B.Sc. Sem V Lecture Notes

1. 1
MATHEMATICS
E – Content
ON ANALYSIS
Vinay M. Raut
Assistant Professor
Shri Shivaji Science College,
Amravati

2. 2
CONFORMAL MAPPING
Content :
• Transformation
• Type of transformation
• Bilinear or Mobius transformation
• Circle under bilinear transformation
• Cross ratio
Transformation :
We know that w f(z)= for complex plane
u iv f (x iy)⇒ + = +
u u(xy)and v v(xy)⇒ = =
These equation define transformation or mapping from z-plane into w-plane the points z and w are called
the image of each after as shown in fig.
Here we write i
z re θ
= and i
w e φ
= ρ
CONFORMAL MAPPING
A mapping that preserves angle between any oriented curves both in magnitude and sense is a
conformal mapping.
Transformation which presence the magnitude of angle but not the sense are called isogonal.
( andθ φ have same magnitude and same sense)
Types of Transformation :
1. Translation : It is defined by w z= + α where α - is complex constant.
By this transformation each point in the z-plane is moved along the direction α through a
distance α by w z .= + α The figure in the z-plane are displaced in the direction of α keeping
their shape, size and orientation unchanged.

3. 3
- EXAMPLES –
Example 1 : Find the image of triangle with vertices 2, 1 i,1 i+ − under w z 1 i= − +
Solution : Let the vertices of the triangle in z-plane are demoted by 1 2 3z 2,z 1 i, z 1 i= = + = −
Here we have to find the vertices of triangle in w-plane and shape given transformation w z 1 i= − +
(translation type)
Now for z1 we have 1 1w z 1 i= − +
1 1w 2 1 i w 1 i⇒ = − + ⇒ = +
Similarly for z2 we have
2 2w z 1 i 1 i 1 i 2i= − + = + − + =
For 3 3z w O⇒ =
The image of triangle in z-plane is transformed to w-plane as shown in figure.
z-plane w-plane
Example 2 : Find the image of circle z 2= under the transformation w z 3 2i= + +
Solution : Given transformation
w z 3 2i= + +
z w (3 2i)⇒ = − +
z w (3 2i)⇒ = − + (taking mod)
2 w (3 2i)⇒ = − + ( z 2)=∵
There in the form of equation of circle where center in at 3 2i+ and radius 2. Thus equation of circle
z 2= in z-plane is transformed to w-plane as shown in figure

4. 4
2. Rotation :
Let the mapping w f(z)= be in the form i 0
w z.e θ
= where 0 realconstantθ =
By this transformation figure in the z-plane are rotated through an angle 0θ in w-plane.
Example 3 : With the help of transformation i /4
w ze π
= determine the region in w-plane corresponding to
the triangular region bounded by the lines x 0,y 0= = and x y 1+ = in z-plane
Solution : Given transformation i /4
w ze π
= (rotation type) put the value of w and z
u iv (x iy) cos isin
4 4
π π 
+ = + + 
 
1 1
(x iy) i
2 2
 
= + + 
 
(By Euler formula i
e cos isinθ
= + θ )
1
(x iy)(1 i)
2
= + +
[ ]
1 1
(x xi iy y) (x y) i(x y)
2 2
= + + − = − + +
Equate real and Imaginary part on both sides
1
u (x y)
2
= − and
1
v (x y)
2
= + …………(1)
Now the region z-plane is bounded by the lines x 0,y 0, x y 1= = + =
For x 0= equation
1
(1) u ( y),
2
⇒ −
1
v (y) u v 0
2
= ⇒ + =
For y 0= equation
x x
(1) u ,v u v 0
22
⇒ = = ⇒ − =
For x y 1+ = equation
1
(1) v
2
⇒ =
Transform region in w-plane is bounded by the lines u v 0,+ = u v 0− = and
1
v
2
= Which in as shown
in figure.

5. 5
Example 4 : Find the image of the rectangular book by x 0,y 0,x 2= = = and y 3= under the
transformation w 2= i /4
e zπ
Solution : Given transformation w 2= i /4
e zπ
i /4
u iv 2. e (x iy)π
⇒ + = +
u iv 2 cos isin (x iy)
4 4
π π 
⇒ + = + + 
 
(by euler)
1 1
u iv 2 i (x iy)
2 2
 
⇒ + = + + 
 
u iv (1 i)(x iy)⇒ + = + +
(x y) i(x y)= − + +
Equate the real and Imaginary part u x y,v x y= − = + Now the region in z-plane in bounded by line
x 0,= y 0,= x 2= and y 3=
For x 0= equation (1) u y,v y u v o⇒ = − = ⇒ + =
For y 0= equation (1) u x,v x,v x u v⇒ = − = = ⇒ =
For x 2= equation (1) u 2 y,v 2 y u v 4⇒ = − = + ⇒ + =
For y 3= equation (1) v x 3 i.e. u v 6⇒ = + − = −
Transformation region in w-plane is bounded by lines u v 0,+ = u v 0− = and u v 6− = − which is as
shown in figure.
Note : To plot the u v 4+ = we put
First u 0= we get v 4=
∴ Intercept in (0,4)
Similarly v 0 u 4= ⇒ =
∴ Intercept is (4,0)
u v 6− = −
3. Stretching or Magnification:

6. 6
This type of mapping is given by w az ,= a is real
4. Inversion :
This type mapping is defined by
1
w
z
=
It’s geometrical transformation is related to the inverse point with respect to the circle
Theorem 1 :
Under the transformation
1
w
z
= a straight line L in the z-plane is mapped into
a) a circle if does not pass through the origin z 0=
b) a straight line if L passes through the origin z 0=
Proof : Rewrite the mapping as
1
z
w
= i.e.
1
x iy
u iv
+ =
+
2 2
1 u iv u iv
x iy
u iv (u iv)(u iv) u v
− −
⇒ + = = =
+ + − +
2 2 2 2
u v
i
u v u v
= −
+ +
Equate real and Imaginary part on both sides
2 2 2 2
u v
x ,y
u v u v
= =
+ +
…………(1)
Any line L in the z-plane is given by the equation Ax By C 0+ + = where A, B, C are real for C 0,≠ it
does not pass through the origin …………(2)
2 2 2 2
u v
Equation (1) A B C 0
u v u v
   
∴ ⇒ + + =   
+ +   
2 2
C(u v ) Au Bv 0⇒ + + = = …………(3)
This represent a circle in the w-plane for C 0≠ hence the straight line L which does not pass
through the origin in mapped into a circle.
If L passes through the origin thenC 0= and then equation (3)⇒
Au Bv 0− =
This is straight line in the w-plane passes through the origin w 0=
Theorem 2 : The transformation
1
w
z
= always maps a circle into a circle with a straight line consider as
a circle of infinite radius.

7. 7
Proof : Now
1
w
z
= on solving we get
2 2
u
x ,
u v
=
+ 2 2
v
y
u v
= −
+
............(1)
Let 2 2
x y 2gx 2fy C 0+ + + + = ............(2)
Be any circle in the z-plane with C 0≠ It means that the circle does not pass through the origin
By equation (1) we get
2 2
2 2 2 2 2 2 2 2
u v u v
2g 2f C 0
u v u v u v u v
−       
+ − + + + =       
+ + + +       
2 2
2 2 2 2 2 2 2
u v u v
2g 2f C 0
(u v ) u v u v
+    
⇒ + + − + =   
+ + +   
2 2
C(u v ) 2gu 2 v 1 0⇒ + + − + + = …………(3)
Since C 0≠ , above represent a circle in the w-plane which also does not pass through the origin w 0.=
If C 0= then the circle equation (2) passes through z 0= and is mapped to equation (3) with C 0.=
2gu 2fu 1 0− + =
This is straight line in the w-plane considering a straight line as a circle of infinite radius above
equation represents a circle with infinite radius.
Hence every circle is mapped into circle under
1
w .
z
=
Example 5 : Under the transformation
1
w
z
= find the image of the circle z 2i 2− =
Solution : We know that
1
w
z
=
2 2 2 2
u v
x ,y
u v u v
⇒ = = −
+ +
…………(1)
The given circle is
z 2i 2− = i.e. x i(y 2) 2+ − =
2 2
x (y 2) 2⇒ + − =
2 2 2
x (y 2) 2⇒ + − = (squaring)
2 2
x y 4y 0⇒ + − = …………(2)
Under equation (1) equation (2) transform to
2 2
2 2 2 2 2 2
u v v
4 0
u v u v u v
−     
+ − − =     
+ + +     
1 4v 0⇒ + = or
1
v
4
= −
This is a straight line parallel to the u-axis in the w-plane.

8. 8
Example 6 : Under the transformation
1
w
z
= find the image of the infinite strips
(a)
1 1
y
4 2
≤ ≤ and (b)
1
o y
2
< <
Solution : We know that from
1
w
z
= we get, 2 2
u
x ,
u v
=
+ 2 2
v
y
u v
= −
+
a) Under this equation the strip
1 1
y
4 2
≤ ≤ transform to 2 2
1 v 1
4 u v 2
−
≤ ≤
+
2 2
u v 4v 0⇒ + + ≤ and 2 2
u v 2v 0+ + ≥
2 2 2
(u 0) (v 2) 2⇒ − + + ≤ and 2 2 2
(u 0) (v 1) 1− + + ≥
Hence the transformed region in the w – plane consists of
1. Boundaries of the two circle
2 2 2
(u 0) (v 2) 2− + + ≥
This is circle with centre (0, 2)− and r 2= and 2 2 2
(u 0) (v 1) 1− + + = this is circle with center
(0, 1)− and radius 1
2. The area enclosed by the boundaries of these circle
The required region is shaded region
b) The region
1
0 y
2
< < is mapped into 2 2
v 1
0
u v 2
< − <
+
i.e. v 0< and 2 2 2
(u 0) (v 1) 1− + + > It
consists of the lower half of the w-plane without interior and the boundary of the circle
2 2 2
(u 0) (v 1) 1− + + = It is as shown in figure.
The Bilinear or Mobius Transformation :
The transformation
z
w ,
z
α + β
=
γ + δ
oαδ −βγ ≠ is called linear or fractional transformation where
, , andα β γ δ are complex constant.

9. 9
Theorem : A General linear transformation is a combination of the transformation of translation,
rotation and stretching.
Proof : Consider linear transformation
W z= α +β …………(1)
Where α and β are complex constant
Let i 0
a e 0θ
α = ≠ …………(2)
We define the following transformation
3f (z) azλ = = is magnification
i 0
2f ( ) e θ
ρ = λ = λ is rotating
1w f ( )= ρ = ρ +β is translation
Consider,
1 2 3 1 2 3(f f f )(z) f f [f (z)]=
1 2(f f )(az)=
1 2f [f (az)]=
i 0
1f (e az)θ
=
i 0
e az zθ
= +β =α +β [By equation (2)]
f(z)= [By equation (1)]
w=
Thus the L. T. is combination of translation rotation and magnification.
Theorem : The bilinear transformation is a combination of translation, rotation stretching and inversion.
Proof : The bilinear transformation is given by
z
w ,
z
α + β
=
γ + δ
oαδ −βγ ≠ …………(1)
Where , , andα β γ δ are complex constant
∴ equation (1) can be written as
(z / )
W
(z / )
α + β γ
=
γ + δ γ
Add /±δ γ
[(z / ) ( / / )]
[z / ]
α + δ γ + β α −δ γ
=
γ + δ γ
/ /
/ 1
(2 / )
 β α −δ γ
=α γ + 
+δ γ 
W
z
µ
= λ +
+ η
…………(2)

10. 10
Where,
α
λ =
γ
( / / )α β α −δ γ
µ =
γ
/η =δ γ
We define
1 2 3 1 2 3f f f (z) f f [f (z)]=
1 2f f (z )= + η
1
1
f
z
 
=  
+ η 
z
µ
= + λ
+ η
W=
f(z)=
Thus the B.T. is combination of L.T., Inversion and translation. But L.T. is combination of
magnification, translation and rotation.
∴ B. T. in combination of translation, rotation magnification and inversion.
Fixed Point of Invariant Point :
The fixed or invariant point of the transformation w f(z)= are the points given by z f (z)=
Theorem : Every bilinear transformation with a single non-infinite fixed point can be put in the form.
1 1
k
W P Z P
= +
− −
(Proof in Book)
Theorem 2 : Every B. T. with two non-infinite fixed point P and Q is of the form
w P z P
k
w Q z Q
− −
=
− −
(Proof in Book)
Where ‘k’ is constant
Note : The B. T. having two fixed point P and Q is given by
w P z P
k
w Q z Q
− −
=
− −
where ‘k’ is constant then B. T. is
1. Parabolic : Fixed point one
2. Elliptic : Fixed point two, k 1=
3. Hyperbolic : Fixed point two, k 1≠ and
4. Loxodromic : Other than above
Example 7 : Find the fixed point and normal form of the B. T.
3z 4
w
z 1
−
=
−

11. 11
Solution : The give transformation is
3z 4
w
z 1
−
=
−
The fixed point by the transformation in is given by
3z 4
z
z 1
−
=
−
(by putting w z= )
2
z 4z 4 0⇒ − + =
2
(z 2) 0⇒ − =
z 2 P(say)⇒ = =
This has only on fixed point
∴ form of the transformation is
1 1
k
w P z P
= +
− −
1 1
3z 4w P P
z 1
=
−− −
−
1 1 z 1
3z 4w 2 3z 4 2z 22
z 1
−
⇒ = =
−− − − +−
−
z 1 z 2 1 1
1
z 2 z 2 z 2
− − +
= = = +
− − −
1 1
1
w 2 z 2
∴ = +
− −
Example 8 : Find the fixed point of B. T.
z
w
2 z
=
−
write its normal form state which is of type.
Solution : The given transformation is
z
w
2 z
=
−
…………(1)
The fixed point of the transformation (1) are given by
2z
z 2z z z
2 z
= ⇒ − =
−
2
z z 0⇒ − =
z(z 1) 0⇒ − =
z 0 or z 1⇒ = =
Let P 0= and Q 0= thus the given transformation has two fixed point. Thus the required transformation
must be of the form
w P z P
k
w Q z Q
− −
=
− −
Consider
w P z z
P Q
w Q 2 z 2 z
−
= − −
− − −

12. 12
z
0
w 0 2 z
zw 1 1
2 z
−
− −=
− −
−
Z 0 Z 0
2 z z 2z 2
− −
= =
− + + −
w 0 1 z 0
w 1 2 z 1
− − 
⇒ =  
− − 
Here
1 1 1
K K
2 2 2
= ⇒ = =
Then the given transformation in hyperbolic
Ex. Find the fixed point of the transformation find the normal form and find its type.
Ex. Find the fixed point of the transformation
z 1
w
z 1
−
=
+
show that the transformation in elliptic
CIRCLE UNDER BILINEAR TRANSFORMATION
We have different from of equation representing a circle and they are as follows
C1 : Real equation of circle can be written as
2 2
x y 2gx 2fy C 0,+ + + + = g,f,C are real center ( g, f )= − − and radius 2 2
g f C= + −
C2 : Complex equation of circle are on follows
1. Z r,− α = α is complex, centre at z =α and radius = r
2. z 1= is unit circle
3. General equation zZ Bz Bz C 0+ + + = where ‘C’ is real center Z B= − and radius
BB C= −
4. 1
2
Z Z
k,
Z Z
−
=
−
K 0≥ this is family of equation of circle such that for k 1= it gives a straight line of
which 1 2z ,z are symmetric point.
Theorem : If 0z is in the upper half of z-plane then the bilinear transformation
i 0
0
z z
w e
z z
θ  −
=  
− 
Maps the upper half of the z-plane into the interior of the unit circle in the w-plane
Proof : Proof in book.
Example 9 : Show that the map of real areas of z-plane onto w-plane by transformation
1
w
z i
=
+
is a
circle find its center and radius

13. 13
Solution : The real axis of z-plane is y 0= (z x iy)= +
If becomes z z 0− = …………(1)
z z 2iy
1
y (z z)
2i
− =


⇒ = −
∵
The given mapping
i.e.
1
w
z i
=
+
1
z i
w
⇒ + =
1
z i
w
⇒ = − + and
1
z i
w
= +
Then equation (1)
iw w w iw w w 0− + − − =
2i w w (w w) 0⇒ + − =
2 2
2i(u v ) 2iv 0⇒ + + =
w u iv
w u iv
w w 2iv
= +

= −
⇒ − =
2 2
u v v 0⇒ + + =
2 2
2 1 1
(u o) v
2 2
   
⇒ − + + =   
   
This represent a circle in w-plane with center
1
o,
2
− 
 
 
and radius
1
2
Example 10 : Prove that under the transformation
z i
w
iz 1
−
=
−
the region Imz o≥ is mapped into the
region w 1≤
Solution : For complex number z x iy= +
Imaginary of z i.e. Imz y=
Given transformation
z i
w
iz 1
−
=
−
Solving for Z we obtain
w i
z
iw 1
−
=
−
…………(1)
Given Imz 0 y 0≥ ⇒ ≥
i.e.
1
(z z) 0
2i
− ≥ then equation (1) ⇒

14. 14
1 w i w 1
0
2i iw 1 iw 1
− + 
− ≥ − − − 
ww 1
0
(ww 1)
−
⇒ ≥
− +
ww 1 0⇒ − ≤
ww 1⇒ ≤ ( ww 1 real 0)+ = >∵
Or w 1≤
Hence Imz 0≥ maps into the region w 1≤
Example 11 : Show that the transformation
2z 3
w
z 4
+
=
−
maps the circle 2 2
x y 4x 0+ − = into straight
line 4u 3 0+ =
Solution : As z x iy= + and z x iy= −
1
z z 2x x (z z)
2
∴ + = ⇒ = +
and 2 2i
y (z z), zz x y
2
= − − = +
Similarly for w u iv= + and w u iv= −
1
u (w w)
2
⇒ = +
and
i
v (w w)
2
= − − also 2 2
ww u v= +
putting this value in given equation of circle i.e. 2 2
x y 4x 0+ − =
we get z z 2(z z) 0− + = …………(1)
Now given transformation is
2z 3
w
z 4
+
=
−
Solving for Z we get
4w 3
z
w 2
+
=
−
Equation (1) ⇒
4w 3 4w 3 4w 3 4w 3
. 2 0
w 2 w 2 w 2 w 2
+ + + + 
− + = − − − − 
(4w 3)(4w 3) 2(4w 3)(w 2) 2(4w 3)(w 2) 0⇒ + + − + − − + − =
2(w w) 3 0⇒ + + =
2(2u) 3 0⇒ + = (asw w 2u)+ =
4u 3 0⇒ + =
4u 3 0+ =
This is equation of straight line thus circle is mapped onto the straight line 4u 3 0+ =

15. 15
Example 12 : Show that the transformation
5 4z
w
4z 2
−
=
−
transform the circle Z 1= into the circle of
radius unity in the w-plane.
Example 13 : Show that the relation
iz 2
w
4z i
+
=
+
transform the real axis in the z-plane into a circle in the
w-plane find the center and radius of circle.
Cross Ratio : If Z1, Z2, Z3, Z4 are district then the quantity
4 1 2 3
1 2 3 4
2 1 4 3
(Z Z )(Z Z )
(Z ,Z ,Z ,Z )
(Z Z )(Z Z )
− −
=
− −
…………(1)
is called the cross ratio of Z1, Z2, Z3, Z4
Easy way to Remember : Form the 4 cyclic difference of Z1, Z2, Z3, Z4
1 2Z Z− 2 3Z Z− 3 4Z Z− 4 1Z Z−
(1) (2) (3) (4)
Take (2) and (4) in numerator and (1) and (2) in denominator. This gives R.H.S. of equation (1)
Theorem : The cross Ratio remains invariant under a bilinear transformation
Proof : Consider a bilinear transformation
az d
w ,
cz d
+
=
+
ad bc 0− ≠
Assume that w1, w2, w3, w4 are the transform of Z1, Z2, Z3, Z4 respectively
then 4 1
4 1
4 1
az b az b
w w
cz d cz d
+ +
− = −
+ +
4 1 1 4
4 1
(az b)(cz d) (az b)(cz d)
(cz d) (cz d)
+ + − + +
=
+ +
4 1
4 1
z (ad bc) z (bc ad)
(cz d)(cz d)
− + −
=
+ +
4 1
4 1
4 1
(ad bc)(z z )
w w
(z d)(z d)
− −
⇒ − =
+ +
…………(1)
Similarly 2 3
2 3
4 1
(ad bc)(z z )
w w
(cz d)(cz d)
− −
− =
+ +
…………(2)
2 1
2 1
2 1
(ad bc)(z z )
w w
(cz d)(cz d)
− −
− =
+ +
…………(3)
and 4 3
4 3
4 3
(ad bc)(z z )
w w
(cz d)(cz d)
− −
− =
+ +
…………(4)

16. 16
4 1 2 3
1 2 3 4
2 3 4 3
(w w )(w w )
(w w w w )
(w w )(w w )
− −
∴ =
− −
4 1 2 3
2 1 4 3
(z z )(z z )
(z z )(z z )
− −
=
− −
1 2 3 4(z ,z ,z ,z )=
Thus the cross ratio is invariant under bilinear transformation
Example 14 : Find the bilinear transformation which maps the points z 1,= 1i 1− into the points
w 0,1,= ∞
Solution : Put the values of z and w in
az b
w
bz d
+
=
+
we get.
a b ai b a b
0 ,1 ,
C d Ci d C d
+ + − +
= = ∞ =
+ + − +
a b 0,Ci d ai b, C d 0⇒ + = + = + − + =
Solving these equation for a, c, d in terms of b we obtain
a b= − c d bi= = −
With these values the transformation
az b
w
Cz d
+
=
+
becomes
z 1
iz i
− +
=
− −
(We can try this by cross ratio method).
Example 15 : Find a bilinear transformation which maps points z o, i,= − − l into w i, , o= l respectively
Solution : Let the B.T. be
az b
w
cz d
+
=
+
…………(1)
For z 0,w i,= = equation
a(0) b b
(1) i
C(0) d d
+
⇒ = =
+
b id⇒ = …………(2)

17. 17
z i,w 1= − = equation
a( i) b
(1) i
C( i) d
− +
⇒ =
− +
…………(3)
z 1,w 0= − = , equation
a( 1) b
(1) o
(C 1) d
− +
⇒ =
− +
…………(4)
b a⇒ =
(2) and (4) id a⇒ = i.e. d ia= −
Then (3) C( i) ia ia a⇒ − − = − +
C ia⇒ =
Putting the values of b, c, d in terms of a in equations (1) we get
az a z 1 z 1
w i
ia z 1a i(Z 1) z 1
+ + +
= = = −
− − −
This is required transformation.
Objective Question :
1. A mapping that preserves angles between any oriented curves both in ………… and …………
is a conformal mapping.
2. Translation transformation is defined by the equation …………
3. Rotation transformation is defined by the equation …………
4. Stretching transformation is defined by the equation …………
5. Inversion transformation is defined by the equation …………
6. Bilinear Transformation given by …………
7. The fixed or invariant points of the transformation is given by …………
8. Normal form of B. T. with one fixed point say α is given by …………
9. Normal form of B. T. with two fixed point say P and Q is given by …………
10. The bilinear transformation with only one fixed point is called ………… transformation.
11. A bilinear transformation which is not parabolic, elliptic or hyperbolic is called …………
transformation.
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