MAT341E Theory of Complex Functions
Worksheet 3
6 November 2007
Beycan Kahraman
040020337
1. Show that each of the following functions are differentiable in the indicated domain of
definition and find f ' ( z ) .
1
( z  0)
a) f ( z )  4
z
1
f ( z )  4 4i  r  4e  4i  r  4 (cos 4  i sin 4 )
r e
4
1
u  r cos 4
v  r  4 sin 4 
 ur  r v
5
5
ur  4r cos 4
vr  4r sin 4 
1
 vr   u
4
4
u  4r sin 4
v  4r cos 4 
r
By Couchy-Riemann equations for polar coordinates, the function is differentiable and
f ' ( z )  e i u r  iv r   e i  4r 5 cos 4  4ir 5 sin 4  4r 5 e i e 4i  4r 5 e 5i

b) f ( z )  r e i 2

(r  0,       )



f ( z )  r  cos  i sin 
2
2



2
2 
1

u

v
r


1

1

r
ur 
cos
vr 
sin 
 differentiable
2
2 v   1 u 
2 r
2 r
r

r 
r

r
 
u  
sin
v 
cos 
2
2
2
2 
i
i

i

1  i 2
1 2
 1
f ' ( z )  e i u r  iv r   e i 
cos 
sin  
e e 
e
2 2 r
2 2 r
2 r
2 r
u  r cos

v  r sin

c) f ( z )  e  cos(ln r )  ie  sin(ln r )
f ( z)  e

cos(ln r )  i sin(ln r ) 
u  e  cos(ln r )
(r  0, 0    2 )
v  e  sin(ln r )

1 

e  sin(ln r )
e  cos(ln r )  u r  r v 
ur  
vr 
differentiable

1 
r
r
v r   u 
r 
u  e  cos(ln r )
v  e  sin(ln r )
e  sin(ln r ) e  cos(ln r )  e  e i
i
i 

 
 sin(ln r )  i cos(ln r )
f ' ( z )  e u r  iv r   e  
i
r
r
r


2. Suppose that a function f ( z )  u  iv is differentiable at a nonzero point z0  r0ei 0 . Show
that f ' ( z 0 ) can be written as f ' ( z0 )  e i (ur  ivr ) where ur and vr are evaluated at (r0 ,0 ) .
f ( z0  z )  f ( z0 )
z  0
z
v 
 u
f ' ( z0 )    i 
x  ( x0 , y0 )
 x
f ' ( z0 )  lim
x  r cos 
y  r sin 
 u r u    v r v  
f ' (r0ei 0 )  


  i

 r x  x   r x  x  ( r0 ,0 )
 u r
v    v r
u  
f ' (r0ei 0 )  
 r
r
  i

r x   r x
r x  ( r , )
 r x
0 0
 u  r
  v  
r 
f ' (r0ei 0 )     ir
 i 
   r
x  r 
x
x  ( r , )
 r  x
0 0
 u  r
 u
  v  r
 
v  r
 
f ' (r0ei 0 )     ir
   i   ir
  i   ir


x  r  x
x  ( r , )  r
r  x
x  ( r , )
 r  x
0 0
0 0
 u
 u

v 
 sin  
v 
f ' (r0ei 0 )    i  cos   ir
   i cos   i sin  

r 
r  ( r , )  r
r 
 r
 ( r0 ,0 )
0 0
v 
 u
f ' (r0ei 0 )  e  i   i 
 e  i ur  ivr  ( r , )
0 0
r  ( r0 ,0 )
 r
3. Show that the following functions are entire
a) f ( z )  3x  y  i (3 y  x)
f ( z )  u ( x, y )  iv ( x, y )
u ( x, y )  3 x  y
v ( x, y )  3 y  x
ux  3
vy  3
uy  1
vx  1
ux  vy
u y  v x

 Couchy  Riemann are satisfied

 at every poınt. Therefore, entire.

b) f ( z )  2 xy  i( x 2  y 2 )
u ( x, y )  2 xy v( x, y )  x 2  y 2
ux  2 y





v y  2 y
They are not equal.
Therefore, not entire.
c) f ( z )  e y eix
f ( z )  e y cos x  i sin x 
u ( x, y )  cos x.e y
v( x, y )  sin x.e y
u x   sin x.e
v y  sin x.e
y
y





They are not equal.
Therefore, not entire.
4. Show that u ( x, y ) is harmonic in some domain & find a harmonic conjugate v ( x, y ) when
a) u ( x, y )  sinh x. sin y
u x  cosh x.sin y
u y  sinh x. cos y

 u xx  u yy  0 (harmonic )
u xx  sinh x.sin y
u yy   sinh x.sin y

ux  vy

v( x, y )   cosh x. cos y  c
b) u ( x, y ) 
ux 
u xx 
y
x  y2
2
x
 2 xy
2
 y2

x2  y2  2 y2
uy 

2


2
x
2
 y2
 2 y x2  y 2  8x2 y x2  y 2
u xx  u yy
x
 2 y x

u xx  u yy 
y
2
 y2
2

2
2

2
u yy 
2 4
2


  8 x yx



x
x2  y 2
x
2
 y2


2

x


2

ux  vy

2
u y  6 xy
u xx  6 x
u yy  6 x

ux  vy

y


y


y


2 4


2 4

2 3
v ( x, y ) 
c) u( x, y)  2 x  x3  3xy2
u x  2  3x 2  3 y 2
u xx  u yy  0
2
2
 y2  2 y x2  y2  4 y x2  y2 x2  y2
 2 y x2  y 2  8x2 y  2 y x2  y 2  4 y x2  y 2
x

2
 2 y x2  y 2  4 y x2  y 2 x2  y 2
0

harmonic
x
c
x  y2
2
harmonic

v( x, y )  2 y  3 yx 2  y 3  c
5. Let the function f ( z )  u (r , )  iv (r , ) be analytic in a domain D that does not include
the origin. Using the Couchy-Riemann equations in polar coordinates and assuming
continuity of partial derivatives, show that, throughout D, the function u ( r ,  ) satisfies the
partial differential equation
r 2 u rr (r , )  ru r (r , )  u (r , )  0
which is the polar form of Laplace’s equation.
Since analytic :
1
1
ur  v
vr   u
r
r
1
1
1
urr   2 v  vr
vr   u
r
r
r
2
r urr  v  u  0
r 2urr  ru r  u  0

urr  
1
1
v  2 u
2 
r
r
6. Show that
a) sinh( z  i )   sinh z
e z i  e z i e z cos   i sin    e z cos   i sin   e z  e z


 sinh z
2
2
2
sinh( z  i) 
b) cosh( z  i)   cosh z
cosh( z  i) 
7. Compute
e z i  e z i e z cos   i sin    e z cos   i sin   e z  e z


 cosh z
2
2
2
 f ( z )dz for
f ( z )  ez and C is the boundary of the square with vertices at the
C
points 0, 1, i  1, i and the orientation of C is counterclockwise.
 e
z
1
1
dz   e dx  i  e
C
x
0
 (1 iy )
0
dy   e
0
x 1
e
0
 ( x i )
dx  i  e  iy dy
1
1
 ie  sin y  i cos y  0  e e
 i x 0
1


 e  1  ie   i  i   1  e  i i  i 





 
1
 i sin y  i cos y 1
0
 

 2 e  1  2e  2  2 e  1  2 e  1  4 e  1
8.

0
and C is the arc form z  1  i to z  1  i along the
curve y  x 3 . Find  f ( z )dz  ?
y0
1,
f ( z)  
4 y,
y0
C
yt , xt

1  t  1
z  t  it

dz  1  3it 2 dt
3

 f ( z )dz   f ( z )dz   f ( z )dz   1  3it dt   4t 1  3it dt
3

0
1
0
1
2
C

1
 t  it

3 0
1

0
 t 4  2it

1
6 1
0
3
2
0
 (1  i )  (1  2i )  2  3i
z2
, find the contour integral of f (z ) over C , where C is
z
a) the semicircle z  2e i ,
0  
9. f ( z ) 

C


 2ei  2 i 
 z  2
i
f ( z )dz   
2ie d  2i  e  1 d  2i  (cos   i sin  )  1d
dz   
z 
2ei

C
C
0
0


 2isin   i cos    0  2isin   i cos     sin 0  i cos 0  2ii    i   2i  4

b) the semicircle z  2e i ,

C
    2
2
 2e  2 i 
2
 z 2


f ( z )dz   
2ie d  2i  ei  1 d  2isin   i cos     
dz   
i
z 
2e

C
C

 2isin 2  i cos 2  2  sin   i cos      2i i  2  i     2i  4
i


c) the semicircle z  2e i ,

C
0    2
2
 2ei  2 i 
2
 z  2


f ( z )dz   
dz

2
ie
d


2
i
ei  1 d  2isin   i cos    0

i




z 
2e

C
C
0
 2isin 2  i cos 2  2  sin 0  i cos 0  0  2i i  2  i   4i


OR
 f ( z) 
dz
By the theorem given in lesson ( for closed contours) : f ( z0 ).2i   
z

z
0 
C
 z  2
C f ( z )dz  C  z dz  22i   4i
x  r cos
y  r sin 
Couchy-Riemann equations for rectangular coordinates:
ux  vy
v x  u y
Let’s calculate these in polar form:
u u  u  1
i.


x  x  r sin 
Since these are equal:
1
vr   u
r
u u r u 1
ii.


y r y r sin 
Since these are equal:
1
u r  v
r
v v r v 1


y r y r sin 

v
v  v 1


x
 x r r sin 