21MAB102T- ADVANCED CALCULUS
AND COMPLEX ANALYSIS
UNIT- IV
ANALYTIC FUNCTIONS
Contents
Definition
Cauchy
of Analytical Function
Riemann Equations
Properties
of analytical functions
Determination
of Harmonic Conjugate
Milne-Thomson’s
method
Analytical functions :- ( Regular functions or Holomorphic
functions)
DEFINITION:A single valued function f(z) is said to be analytic at a point z0
,if it has a derivative at z0 and at every point in some
neighbourhood of z0 .
NOTE :
If it is analytical at every point in a region R, then it is said to
be analytic in the region R.
Necessary condition for a complex
function f(z) to be analytic:Derivation of Cauchy-Riemann
equations:STATEMENT:If f(z) = u(x,y) + i v(x,y) is analytic in a region
R of the z-plane then
I) ux , uy, vx , vy exist and
Ii) ux = vy and uy = -vx at every point in that
region.
Necessary condition for a complex
function f(z) to be analytic:Derivation of Cauchy-Riemann
equations:PROOF:Let f(z) = u(x,y) + i v(x,y)
We first assume f(z) is analytic in a region R. Then by the
Definition, f(z) has a derivative f’(z) everywhere in R.
Now
Derivation of Cauchy-Riemann
equations:Let z = x+iy
Δz = δx + i δy

( z+δz) = (x+δx) + i (y+δy)
 f ( z  z )  u ( x  x, y  y)  i v ( x  x, y  y)
We know that, f(z) = u(x,y) + i v(x,y)
Now
Derivation of Cauchy-Riemann
equations:Case (i) :- if δz→0 , first we assume that δy=0 and δx→0
∴
u
v

i
=
x
x
∴
f ΄(z) = ux+ i vx
-------→(1)
Derivation of Cauchy-Riemann
equations:CASE (II) :- If δz→0 , now we assume δx=0 and δy→0
∴
1 u
v
= i y  y
∴ f ΄(z) = uy+ i vy-------→(2) (Since 1/i = -i)
Derivation of Cauchy-Riemann
equations:From (1) & (2) , we get
ux + i vx
=
-i uy+ vy
Equating real and imaginary parts weget,
Ux = vy and uy = - vx
The above equations are called Cauchy-Riemann
equations
(or) C-R equations .
therefore the function f(z) to be analytic at the point
z, it is necessary that the four partial derivatives ux , uy,
vx, vy should exist and satisfy the C-R equations.
Sufficient condition for f(z) to be
analytic
STATEMENT:The singled valued continuous function f(z) = u + i v is
analytic in a region R of the z-plane, if the four partial
derivatives ux , uy , vx , vy , (i) exist , (ii) continuous
, (iii) they satisfy the C-R equations ux = vy and uy =
- vx
at every point of R.
Note:- all polynomials, trigonometric, exponential
functions are continuous.
Cauchy-Riemann Equations in Polar
form
STATEMENT:z = reiθ , then
If f(z) = u(r,θ) + i v(r,θ) is differential and
u  1  v
1 
 
 u   v
r  r  
r 
v
 1  u
1
  
 v  
r
 r  
r

r
r

u


Proof:- let z = reiθ
and f(z) = u+iv
i.e.,
u+iv = f( reiθ )
(1)
Cauchy-Riemann Equations in Polar
form
Differentiating partially w.r.t.
u
v
i
 f   re   e 
r
r
i
i
Differentiating partially w.r.t.
u
v
i
 f   re  re  i 


  ri  f   re  e 
i
̀r ́ we get,
(2)
̀θ ́ weget,
i
i
i
v 
 u
  ri    i  ( from eqn. (2))
r 
 r
 u   v 
 ir 
  r    (3)
 r   r 
Cauchy-Riemann Equations in Polar
form
Equating real and imaginary parts in eqn. (3) , we get,
u
v
r

r
and
i.e., u   r v and

r
v
u
r

r
v  ru

 1 
(or ) v  
 u and
 r 
r

r
1
u 
r
r

v


EXAMPLES
1) Show that f(z) = z3 is analytic.
Proof:-
given f(z) = z3 =(x+iy)3 = x 3 + 3 x 2(iy) +
3x(iy)2 + (iy)3
= (x 3 - 3xy2) + i ( 3x2y – y3 )
We know that f(z) = u+iv
u  x  3xy
, v  3x y  y
3
2
u
So ,
 3x  3 y
x
u
  6 xy
y
2
2
2
,
,
3
v
 6 xy
x
v
 3x  3 y
y
2
2
EXAMPLES
from the above equations weget ,
u  v and u   v
x
y
y
x
 C  R equations are satisfied .
Here ux , uy , vx , vy exists and continuous.
Hence the given function f(z) is analytic.
2) Examine the analyticity of the following functions and
find its derivatives.
i)
f ( z)  e
z
ii ) f ( z )  cos z
iii ) f ( z )  sinh z
EXAMPLES
i) SOLUTION:-
f ( z)  e  e
z
x  iy
 e e  e  cos y  i sin y 
x
Here u  e cos y
x
iy
x
and
u  e cos y
v  e sin y
u   e sin y
v  e cos y
x
x
x
x
x
x
y
 u  v and
x
v  e sin y
x
y
u  v
y
y
x
 C  R equations are satisfied .
 f ( z ) is analytic everywhere in the complex plane.
EXAMPLES
Now f ( z )  u x  i vx
 e cos y  i e sin y
x
e
x
x
 cos y  i sin y 
e e
x
e
e
x  iy
z
iy
EXAMPLES
II) SOLUTION:-
f ( z )  cos z
 cos( x  iy )
 cos x cos(iy )  sin x sin(iy )
 cos x cosh y  i sin x sinh y ( cos(ix)  cosh x
sin(ix)  i sinh x)
 u  cos x cosh y
v   sin x sinh y
u   sin x cosh y
v   cos x sinh y
u  cos x sinh y
v   sin x cosh y
x
y
 u  v and
x
y
x
y
u v
y
x
EXAMPLES
 C  R equations satisfied
 It is analytic
Also
f ( z )  u  i v
x
x
   sin x cosh y   i   cos x sinh y 
1
  sin x cos iy  i (  cos x   sin(iy ))
i
  sin x cos  iy   cos x sin(iy )
  sin( x  iy ) 
  sin z
EXAMPLES
SOLUTION:-
1
f ( z )  sinh z  sin(iz )
i
  i (sin i ( x  iy ))
  i (sin(ix) cos y  cos(ix) sin y )
  i (i sinh x cos y  cosh x sin y )
 sinh x cos y  i cosh x sin y
 u  sinh x cos y
,
v  cosh x sin y
u  cosh x cos y ,
v  sinh x sin y
u   sinh x sin y ,
v  cosh x cos y
x
y
 u  v and
x
y
x
u  v
y
y
x
EXAMPLES
 C  R equations are satisfied
 f ( z ) is analytic.
Now
f ( z )  u  i v
x
x
  cosh x cos y   i  sinh x sin y 
 1

  cos(ix) cos y   i    sin(ix) sin y 
 i

 cos(ix  y )
 cos i ( x  iy )
 cos i z
 cosh z
 1/ i    i 
TRY IT
Examine the analyticity of the following functions and
find its derivatives.
i)
f ( z )  e x ( cos y  i sin y )
ii ) f ( z )  e  x ( cos y  i sin y )
iii ) f ( z )  sin x cosh y  i cos x sinh y
EXAMPLES
3) Show that the function f ( z )  xy is not regular (analytic)
at the origin, although C  R equations are satisfied
at the origin.
Solution : 
Given f ( z )  xy
Hence u 
xy
and
v0
u
u ( x  x, y )  u ( x, y )
Now, u x 
 lim
x x0
x
EXAMPLES
u x (0,0)  lim
x 0
llly
u (x,0)  u (0,0)
0
x
u y (0,0)  0
vx (0,0)  0
v y (0,0)  0
 u x  v y and u y   vx at the origin.
 C  R equations are satisfied at the origin.
EXAMPLES
But
f (0  z )  f ( z )
f (0)  lim
z 0
z
xy  0
  xlim
 iy 0
z
Along the curve y = mx
m x
m
f (0)  lim

y  mx
1  im
x 0 x 1  im 
2
 The limit is not unique, since it depends on 'm'.
 f (0) does not exist.
Hence f(z) is not regular at the origin.
C-R equations in polar form
EXAMPLES
1) Check for analyticity of logz
(Or)
Show that f(z) = logz is analytic everywhere except at
the origin and find its derivatives.
Solution:- f ( z )  log z
 log(r ei )
( z  r ei )
 log r  log ei
 log r  i
w.k .t. f ( z )  u  iv
Here u  log r and v  
EXAMPLES


1
r
u  0
ur 
vr  0
v  1
ur , u , vr , v exist, are continuous and satisfy C-R equations
1
ur  
r

1
 v and v r   

r

 u  everywhere except at r  0  i.e. z  0.


f ( z ) is analytic everywhere except at z  0.
EXAMPLES
Prove that f(z) = zn is analytic function and find its
derivatives.
n
i n
f
(
z
)

z

(
re
)
PROOF: r n ein
 r n  cos n  i sin n 
 u  r n cos n
;
v  r n cos n
;
vr  nr n1 sin n
u   nr n sin n ;
v  nr n cos n
ur  nr n1 cos n
1
 ur    v
r
and
1
vr     u
r
EXAMPLES
Thus ur , u , vr , v exist , are continuous and
satisfy C  R equations everywhere.
 f ( z ) is analytic.
Also
 ur  i vr 

f ( z)  

i
e


nr n1 cos n   i  nr n1 sin n 


ei
nr n1  cos n  i sin n 

ei
nr n1ein
i n 1
n 1


n
re

nz


ei
Laplace Equations
In Cartesian form :
 2
 2

0
2
2
x
y
i.e.,  2  0
In Polar form :
 2
1 
1  2

 2
 0
2
2
r
r r
r 
HARMONIC FUNCTIONS
A real valued function of two real variables x and
y is said to be harmonic, if
i)
The second order partial derivatives uxx , uxy , uyx ,
uyy exist and they are continuous.
And
ii)
2
2

u

u
The laplace equation
 2  0 satisfies.
2
x
y
CONJUGATE HARMONIC FUNCTIONS:If u+iv is an analytic function of z then v is called a
conjugate harmonic function of u; (or) u is called a
conjugate harmonic function of v; (or) u and v are called
conjugate harmonic functions.
Properties of Analytic functions
PROPERTY (1) :- The real and imaginary parts of an analytic
function f(z) = u+iv satisfy the laplace equation (or) real part
“u” and imaginary part “v” of an analytic function f(z) = u+iv
are harmonic functions.
Proof:-
Given f(z) = u+i v is an analytic function.
I.E.,
U and v are continuous, ux , uy , vx , vy are exist and
they satisfy the C-R equations ux = vy
and
(1)
uy = - vx
(2)
Properties of Analytic functions
Diff . eqn.(1) partially w.r .t . x, weget ,
u xx  v yx  (3)
Diff . eqn.(2) partially w.r .t . y , weget ,
u yy   vxy  (4)
Adding (3) & (4) weget ,
u xx  u yy  v yx  vxy  0
 v yx  vxy 
 u satisfies Laplace equation.
Hence u is a Harmonic function.
Properties of Analytic functions
NOW ,
Diff . eqn.(1) partially w.r.t . y , weget ,
u xy  v yy  (5)
Diff . eqn.(2) partially w.r.t. x, weget ,
u yx   vxx  (6)
subracting (5) & (6) weget ,
v yy  vxx  u xy  u yx  0
 u xy  u yx 
 v satisfies Laplace equation.
Hence v is a Harmonic function.
Thus u and v are harmonic functions.
NOTE:- The converse of the above result need not be true.
Properties of Analytic functions
TRY IT
Prove that the real and imaginary parts of an analytic
function f(z) = u(r,θ) + i v(r, θ) satisfy the laplace
equation in polar coordinates.
i.e., To prove that
u rr
1

r

1
 ur   2  u

r 
vrr
1


r
v   1  v
 r  2  

r 
and
Properties of Analytic functions
ORTHOGONAL CURVES:Two curves are said to be orthogonal to each other then they
intersect at right angles. [ Product of slopes m1 m2 = -1]
PROPERTY (2) :If f(z) = u+ iv is an analytic function then the family of curves
U(x,y) = a and v(x,y) = b (where a&b are constants) cut each
other orthogonally.
Proof:- Given :
u ( x, y )  a and v( x, y )  b
Taking differentials on both sides, weget ,
du  0
Properties of Analytic functions

u
u
dx 
dy  0
x
y

llly
u x
dy

 m1
dx
uy
v ( x, y )  b


v
v dy

 0
x
y dx
vx
dy

 m2
dx
vy
Properties of Analytic functions
 u x  vx 
Pr oduct of slopes, m1 m2  


 u y  v y 
(u x ) (u y )

(u y ) (u x )
 ux  vy



 and u y   vx 
 1
Hence the two curves in eqns. are orthogonal curves.
Properties of Analytic functions
RESULT :- (1) An analytic function with constant
modulus is constant.
PROOF:Let f(z) = u + i v be an analytic function
 f ( z)  u 2  v2
f ( z)  c
Given :
i.e.,
u v c
2
2
 u 2  v 2  c 2  (1)
Properties of Analytic functions
Diff . eqn.(1) partially w.r .t . x, weget ,
2uu x  2vvx  0
 uu x  vvx  0  (2)
Diff . eqn.(1) partially w.r .t . y , weget ,
2uu y  2vv y  0
 uu y  vv y  0  (3)
Since f ( z ) is analytic, it satisfies C  R equations
i.e., u x  v y and u y   vx

(2)  uu x  v( u y )  0  uu x  vu y  0
(3)  uu y  v (u x )  0
 uu y  vu x  0
Properties of Analytic functions
Squaring and adding the above equations, we get,
 uu
 vu y    uu y  vu x   0
2
x
2
 u 2u x2  v 2u y2  2uvu x u y  u 2u y2  v 2u x2  2uvu y u x  0
 u 2 u x2  u y2   v 2 u x2  u y2   0
  u 2  v 2   u x2  u y2   0
But u  v  c  0 ( from eqn. (1))
2
2
2
 ux  u y  0
2
2
 (4)
Properties of Analytic functions
Since
f ( z)  u  i v
f ( z )  u x  i vx
 ux  i u y

( by C  R eqns.)
f ( z )  u x2  u y2
 f ( z )  u x2  u y2
2
0
 from (4) 
 f ( z )  0
 f ( z ) is a constant
 An analytic function with constant modulus is constant.
Properties of Analytic functions
RESULT :- (2)
x+iy then
If f(z) = u+iv is a regular function of z =

2

f ( z )   4 f ( z )
2
2
Pr oof : 
2 
 2
2
2

To prove that 

f
(
z
)

4
f
(
z
)
2
2 

x

y


Let
f ( z )  u  iv
f ( z )  u  iv

f ( z ) f ( z )   u  iv  u  iv   u 2  v 2

f ( z)
2
 u 2  v2
Properties of Analytic functions
Now,
2 
2  2
 2
 2
2
2

f
(
z
)


u

v

 x 2 y 2  
 x 2 y 2 




2 2
2 2
2 2
2 2
 2 (u )  2 (v )  2 (u )  2 (v )
x
x
y
y
 (1)
 2
Now, consider ,
 u   2uu x
x
2 2

2

u

2
uu

2
uu

2
u


 
x
xx
x
2
x
x
2 2
2
llly
u

2
uu

2
u


yy
y
y 2
Properties of Analytic functions
 2u 2  2u 2
2
2



2
u
u

u

2
u

u



xx
yy
x
y 
2
2
x
y
 f ( z ) is analytic 
 2 u (0)  u x  u y  

u
is
harmonic


 f ( z ) is analytic,

2
2
 2  u x  ( vx )  


C

R
eqns
.
satisfied


2
 2 u x2  vx2
 2 f ( z )
2
2

 f ( z )  u x  i vx



2
2
  f ( z )  u x  vx 
Properties of Analytic functions
llly
 2v2  2v2
2
 2  2 f ( z )
2
x
y
2 
 2
2
2
2
 (1)   2  2  f ( z )  2 f ( z )  2 f ( z )
y 
 x
 4 f ( z )
Thus proved
2
EXAMPLES
1) If f ( z )  e x (cos y  i sin y ) is analytic function
prove that u , v are harmonic functions.
Solutions : 
To prove that u and v are Harmonic functions
i.e., T .P.T .
Here
u xx  u yy  0 and vxx  v yy  0
u  e x cos y
v  e x sin y
u x  e x cos y
vx  e x sin y
u xx  e x cos y
vxx  e x sin y
u y   e x sin y
v y  e x cos y
u yy   e x cos y
v yy   e x sin y
EXAMPLES
 u xx  u yy  e x cos y  e x cos y  0
and
vxx  v yy  e x sin y  e x sin y  0
 Both u & v satisfies Laplace equation
Hence u & v are Harmonic functions.
EXAMPLES
EXAMPLES
CONSTRUCTION OF ANALYTIC
FUNCTION
MILNE-THOMSON METHOD :-
To find the analytic function f(z):
i)
When u(x , y) is given ( i.e., Real part is given)
f ( z)   ux ( z, 0) dz  i  u y ( z, 0) dz
Ii) when v(x ,y ) is given ( i.e., Imaginary part is given)
f ( z )   v y ( z, 0) dz  i  vx ( z, 0) dz
CONSTRUCTION OF ANALYTIC
FUNCTION
METHOD TO FIND OUT THE HARMONIC
CONJUGATE:
Let f(z) = u + i v be an analytic function.
GIVEN: u(x ,y )
 v   u y dx   u x dy


 treating y   Integrating the terms 
 as constant   independent of x 

 

EXAMPLES
If u(x ,y ) = x2 + y2 , find v(x ,y ) and hence find
f(z).
Solution : 
Given : u  x 2  y 2

ux  2 x , u y   2 y
we know that ,
v   u y dx   u x

dy

 treating y   Integrating the terms 

 

 as constant   independent of x

EXAMPLES
 v   (2 y ) dx   2 x dy


 IInd integral is zero since 
 2 xy  0 

 there is no term indep. of "x " 
v  2 xy
f ( z)  u  i v
 f ( z )   x 2  y 2   i  2 xy 
 x 2  i 2 y 2  2 x(iy )
  x i y

f ( z)  z 2
2
EXAMPLES
Find f(z),
when u(x ,y ) = x2 + y2 .
(same example, using Milne-Thomson method, finding
f(z) )
Solution : 
Given : u  x 2  y 2

ux  2 x ,

u x ( z ,0)  2 z , u y ( z ,0)  0
uy   2 y
By Milne-Thomson method ,
f ( z )   u x ( z , 0) dz  i  u y ( z, 0) dz
  2 z dz  i  0 dz

f ( z)  z 2
EXAMPLES
Show that the function u(x ,y ) = sinx coshy is harmonic.
Find its harmonic conjugate v( x , y ) and the
analytic function f(z) =u + i v .
Solution : 
Given :

u  sin x cosh y
u x  cos x cosh y
u y  sin x sinh y
u xx   sin x cosh y
u yy  sin x cosh y
u xx  u yy  0
 u is harmonic.
EXAMPLES
To find v ( x, y ) : 
we know that , v   u y dx   u x dy


 treating y   Integrating the terms 
 as constant   independent of x 

 

 V    sin x sinh y  dx    cos x cosh y  dy
  sinh y  sin x dx  0
  sinh y (  cos x)

V  cos x sinh y
since no term is 


independent of x 
EXAMPLES
Now,
f ( z )  u  i v  sin x cosh y  i cos x sinh y
 sin(iy ) 
 sin x cos(iy )  i cos x 

i


 sin x cos(iy )  cos x sin(iy )
 sin( x  iy )
 sin z

f ( z )  sin z
EXAMPLES
Construct analytic function f(z) of which imaginary part
v(x,y) = - 2 sinx (ey - e-y ).
Solution : 
Given :
v( x, y )   2 sin x  e  e
y

v   4 cos x sinh y
v   4 sin x cosh y
 v ( z , 0)  0
x
,
,
e e
y
 2sinh y 
y
y
f ( z )   v ( z , 0) dz  i  v ( z , 0) dz
y
f ( z )  4 cos z  c
y
v ( z , 0)   4sin z
  4 sin z dz


i.e., v   4 sin x sinh y
x

y
x
EXAMPLES
Find the analytic function f(z) = u+iv such that,
u+v = x3 + 3x2 y - 3xy2 - y2 + 4x + 5 and f(0) = 2+3i .
Solution : 
we knowthat , f ( z )  u  i v
i f ( z )  iu  v
 f ( z )  i f ( z )  u  i v  iu  v
 f ( z ) (1  i )   u  v   i  u  v 
F ( z)

U  iV
where F ( z )  f ( z ) (1  i )
U   u  v  , V  u  v  x  3x y  3xy  y  4 x  5
3
2
2
2
EXAMPLES
By Milne-thomson method,
F(z) =  v ( z , 0) dz  i  v ( z , 0) dz
y
Now,
v
v
x
x
 3 x  6 xy  3 y  4
y
 3 x  6 xy  2 y
2
2
2
v ( z , 0)  3 z
2
v ( z , 0)  3 z
2
x
 4
y

F ( z )   3 z dz  i   3 z  4  dz
2
3z

3
3
 3z
i
 3
2
3

 4z 

EXAMPLES

F ( z )  z  i ( z  4)  c

(1  i ) f ( z )  z (1  i )  i 4 z  c
3
3
3

i4z
c
f ( z)  z 

(1  i ) (1  i )
i 4 z (1  i )
 z 
c
(1  i ) (1  i )
3
3
1
4 z (i  1)
c
2
f ( z )  z  2 z (1  i )  c
 z 
3

1
3
1
 (1)
EXAMPLES
Given : f (0)  2  3i
put z  0 in (1), weget , f (0)  c
1

c  2  3i
1

f ( z )  z  2 z (1  i )   2  3i 

f ( z )   z  2 z  2   i (2 z  3)
3
3
EXAMPLES
EXAMPLES
EXAMPLES
CONFORMAL MAPPING
INTRO.:
Suppose two curves c1 , c2 in the Z-plane intersect at
z0 and the corresponding curves ϒ1 , ϒ2 in the w-plane
intersect at w0 by the transformation w= f(z).
if the angle between the two curves in the z-plane is
same as the angle between the curves in the w-planes
both in magnitude and in direction, then the
transformation w = f(z) is said to be conformal mapping.
DEFINITION:A transformation that preserves angles between every
pair of curves through a point both in magnitude and
sense of rotation is said to be conformal at that point.
CONFORMAL MAPPING
ISOGONAL TRANSFORMATION:-
The transformation which preserves angle between
every pair of curves in magnitude and not in
direction(sense) is called an isogonal transformation.
THEOREM:If f(z) is analytic and f'(z)≠ 0 in a region R of the z-plane
then the mapping performed by w=f(z) is conformal at all
points of R.
CONFORMAL MAPPING
CRITICAL POINTS:The point at which the mapping w=f(z) is not
conformal, i.e., f ‘(z) = 0 is called a critical point of the
mapping.
Eg . :
Consider


w  f ( z )  sin z
f ( z )  cos z
f (0)  0 , when z  

3
,
,.........
2
2
2n  1 

i.e., z 
, where n is an int eger ,
2
which are the critical points of the given transformation.
Standard Transformations

TRANSLATION
Maps of the form z → z + k, where k є C

MAGNIFICATION AND ROTATION
MAPS OF THE FORM Z → K Z , WHERE K Є C

INVERSION
MAPS OF THE FORM Z → 1 / Z
EXAMPLE FOR TRANSLATION
Find the region of the w-plane into which the rectangular region in the zplane bounded by the lines x=0, y=0, x=1, y=2 is mapped under the
transformation w=z+2-i .
Solution:-
given : w = z + 2 - i
→ (u+ iv) = (x+iy) + (2-i)
= (X+2) + i (y-1)
Equating real and imaginary parts, we get,
U = x+2
and
v = y -1
EXAMPLE FOR TRANSLATION
Given boundary lines are:
transformal boundary lines are:
X=0
U=2
Y=0
V = -1
X=1
U=3
Y=2
V=1
v
Y
U=2
U=3
V=1
Y=2
u
Y=0
V=-1
X=0
X=1
X
MAGNIFICATION AND ROTATION
Let w = a z, where a ≠ 0
if a = │a│ ei α and, z = │z │ ei θ, then
w = │a│ │z│ ei(θ + α )
the image of z is obtained by rotating the vector z through the angle
α and magnifying or contracting the length of z by the factor │a│.
Thus the transformation w = a z is referred to as a rotation or
magnification.
EXAMPLE FOR MAGNIFICATION
Determine the region R of the w plane into which the triangular
region D enclosed by the lines x = 0, y = 0, x + y = 3 is transformed
under the transformation w = 2z.
Solution:
let w = u +i v ; z = x + i y
given:
w=2z
i.e.,
u +i v = 2 (x + i y)
i.e.,
u=2x ;
v = 2 y and
u+v = 2(x+y)
EXAMPLE FOR MAGNIFICATION
When X = 0
,
u=0
Y=0
,
v=0
X+Y=3 ,
u+v=6
Thus the transformation w = 2z maps a triangle in the zplane into a 2-times magnified triangle in the w-plane.
EXAMPLE FOR ROTATION
Consider the transformation w = eiπ/4 z and determine the
region in the w-plane corresponding to triangle region bounded
by the lines x=0 , y=0 , x+y=1.
Solution : 
Given : w  e  z
i /4


u

iv

e


i /4
 x  iy 

 
 
  cos    i sin     x  iy 
4
 4 

1 
 1

i
  x  iy 
2
 2
x y x y

 i

 2   2 
EXAMPLE FOR ROTATION

x y
u
and
2
x y
v
2
Y
y
y
u
and v 
2
2
when x  0,
X+Y=1
X
 y   2 u and y  2 v
  2u  2v
V
 u  v
when y  0,
x
x
u
and v 
2
2
 uv
when x  y  1
1
 v
2
V=1/ 2
U=-V
U=V
U
EXAMPLE FOR ROTATION
The region in the z-plane is mapped on to the region
1
bounded by u = -v, u = v, v 
in the w-plane.
2
 The mapping w  ze
i  /4
performs a rotation
of R through an angle  / 4.
INVERSE TRANSFORMATION
INVERSE TRANSFORMATION
INVERSE TRANSFORMATION
EXAMPLE OF INVERSE
TRANSFORMATION
EXAMPLE OF INVERSE
TRANSFORMATION
EXAMPLE OF INVERSE
TRANSFORMATION
EXAMPLE OF INVERSE
TRANSFORMATION
Find the images of the finite strips,
1
1
1
 y
under the transformation w  .
4
2
z
Solution : 
1
Given : w 
z
1
 z
w
1
u  iv
i.e., x  iy 

u  iv u  v
u
v
 x
and y 
u v
u v
 (1)
 (2)
2
2
2
2
2
2
EXAMPLE OF INVERSE
TRANSFORMATION
1
1
Given :  y 
4
2
1
when y 
eqnuation (2) becomes ,
4
1
v

4
u v
 u  v   4v
2
2
2
2
 u  v  4v  4  4  0
2

2
u   v  2  4
2
2
which is a circle whose centre at (0, 2) and
radius is 2 in w  plane.
EXAMPLE OF INVERSE
TRANSFORMATION
1
, equation (2) becomes ,
2
1
v

2
u v
 u  v   2v
 u  v  2v  1  1  0
when y 
2

2
2
2
2
2
u   v  1  1
2
2
which is a circle whose centre is at (0, 1) and
radius is 1.
Y
V
Y=1/2
Y=1/4
Y=0
X
BILINEAR TRANSFORMATION
Def . : 
az  b
The transformation w 
, where a, b, c, d
cz  d
are complex cons tan ts and ad  bc  0 is known
as bilinear transformation.
Note : 
(i ) A bilinear transformation is also called
as Mobius transformation or a linear fractional
transformation.
az  b
 wd  b
(ii ) The inverse mapping of w 
is z 
cz  d
cw  a
is also called as a bilinear transformation.
BILINEAR TRANSFORMATION
FIXED POINTS (OR) INVARIANT POINTS :-
If the image of a point z under a transformation w=f(z)
is itself, then the point is called a fixed point or an
invariant point of the transformation.
Thus fixed point of the transformation w=f(z) is given
by z = f(z).
z
Eg .: Let w 
, find the fixed po int (or )
z2
in var ient po int .
Solution : 
z
put w  z then z 
z2
 z ( z  3)  0
 z  0, z  3 are two fixed po int s.
 z  2z  z
2
BILINEAR TRANSFORMATION
Definition of cross ratio:If z ,z ,z ,z are four points in the z-plane then
1
2
3
4
z - z  z - z 

the ratio
is called the cross ratio of these points.
 z - z  z - z 
1
2
3
4
1
4
3
2
Cross Ratio Property of a bilinear transformation:The cross ratio of four points is invariant under a
bilinear transformation.
i.e., If w ,w ,w , w are the images of z ,z ,z , z
1
2
3
4
1
respectively under a bilinear transformation then
 w - w

 w - w
1
2
1
4
 w - w   =  z - z  z - z  
  w - w     z - z   z - z  
3
4
1
2
3
4
3
2
1
4
3
2
2
3
4
BILINEAR TRANSFORMATION
Note:the bilinear transformation which transforms the points
Z 1 , z2 , z3 of z-plane respectively into the points w 1,w2 ,w3
of W-plane is given by
  w - w  w - w    z - z  z - z  

 = 

  w - w  w - w    z - z  z - z 
1
1
2
2
3
3
1
1
2
2
3
3
BILINEAR TRANSFORMATION
EXAMPLES
1) Find the bilinear transformation which maps the points
z = 0,-i,-1 into w = i,1,0.
Solution:Given : z  0, z   i, z   1
1
2
1
and w  i , w  1, w  0 .
1
2
3
The bilinear transformationion is got by using the relation
 w  w  w  w    z  z  z  z 
 w  w  w  w   z  z  z  z 
1
1
2
2
3
3
1
1
2
2
3
3
w  i 1  0   z  0  i  1



 i  1 0  w   0  i  1  z 
BILINEAR TRANSFORMATION
EXAMPLES
 ( i ) ( w  i ) (1  z )  z (1  i ) (  w) (i  1)
  i  iwz  1  z   2iwz
  iw  iwz 1  z
 w ( zi  i )  (1  z )
1 z
 w
zi  i

1 z
w
( i ) (1  z )
BILINEAR TRANSFORMATION
EXAMPLES
2) Find the bilinear transformation which transforms the po int s
z   , i ,0 int o the po int s w  0, i,  respectively.
Solution : 
Given : z   , z  i, z  0
1
2
3
and w  0, w  i , w   .
1
2
3
The bilinear transformationion is got by using the relation
 w  w   w  w    z  z  z  z 
 w  w  w  w   z  z  z  z 
1
1
2
2
3
3
1
1
2
2
3
3
BILINEAR TRANSFORMATION
EXAMPLES
w

 z

 w  w  w   1  z   1  z  z 
w
 
z



z 
w
 w  w  w  1    z   1    z  z 
z 
w 


( w  0)(0  1)
(0  1) (i  0)


(0  1)(i  0)
( z  0) (0  1)
2
1
3
1
3
2
1
2
1
2
1
3
3
(  w)
( i )


( i )
( z )

1
w
z
3
1
3
BILINEAR TRANSFORMATION
TRY IT
Find the bilinear transformation which maps the points,
I) 1, -i , 2
onto 0, 2, i respectively.
Ii) -i,0,i into -1, i, 1 respectively.
Iii)
0, 1,  into i, -1, -i respectively.