vtu m4 notes

1. 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
C1
1
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
-1
z)f 2
1
≠= ,(
φθ ii
Rewandrez ==
( ) ( )θφ r,-R, /1=
( )θr,
22
vu
iv-u
ivu
1
iyx
+
=
+
=+
,22
vu
u
x
+
= ,22
vu
-v
y
+
=
0c2fy2gxyx 22
=++++

2. 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-2guvuc 22
=+++
0c ≠
2
z
2
z
0z ≠
( ) i(2xy)yxiyx 22
+−=+ 2
2xyvyxu 22
=−= ,
Aui.eAyx 22
==−
lyorthogonalintersectBxyandAyx 22
==−
2
z
BxyandAyx 22
==−
uyaoruyx 2222
=−=−
v2ayorv2xy ==
)1()(4 −−−−−= 222
a-uav

3. 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
2
bor −=−= 222
xuyxu
)(4) 2222
bubbb4(u +=+=
( )0,2
b-
2
z
z
ew =
0( ≠= z1
ez)f
isiny)cosyeeivu xiyx
+==+ +
(
sinyevcosyeu xx
== 2x22
evu =+

4. 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 =+
tany
u
v
=
( ) 2c2c22
reevu ===+
2
muvormtanc
u
v
===
z
a
zw
2
+=
( )
z
a
zf
2
1
−=1
a±
θi
rezLet = θθ i-
2
i
e
r
a
reivu +=+
( )θθθθ isin-cos
r
a
)isinr(cos
2
++=

5. -------(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
v
sin
rar
u
22
/
:
/
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
a
sin
v
cos
u
4=−
θθ
1
)) 22
=−
θθ (2asin
v
(2acos
u 22
θi
re=z
θθ tan
x
y
x
y
tan-1
=





= and
1=− 2
2
2
2
B
v
A
u

6. 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
baz
w
)2(−−−
−
+
=
acw
b-dw
z
0b-a)z(dcz
dcz
baz
w 2
=−+
+
+
= or
321 zzz ,,
yrespectivlwww 32,1,

7. 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
−
−−
toonzzz 321 ,, yrespectivlwww 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
iz1
w
+
=
1<z
z)(1
w)-i(1
z
+
=
1<z
1<z

8. 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 ∞→∑=
naszf( k
n
1k
δα )k where ∞→→ naszk 0δ is defined as
the complex line integral along the path C, and is denoted as ∫C
dzf(z)
and curve.closedsimpleforusedisdzf(z)∫C
Properties:
I) If C- denotes the curve traversed from Q to P then, ∫−C
dzf(z) = -
∫C
dzf(z)
II) If C- is split into a number of parts C1, C2,C3, -- then
∫C
dzf(z) = ∫
1C
dzf(z) + ∫
2C
dzf(z) +---
III) [ ]∫ ± zz)fz)f 221 d((1 λλ = ∫ ∫±
C
221 z)dzfdzz)f
C
((1 λλ
Line Integral of a Complex Valued function:
( )[ ]i6
2
1
±+−= i1z

9. 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
dzf(z) = ∫ ++
C
idy)iv)(dx(u = ∫ +
C
idy)v-dx(u ∫ +
C
dyudxv .
EXAMPLES:
Evaluate: ∫C
dzz2
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
dzf(z) = ∫
+=
=
iz
z
3
0
dzz2
.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
= x2
– y2
+ i(2xy) and dz = dx + idy .
∫C
dzf(z) = [ ]
( )
( )
( )idydxi2xyyx
0,0
22
++−∫
1,3
= [ ]
( )
( )
∫ −−
1,3
)
0,0
22
dy2xydxy(x + i [ ]
( )
( )
∫ +−
1,3
)
0,0
22
dx2xydyy(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
dzf(z) = ∫
1
2
dyy18
0
+ i ∫
1
2
dyy26
0
= 



+
3
26
i6 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 = x2
dx
Along C2 : x= 3. dx = 0 and y- changes from 0 to 1.
Z2
dz = (3 + iy)2
i dy.

10. ∫C
dzz2
= ( ) dyiy3idxx
3
0x
2
∫∫ ==
++
1
0
2
y
=
3
8
19 i+
Ex: Evaluate the integral
I= ∫C
dzz
2
, 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
dzz
2
= ∫op
2
dzz + ∫pQ
2
dzz + ∫QR
2
dzz + ∫RO
2
dzz
On OP: y= 0 , z = x , 0≤ x≤1 , ∫C
dzz
2
= ( ) ( )
3
1
0
2
=++∫=
1
0x
2
i.0dxx ---------(I)
On PQ; x=1 , z = 1+iy , 0≤ y ≤1 , ∫C
dzz
2
= ( )( )
3
42
ii.dy0y1
1
0y
2
=++∫=
------(II)
On QR: y=1, z = x+i, x- decreases from 1 to 0 .
∫C
dzz
2
= ( )( )
3
4
-i.0dx1x
1
0
2
=++∫=x
------- (III)
On RO, we have x = 0 , z = iy , y decreases from 1 to 0.
∫C
dzz
2
= ( )( )
3
2 i
-i.dy0y0
0
1y
=++∫=
--------(IV)
∫C
dzz
2
= i
3
i
3
4
i +−=−−+ 1
3
4
3
1
.
Ex: Evaluate the Integral ( )∫
+
=
−
i1
z
2
dziyx
0
along i) the straight line y= x
ii) the parabola y = x2
.

11. 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(xdziy)x(
1
0z
i1z
0z
22
+∫ ∫ −=−
+=
=
+=
=
z
i
dt)t(t)1(idt)(dtit)t(
1
0
1
0
22
∫ ∫ −+=+−
=i
ii
= )5(
6
1
i−
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(xdziy)x(
1
0z
i1z
0z
22
+∫ ∫ −=−
+=
=
+=
=
z
i
∫ ∫ +−=+−=
=
1
0
1
0
22
dt)21(t)1(dt)ti2(dtit)t(
t
iti
= )5(
6
1
i+