UNIT-IV

Transformations in the complex plane:

Consider the complex valued function

W=f(z)---------(1)

A complex number z=x+iy determines a point P(x,y) in the complex plane

and is referred as the point z. The point w= u+iv, is represented by a point

Q(u,v) in theu-v plane. Thus w= f(z) represents a transformation and

transforms a point P(x,y) to a unique point Q(u,v) in the complex plane.

Conformal Transformation:

The transformation w= f(z) transforms the curves C1 and C2 to the curves

C11 and C2

1 and intersects at a given point then the transformation is said

to be a conformal transformation.

I):The transformation w=1/z

f(z)= 1/z is analytic with

As such, this transformation is conformal at every point . The transformation

is carried by taking in polar form. Then from

the above Equation so that a point is transformed to the point

Ex: show that the transformation

w = 1/z transforms a circle to a circle or a straight line.

Using W = u+iv, z= x+iy. And w = f(z) gives

equating real and imaginary parts gives.

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

Let us consider any circle in z-plane . It’s Cartesian equation is of the form

0.zfor z

-1z)f

2

1 ≠= ,(

φθ iiRewandrez ==

( ) ( )θφ r,-R, /1=

( )θr,

22 vu

iv-u

ivu

1iyx

+=

+=+

,22

vu

ux

+= ,

22vu

-vy

+=

0c2fy2gxyx 22 =++++

Substituting for x and y from equation (I), we get

-------(II)

The above equation represents a circle in the w- plane if ,

and a straight line if c =0.

(II) :Transformation w =

1.Consider the transformation w= --(I)

The transformation is conformal for .

Now u+iv =

So, that

Let . .Which represents a rectangular

Hyperbola. V= constant. V=2B, Which is also a rectangular hyperbola.

The two families of curves

Under the given transformation w = the rectangular hyperbolas

in the z-plane transforms to the st-lines u=A, and v=2B, in the w-plane.

2.Now consider a line parallel to y-axis. The equation Of this is of the form

x=a, where a- is a constant.

Then

The equation represents equation of parabola in the w-plane having vertex at

( ) 012fu- 2guvuc22 =+++

0c ≠

2z

2z0z ≠

( ) i(2xy)yxiyx 22 +−=+ 2

2xyvyxu 22 =−= ,

Au i.eAyx 22 ==−

lyorthogonalintersect Bxy and Ayx 22 ==−

2z Bxy and Ayx

22 ==−

uyaor uyx 2222 =−=−

v 2ayor v2xy ==

)1()(4 −−−−−= 222a-uav

the point , and its axis is the negative u-axis.

3.Again consider a line parallel to the x-axis. It’s equation is of the form

y=b, where b –is constant.

V=2xy or v=2xb

Which represents a parabola in the w-plane having vertex at the point

and its axis is positive u-axis.

Hence the transformations W= transforms st-lines parallel to y-axis

to parabolas having the negative u-axis as their common axis and the straight

lines parallel to x- axis to parabolas having the positive u-axis as their

common axis.

(III) :Transformation:

Here for any z. Therefore the transformation is

conformal for all z.

-----(1)

We shall find the image in the w-plane corresponding to the straight

lines parallel to the co-ordinate axes in the z-plane. Let x = constant

y= constant.

Squaring and adding equations (1) we get

)0,2(a

2bor −=−= 222 xu yxu

)(4) 2222 bubbb 4(u +=+=

( )0,2b-

2z

zew =

0( ≠= z1 ez)f

isiny)cosyee ivu xiyx +==+ + (

sinyevcosyeu xx == 2x22 evu =+

and by dividing

Case-1:Let x = c Where c is a constant.

This represents a circle with center origin and radius r, in the w-plane.

Case-2: Let y= c where c- is a constant

This represents a st-line passing through origin in the w-plane.

Conclusion : The st-line parallel to the x-axis in the z-plane maps onto

a st-line passing through the origin in the w-plane. The st-line parallel to

y-axis in the z-plane maps onto a circle with center origin and radius r.

A tangent is drawn at the point of intersection of these two curves in the w-

plane, the angle subtended is 90. Hence the two curves are orthogonal

trajectories of each other.

(IV):The Transformation :

Here The transformation is conformal at all points except

at 0 and . The transformation is also known as the “Joukowski’s “

transformation.

Equating real and imaginary parts

After simplifying

2x22 evu =+

tanyu

v=

( ) 2c2c22 reevu ===+2

muvor mtancu

v===

z

azw

2

+=

( )z

azf

21 −= 1

a±

θirezLet = θθ i-2

i er

areivu +=+

( )θθθθ isin-cosr

a)isinr(cos

2

++=

-------(1)

Squaring and adding

Case-1: When r = constant then above eqn becomes

Which represents an ellipse in w-plane with foci

Hence the circle in the z-plane maps onto an

ellipse in the w-plane with foci

2. Eliminating r in the equation (1).

or

Now represents a circle with centre origin and radius r

in the z-plane.

This represents a st-line in the z-plane passing through origin.

Now the equation (2) becomes

Where A=2 acos θ , B=2a sin θ

( ) ( )ra-r

vsin

rar

u22 /

:/

cos =+

= θθ

( ) ( )1

//22

=+

+ ra-r

v

rar

u

2

2

2

2

1=+2

2

2

2

b

v

a

u

)0,( 22 ba −±

constant,rz ==

a,0)2(±

2

2

2

2

2

asin

v

cos

u4=−

θθ1

)) 22=−

θθ (2asin

v

(2acos

u22

θire=z

θθ tanx

y

x

ytan-1 =

= and

1=−2

2

2

2

B

v

A

u

This represents a hyperbola in the w-plane with foci.

The both conics (ellipse & hyperbola) have the same foci, independent of

r and θ and they are called confocal conics .

(V):Bilinear Transformation:

Let a,b,c, and d be complex constants such that

ad – bc 0. Then the transformation defined by,

is called bilinear transformation. Solving for z, we find

Which is called the inverse bilinear transformation.

The transformation (1) establishes one-one correspondence between the

points in the z-and w- plane.

Now from equation (1)

Since the above equation is a quadratic equation there exists exactly two

such points for a given transformation. These are called the fixed points

or invariant points of the transformation.

Note 1:

There exists a bilinear transformation that maps three given distinct

points Onto three given distinct points

a,0)BA( 22 2()0, ±≡−±

≠

)1(−−−+

+=

dcz

bazw

)2(−−−−

+=

acw

b-dwz

0b-a)z(dczdcz

bazw 2 =−+

+

+= or

321 zzz ,,

yrespectivl www 32,1,

Solving this equation for w in terms of z,

we obtain the bilinear transformation that

transforms

Ex: Find bilinear transformation that maps the points 1,i,-1 on to the

points i,0,-1 respectively.

Under this transformation find the image of

Also find the invariant points of this transformation.

Using the formula

We get

To find the image of

we rewrite the above equations as

If and using this condition we get u>0

Under this transformation the image of is u>0

Which is right half of w plane

( )( )( )( )

=−

−−

123

321

www-w

wwww ( )( )( )( )123

321

zzw-z

zzzz

−

−−

to onzzz 321 ,, yrespectivl www 32,1,

1z <

1,,1 −=== 321 zzz i i−=== 32 wwi,w ,0

( )( )( )( )

=−

−−

123

321

www-w

wwww ( )( )( )( )123

321

zzw-z

zzzz

−

−−

iz-1

iz1w

+=

1<z

z)(1

w)-i(1z

+=

1<z

1<z

To find invariant points set w = z. The quadratic equation is obtained

These two are invariant under the transformation

COMPLEX INTEGRATION.

Complex Line integrals:

Consider a continuous function f(z) of the complex variable z= x+iy defined

at all points of a curve c. Divide the curve C- into n-parts by arbitrarily

taking points P0(z0), P1(z1),---

Now δzk = zk- zk-1 k=1,2,3,---.

The concept of line integral is that the curve is divided into smaller units and

the smallest part of the curve is a st-line.

Then ∞→∑=

n aszf(k

n

1k

δα )k

where ∞→→ n aszk

0δ is defined as

the complex line integral along the path C, and is denoted as ∫C

dz f(z)

and curve. closed simplefor used isdz f(z)∫C

Properties:

I) If C- denotes the curve traversed from Q to P then, ∫−C

dz f(z) = -

∫C

dz f(z)

II) If C- is split into a number of parts C1, C2,C3, -- then

∫C

dz f(z) = ∫1C

dz f(z) + ∫2C

dz f(z) +---

III) [ ]∫ ± zz)fz)f221

d((1

λλ = ∫ ∫±C

221z)dzf dz z)f

C

((1

λλ

Line Integral of a Complex Valued function:

( )[ ]i62

1±+−= i1z

Let f(z) = u(x,y) + i v(x,y) be a complex valued function defined over a

region R , and C- be a curve in the region, Then

∫C

dz f(z) = ∫ ++C

idy)iv)(dx(u = ∫ +C

i dy) v -dx (u ∫ +C

dy u dx v .

EXAMPLES:

Evaluate: ∫C

dz z 2 i) Along the st-line from z = 0 to z = 3+i..

ii) Along the curve made up of two line segments, one from z = 0 to z= 3,

and another z = 3 to z = 3+i.

Soln: ∫C

dz f(z) = ∫+=

=

iz

z

3

0

dz z 2 .z- varying from 0 to 3 + i .

(x,y) varies from (0,0) to (3,1). Equation of line joining the points (0,0)

and (3,1).

03

01

−

−=

0-x

0-y y = x/3.

And z2 = (x+iy)

2 = x

2 – y

2 + i(2xy) and dz = dx + idy .

∫C

dz f(z) = [ ]( )

( )

( )idydxi2xyyx0,0

22 ++−∫1,3

= [ ]( )

( )

∫ −−1,3

)0,0

22 dy 2xydxy(x + i [ ]( )

( )

∫ +−1,3

)0,0

22 dx 2xydyy(x

Now y = x/3, x = 3y. we convert these integrals into the variable

defined and integrate w.r.t y from 0 to 1.And dx = 3 dy.

∫C

dz f(z) = ∫1

2dyy 180

+ i ∫1

2dyy 260

=

+

3

26i6 along the given path.

iii) Along the curve made up of two line segments one from

z = 0 to z = 3, and another z = 3 to z = 3 + i.

z = 0 to z = 3 -� (x,y) varies from (0,0) to (3,0)

z = 3 to z = 3 + i � (x,y) varies from (3,0) to (3,1).

Along C1 y = 0: dy = 0.and x- varies from 0 to 3 ,Z2 dz = x

2 dx

Along C2 : x= 3. dx = 0 and y- changes from 0 to 1.

Z2 dz = (3 + iy)

2 i dy.

∫C

dz z2 = ( ) dyiy3idxx

3

0x

2

∫∫==

++1

0

2

y

= 3

819 i+

Ex: Evaluate the integral

I= ∫C

dz z2

, where C- is the square region having vertices at origin O and

the points P(1,0), Q(1,1), and R(0,1).

Here the given curve C- is made up of the line segments OP, PQ, QR & RO.

∫C

dz z2

= ∫op

2

dz z + ∫pQ

2

dz z + ∫QR

2

dz z + ∫RO

2

dz z

On OP: y= 0 , z = x , 0≤ x≤1 , ∫C

dz z2

= ( ) ( )3

10

2

=++∫=

1

0x

2 i.0dxx ---------(I)

On PQ; x=1 , z = 1+iy , 0≤ y ≤1 , ∫C

dz z2

= ( )( )3

42 ii.dy0y11

0y

2 =++∫=

------(II)

On QR: y=1, z = x+i, x- decreases from 1 to 0 .

∫C

dz z2

= ( )( )3

4-i.0dx1x

1

0

2 =++∫=x

------- (III)

On RO, we have x = 0 , z = iy , y decreases from 1 to 0.

∫C

dz z2

= ( )( )3

2 i-i.dy0y0

0

1y

=++∫=

--------(IV)

∫C

dz z2

= i3

i

3

4i +−=−−+ 1

3

4

3

1.

Ex: Evaluate the Integral ( )∫+

=

−i1

z

2 dziyx0

along i) the straight line y= x

ii) the parabola y = x2.

Proof:: The parametric equations of the given st-line is given by x=t , y= t,

so that z = x+iy = t+it. As z varies from 0 to 1+I, the parameter increases

from 0 to 1.

Hence along the given line, the given integral is,

idy)dx()y(x dz iy)x(1

0z

i1z

0z

22 +∫ ∫ −=−+=

=

+=

=

z

i

dt)t(t )1( idt)(dt it)t(1

0

1

0

22

∫ ∫ −+=+−=i

ii

= )5(6

1i−

ii) The parametric equations of the given parabola are x= t, y = t2 .

z= x+i y = t (1+ i t) as z varies from 0 to 1+I , the parameter t increases

from 0 to 1.

The given integral is,

I = idy)dx()y(x dz iy)x(1

0z

i1z

0z

22 +∫ ∫ −=−+=

=

+=

=

z

i

∫ ∫ +−=+−==

1

0

1

0

22 dt)21(t )1( dt) t i 2 (dt it)t(t

iti

= )5(6

1i+