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.

Complex numbers

The domain of definition of f The range of f

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

( )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.

z-plane w-plane

domain of definition

w = f(z)

• Example: If f (z) = z2, then

case #1: case #2:

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

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

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

δz-z-wzf

00 0)(

whenever

that such a given for if

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)(),(

)(lim)(

vx,yvii

ux,yui

Then .

W F(z)

.)]()(lim 000

WwzFzfzz

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

wzFzfiizz

.0 Wif ,W)(

)( lim)( 0

BITS Pilani, Pilani Campus 14

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

Proof:

( ) / 2f z iz

1lim ( )

| || 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 | ( ) |

z if z

• Example If then the limit does not exist.

( ,0)z x

lim ( )z

0lim 1

(0, )z y0

0lim 1

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

• The -Neighborhood of Infinity

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(∞)

1lim)(lim.1

zfzzzz

1lim)(lim.2 w

1lim)(lim.3

δz-zzf

z-z zf

,0)(lim)10

whenever

that such

eachfor

1lim)(lim

zfzzzz

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

The function f(z) is said to be

continuous in a region R if it is

continuous at all points of the

region R.

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.

.continuousallare

then ,continuous are and If 3.Theorem

0)(,)(

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, . .

f zs t

0| |z z When 0

| ( ) || ( ) ( ) |

f zf z f z

f(z0) f(z)

If f (z) = 0, then 00

| ( ) || ( ) |

f zf z

Contradiction!

0| ( ) |f z

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

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

zfzfzf

exists. at derivative its if

at abledifferenti be to said is function The

zfzzfzf

)()(lim)(

0 to reduces (1) then If

ablityDifferenti Continuity

Continuityablity Differenti :Problem

)()(lim)(

zfzfzf

at abledifferenti is Let :Proof

)()(lim 00

zfzfzz

)()(lim 0

zfzfzz

)(lim)()(

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

yxvyxyxu

have we For

0000 zz

zzzzzzzz

000 )()(

zzzzzz

paththealong

unique.notis

then if Thus,

)()(lim

else. whereno

and origin the at abledifferenti is

then When

.0)()(

)()( zfdz

d zf 1.

,0.2 cdz

,1.3 zdz

,.4 1 nn nzzdz

)()()()(.6 zgzfzgzfdz

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

)()()()(

zfzgzfzg

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

w gW zfw

rule Chainby hence

and Let Ex.

. pointany at

exist not does thatshow (a)8Q

zfzzf )(,)(:

then Let :Solution

)()( 00

any whereexist not does)(

Calong

exist. NOT does thatShow

by defined function a be Let Q.9

)()()(

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

).,(),()('

yxivyxuzf

as writtenbe can

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

zfzzfzf

at abledifferenti is Since

),(),()(

yxviyxuzf

iyxziyxz

that Note

yyxxiv

yyxxuzzf

yxvyyxxvi

yxuyyxxu

),(),(

)0,0(),(

givesEq. )1(

),,(),(

yxvyxui

yxviyxu

???WHY

yxatviuzfand

yxatvuvu

:thatsuch

point the of oodneighbourh

some in throughout defined function

any be Let

)()()(

x, yi v x,y u zf

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

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

00 yxvuvu

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

θ,rvvuu

θr,vir,θ uf(z)

oorrri viuezf

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

,z zf 2)(

function the For1: Example

ivuxyiyx z zf 2)( 222

Consider:Solution

xyyxvyxyxu 2),(,),( 22

xyyx vuvu &

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

vvuu yxyx

for continuous are and (ii) ,,,

pointany at abledifferenti is z,zzf 2)(

zyixivuzf xx 222)(

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

function the For:2 Example

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

Further else. whereno and

atonly abledifferenti is

by defined function

the of partsimaginary & real

the denote Let :Q.672 Page

. at abledifferenti NOT is although

at satisfied are equations-CR thatShow

z-fzzf zf

zf.z zf

limlim

)()(lim)(

have We

IMethod

exist?Does Let Q.

point.any at exist not does

paththealong

)0,0(,

z = x + i y

-yx, vu

x-iyz zf

II-Method

point.any at abledifferenti not is

satisfied not areequationsC.R.

zfezf z

find and abledifferenti is

thatShow Let 2b.Q

yevyeu

sin,cos

)sin(cos

:Solution

. point

any at continuous are

Clearly

,,,)2(

sincos

and exists

yeiy-e

sincos

: find To

any at continuous

,,,)2(

iVUzf zF

& exists

)sin(cos

if point a at

analytic be to said is function

and at abledifferenti is ,)()( 0zzfi

of oodneighbourh

some in abledifferenti is

. set the of point each

at abledifferenti is f if set open

an inanalytic is functionA

y.analyticitimply not doesablity Differenti

:Remark

)( 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

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.

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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