Complex Variable
Theorem:
The continuous single valued function is 𝑓 (𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) will be analytic in a region
R if and only if four partial derivatives 𝑢𝑥 , 𝑣𝑥, 𝑢𝑦, 𝑣𝑦, exists, continuous and Cauchy Riemann
equations 𝑢𝑥 = 𝑣𝑦 , 𝑢𝑦 = −𝑣𝑥 are satisfied.
Proof: If f ( z ) = u ( x, y) + iv( x, y) is analytic then
f ( z ) = lim z→0
f ( z + z ) − f ( z )
z
{u ( x + x, y + y ) + iv( x + x, y + y )} − {u ( x, y ) + iv( x, y )}
x + iy
y →0
= lim x→0
Exists and unique. Two cases arise.
Case 1. Along x-axis y = 0 and x → 0
f ( z ) = lim x→0
{u ( x + x, y ) − u ( x, y )} + i{v( x + x, y ) − v( x, y )}
x
f ( z ) = u x + ivx
……………………………………………..(1)
Case 2. Along y-axis x = 0 and y → 0
f ( z ) = lim y→0
{u ( x, y + y ) − u ( x, y )} + i{v( x, y + y ) − v( x, y )}
iy
1
f ( z ) = u y + v y
i
f ( z ) = −iu y + v y
From equation (1) and (2) we get
u x + ivx = −iu y + v y
Therefore 𝑢𝑥 = 𝑣𝑦 , 𝑢𝑦 = −𝑣𝑥
……………………………………………….(2)
Since u x and u y are continuous then
u = u ( x + x, y + y) − u ( x, y)
u = {u ( x + x, y + y) − u ( x, y + y)} + {u ( x, y + y) − u ( x, y)}
=(
=
u
u
+  1 )x + ( + 1 )y
x
y
u
u
x +
y +  1x + 1y
x
y
…………………………..(3)
Where  1 → 0 and 1 → 0 Since x → 0 and y → 0
Similarly v x and v y are continuous then
v =
v
v
x + y +  2 x +  2 y
x
y
Where  2 → 0 and  2 → 0 Since x → 0 and y → 0
Now w = u + iv = (
u
v
u
v
+ i )x + ( + i )y + x + y
x
x
y
y
Where  =  1 + i 2 → 0 and  = 1 + i 2 → 0 Since x → 0 and y → 0
Now using Cauchy Riemann equations
w = u + iv = (
u
v
v
u
+ i )x + (− + i )y + x + y
x
x
x
x
w = (
u
v
v
u
+ i )x + (i 2
+ i )y + x + y
x
x
x
x
w = (
u
v
u
v
+ i )x + i( + i )y + x + y
x
x
x
x
w = (
u
v
+ i )(x + iy ) + x + y
x
x
w = (
u
v
+ i )z + x + y
x
x
Dividing by z and Taking limit z → 0
dw
w
u
v
= f ( z ) = lim z→0
=( +i )
dz
z
x
x
Hence f (z ) is analytic.
Problem: Show that the function u = x 2 − y 2 − 2 xy − 2 x + 3 y is harmonic. And find the
conjugate harmonic function v .
Problem: Show that the function u = x 2 − y 2 − 2 xy − 2 x + 3 y is harmonic. And find the
conjugate harmonic function of u
Solution: Given that u = x 2 − y 2 − 2 xy − 2 x + 3 y
u
= 2x − 2 y − 2
x
u
= −2 y − 2 x + 3
y
 2u
 2u
= −2
=
2
and
y 2
x 2
 2u  2u
+
=0
Therefore
x 2 y 2
Implies that u is harmonic
From CR equations
u v
=
= 2x − 2 y − 2
x y
and
u
v
=−
= −2 y − 2 x + 3
y
x
…………………. (1)
…………………..(2)

Integrating (1) on both sides v = (2 x − 2 y − 2)dy
 v = 2 xy − y 2 − 2 y + F ( x)
…………………(3)
Differentiate
v
= 2 y + F ( x)
x
 2 y + 2 x − 3 = 2 y + F ( x)
 F ( x) = 2 x − 3
 F ( x) =  (2 x − 3)dx = x 2 − 3x + c
Putting this value in equation (3)
 v = 2 xy − y 2 − 2 y + x 2 − 3x + c
Conjugate harmonic function
Problem: Show that u = e x ( x cos y − y sin y) is harmonic. Also find the conjugate harmonic
function of u
Problem: Find the conjugate harmonic function of u = e − x ( x sin y − y cos y)
Cauchy’s Integral formula:
If f (z ) is analytic for all points inside of C and connected a simple closed curve C. a is
any point inside C. Then f (a) =
1
f ( z)
dz .

2i c z − a
Proof: Since f (z ) is analytic for all points inside of C
f ( z ) = f (a) + ( z − a) f (a) + ( z − a)
Now
where z → a , then  → 0
1
f ( z)
1
f (a) + [ f ( z ) − f (a)]
dz =
dz .


2i c z − a
2i c
z−a
f (a) dz
1 [ f ( z ) − f (a)]
+
dz .

c
2i z − a 2i c
z−a
=
=
=
f (a) dz
1 [ f (a) + ( z − a) f (a) + ( z − a) − f (a)]
+
dz .

2i c z − a 2i c
z−a
f (a)
1 ( z − a)[ f (a) +  ]
2i +
dz .
2i
2i c
z−a
= f (a) +
f (a)
1
dz +
dz .

c
2i
2i c
= f (a) + 0 +
1
dz .
2i c
1
f ( z)
1
dz − f (a) =
dz .

2i c z − a
2i c
Considering C is very small so that    for all points on C.
1
f ( z)
1

dz − f (a) =
dz 
L.


2i c z − a
2i c
2
1
f ( z)
dz − f (a) = 0 .

2i c z − a
1
f ( z)
dz − f (a) = 0 .

2i c z − a
f (a) =
1
f ( z)
dz .

2i c z − a
Cauchy’s Integral formula for nth order derivative:
f n (a) =
Evaluate:
n!
f ( z)
dz .

2i c ( z − a) n+1
z
1.
 z − 1dz
2.
z2
c ( z − i)dz
3.
sin z 2 + cos z 2
c ( z − 1)( z − 2) dz where C is the circle z = 3
4.
5.
6.
where C is the circle z = 2
c

sin 3z
c
(z +

2
where C is the circle z = 2
dz where C is the circle z = 5
)
e tz
c ( z 2 + 1)dz

c
sin 6 z
(z −

)
6
where C is the circle
dz
z =3
where C is the circle
3
Solution 1:
We know
f (a) =
1
f ( z)
dz .
2i c z − a
Here f ( z ) = z and a = 1
f (1) = 1
then
Now
z
 z − 1dz = 2if (1) = 2i  1 = 2i
c
Solution 2:
We know
f (a) =
1
f ( z)
dz .

2i c z − a
2
Here f ( z ) = z and a = i
then
f (i ) = i 2
z =1
Now
z2
2
c z − i dz = 2if (i) = 2i  i = 2i(−1) = −2i
Solution 3:
We know
f (a) =
1
f ( z)
dz .
2i c z − a
Now
sin z 2 + cos z 2
c ( z − 1)( z − 2) dz Here f ( z) = sin z 2 + cos z 2
sin z 2 + cos z 2
c ( z − 1)( z − 2) dz
f ( z)
=
dz
c ( z − 1)( z − 2)
1
1
= (
−
) f ( z )dz
c z −2
z −1
f ( z)
f ( z)
=
dz − 
dz
cz −2
c z −1
= 2if (2) − 2if (1)
= 2i (sin  2 2 + cos  2 2 ) − 2i (sin  12 + cos  12 )
= 2i (0 + 1) − 2i (0 − 1)
= 2i + 2i
= 4i
Solution 4.

c
sin 3z
(z +

2
dz where C is the circle z = 5
)
We know
f (a) =
1
f ( z)
dz .

2i c z − a
f ( z ) = sin 3z and a = −

2
then f (−



) = sin 3(− )
2
2

sin 3z
c

dz = 2i  f (− ) = 2i  sin 3(− ) = 2i

2
2
(z + )
2
e tz
dz
Solution 5. c 2
( z + 1)
where C is the circle
z =3
We know
1
f ( z)
dz

c
2i z − a
f (a) =
f ( z ) = e tz
Now
e tz
f ( z)
c ( z 2 + 1)dz = c ( z + i)( z − i)dz
=
1
1
1
(
−
) f ( z )dz

2i c z − i z + i
=
1
f ( z)
f ( z)
(
dz − 
dz )
2i c z − i
z
+
i
c
1
[2if (i ) − 2if (−i )]
2i
1
=  2i (e ti − e −ti )
2i
=   2i sin t
= 2i sin t
=
Solution 6.

c
sin 6 z
(z −

6
)
dz
where C is the circle
3
We know
f n (a) =
n!
f ( z)
dz .

c
2i ( z − a) n+1
z =1
f ( z ) = sin 6 z , a =

6
and n = 2
Now

c
sin 6 z
(z −

6
)
dz =
3
2i

f ( )
2!
6
……………………(1)
5
Here f ( z ) = 6 sin z cos z ,
f ( z ) = 6[5 sin 4 z cos z cos z + sin 5 z (− sin z )]

21
f ( ) =
6
16
From (1)

c
sin 6 z
(z −

6
)
dz
3
2i

f ( )
2!
6
21
= i 
16
=
e tz
dz
Problem 7. c
( z + 1) 3
ze tz
dz
Problem 8. c
( z + 1) 3
Theorem: if f (z ) is analytic inside and on a simple closed curve C except at the pole z = a
of order m then the residue of f (z ) at z = a is
Re s(a) = lim z→a
1
d m−1
[( z − a) m f ( z )]
m−1
(m − 1)! dz
If z = a is a simple pole then Re s(a) = lim z→a ( z − a) f ( z )
Cauchy’s Residue Theorem:
Let f (z ) be analytic inside and on a simple closed curve C except at a finite number of singular
points a1 , a2 , a3 ,.................., an , then
 f ( z)dz = 2i[Re s(a ) + Re s(a ) + .................. + Re s(a )]
1
2
n
c
Proof: Let a1 , a2 , a3 ,.................., an be the center of the circle C1 , C2 , C3 ,.................., Cn
respectively. f (z ) be analytic inside and on a simple closed curve C.
 f ( z)dz =  f ( z)dz +  f ( z)dz + ....................... +  f ( z)dz
C
C1
C2
But Re s(a1 ) =
Re s(a2 ) =
(1)
Cn
1
f ( z )dz   f ( z )dz = 2i Re s(a1 )
2i C1
C1
1
f ( z )dz   f ( z )dz = 2i Re s(a2 )
2i C2
C2
…………………………………………………………………………………………
………………………………………………………………………………………………………
Re s(an ) =
1
f ( z )dz   f ( z )dz = 2i Re s(an )
2i Cn
Cn
From (1)
 f ( z)dz = 2i Re s(a ) + 2i Re s(a ) + .................. + 2i Re s(a )
1
2
n
c
 f ( z)dz = 2i[Re s(a ) + Re s(a ) + .................. + Re s(a )]
1
c
2
n
ez
1
dz =
Ex: Show that c 2
2 2
(z +  )

Solution: let
ez
ez
f ( z) = 2
=
( z +  2 ) 2 ( z + i ) 2 ( z − i) 2
2
2
To find the pole, ( z + i) ( z − i) = 0
z = i,−i is a pole of order 2.
Now
Re s(i ) = lim z →i
1
d
[( z − i ) 2 f ( z )]
(2 − 1)! dz
= lim z →i
1
d
ez
[( z − i ) 2
]
(2 − 1)! dz
( z + i ) 2 ( z − i ) 2
= lim z →i
d
ez
[
]
dz ( z + i ) 2
= lim z →i
( z + i) 2 e z − e z .2( z + i)
( z + i ) 4
= lim z →i
( z + i )e z − e z .2
( z + i ) 3
=
2(i − 1)ei
− 8i 3
=
2(i + i 2 )ei
− 8i 3
=
− ( + i )ei
4 3
Re s(−i) = lim z →−i
1
d
[( z + i) 2 f ( z )]
(2 − 1)! dz
= lim z →−i
1
d
ez
[( z + i ) 2
]
(2 − 1)! dz
( z + i ) 2 ( z − i ) 2
= lim z →−i
d
ez
[
]
dz ( z − i) 2
= lim z →−i
( z − i ) 2 e z − e z .2( z − i )
( z − i ) 4
= lim z →−i
( z − i )e z − e z .2
( z − i ) 3
=
=
2(−i − 1)e −i
8i 3
2(−i + i 2 )e −i
8i 3
− ( − i )e −i
=
4 3
By Cauchy residue theorem
ez
c ( z 2 +  2 ) 2 dz = 2i[Re s(i) + Re s(−i)]
= 2i[
=
1
2 2
=
=
− ( + i )ei − ( − i )e −i
+
4 3
4 3
1
2 2
1
2 2
.2
[− (ei + e −i ) − i (e i − e −i )]
[−2 cos  − i.2i sin  ]
=
1
