complex variable & numerical method

1. Subject :- Complex Variable & Numerical Method
(2141905)
Topic :- Transformations
Branch :- Mechanical
Semester :- 4th
•Made By :-
-Digvijasinh Gohil(150990119008)
-Jaimin Prajapati (150990119010)
-Abhijitsinh Kher (150990119012)
Guided By :-
Dr. Purvi Naik

2. Inversion
A transformation of the form 𝑤 =
1
𝑧
is called inversion.
This transformation maps circle & line into circle & line.
Ex.1:- Find the image of the line 𝑥 − 𝑦 = 1 under the transformation 𝑤 =
1
𝑧
.
Solution:
z
w
1

w
z
1


3. Comparing real & imaginary parts,
Given 𝑥 − 𝑦 = 1,
Substituting 𝑥 & 𝑦,
22
vu
ivu
iyx



22
vu
v
y



22
vu
u
x


12222









 vu
v
vu
u

4. vuvu  22
Which represents a circle with the centre at
1
2
,
1
2
and radius
1
2
.
022
 vuvu
2
1
2
1
2
1
22












 vu

5. Ex.2 :- Find the image of 𝑧 + 1 = 1 under the mapping 𝑤 =
1
𝑧
.
Solution:
z
w
1

w
z
1

ivu 

1
22
vu
ivu
iyx




6. Comparing real and imaginary parts,
Given
,22
vu
u
x


22
vu
v
y



11 z
11  iyx
1)1(
2
 iyx
1)1( 22
 yx
1222
 xyx

7. Substituting 𝑥 & 𝑦,
02
2
2222
2
22





















 vu
v
vu
u
vu
u
 
0
2
22222
22





vu
u
vu
vu
0
21
22



vu
u
021  u
2
1
u

8. Which represent a straight line.
Hence, the image of the circle 𝑧 + 1 = 1 in z-plane is the straight
line 𝑢 = −
1
2
in the w-plane .

9. Ex.3 :- Find the image of 𝑧 + 2𝑖 = 2 under the transformation 𝑤 =
1
𝑧
.
Solution:
z
w
1

w
z
1

ivu 

1
22
vu
ivu
iyx




10. Comparing real and imaginary parts,
Given,
,22
vu
u
x

 22
vu
v
y



22  iz
22  iiyx
4)2(
2
 yix
4)2( 22
 yx
0422
 yyx

11. Substituting 𝑥 & 𝑦,
04
2
2222
2
22





















 vu
v
vu
u
vu
u
 
0
4
22222
22





vu
v
vu
vu
041  v
4
1
v
Which represent a straight line.
Hence, the image of the circle 𝑧 + 2𝑖 = 2 in z-plane is the straight line 𝑣 =
1
4
in the w-plane.

13. Square Transformation
 Let z=r𝑒 𝑖𝛉 & w=R𝑒 𝑖𝜙 than the square transformation is defined as w=𝑧2 .
R𝑒 𝑖𝜙=(r𝑒 𝑖𝛉)2
R=𝑟2, 𝜙=2𝛉
Ex.1:- Find the image of the triangular region bounded by the lines 𝑥 = 1, 𝑦 =
1, 𝑥 + 𝑦 = 1 under the transformation w=𝑧2
.
Solution: w=𝑧2
𝑢 + 𝑖𝑣 = (𝑥 + 𝑖𝑦)2
𝑢 + 𝑖𝑣 = 𝑥2
− 𝑦2
+ 2𝑖𝑥𝑦

14. Comparing real & imaginary parts,
𝑢 = 𝑥2
− 𝑦2
, 𝑣 = 2𝑥𝑦
1) When𝑥 = 1,
𝑢 = 1 − 𝑦2, 𝑣 = 2𝑦
∴y =
v
2
Substituting 𝑦 in 𝑢,
𝑢 = 1 −
𝑣2
4
4𝑢 = 4 − 𝑣2
∴ 𝑣2 = −4(𝑢 − 1)
Which represents a parabola with vertex (1,0) and opening to left of the vertex.
The image of the line 𝑥 = 1 is the parabola 𝑣2 = −4(𝑢 − 1).

15. 2) When 𝑦 = 1,
𝑢 = 𝑥2
− 1, 𝑣 = 2𝑥
∴ 𝑥 =
𝑣
2
Substituting 𝑥 in 𝑢,
𝑢 =
𝑣2
4
− 1
4𝑢 = 𝑣2 − 4
∴ 𝑣2
= 4(𝑢 + 1)
Which represents a parabola with vertex (-1,0) and opening to right of the
vertex.
The image of the line y= 1 is the parabola 𝑣2 = 4(𝑢 + 1).

16. 3) When 𝑥 + 𝑦 = 1,
𝑥 + 𝑦 2 = 1
𝑥2 + 𝑦2 + 2𝑥𝑦 = 1
𝑥2
+ 𝑦2 2
= 1 − 2𝑥𝑦 2
𝑥2 − 𝑦2 2=1 − 4𝑥𝑦
Substituting 𝑢 = (𝑥2
− 𝑦2
) and 𝑣 = 2𝑥𝑦,
𝑢2 = 1 − 2𝑣 = −2 𝑣 −
1
2
Which represents a parabola with vertex 0,
1
2
opening below the vertex.
The image of the line 𝑥 + 𝑦 = 1 is the parabola 𝑢2 = 1 − 2𝑣.

18. Ex2:- Find the image of the square with vertices (0,0), (2,0), (2,2), (0,2) in the z-
plane under the transformation 𝑤 = 𝑧2 .
Solution:
let ABCD be the square in the z-plane with vertices A(0,0), B(2,0), C(2,2), D(0,2) .
1) For AB y=0,
u = 𝑥2
, 𝑣 = 0
Which represents the right part of the right part of the u-axis where u>0.
The image of the line y=0 is the right part of the u-axis (u>0).
w=𝑧2
𝑢 + 𝑖𝑣 = (𝑥𝑖𝑦)2
𝑢 + 𝑖𝑣 = 𝑥2
− 𝑦2
+ 2𝑖𝑥𝑦
Comparing real & imaginary parts,
𝑢 = 𝑥2 − 𝑦2, 𝑣 = 2𝑥𝑦

19. 2) For BC x=2,
𝑢 = 4 − 𝑦2 , 𝑦 = 4𝑦
𝑦 =
𝑣
4
Substituting y in u,
𝑢 = 4 −
𝑣2
16
∴ 𝑣2 = −16(𝑢 − 4)
Which represents a parabola with vertex (4,0) and opening to left of the vertex.
The image of the line x=2 is the parabola 𝑣2 = −16(𝑢 − 4).

20. 3) for CD y=2,
𝑢 = 𝑥2
− 4, 𝑣 = 4𝑥
𝑥 =
𝑣
4
Substituting x in u,
𝑢 =
𝑣2
16
− 4
∴ 𝑣2
= 16(𝑢 + 4)
Which represents a parabola with vertex (-4,0) and opening to right of the
vertex.
The image of the line y=2 is the parabola 𝑣2
= 16(𝑢 + 4).

21. 4) For DA x=0,
u = −𝑦2
, 𝑣 = 0
Which represents the left part of the right part of the u-axis where u<0.
The image of the line x=0 is the right part of the u-axis (u<0).

22. Ex.3:- Find the image of the region bounded by 1 ≤ r ≤ 2 and
π
6
≤ 𝛉 ≤
𝜋
3
in the z-
plane under the transformation 𝑤 = 𝑧2
. Show the regions graphically.
Solution:
𝑤 = 𝑧2
Comparing both the sides,
Given 1 ≤ r ≤ 2 ,
1 ≤ 𝑟2 ≤ 2
1 ≤ R ≤ 2
R𝑒 𝑖𝜙=(r𝑒 𝑖𝛉)2
R=𝑟2, 𝜙=2𝛉

23. And
π
6
≤ 𝛉 ≤
𝜋
3
π
6
≤
𝜙
2
≤
≤ 𝜙 ≤
2𝜋
3
Hence, the image of the region bounded by 1 ≤ r ≤ 2 and
π
6
≤ 𝛉 ≤
𝜋
3
in the z-plane is
the region 1 ≤ R ≤ 2 and
π
3
≤ 𝜙 ≤
2𝜋
3
in the w-plane.

24. Example1 :- Find the image of 2x+y-3=0 under the transformation w=z=2i
Solution
W=z+2i
u+iv=x+iy+2i
=x+i(y+2)
Comparing real and imaginary parts,
u=x, v=y+2
x=u, y=v-2
Given
2x+y-3=0
2u+(v-2)-3=0
2u+v-5=0

25. Hence, the 2x+y-3=0 in the z-plane is mapped onto a line 2u+v-5=0 in
the w-plane

26. Example 2:- Detrmine and sketch the image of 𝑧 =1 under the transformation
w=z+i.
Solution :-
W=Z+I
u+iv = x+iy+I
= x+I (y+i)
Comparing real and imaginary parts,
u=x, v=y+1
x=u, y=v-1
𝑧 =1
𝑥2 +𝑦2 =1
𝑥2 +𝑦2 =1
𝑢2
+(𝑣 − 1)2

27. Hence the image of the circle 𝑧 =1 in the z- plane is the
circle 𝑢2
+(𝑣 − 1)2
with the centre at (0,1) and radius of 1 in the w-plane

28. Example 3:- Find the image of 𝑧 =2 under the mapping w=z+3+2i.
W=z+3+2i
u+iv = x+iy+3+2i
=(x+3i)+i(y+2)
Comparing real and imaginary parts,
u=x+3, v=y+2
x=u-3, y=v-2
𝑧 =2
𝑥2 +𝑦2 =2
𝑥2 +𝑦2 =4
(𝑢 − 3)2 +(𝑣 − 2)2=4

29. Hence, the circle 𝑧 =2 in the z- plane is the
circle with the centre at (3,2) and radius of 2 in the w-plane

31. Example 1:- Find the fixed points of the transformation W=
𝟔𝒁−𝟗
𝒁
.
Solution :-
The fixed points are obtained by putting w=z.
z=
6𝑍−9
𝑍
𝑧2=6z-9
𝑧2-6z+9=0
(𝑧 − 3)2
=0
Z=3,3

32. Example2 :- Find the invariant of the transformation W=
𝟏+𝒛
𝟏−𝒛
.
Solution:-
The invariant points are obtained by putting w=z.
Z=
1+𝑧
1−𝑧
z-𝑧2
=1+z
𝑧2
=-1
(Z=+i) and( z=-i)

33. Example 3:- find the invariant points of the transformation w=
𝟐𝒛+𝟔
𝒛+𝟕
.
Solution :-
The invariant points are obtained by putting w=z.
W=
2𝑧+6
𝑧+7
𝑧2+7z=2z+6
𝑧2+5z-6=0
(z+6)(z-1)=0
Z=-6,1

34. Examples 4:- Find the fixed points of the mapping w=
𝟏
𝒛+𝟐𝒊
.
Solution :-
The fixed point are obtained by putting w=z
Z=
1
𝑧+2𝑖
𝑧2+zi=1
𝑧2+2zi-1=0
Z=
−2𝑖± −4+4
2
=-i