### 181202(MA 2161) – MATHEMATICS - II UNIT 3 notes

Unit.3 Analytic Functions
1
UNIT III ANALYTIC FUNCTIONS
Part-A
Problem 1 State Cauchy – Riemann equation in Cartesian and Polar coordinates.
Solution:
Cartesian form:
u v
x y
 

 
, u v
y x
 
 
 
Polar form:
u 1 v , v 1 u
r r  r r 
   
  
   
Problem 2 State the sufficient condition for the function f(z) to be analytic.
Solution:
The sufficient conditions for a function f z  u iv to be analytic at all the points in a
region R are
(1) , x y y x u  v u  v
(2) , , , x y x y u u v v are continuous functions of x and y in region R .
Problem 3 Show that f z  ez is an analytic Function.
Solution:
f z  u iv ez
 exiy
 exeiy
 ex cos y i sin y
u  ex cos y, v  ex sin y
x , x
x x u  e cosy v  e siny
x , x
y y u  e siny v  e cosy
i.e., , x y y x u  v u  v
Hence C-R equations are satisfied.
 f z  ez is analytic.
Problem 4 Find whether f (z)  z is analytic or not.
Solution:
Given f z  z  x iy
i.e., u  x, v   y
Unit.3 Analytic Functions
2
u 1, v 0
x x
 
 
 
u 0, v 1
y y
 
  
 
x y u  v
C-R equations are not satisfied anywhere.
Hence f z  z is not analytic.
Problem 5 State any two properties of analytic functions
Solution:
(i) Both real and imaginary parts of any analytic function satisfy Laplace equation.
i.e.,
2 2
2 2 u u 0
x y
 
 
 
or
2 2
2 2 v v 0
x y
 
 
 
.
(ii) If w  u  iv is an analytic function, then the curves of the family ux, y  c, cut
orthogonally the curves of the family vx, y  c .
Problem 6 Show that   2 f z  z is differentiable at z 0 but not analytic at z 0 .
Solution:
2
0 0 0 0
( ) (0) 0
z 0 z z z
f z f z zz Lim Lim Lim Limz
 z  z  z 

   

 f z is differentiable at z  0.
Let z  x  iy
z  x  iy
z 2  z z   x  iy x  iy  x2  y2
f z  x2  y2 i0
u  x2  y2 , v  0
2 , 0 x x u  x v 
2 , 0 y y u  y v 
The C-R equation x y u  v and y x u  v are not satisfied at points other than z = 0.
Therefore f z is not analytic at points other than z  0. But a function can not be
analytic at a single point only. Therefore f z is not analytic at z = 0 also.
Problem 7 Determine whether the function 2xy  i(x2  y2 ) is analytic.
Solution:
Given f  z   2xy  i x2  y2 
i.e., u  2xy , v  x2  y2
Unit.3 Analytic Functions
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u 2y, v 2x
x x
 
 
 
u 2x, v 2y
y y
 
  
 
x y u  v and y x u  v
C-R equations are not satisfied.
Hence f  z is not analytic function.
Problem 8 Show that v  sinhx cosy is harmonic
Solution:
v  sinhxcosy
v coshxcosy, v sinhxsiny
x y
 
  
 
2 2
2 2 v sinhxcosy, v sinhycosy
x y
 
  
 
2 2
2 2 v v sinhxcosy sinhycosy 0
x y
 
   
 
Hence v is a harmonic function.
Problem 9 Construct the analytic function f (z) for which the real part is excosy .
Solution:
u  ex cos y
u ex cos y
x

Assume     1 u x, y z,0
x

  1  z,0  ez
u exsiny
y

 

Assume
    2
,
,0
u x y
z
y

  2  z,0  0
  '      
1 2 f z   f z dz   z,0 dz  i z,0 dz
 ez dz  i 0
f z  ez C .
Unit.3 Analytic Functions
4
Problem 10 Prove that an analytic function whose real part is constant must itself be a
constant.
Solution:
Let f z  u iv be an analytic function
, .............(1) x y y x u  v u  v
Given
u  ca constant
0 x u  , 0 y u 
0 & 0 1 y x v  v  by
We know that f z uiv
  x x f  z  u iv
f z  0i0
f z  0
Integrating with respect to z , f z  C
Hence an analytic function with constant real part is constant.
Problem 11 Define conformal mapping
Solution:
A transformation that preserves angle between every pair of curves through a
point both in magnitude and sense is said to be conformal at that point.
Problem 12 If w  f (z) is analytic prove that dw w i w
dz x y
 
  
 
where w  u  iv and
prove that
2
w 0
z z

 
Solution:
w  u x, y ivx, y is an analytic function of z .
As f z is analytic we have , x y y x u  v u  v
Now '    
x x y y y y
dw f z u iv v iu i u iv
dz
       
u i v i u i v
x x y y
     
           
u iv i u iv
x y
 
    
 
w i w
x y
 
  
 
Unit.3 Analytic Functions
5
W.K.T. w 0
z

2
w 0
z z

 
 
Problem 13 Define bilinear transformation. What is the condition for this to be
conformal?
Solution:
The transformation w az b , a d bc 0
cz d

  

where a, b, c, d are complex
numbers is called a bilinear transformation.
The condition for the function to be conformal is dw 0
dz
 .
Problem 14 Find the invariant points or fixed points of the transformation w 2 2
z
  .
Solution:
The invariant points are given by z 2 2
z
 
i.e., z 2 2
z
 
z2  2z  2
z2  2z  2  0
  
 
2 4 4 1 2
2 1
z
 

2 4 8 2 2
2 2
   i
 
1 i
The invariant points are z 1 i, 1 i
Problem 15 Find the critical points of (i) w z 1
z
  (ii) w  z3 .
Solution:
(i). Given w z 1
z
 
For critical point dw 0
dz

2
dw 1 1 0
dz z
  
z  i are the critical points
Unit.3 Analytic Functions
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(ii). Given w  z3
dw 3z2 0
dz
 
z  0
 z  0 is the critical point.
Part-B
Problem 1 Determine the analytic function whose real part is
u  x3 3xy2  3x2 3y2 1.
Solution:
Given u  x3 3xy2  3x2 3y2 1
  2 2
1 x, y u 3x 3y 6x
x

   

  2
1  z,0  3z  6z
  2 x, y u 6xy 6y
y

  

  2  z,0  0
By Milne Thomason method
      1 2 f z   z,0 dz i z,0 dz
 3z2  6z dz  0
3 2
3 6 3 3 2
3 2
 z  z C  z  z C
Problem 2 Find the regular function f z whose imaginary part is
v  ex x cos y  y sin y
Solution:
v  ex x cos y  y sin y
       2 x, y v e x cos y x cos y y sin y e x
x
   
    

       2  z,0  e z  z ez  e z  zez  ez 1 z
    1 x, y u e x x sin y y cos y sin y 1
y
  
      
    1  z,0  ez 0  0  0  0
By Milne’s Thomson Method
      1 2 f z   z,0 dz  i z,0 dz
Unit.3 Analytic Functions
7
  0 dz  i1 z e zdz
   
 2 1 1
1 1
e z e z i z C
      
                
 i 1 ze z  ez   C
 i e z  ze z  e z   C  i zez  C    
Problem 3 Determine the analytic function whose real part is sin 2 .
cosh 2 cos 2
x
y  x
Solution:
Given sin 2
cosh 2 cos 2
u x
y x

      
  1 2
cosh 2 cos 2 2cos 2 sin 2 2sin 2
,
cosh 2 cos 2
u y x x x x x y
x y x

  
 
 
    
 
2
1 2
1 cos 2 2cos 2 2sin 2
,0
1 cos 2
z z z
z
z

 

    
 
2
2
1 cos 2 2cos 2 2 1 cos 2
1 cos 2
z z z
z
  

     
 2
1 cos 2 2cos 2 2 1 cos 2 1 cos 2
1 cos 2
z z z z
z
   

2 cos 2 21 cos 2  2 cos 2 2 2 cos 2
1 cos 2 1 cos 2
z z z z
z z
   
 
 
2 1
1 cos 2 1 cos 2
2
z z

  
   
 
 
2
2
1 cos
sin
ec z
z
   
      
  2 2
cosh 2 cos 2 0 sin 2 2sinh 2
,
cosh 2 cos 2
u y x x y x y
y y x

  
 
 
 2
2sin 2 sinh 2
cosh 2 cos 2
x y
y x

  2  z,0  0
By Milne’s Thomson method
      1 2 f z   z,0 dz  i z,0 dz
  cosec2 z dz  0  cot z  C
Unit.3 Analytic Functions
8
Problem 4 Prove that the real and imaginary parts of an analytic function w  u  iv
satisfy Laplace equation in two dimensions viz 2u  0 and 2v  0 .
Solution:
Let f z  w  u  iv be analytic
To Prove: u and v satisfy the Laplace equation.
i.e., To prove:
2 2
2 2 u u 0
x y
 
 
 
and
2 2
2 2 v v 0
x y
 
 
 
Given: f z is analytic
 u and v satisfy C-R equations
u v
x y
 

 
….. (1)
and u v
y x
 
 
 
….. (2)
Diff. (1) p.w.r to x we get
2 2
2
u v
x x y
 

  
….. (3)
Diff. (2) p.w.r. to y we get
2 2
2
u v
y y x
 
 
  
….. (4)
The second order mixed partial derivatives are equal
i.e.,
2v 2v
x y y x
 

   
(3) + (4)
2 2 2 2
2 2 u u v v 0
x y x y y x
   
    
     
 u satisfies Laplace equation
Diff. (1) p.w.r to y we get
2 2
2
u v
y x y
 

  
…(5)
Diff. (2) p.w.r. to xwe get
2 2
2
u v
x y x
 
 
  
… (6)
(5) + (6)
2 2 2 2
2 2 v v v v 0
x y y x x y
   
    
     
i.e.,
2 2
2 2 v v 0
x y
 
 
 
 v Satisfies Laplace equation
Problem 5 If f(z) is analytic, prove that 2 2
2
2
2
2
f (z) 4. f (z)
x y
   

 

Solution:
Let f(z) = u + iv be analytic.
Unit.3 Analytic Functions
9
Then ux = vy and uy = -vx (1)
Also uxx + uyy = 0 and vxx + vyy = 0 (2)
Now |f(z)|2 = u2 + v2 and f(z) = ux + ivx
x x
| f (z) |2 2u.u 2v.v
x
 

and   xx
2
xx x
2
x
2
2
2
| f (z) | 2 u u.u v v.v
x
   

(3)
Similarly   yy
2
yy y
2
y
2
2
2
| f (z) | 2 u u.u v v.v
y
   

(4)
| f (z) | 2u u u(u u ) v v v(v v )
x y xx yy
2
y
2x
xx yy
2
y
2
x
2
2
2
2
2
         

 

  ) 0 ( v u v ) 0 ( u v u 2 2x
2
x
2
x
2
x      
 2 
x
2
x  4 u  v
2  4.f (z)
Problem 6 Prove that     2 Re f z 2  2 f  z 2
Solution:
Let f z  u  iv
  Re f z 2  u2
 2  2 x u uu
x

   
2
2
2 2 x u uu
x x
 

 
2  xx x x  uu u u
2 2 xx x  uu  u 
  2
2 2
2 2 yy y u uu u
y
       
    2 2
2 2 2
2 2 2 xx yy x y u u u u u u
x y
                  
2 0 2 2 x y  u  u  u 
  2  2 f  z
Unit.3 Analytic Functions
10
Problem 7 Find the analytic function f z  u  iv given that
2u  v  ex cos y sin y
Solution:
Given 2u  v  ex cos y sin y
f z  u iv...............1
if z  iu v...............2
122 f z  2u i2v...........(3)
3 2(2 i) f z  2u  v i 2v u............(4)
F  z U  iV
2u  v U  ex cos y sin y
  1 x, y U ex cos y ex sin y
x

  

  1  z,o  ez
  2 x, y u ex sin y ex cos y
x

   

  2  z,o  ez
By Milne Thomson method
      1 2 F z  z,o i z,o
 F z dz  ezdz  i ezdz
F z  1 iez C     (5)
From (4) & (5)
1iez C  2 i f z
  1
2 2
f z i ez C
i i

 
 
  1 3
5 2
f z i ez C
i

 

Problem 8 Find the Bilinear transformation that maps the points 1  i,  i, 2  i of the
z-plane into the points 0, 1, i of the w-plane.
Solution:
Given 1 1 z 1 i, w  0
2 2 z  i, w 1
3 3 z  2 i, w  i
Cross-ratio
  
  
  
  
1 2 3 1 2 3
1 2 3 1 2 3
w w w w z z z z
w w w w z z z z
   

   
Unit.3 Analytic Functions
11
  
  
   
     
0 1 1 2
0 1 1 2
w i z i i i
i w i i i z
          
          
 
  
  
  
1 1 2
1 1 2
w i z i i i
i w i i i z
     

     
 
 
  
  
1 1 2
1 2 2
w i z i
w i i i z
   

   
 
 
  
  
1 2 1
1 2 2
w i z i
w i i i z
   

   
 
  
 
 
2 1
1 2 1 2
w z i
w i i i i z
  

    
 
 
 
 
2 1
1 2 2 2
w z i
w i i i i z
  

     
 
 
 
 
2 1
3 2
w z i
w i i i z
  

   
  
  
3 2
2 1
w i i i z
w z i
   

  
  
  
3 2
1
2 1
i i i z
w z i
  
 
  
 
 
 
3 2 1
2 1
i i i z
w z i
  
 
  
 
 
3 2 1
2 1
i i i z
w z i
  
 
 
    
 
2 1 3 2
2 1
i z i i i z
w z i
     

 
 
    
2 1
2 1 3 2
w z i
i z i i i z
 

     
 
    
2 1
2 1 3 2
i z i
w
z i i i z
 

     
2  1 
2 2 2 6 3 3 2 1
i z i
w
z i i z i zi
 

       
2  1 
.
5 3
i z i
w
z i zi
 

   
Problem 9 Prove that an analytic function with constant modulus is constant.
Solution:
Let f z  u iv be analytic
Unit.3 Analytic Functions
12
By C.R equations satisfied
i.e., ux  vy , uy  vx
 f z  u iv
 f  z   u2  v2  C
   f z 2  u2  v2  C2
u2  v2  C2...............(1)
Diff (1) with respect to x
2u u 2v v 0
x x
 
 
 
0.............(2) x x uu  vv 
Diff (1) with respect to y
2u u 2v v 0
y y
 
 
 
0............(3) x x uv  vu 
(2) (3)  2 2  0 x u  v u  v u 
0 x u 
(2) (3)  2 2  0 x  v  u  u  v v 
0 x v 
W.K.T   0 x x f  z  u  iv 
f z  0
Integrate w.r.to z
f z  C
Problem 10 When the function f z  u iv is analytic show that   1 u x, y  C and
  2 v x, y  C are Orthogonal.
Solution:
If f z  u  iv is an analytic function of z , then it satisfies C-R equations
, x y y x u  v u  v
Given   1 u x, y  C .............(1)
  2 v x, y  C .............(2)
By total differentiation
du u dx u dy
x y
 
 
 
dv v dx v dy
x y
 
 
 
Unit.3 Analytic Functions
13
Differentiate equation (1) & (2) we get du  0 , dv  0
u dx u dy 0
x y
 
  
 
v dx v dy 0
x y
 
 
 
1
/ ( )
/
dy u x m say
dx u y
 
  
 
2
/ ( )
/
dy v x m say
dx v y
 
 
 
1 2
/ / ( )
/ / x y y x
m m u x v x u v u v
u y v y
   
       
   
1 2 m m  1
The curves   1 u x, y  C and   2 v x, y  C cut orthogonally.
Problem 11 Show that the function 1 log  2 2 is
2
u  x  y harmonic and determine its
conjugate.
Solution:
Given 1 log  2 2 
2
u  x  y
  2 2 2 2
1 . 1 2
2
u x x
x x y x y

 
  
  2 2
1 . 1 2
2
u y
y x y

  2 2
y
x y

   
   
2 2 2 2 2
2 2 2 2 2 2 2
u x y x 2x y x
x x y x y
   
 
  
  
   
2 2 2 2 2 2
2 2 2 2 2 2 2
u x y 1 2y x y
x x y x y
   
 
  
 
2 2 2 2 2 2
2 2 2 2 2
u u y x x y 0
x y x y
    
   
  
Hence u is harmonic function
To find conjugate of u
  1 2 2 x, y u x
x x y

 
 
  1
z,o 1
z
 
Unit.3 Analytic Functions
14
2 x, y u 2 y 2
y x y

 
 
  2  z,o  0
By Milne Thomson Methods
      1 2 f  z  z,o i z,o
f  z dz 1 dz 0
z
    
 log z  c
f z log rei
f z  u iv  log r  i
u  log r , v 
u log x2 y2 r2 x2 y2 , tan 1 y
x
               
v tan 1 y
x
   
 
 Conjugate of u is tan 1 y
x
  
 
 
.
Problem 12 Find the image of the infinite strips 1 1
4 2
 y  under the
transformation w 1
z
 .
Solution: w 1
z
 z 1
w
 
2 2
z 1 u iv
u iv u v

 
 
2 2 x u ........(1)
u v
 

2 2 y v ........(2)
u v
 

Given strip is 1 1
4 2
 y  when 1
4
y 
2 2
1 ( 2)
4
v by
u v
 

u2  (v  2)2  4..........(3)
which is a circle whose centre is at 0,2 in the w-plane and radius 2.
When 1
2
y 
2 2
1 ( 2)
2
v by
u v

u2  v2  2v  0
u2  (v 1)2 1..........(4)
Unit.3 Analytic Functions
15
which is a circle whose centre is at 0,1 and radius is 1 in the w-plane.
Hence the infinite strip 1 1
4 2
 y  is transformed into the region between circles
u2  (v 1)2 1and u2  (v  2)2  4 in the w-plane.
Problem 13 Obtain the bilinear transformation which maps the points z 1, i, 1 into
the points w  0,1, .
Solution: We know that
  
  
  
  
1 2 3 1 2 3
1 2 3 1 2 3
w w w w z z z z
w w w w z z z z
   

   
  
  
  
  
0 1 1 1
0 1 1 1
w z i
w i z
    

     
   
1 1. 1
1 1 1
w z i
i z
 
 
   
1.1
1 1
w z i
z i
 
 
 
  1
1
w i z
z

 

Problem 14 Find the image of z  2i  2 under the transform w 1
z

Solution:
Given w 1
w
 z 1
w
 
Now w  u  iv
1
2
y 
1
4
y 
x
y
z  plane
u
v
0,1
0,2
w plane
Unit.3 Analytic Functions
16
   2 2
z 1 1 u iv u iv
w u iv u iv u iv u v
 
   
   
i.e., 2 2
x iy u iv
u v

 

2 2 x u ..........(1)
u v
 

2 2 y v ..........(2)
u v

Given z  2i  2
x  iy  2i  2
x  i( y  2)  2
x2   y  22  4
x2  y2  4y  0...................(3)
Sub (1) and (2) in (3)
2 2
2 2 2 2 2 2 u v 4 v 0
u v u v u v
                         
   
2 2
2 2 2 2 2 2 2 2
u v 4 4v 0
u v u v u v
         
   
 
2 2 2 2
2 2 2
4
0
u v v u v
u v
  

  
 
2 2
2 2 2
1 4
0
v u v
u v
 

1 4 0 1 ( 2 2 0)
4
 v  v   u  v  This is straight line in w-plane.
x2   y  22  4
0,2
v
u
x
y
z  plane
w plane
1
4
v  
v  0
Unit.3 Analytic Functions
17
Problem 15 Prove that
1
w z
z

maps the upper half of the z-plane onto the upper half
of the w-plane.
Solution:
(1 )
1
w z w z z
z
   

w wz  z
w  w1 z
w  (w1)z
1
z w
w

Put z  x  iy, w u  iv
1
x iy u iv
u iv

 
 
  
  
1
1 1
u iv u iv
u iv u iv
  

   
   
 
2
2 2
1 1
1
u u iuv iv u v
u v
    

 
 
 
2 2
1 2 2
u v u iv
u v
  

 
Equating real and imaginary parts
   
2 2
2 2 2 2 ,
1 1
x u v u y v
u v u v
 
 
   
 2 2 0 0
1
y v
u v
  
 
 2 2 0 0
1
y v
u v
  
 
v  0
Thus the upper half of the z plane is mapped onto the upper half of the w plane.
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