1: Example

)(zf

ivuxyiyx z zf 2)( 222

Consider:Solution

xyyxvyxyxu 2),(,),( 22


xvyv

yuxu

yx

yx

2,2

,2,2

xyyx vuvu &

x, y all for satisfied are equations- CR (i)

x, y

vvuu yxyx

all

for continuous are and (ii) ,,,


and

pointany at abledifferenti is z,zzf 2)(

zyixivuzf xx 222)(


find out the points where the function is differentiable. Also find

,)(2

zzf

function the For:2 Example

)(zf

222)( yxz zf Consider

0),(&),( 22 yxvyxyxu


,0,0,2,2 yxyx vvyuxu

y. x 0

have must wethen

satisfied, are equations- CR If

0)0,0()0,0()0(

)00()(

xx ivu f

,zf

Further else. whereno and

atonly abledifferenti is


by defined function

the of partsimaginary & real

the denote Let :Q.672 Page

f

u & v

0,0

0,)(

2

z

zz

zzf

. at abledifferenti NOT is although

at satisfied are equations-CR thatShow

)00(

)00(

,f

,


z

z

z

zzz

z

z-fzzf zf

zf.z zf

zz

z

00

0

limlim

)()(lim)(

)()(

have We

IMethod

exist?Does Let Q.


point.any at exist not does

paththealong

paththealong

)(

1

1

lim

2

1

)0,0(,

zf

C

C

yix

yixyx

z = x + i y

c1

c1

c2

c2


xyyx

yx

yx

vuvu

vv

uu

-yx, vu

x-iyz zf

,

1,0

,0,1

)(

Thus

II-Method


point.any at abledifferenti not is

satisfied not areequationsC.R.

)(zf


).()(

)(.)(

zfzf

zfezf z

and

find and abledifferenti is

thatShow Let 2b.Q

yevyeu

yiye

eezf

xx

x

iyxz

sin,cos

)sin(cos

)(

:Solution


.cos

,sin

,sin

,cos

yev

yev

yeu

yeu

xy

xx

xy

xx


. point

any at continuous are

(1)

Clearly

)(

,,,)2(

&

x,y

vvuu

vuvu

yxyx

xyyx


ziyx

xx

xx

eee

yeiye

ivuzf

zf

.

sincos

)(

)(

and exists


yeV

y-eU

ViU

yeiy-e

zf zF

zf

x

-x

x-x

sin

,cos

sincos

)()(

)(

(say)

Let

: find To


.cos

,sin

,sin

,cos

yeV

yeV

yeU

yeU

xy

xx

xy

xx


point

any at continuous

are

Thus,

)(

,,,)2(

&)1(

yx,

VVUU

VUVU

yxyx

xyyx


z

iyx

xx

xx

e

ee

yeiye

iVUzf zF

zF

& exists

.

)sin(cos

)()(

)(


if point a at

analytic be to said is function

0

)(

z

zfA

and at abledifferenti is ,)()( 0zzfi

.

)()(

0z

zfii

of oodneighbourh

some in abledifferenti is


. set the of point each

at abledifferenti is f if set open

an inanalytic is functionA

S

S

zf )(

y.analyticitimply not doesablity Differenti

:Remark


:Ex2

)( z zf

f (z) is differentiable at origin and no where else.

But f (z) is not analytic at the origin as it is not differentiable in any neighborhood of origin.


. in throughout constant is then

, domain a in everywhere If

:Theorem

Dzf

Dzf

)(

0)(

Entire function: A function f(z) is said to

be an Entire function if f(z) is analytic

at each point in the entire finite plane.


Singular Point: Let a function f (z) is, not analytic at a point z0, but analytic at some point in every neighbourhood of z0.

Then z0 is called a singularity of f(z).

Example: Every polynomial is an entire

function.


z

zf 1

)()1(

Examples

. ofy singularit a is )(0 zf z


2)()2( z zf

anywhereanalytic not is zf )(

point singular no has )(zf

lecture 23 25

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