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UNIT –IV
COMPLEX INTEGRATION
CAUCHY’S INTEGRAL THEOREM
Statement:
If f(z) is analytic and f’(z) is continuous at all points inside and
on a simple closed curve c, then ∫ f ( z )dz = 0
...
(1)
Since f’(z) is continuous, the four partial derivatives

∂u ∂u ∂v ∂v
, , ,
∂x ∂y ∂x ∂y

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are also exists and continuous in R and on c
By Green’s theorem in the plane
 ∂N ∂M 
∫ Mdx + Ndy = ∫∫  ∂x − ∂y dxdy



c
R 
∴ (1) Becomes,
 ∂v ∂u 
 ∂u ∂v 
∫ f ( z )dz = ∫∫  − ∂x − ∂y dxdy + i ∫∫  ∂x − ∂y dxdy ……(2)






c
R 
R 
Since f(z) = u+iv is analytic by the CR – Equation,

∴ From (2)

∫
c

 ∂u ∂u 
 ∂u ∂u 
f ( z )dz = ∫∫  − dxdy + i ∫∫  − dxdy
 ∂y ∂y 
∂x ∂x 

R 
R 
= 0 + i0 =0

CAUCHY’S INTEGRAL FORMULA (OR) CAUCHY’S
FUNDAMENTAL FORMULA
Statement:
If f(z) is analytic inside and on a simple closed curve c and if ‘a’ is
any point within c, then,
1
f ( z)
f(a) =
∫ z − a dz , the integration around c being taken in the
2πi c
positive direction
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Now

f ( z)
is analytic inside
z−a

and on c except at z =a
Draw a circle c1 : z − a = r with center at z = a and radius r units such
that c1 lies entirely inside c
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Evaluate

z2 +1
∫ z 2 − 1 dz where c is circle
c

(i) z − 1 = 1 (ii ) z + 1 = 1 (iii ) z − i = 1
Solution:
z2 +1
Given ∫ 2
dz =
c z −1

z2 +1
∫ ( z + 1)( z − 1) dz
c

z2 +1
A
B
=
+
( z + 1)( z − 1) z + 1 z − 1
z 2 + 1 = A( z − 1) + B( z + 1)
Put z = 1,B = 1, put z =-1 , A = -1

Consider

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2

z +1
1
1
dz = − ∫
dz + ∫
dz
( z + 1)( z − 1)
z +1
z −1
c
= 2πi[− f (a ) + f (a)]
Here the point are z = -1 and z = 1
(i) c is the circle z − 1 = 1
...
The point z = -1 lies inside and z = 1 lies outside the
circle z + 1 = 1

z2 +1
dz = 2πi[− f (a) + 0]
( z + 1)( z − 1)
c
= − 2πi [ since f(z) = 1]
(iii) c is the circle z − i = 1
∴∫

When Z = 1, 1 − i = 2 > 1 lies outside c
When z = -1, − 1 − i = 2 > 1 lies outside c
∴ ∫ f ( z )dz = 0
c

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Using Cauchy’s Integral formula, evaluate

zdz

∫ (z − 1)(z − 2)

2

where c is the circle

c

1
2

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z−2 =

Solution: Consider

z
(z − 1)(z − 2)2

Hence the point z – 1 lies outside circle z − 2 =

1
and the point z = 2 lies outside
2

1

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1!
2!

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This is known as Taylors series of f(z) about z = a
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Having its centre at z = a and within which the
function is analytic
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The radius of this circle is
called the redius of convergence
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1!
2!
3!
The series is called Maclaurin’s series of f(z)
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Some Important Results: If z < 1, we have

∞

•

(1 + z )−1 = 1 − z + z 2 − z 3 +
...
= ∑ z n
n =0

∞

•

(1 + z )−2 = 1 − 2 z + 3z 2 − 4 z 3 +
...
= ∑ (n + 1) z n
n =0

1
at x = 1 in Taylor series
z−2
1
Solution: Given f (z) =
f(1) = -1
z−2
−1
f ' ( z) =
, f ' (1) = −1
( z − 2) 2
1
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1!
2!
( z − 1) 2
( z − 1) 3
= − 1 + ( z − 1)(−1) +
(−2) +
(−6) +
...


f ( z ) = f (a) + (z − a )

[

]

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2
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 − 8 1 − z +  z  −
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 − 8 1 − 3 +  3  −
...
 − 1 − +   −
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Find Laurent’s series expansion of f (z) =
in 1 < z + 1 < 3
z ( z − 2)( z + 1)
7z − 2
Solution: Let f (z) =
z ( z − 2)( z + 1)

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7z − 2
A
B
C
= +
+
z ( z − 2)( z + 1) z z − 2 z + 1
7 z − 2 = A( z − 2)( z + 1) + Bz ( z + 1) + Cz ( z − 2)
Put z = 0, A = 1
z = 2, B = 2
z = −1, C = −3
7z − 2
z ( z − 2)( z + 1)
1
2
3
f ( z) = +
−
z z − 2 z +1
F (z) =

Given 1 < z + 1 < 3

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Now put t = z + 1, then 1 < t < 3

⇒1< t, t < 3

t
1
⇒ < 1, < 1
t
3

1
2
3
+
−
t −1 t − 3 t
3
1
2
= − +
−
t
 1
 t
t 1 −  31 − 
 t
 3

∴ f ( z) =

−1

−1

3 1 1
2 t 
= − + 1 −  − 1 − 
t t t
3  3
2
2



3 1 1 1
1 + +   +
...

=− +
 
 3  3 3

t t  t t 




2
2



3
1 
1 + 1 +  1  +
...

=−
+




 3

z +1 z +1
z + 1  z +1
3
 3 





CALCULUS OF RESIDUES
Definition: Zero of an analytic function
A point z = a is said to be a zero of an analytic function f (z) if f(z) is
zero at z = a
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If f (a) = f ' (a) =
...

Example; Let f (z) = z 2
Then f ' ( z ) = 2 z , f ' ' ( z ) = 2
f (0), f ' (0) = 0, f ' ' (0) = 2 ≠ 0
∴ z = 0 is a zero of order 2
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Types of Singular Point:
Isolated Singular Point:

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A point z = a is said to be an isolated singular point of f (z) if
(i) f (z) is not analytic at z = a
(ii) f (z) is analytic at all points for some neighbourhood of z = a

z
(z − 1)( z − 2)
Then z = 1, 2 are isolated points
...


Example
Let f ( z ) = e1 2
Clearly z = 0 is a singular point
2
1
1 z
1z
z +
...

z 2!  z 
∴ z = 0 is an essential singular point

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Removable singular point:
A singular point z = a is said to be a removable singular point of f(z) if the
Laurent’s series of f(z) about z = a does not contain the principal part
...

z
z
3! 5!

2
4
z
z
=1 −
+
−
...


f ( z) =

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1
...



z
3


z2
+
...

= 1+

∴ Z=0 is a removable singular point
...

⇒ z = 0, nπ are simple poles (pole of order 1)

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2
...
Find the pole and its order
...

z −
3! 5!



1
1
z
−
+
−
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Definition
A function f (z ) is said to be an entire function or integral function if it is
analytic everywhere in the finite plane except at infinity
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Example

Consider f ( z ) =

cos πz 2
(z − 1)(z + 2)

Then f (z ) is not analytic at z=1,-2
∴ f (z 0 is a meromorphic function
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Suppose z = a is a pole of order 1
Then {Re sf ( z ) }z = a =

Lim

(z − a) f (z)

z→ a

2
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Suppose z =a is a pole of order 1 and f ( z ) =

Then { Res f ( z ) } z = a =

P(z )
Q( z )

P(a )
Q ' (a )

CAUCHY’S RESIDUE THEOREM

If f (z ) is analytic at all point inside and on a simple closed curve C
Except at a finite number of point z1, z 2, z3
...
zn ]

1, 2, 3,

c

Proof
Given that f (z ) is not analytic
Only at z1 , z 2 , z3,
...
cn with centre at
z1 , z 2 , z3 ,
...
ρ n
Then f (z ) is analytic in the region
Between c and c1 , c2 , c3
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+ ∫ f ( z )dz
c

c1

c2

…
...
z n are the singular points of f (z )
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(2)

i

From (1) and (2)
∫ f (z )dz = 2π Re sf (z ) z = z + 2π Re sf (z ) z = z + 2π Re sf (z ) z = z
1

c1

2

= 2π Re sf ( z ) z = z + Re sf ( z ) z = z +
...
zn }

z+2
about each singularity
...

∴ [Re sf ( z )]z = 2 = lim( z − 2) f ( z )
Example 1 Find the residue of f(z) =

z →2

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= lim( z − 2)
z →2

z+2
4
=
2
(z − 2)(z + 1) 9

d
2
[( z − 2) f ( z )]
z → −1 dz
d
z+2
2
= lim [(z − 2)
]
z → −1 dz
( z − 2)( z + 1) 2
d  z + 2
= lim 
z → −1 dz z − 2 


 ( z − 2 )(1) − ( z + 2 )(1) 
4
= lim 
=−
2
z → −1
9
(z − 2)



∴ [Re sf ( z )]z = −1 = lim

Example 2
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d
2
∴ [Re sf ( z )]z = zi = lim ( z − 2i ) f ( z )
dz
z →2i

= lim

d
1
(z − 2i)2
2
dz
(z − 2i) (z + 2i)2

z→2i

= lim

d
(z − 2i)2
dz

z →2i

(z + 2i)2 (0) − 2(z + 2i)
= lim
z →2i
(z + 2i)4
=−

8i
256

∴By Residue theorem
dz

∫ (z
c

 − 8i  π
= 2πi
=
2
 256 6
+4
4

)

CONTOUR INTEGRATION

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TYPE: I
2π

Integrals of the type

∫ f (cos θ , sin θ )dθ

where f (cos θ , sin θ ) is

0

A rational function of sin θ & cos θ
...
On z = 1
...


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2π

Example 1
...

0

2π

Solution: I =

dθ

∫ 13 + 5 sin θ

limit : 0 to 2 π

0

Contour: z = 1

Put z = e iθ

dz = ie iθ dθ , sin θ =

⇒ dθ =

1 iθ
1
1
e − e − iθ =  z − 
2i
2i 
z

(

)

dz
iz

dz

I =∫


 z 2 −1  
iz 13 + 5
 2iz  





dz
= 2∫ 2
5 z + 26iz − 5
c
c

1
5 z + 26iz − 5
∴ I = 2 ∫ f ( z )dz …………
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0

sin 2 θ
dθ ,
I = ∫ a + b cos θ
2π

Solution: Let

0

2π

=

1 − cos 2θ

∫ 2a + 2b cosθ dθ
0

We can write cos 2θ = Real part of e 2iθ ,Q e 2iθ = cos 2θ + i sin 2θ
2π

∴ I = R
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P ∫
2
 z + 1  iz
0
2a + b
 z 



dθ =

2π

= R
...
P

1
f ( z )dz ………(1)
i ∫
0

2π

1− z2
bz 2 + 2az + b
To find Residues:
Poles of f(z) are given by bz 2 + 2az + b = 0
Where f ( z ) =

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− 2a ± 4a 2 − 4b 2
2b

Z=

− a ± a2 − b2
=
b

Let α =

− a + a2 − b2
− a − a2 − b2
and β =
, since a>b,
b
b

α < 1& β > 1
∴ The simple pole z = α lies inside C,

∴ [Re sf ( z )]z =α = lim(z − α ) f ( z )
z →α

= lim( z − α )
z →α

=

1− z2
b(z − α )( z − β )

1−α 2
a − a2 − b2
=
b(α − β )
b2

a − a2 − b2
∴∑ R =
b2

∴ By Cauchy’s Residue theorem

∫ f ( z )dz = 2πi∑ R
c

 a − a2 − b2
= 2πi

b2







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(1) becomes
2π

1
I= R
...
P 2π 


b2


2π
= 2 a − a2 − b2
b
Type II Integration around semi-circular contour
Consider the integral
∞
P( x)
Improper integrals of the form ∫
dx ,where P(x) and Q(x)
Q( x)
−∞
Are polynomials in x such that the degree of Q exceeds that of P atleast by two
and Q(x) does not vanish for any x
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−
∞

Example 1: Prove that

2

z −z+2
z + 10 z 2 + 9
f ( z )dz where C is the closed contour consisting of Γ, semi- large

Solution: Let f ( z ) =
Consider

∫

4

C

circle of radius R and the real axis from –R to R
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Then

∫
c

f ( z )dz = ∫ f ( z )dz +
Γ

R

∫ f ( x)dx

−R

When R → ∞, ∫ f ( z )dz
Γ

∞

Hence

∫ f ( x)dx = ∫ f ( z )dz
C

−∞

By using residue theorem,

∫ f ( z )dz = 2πi∑ R

C
∞

Hence

∫ f ( x)dx = 2πi∑ R

−∞

The pole of f(z) are given by z 2 + a 2 = 0 ⇒ z = ± ai
The simple pole z = ai lies in the upper half plane

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[Re sf ( z )]z =ai = lim (z − ai ) f ( z )
z → ai

ze imz
= lim ( z − ai )
z → ai
(z + ai )(z − ai )

=

∞

∴ ∫ f ( x)dx = 2πi
−∞

aie − ma e − ma
=
2ai
2

e − ma
= πie − ma
2

xeimx
⇒ ∫ 2
dx = πie − ma
2
x +a
−∞
∞

∞

x(cos mx + i sin mx)
dx = πie − ma
2
2
x +a
−∞
Equating imaginary parts we get

∫

∞

∫

−∞

x sin mx
dx = πie − ma
x2 + a2
∞

⇒∫
0

x sin mx
π
dx = e − ma
2
2
x +a
2

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