Unit_III

Complex Numbers:

In the system of real numbers R we can solve all quadratic equations of the

form

0,0x 2 ≠=++ acbxa , and the discriminant 042 ≥− acb . When the discriminant

042 <− acb , the solution of this quadratic equation do not belong to the

system of .

In fact , a simple quadratic equation of the form 012 =+x , does possesses

solution in real. This difficulty was overcame by introducing the imaginary

part unit i, where 12 −=i .

Thus the set of complex numbers defined as . }{ 1,:)( −=∈+= iandRyxiyxC .

Some Basic Results: 1. If z = x +iy is a complex number, then the complex number iyxz −= is

called the complex conjugate of z , and 22))((z z yxiyxiyx +=−+=

2.If z= x+iy is a complex number , then the modulus of z, denoted by 22 y z += x

3. A complex number z = x + iy is represented by a point p(x ,y) in the

Cartesian plane with abscissa x and ordinate y. Then the x-axis is called real

axis and the y-axis is called the imaginary axis.The point p(x, y) is referred

to as the point z.

Let OP= r and .θ=∠XOP Then θθ rsin y , cosr x ==

Every Complex number can z = x +iy be expressed in the form as given

below

)isinr(cosz θθ += polar form θi

rez = exponential form

We observe that 22x r y+= the modulus of z and it represents the

distance of the point z from the origin. Also

.arg tan θ1-

zofumentthecalledisanglethex

yθ

=

4. Let 000z iyx += then )()(z-z 000 yyixx −+−=

( ) ( )2

0

2

0 yyxx −+−= .

Now 0z-z may be represented as

)sin(cosz-z 0 θθ iR += and Rzz =− 0 serves as the complex equation

of the circle C with ( )00 , yx and radius R. In particular 1z = represents the

circle with center at the origin and radius equal to 1.

Functions of a complex variable:

Let C be a set complex numbers. If to each complex number z in C there

corresponds a unique complex number w ., then w is called a complex

function of z defined on C, and we write w = f(z). Hence ,w has a real part ,

say u and an imaginary part , say v. Then , w has the representation

W= f(z) = u(x,y) + i v(x,y) (Cartesian form) W = u(r,θ) + i v(r,θ) (polar form)

Continuity :

A complex valued function f(z) is said to be continuous at a point 0

z if f(z)

is defined at 0

z and 0z z

zf(z)lim0

=→

.

Note:

1. If a complex valued function f(z) is differentiable at a point 0

z , then

it is continuous at 0

z .

2. The converse of the above result is not always true. The continuity of

a complex function need not imply its differentiability.

Derivative of a complex function:

The derivative of a complex function f at a point 0z z = , denoted by ),(zf 0′

is defined as z

zz

∆

−∆+=

→∆

)f(z)f(z)(f 00

0 z0

1

lim , provided this limit exists.

Substituting havewe,z-zz0

=∆

zz

∆

−=

→∆

)f(z)f(z)(f 0

0 z0

1

lim

We should remember that by the definition of limit f(z) is defined in a

neighborhood of 0

z and z may approach 0

z from any direction in the

complex plane . The derivative of a function at a point is unique if it exists.

Analytic Functions. Cauchy-Riemann equations.

In complex analysis we are interested in the functions, which are

differentiable in some domain, called the analytic functions. A large variety

of functions of complex variables which are useful for applications purpose

are analytic.

A Function f(z) is said to be analytic at a point 0

z , if it is differentiable

at 0

z and, in addition , it is differentiable throughout some

neighborhood of 0

z .

Further a function f(z) is said to be analytic in a domain D if f(z) is defined

and differentiable at all points of D. In fact , analyticity is a ‘global ‘

property while differentiability is a ‘local’ property.

The terms regular and holomorphic are also used in place of analytic.

Cauchy-Riemann Equations : Cauchy –Riemann equations provide a criterion for the analyticity of a

complex function W = f(z) = u(x,y) + i v(x,y) .

Statement: Necessary conditions for a function to be analytic.:

If f(z) = u(x,y) + i v(x,y) is continuous in some neighborhood of a point

z= x+ iy and is differentiable at z , then the first order partial derivatives of

u(x,y) and v(x,y) exist and satisfy the Cauchy-Riemann equations

xyyx

vuvu −== and . At the point z = x + iy.

Proof:

Since f(z) is differentiable at z, we have

z

zz

∆

−∆+=

→∆

)f(z)f(z)(f lim

0 z

1

{ } { }yi

)yv(x,i(x.y)uy)yx,(x v)yy,(u )(f lim

0 z

1

∆+∆

+−∆+∆++∆+∆+=

→∆ x

ixxz

--------(I)

Let us assume z∆ to wholly real and wholly imaginary.

Case I: When z∆ wholly real, then y∆ = 0 , so that z∆ = x∆ .The limit on

the right side of equation (I) becomes,

{ }x

xxz

∆

∆+=

→∆

y)u(x. -y),(u )(f lim

0 x

1

{ }x

xx

∆

∆++

→∆

y)v(x. -y),(vlimi

0z

=

x

v i

x

u

∂

∂+

∂

∂. --------------------------------------(II)

Case II: When z∆ wholly imaginary, then x∆ = 0 , so that z∆ = yi∆ .The

limit on the right side of equation (I) becomes,

{ }y

xz

∆

∆+=

→∆ i

y)u(x. -y)y,(u )(f lim

0 y

1 { }y

x

∆

∆++

→∆ i

y)v(x. -y)y,(vlimi

0y

=

y

v

y

u

i

1

∂

∂+

∂

∂

y

ui

y

v

∂

∂−

∂

∂= ---------------------------------------(III)

Since f(z) is differentiable the value of the limits obtained from (II) and (III)

must be equal.

x

v i

x

u

∂

∂+

∂

∂

y

ui

y

v

∂

∂−

∂

∂=

Comparing the real and imaginary parts, we get

x

v

y

u

y

v

x ∂

∂−=

∂

∂

∂

∂=

∂

∂and

u at the z = (x,y).

These are known as the Cauchy-Riemann equations. Satisfaction of these

equations is necessary for differentiability and analyticity of the function f(z)

at a given point. Thus, if a function f(z) does not satisfy the Cauchy-

Riemann equations at a point, it is not differentiable and hence not analytic

at that point.

Ex 1: If w= logz, find dz

dw , and determine where w is not analytic.

Let us consider z in exponential form , )sin(cosrez i θθθir +==

( )

=+= −

x

ytan,yxr 122 θ

iy)log(xivuw +=+=

( )

++=

x

y taniyxlog

2

1 1-22

Equating real and imaginary parts

( )

=+=

x

ytanv:ylog

2

1u 1-22

x

2222 yx

y

x

v:

y

x

x

u

+−=

∂

∂

+=

∂

∂

x

2222 yx

x

y

v:

y

y

y

u

+=

∂

∂

+=

∂

∂

x

Now from- C-R equations

x

v

y

uand

y

v

x ∂

∂−=

∂

∂

∂

∂=

∂

∂u

Thus w = logz the C-R equations holds good for ( ) 0yx 22 ≠+

Further , ( ) 2222 yx

yi

yx

x

dx

dvi

x

u

dz

dw

+−

+=+

∂

∂=

dz

dw ( )( ) z

1

z

z

yx

iy-x22

==+

=z

Thus every point other than origin ( )0y xi.e. 22 ≠+ w=logz is

differentiable and the function logz is analytic every where except at origin.

Ex2: Show that the function w= sinz is analytic and find the derivative.

iy)sin(x ivu w +=+=

siniycosx cosiysinx += ------------------------(1)

Now 2

eecosxand

2i

eexsin

-ixixix −=

−=

− ix

coshx cosix :isinhxsinix ==

Using these in equation (1)

W ( ) sinhyicosx coshysinx ivu +=+=

Equating real and imaginary parts, we get

u = sinx coshy : v= cosx sinhy

ysinhsinx

v ,coshy cosx

x

ux−=

∂

∂=

∂

∂

-------------------------(2)

coshycosx y

v ,sinhy sinx

y

u=

∂

∂−=

∂

∂

The C-R equations are satisfied

x

v

y

uand

y

v

x ∂

∂−=

∂

∂

∂

∂=

∂

∂u

f(z) = sinz is analytic.

( ) =z1f

dx

dvi

x

u+

∂

∂ ( ) ysinhsinx-icoshycosx +=

yisinsinx cosiycosx −=

( ) cosz iyxcos =+=

Consequences of C-R Equations:

1). If f(z)= u + iv is an analytic function then u and v both satisfy the two

dimensional Laplace equation.

0yx 2

2

2

2

=∂

∂+

∂

∂ φφThis equation is also written as 02 =∇ φ .

Here 2∇ is the two- dimensional Laplacian.

Since f(z) is analytic we have Cauchy-Riemann equations

)(x

v

y

uand )(

y

v

xIII

u−−−

∂

∂−=

∂

∂−−−

∂

∂=

∂

∂

Differentiating (I) w,r,t. x and (II) w.r.t y partially we get

xy

v

y

uand

yx

v

x

2

2

22

2

2

∂∂

∂−=

∂

∂

∂∂

∂=

∂

∂ u

But xy

v

yx

v 22

∂∂

∂=

∂∂

∂ is always true and hence we have

ory

u

x 2

2

2

2

∂

∂−=

∂

∂ u 0

y

u

x

u2

2

2

2

=∂

∂+

∂

∂ ,this implies u is harmonic.

Similarly Differentiating (I) w.r.t . y and (II) w.r.t.y partially we get

ory

v

x

v2

2

2

2

∂

∂−=

∂

∂0

y

v

x

v2

2

2

2

=∂

∂+

∂

∂ , this implies v harmonic.

If f(z) = u + iv is an analytic function, then u and v are harmonic

functions. Here , u and v are called harmonic conjugates of each other.

Consequence II:

If f(z) = u + iv is an analytic function, then the equations 1

c),( =yxu

And 2

c),(v =yx represent orthogonal families of curves.

Soln:

( )1

cyx,u = -------------(i)

( )2

cyx,v = --------------(ii)

Differentiating eqn (i) partially w.r.t x

( )1

m

yu

xu

dx

dy0

dx

dy

y

u

x

u=

∂∂

∂∂

−==∂

∂+

∂

∂or --------- (I)

Differentiating eqn (ii) partially w.r.t. x

( )2

m

yv

xv

dx

dy0

dx

dy

y

v

x

v=

∂∂

∂∂

−==∂

∂+

∂

∂or ---------(II)

The two families are orthogonal to each other , then 1mm21

−= ,

And using C-R equations

( )( )( )( )yvyu

xvxum

21∂∂∂∂

∂∂∂∂=m =

( )( )( )( )

1−=∂∂∂∂

∂∂−∂∂

yvyu

yuyv

Hence the curves intersect orthogonally at every point of intersection.

Note: The converse of the above result is not true. The following example

reveals the property.

22

2

yxvy

xu 2: +==

2

22

1

2

cyxcy

x=+= 2:

( ) ( )( ) 122

mx

2y

yx-

y2x

yu

xu

dx

dy==−=

∂∂

∂∂

−= for curve 1

c

( ) ( )( ) 2

m2y

x

4y

2x

yv

xv

dx

dy=−=−=

∂∂

∂∂

−= for curve 2

c

1mm21

−= . They intersect orthogonally.

But C-R Equations are not satisfied

x

v

y

uand

y

v

x ∂

∂−≠

∂

∂

∂

∂≠

∂

∂u.

Some different forms of C-R Equations:

If w = f(z) = u+ iv , is analytic , then the following results follows.

1. ( ) =z1f

x

v i

x

u

∂

∂+

∂

∂

y

ui

y

v

∂

∂−

∂

∂=

∂

∂+

∂

∂−=

y

vi

ui

y

x

v i

x

u

∂

∂+

∂

∂=

∂

∂+

∂

∂−=

y

vi

ui

y

y

wi

x

w

∂

∂−=

∂

∂

2. ( )22

21

x

v

x

uzf

∂

∂+

∂

∂= ( )

22

21

x

v

x

uzf

∂

∂+

∂

∂=

22

u

x

u

∂

∂+

∂

∂=

y

22

x

v

∂

∂+

∂

∂=

y

u using C-R Equations.

Based on the results above mentioned the following results are valid,

a) ( )2

zfx

∂

∂ ( )2

zfy

∂

∂+ ( )

21f z=

b) ψ is any differential function of x and y then

22

x

∂

∂+

∂

∂

y

ψψ

22

vu

∂

∂+

∂

∂=

ψψ ( )21f z= .

c)

∂

∂+

∂

∂2

2

2

2

yx( )[ ]2

zfRe ( )21f2 z=

d)

∂

∂+

∂

∂2

2

2

2

yx( ) 2

f z ( )21f4 z=

Construction of An Analytic Function When real or

Imaginary part is Given

(Putting in Exact differential M dx + N dy = 0)

The Cauchy-Riemann equations provide a method of constructing

an analytic function f(z) = u+iv when u or v or vu ± is given.

Suppose u is given, we determine the differential dv , since

v = v(x,y),

dyy

vdx

x

vdv

∂

∂+

∂

∂= using C-R equations ,this becomes

dy N dx M. dy x

udx

y

udv +=

∂

∂+

∂

∂−=

And it is clear that y

M

x

N

∂

∂−

∂

∂ = 0y

u

x

u2

2

2

2

=∂

∂+

∂

∂

Because u is harmonic. This shows that M dx + N dy is an exact differential.

Consequently , v can be obtained by integrating M w.r.t. x by treating y as a

constant and integrating w.r.t. y only those terms in N that do not contain x,

and adding the results.

Similarly, if v is given then by using

dyy

udx

x

udu

∂

∂+

∂

∂= dy

x

vdx

y

v

∂

∂−

∂

∂= .

Following the procedure explained above we find u, and hence

f(z) u + iv can be obtained. Analogous procedure is adopted to find

u+iv when vu ± is given.

Milne-Thomson Method:

An alternative method of finding ivu ± when u or v or vu ±

is given.

Suppose we are required to find an analytic function f(z) = u+ iv

when u is given. We recall that

∂

∂−

∂

∂=

y

ui

x

u)(f 1

z ------------------------(I)

Let us we set y)x,(y

u and y)x,(

x

u21

φφ =∂

∂=

∂

∂ -------(II)

Then y)x,(i)yx,( (z)f21

φφ −=′ --------(III)

Replacing x by z and y by 0, this becomes

z,0)(i)z,0( (z)f21

φφ −=′ -------(IV)

From which the required analytic function f(z) can be got.

Similarly , if v is given we can find the analytic function f(z) = u+ iv by

starting with

∂

∂+

∂

∂=

x

vi

y

v)(f 1

z Analogous procedure is used when vu ± is given.

Applications to flow problems: As the real and imaginary parts of an analytic function are the

solutions of the Laplace’s equation in two variable. The conjugate functions

provide solutions to a number of field and flow problems.

Let v be the velocity of a two dimensional incompressible fluid with

irrigational motion, jy

vi

x

vV

∂

∂+

∂

∂= ------------------------------(1)

Since the motion is irrotational curl V= 0.

Hence V can be written as

jy

ix ∂

∂+

∂

∂=∇

φφφ -----------------------------(II)

Therefore , φ is the velocity component which is called the velocity

potential. From (I) and (II) we have

yy

v,

xx

v

∂

∂=

∂

∂

∂

∂=

∂

∂ φφ ------------------------(III)

Since the fluid is incompressible div V = 0.

0yyxx

=

∂

∂

∂

∂+

∂

∂

∂

∂ φφ -----------------------(IV)

0yx 2

2

2

2

=∂

∂+

∂

∂ φφ This indicates that φ is harmonic.

The function ( )yx,φ is called the velocity potential , and the curves

( ) cyx, =φ are known as equi -potential lines.

Note : The existence of conjugate harmonic function ( )yx,ψ so that

( ) ( )yx,iyx,w(z) ψφ += is Analytic.

The slope is Given by

x

y

v

v

x

y

y

x

dx

dy=

∂∂

∂∂

=

∂∂

∂∂

−=φ

φ

ψ

ψ

This shows that the velocity of the fluid particle is along the tangent to the

curve ( ) 1cyx, =ψ , the particle moves along the curve.

( ) 1cyx, =ψ - is called stream lines ( ) cyx, =φ - called equipotential

lines. As the equipotential lines and stream lines cut orthogonally.

( ) ( )yx,iyx,w(z) ψφ +=

yi

xxi

xdz

dw

∂

∂−

∂

∂=

∂

∂+

∂

∂=

φφψφ

yx

vv −=

The magnitude of the fluid velocity ( )dz

dwvv 2

y

2

x=+

The flow pattern is represented by function w(z) known as complex

potential.

Complex potential w(z) can be taken to represent other two-dimensional

problems. (steady flow)

1. In electrostatics ( ) cyx, =φ --- interpreted as equipotential lines.

( ) 1cyx, =ψ --- interpreted as Lines of force

2. In heat flow problems:

( ) cyx, =φ --- Interpreted as Isothermal lines

( ) 1cyx, =ψ --- interpreted as heat flow lines.

Cauchy –Riemann equations in polar form:

Let )iv(r,)u(r,)f(ref(z) i θθθ +== be analytic at a point z, then

there exists four continuous first order partial derivatives ,

θθ ∂

∂

∂

∂

∂

∂

∂

∂ v,

r

v,

u,

r

u and satisfy the equations

, .u

r

1

r

v:

v1

r

u

θθ ∂

∂−=

∂

∂

∂

∂=

∂

∂

r

Proof: The function is analytic at a point θi

rez = .

z

zz

∆

−∆+=

→∆

)f(z)f(z)(f lim

0 z

1 exists and it is unique.

Now ).iv(r,)u(r, f(z) θθ +=

Let z∆ be the increment in z , corresponding increments are

. andr in ,r θθ∆∆

{ } { }

z

)v(r,i)(r.u)r,(r v),(u lim)(f

0 z

1

∆

+−∆+∆++∆+∆+=

→∆

θθθθθθ irrz

{ }z

)u(r. -),r(ru lim)(f

0 z

1

∆

∆+∆+=

→∆

θθθz

{ }z

r

∆

∆+∆++

→∆

).v(r -),(rvlimi

0z

θθθ

------------------------(I)

Now θirez = and z is a function two variables r and θ , then we have

.z

rr

zz θ

θ∆

∂

∂+∆

∂

∂=∆

( ) ( ) θθ

θθ ∆∂

∂+∆

∂

∂=∆ ii rerre

rz

θθθ ∆+∆=∆ ii

erirez

When z∆ tends to zero, we have the two following possibilities.

(I). Let rezthat so,0i ∆=∆=∆ θθ

And 0r implies , 0Z →∆→∆

{ }re

)u(r. -),r(ru lim)(f

i0 r

1

∆

∆+=

→∆θ

θθz

{ }re

)v(r. -),(rvlimi

i0r ∆

∆++

→∆ θ

θθr

The limit exists,

∂

∂+

∂

∂= −

r

vi

r

u)(f 1 θi

ez -----------------(I)

2. Let θier iz that so , 0r =∆=∆

And 0imply 0,z →∆→∆ θ

{ }θ

θθθθθ ∆

∆+=

→∆i

0

1

er i

)u(r. -),(ru lim)(f z

{ }θ

θθθθ ∆

∆++

→∆ i0z eir

)v(r. -),(rvlimi

=

∂

∂+

∂

∂

θθθ

vi

u

re i

1i

∂

∂+

∂

∂−=

θθθ

vui

er

1i

∂

∂−

∂

∂=

θθθ u

r

iv

r

1e z)(f i-1 ------------ (II)

From (I) and(II) we have

θ∂

∂=

∂

∂ v

r

1

r

u

θ∂

∂−=

∂

∂ u

r

1

r

v or

θθurv,vru

rr−==

Which are the C-R Equations in polar form.

Harmonic Function:

A function φ -is said to be harmonic function if it satisfies Laplace’s

equation 02 =∇ φ

Let )iv(r,)u(r,)f(ref(z) i θθθ +== be analytic. We shall show that

u and v satisfy Laplace’s equation in polar form.

01

rr

1

r 2

2

22

2

=∂

∂+

∂

∂+

∂

∂

θ

φφφ

r

The C-R equations in polar form are given by,

θ∂

∂=

∂

∂ v

r

1

r

u ---------(I)

θ∂

∂−=

∂

∂ u

r

1

r

v------------(II)

Differentiating (I) w.r.t r and (II) w.r.θ , partially , we get

r

u

r

ur

2

2

2

∂∂

∂=

∂

∂+

∂

∂

θ

v

r :

2

22 u

r

vr

θθ ∂

∂−=

∂∂

∂

And we have r∂∂

∂=

∂∂

∂

θθ

v

r

v 22

2

2

2

2 u

r

1

r

u

r

ur

θ∂

∂−=

∂

∂+

∂

∂

Dividing by r we, get

0u1

r

u

r

1

r

u2

2

22

2

=∂

∂+

∂

∂+

∂

∂

θr

Hence u-satisfies Laplace’s equation in polar form.

The function is harmonic. Similarly v- is harmonic.

Orthogonal System:

Let )f(r θ= and −= φθ

φ ,dr

dtan r being the angle between

the radius vector and tangent. The angle between the tangents at

the point of intersection of the curves is 21

φφ − . 1TanTan21

−=φφ ,is

the condition for orthogonal.

Consider 1

c)u(r, =θ .

Differentiating w.r.t θ , treating r as a function of θ .

0u

d

dr

r

u=

∂

∂+

∂

∂

θθ

∂

∂

∂

∂

−=

r

u

u

d

dr θ

θ

Thus dr

drTan

1

θφ = = ( )

( )θ∂

∂∂

∂

ur

u

r =

( )( )

θ∂∂

∂∂

ur

u

r- ---------(I)

Similarly for the curve 2

c)v(r, =θ

=2

Tanφ ( )( )

θ∂∂

∂∂

vr

v

r- -----------(II)

( ) ( )( ) ( )

θθ

φφ

∂∂

∂∂

∂∂

∂∂

=vu

rvr

rur

anan21

TT

By C-R Equations θθurv,vru

rr−==

The equation reduces to

( ) ( )( ) ( ) 1

uv

u-v

anan21

−=

∂∂

∂∂

∂∂

∂∂

=

θθ

θθφφTT

Hence the polar family of curves 1

c)u(r, =θ and 2

c)v(r, =θ ,

intersect orthogonally.

Construction of An Analytic Function When real or

Imaginary part is Given(Polar form.)

The method due to Exact differential and Milne-Thomson is

explained in earlier section .

Ex: Verify that ( )θcos2r

1u

2= is harmonic . find also an analytic

function.

Soln: θcos2r

2

r

u2

−=

∂

∂ : θθ

sin2r

2u2

−=

∂

∂

θcos2r

6

r

u42

2

=∂

∂ : θ

θcos2

r

4u22

2

−=∂

∂ .

Then the Laplace equation in polar form is given by,

=∂

∂+

∂

∂+

∂

∂2

2

22

2

u1

r

u

r

1

r

u

θrθcos2

r

64

θcos2r

24

− 0cos2

r

42

=− θ

Hence u-satisfies the laplace equation and hence is harmonic.

Let us find required analytic function f(z) = u+iv.

We note that from the theory of differentials,

θθ

dv

drr

vdv

∂

∂+

∂

∂=

Using C-R equations θθurv,vru

rr−==

θθ

dr

urdr

u

r

1-

∂

∂+

∂

∂=

θθθ dcos2r

2d sin2

r

223

−

−= r

= θsin2

r

1-d

2

From this c sin2r

1-v

2+= θ

c+

+

=+= θθ sin2

r

1-icos2

r

1 ivu f(z)

22

[ ] icisin2-cos2r

12

+= θθ

( )

icer

1e

r

12i

2i-

2+=+=

θ

θic

ic.z

1f(z)

2+=

Ex 2:Find an analytic function f(z)= u+iv given that

θsinr

1-rv

= 0r ≠

Soln: θsinr

1r

r

v2

+=

∂

∂ : θθ

cosr

1r

v

−=

∂

∂

To find u using the differentials

θθ

du

drr

udu

∂

∂+

∂

∂=

Using C-R equations θθurv,vru

rr−==

θθ

dr

v-dr

v

r

1

∂

∂+

∂

∂=

θθθ dsinr

11r-drcos

r

1-r

r

12

+

=

= θθθ d sinr

1r -drcos

r

11

2

+

−

+= θcos

r

1rd

ccosr

1ru +

+= θ

f(z) = u+iv

θθ sinr

1-ricos

r

1r

++

+= c

( ) cisin-cosr

1isinr(cos +++= θθθθ

cz

1z

r

1er f(z) i-i ++=+= θθ

e

Ex: Construction an analytic function given θcos2ru 2= .

(Milne Thomson Method)

θcos2ru 2= -----------(I)

θ2rcos2r

u=

∂

∂ θ

θsin22r

u 2−=∂

∂

∂

∂+

∂

∂= −

r

vi

r

u)(f 1 θi

ez

Using C-R equations θθur v,vur

rr−==

( ) ( )

+= θθθ

sin22r-r

1-ircos2 2ezf

2i-1

( )[ ]θθθsin22r ircos2 2e

-i +=

[ ]θθθisin2cos2er 2

-i +=

Now put r = z , and 0=θ .

( ) 2zzf 1 = on integrating

( ) cz zf 2 += .

COMPLETION OF UNIT-I