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KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
B
...
(II Sem
...

2𝜋

∞

∫0 𝑓(cos 𝜃 , sin 𝜃) 𝑑𝜃, ∫−∞ 𝑓(𝑥) 𝑑𝑥

Course Outcome: Extend the concepts of complex functions for calculating Taylor’s series, Laurent’s series and
definite integrals

Application in Engineering: Complex analysis is a branch of mathematics that studies analytical properties of
functions of complex variables
...
The discipline covers a wide range of different techniques including
solution methods to free-boundary problems such as Hele-Shaw and Stokes flow, conformal mappings, Fourier and
other transform methods and Riemann-Hilbert problems
...
Importantly, there has been a surge of activity in the advancement of complex analysis methods
in recent years, driven by applications in engineering, biology, and medicine
...


1|Page

CONTENTS
Topic

Page No
...
Terminology used in complex integration
...
Complex integrals

3

3
...
Extension of Cauchy’s integral theorem

7

5
...
Cauchy’s integral formula for derivative of an analytic function

10

7
...
Taylor Series

15

9
...
Zero of an analytic function and singularity

17

11
...
The calculus of residues

19

13
...
Contour Integration

26

15
...
E-resources

33

2|Page

Terminology used in complex integration:
Continuous Arc: The set of points (𝑥, 𝑦) defined by 𝑥 = ∅(𝑡), 𝑦 = 𝜓(𝑡), with parameter 𝑡 in the interval (𝑎, 𝑏) ,
define a continuous arc provided ∅(𝑡) & 𝜓(𝑡) are continuous functions
...

Simple Curve: A curve having no self-intersections is known as simple curve
...
e
...


Closed Curve: A curve is said to be closed if end points are coinciding
...

Closed Contour: A piecewise smooth closed curve without points of self-intersection is called closed contour
...

Positive sense: of traversing a contour is the direction such that the interior domain bounded by the given closed
contour remains on the left of the direction of the motion
...


Example
...

Solution
...
Evaluate ∳𝐶 𝑧 𝑑𝑧 , where 𝐶 is the circle 𝑥 = cos 𝑡 , 𝑦 = sin 𝑡 , 0 ≤ 𝑡 ≤ 2𝜋
...
Here, we have
𝑧(𝑡) = cos 𝑡 + 𝑖 sin 𝑡 = 𝑒 𝑖𝑡
𝑓(𝑧) =
1

𝑧′(𝑡) = 𝑖𝑒 𝑖𝑡

1
= 𝑒 −𝑖𝑡
𝑧
2

2

Thus, ∳𝐶 𝑧 𝑑𝑧 = ∫0 𝑒 −𝑖𝑡 𝑖𝑒 𝑖𝑡 𝑑𝑡 = ∫0 𝑖 𝑑𝑡 = 𝟐𝝅𝒊
...


Example
...
We have ∫𝐶(𝑥 2 + 𝑖𝑦 2 )𝑑𝑧 = ∫𝐶 (𝑥 2 + 𝑖𝑦 2 )𝑑𝑧 + ∫𝐶 (𝑥 2 + 𝑖𝑦 2 )𝑑𝑧
1

2

on 𝐶1 : 𝑦 = 𝑥 ⇒ 𝑑𝑦 = 𝑑𝑥 and 𝑥 → 0 𝑡𝑜 1
1

Now, ∫𝐶 (𝑥 2 + 𝑖𝑦 2 )𝑑𝑧 = ∫0 (𝑥 2 + 𝑖𝑥 2 )(1 + 𝑖)𝑑𝑥
1

1

𝟐

= (1 + 𝑖)2 ∫0 𝑥 2 𝑑𝑥 = 𝟑 𝒊
on 𝐶2 : 𝑥 = 1 ⇒ 𝑑𝑥 = 0 and 𝑦 → 1 𝑡𝑜 2
2

∫𝐶 (𝑥 2 + 𝑖𝑦 2 )𝑑𝑧 = ∫1 (1 + 𝑖𝑦 2 )𝑖𝑑𝑦
2

2

2

𝟕
𝟑

= − ∫1 𝑦 2 𝑑𝑦 + 𝑖 ∫1 𝑑𝑦 = − + 𝒊
2

7

𝟕

𝟓

Finally, ∫𝐶(𝑥 2 + 𝑖𝑦 2 )𝑑𝑧 = 3 𝑖 + (− 3 + 𝑖) = − 𝟑 + 𝟑 𝒊
...
Evaluate ∫0 (𝑥 − 𝑦 − 𝑖𝑥 2 ) 𝑑𝑧, along real axis from 𝑧 = 0 𝑡𝑜 𝑧 = 1 and then along a line parallel to
imaginary axis from 𝑧 = 1 𝑡𝑜 𝑧 = 1 + 𝑖
...
Let 𝐼 = ∫0 (𝑥 − 𝑦 − 𝑖𝑥 2 ) 𝑑𝑧
= ∫𝐶 (𝑥 − 𝑦 − 𝑖𝑥 2 )𝑑𝑧 + ∫𝐶 (𝑥 − 𝑦 − 𝑖𝑥 2 )𝑑𝑧
1

2

on 𝐶1 : 𝑦 = 0 ⇒ 𝑑𝑦 = 0 and 𝑥 → 0 𝑡𝑜 1
1

1

1

Now, ∫𝐶 (𝑥 − 𝑦 − 𝑖𝑥 2 )𝑑𝑧 = ∫0 (𝑥 − 𝑖𝑥 2 ) 𝑑𝑥 = 2 − 3 𝑖
1

on 𝐶2 : 𝑥 = 1 ⇒ 𝑑𝑥 = 0 and 𝑦 → 0 𝑡𝑜 1
1

1

∫𝐶 (𝑥 − 𝑦 − 𝑖𝑥 2 )𝑑𝑧 = ∫0 (1 − 𝑖 − 𝑦)𝑖𝑑𝑦 = 1 + 2 𝑖
2

1
2

1
3

1
2

3
2

1
6

∴ 𝐼 = − 𝑖 + 1 + 𝑖 = + 𝑖
...
Evaluate ∫𝐶 (𝑧 − 𝑧 2 ) 𝑑𝑧, where 𝐶 is the upper half of the circle
...
What is the value of integral if
𝐶 is the lower half of the circle?

4|Page

Solution
...


Example
...

Solution
...
𝑒 2𝜋𝑖 + 𝑒 2𝜋𝑖 + 0 − 1]
= −[−2𝜋𝑖 + 1 + 0 − 1] = 𝟐𝝅𝒊
...

𝜕𝑁

By Green’s Theorem, ∮𝐶 (𝑀𝑑𝑥 + 𝑁𝑑𝑦) = ∬𝑅 ( 𝜕𝑥 −
𝜕𝑣

𝜕𝑢

𝜕𝑢

𝜕𝑣

𝜕𝑣

𝜕𝑢

𝜕𝑢

𝜕𝑣

𝜕𝑀
) 𝑑𝑥𝑑𝑦
𝜕𝑦

∮𝐶 𝑓(𝑧)𝑑𝑧 = ∬𝑅 (− 𝜕𝑥 − 𝜕𝑦) 𝑑𝑥𝑑𝑦 + 𝑖∬𝑅 (𝜕𝑥 − 𝜕𝑦) 𝑑𝑥𝑑𝑦
∮𝐶 𝑓(𝑧)𝑑𝑧 = ∬𝑅 (− 𝜕𝑥 − 𝜕𝑦) 𝑑𝑥𝑑𝑦 + 𝑖∬𝑅 (𝜕𝑥 − 𝜕𝑦) 𝑑𝑥𝑑𝑦
𝜕𝑢

𝜕𝑢

𝜕𝑢

𝜕𝑢

= ∬𝑅 (𝜕𝑦 − 𝜕𝑦) 𝑑𝑥𝑑𝑦 + 𝑖∬𝑅 (𝜕𝑥 − 𝜕𝑥 ) 𝑑𝑥𝑑𝑦
=0
Hence, ∮𝐶 𝑓(𝑧)𝑑𝑧 = 0
5|Page

Note: Cauchy’s Integral theorem without assumption that 𝑓 ′ (𝑧) is continuous is known as Cauchy-Goursat Theorem
...
e Cauchy-Goursat Theorem can be stated as:
If f is analytic at all points within and on a simple closed contour C,
Then, ∮𝐶 𝑓(𝑧)𝑑𝑧 = 0

C-R equations:

𝜕𝑢
𝜕𝑥

=

𝜕𝑣
𝜕𝑦

𝑎𝑛𝑑

𝜕𝑢
𝜕𝑦

=−

𝜕𝑣
𝜕𝑥

Example
...


Solution
...
= ∫𝐶 𝑒 𝑖𝑧 𝑑𝑧 + ∫𝐶 𝑒 𝑖𝑧 𝑑𝑧 + ∫𝐶 𝑒 𝑖𝑧 𝑑𝑧
1

2

1+𝑖

3

−1+𝑖

−1−𝑖

= ∫−1−𝑖 𝑒 𝑖𝑧 𝑑𝑧 + ∫1+𝑖 𝑒 𝑖𝑧 𝑑𝑧 + ∫−1+𝑖 𝑒 𝑖𝑧 𝑑𝑧
1

= [𝑒 𝑖(1+𝑖) − 𝑒 𝑖(−1−𝑖) + 𝑒 𝑖(−1+𝑖) − 𝑒 𝑖(1+𝑖) + 𝑒 𝑖(−1−𝑖) − 𝑒 𝑖(−1+𝑖) ]
𝑖
1

= 𝑖 [𝑒 −1+𝑖 − 𝑒 1−𝑖 + 𝑒 −1−𝑖 − 𝑒 −1+𝑖 + 𝑒 1−𝑖 − 𝑒 −1−𝑖 ]
= 𝟎 = 𝑹
...
Evaluate ∳𝐶 𝑒 𝑧 𝑑𝑧 , where C is shown in given figure
...

The function 𝑒 𝑧 is entire and C is a simple closed contour
...

Hence, ∳𝐶 𝑒 𝑧 𝑑𝑧 = 0

𝑑𝑧

Example
...

1

Solution
...

Thus, ∫𝐶

𝑑𝑧
𝑧2

= 0
...
If
a domain is not simply connected, then it is called multiply connected domain
...
a)
...
b)
...


6|Page

A region which is not simply connected is called a multiply connected region
...

A multiply connected region enclosed between two or more separate curves
...


Simply connected region

Doubly connected region

Triply connected region

Extension of Cauchy’s integral theorem
Statement: If 𝑓(𝑧) is analytic in the region 𝑅 between two simple closed curves 𝐶1 and 𝐶2 , then
∳𝐶 𝑓(𝑧)𝑑𝑧 = ∳𝐶 𝑓(𝑧)𝑑𝑧, when integral along each curve is taken in anti-clockwise direction
...
, 𝐶𝑛 are simple closed curves with a
positive orientation such that 𝐶1 , 𝐶2 , …
...

If 𝑓 is analytic on each contour and at each point interior
to 𝐶 but exterior to all the 𝐶𝑘 , 𝑘 = 1,2,3, … … , 𝑛
∳𝐶 𝑓(𝑧)𝑑𝑧 = ∑𝑛𝑘=1 ∳𝐶 𝑓(𝑧)𝑑𝑧
𝑘

Cauchy’s integral formula
Statement: If 𝑓(𝑧) is analytic within and on a closed curve 𝐶 and 𝑧0 is any point within 𝐶, then
1

𝑓(𝑧)

𝑓(𝑧)

0

0

𝑓(𝑧0 ) = 2𝜋𝑖 ∮𝐶 𝑧−𝑧 𝑑𝑧 OR ∮𝐶 𝑧−𝑧 𝑑𝑧 = 2𝜋𝑖
...
By Cauchy’s theorem, we have
𝑓(𝑧)

∮𝐶 𝑧−𝑧 𝑑𝑧 + ∮𝐿
0

1

𝑓(𝑧)
𝑑𝑧
𝑧−𝑧0

𝑓(𝑧)

∮𝐶 𝑧−𝑧 𝑑𝑧 = − ∮𝐶

0

0

+ ∮𝐶

0

𝑓(𝑧)
𝑑𝑧
𝑧−𝑧0

+ ∮𝐿

2

𝑓(𝑧)
𝑑𝑧
𝑧−𝑧0

=0

L

𝑓(𝑧)
𝑑𝑧
𝑧−𝑧0

Now,

z
L C

C

The equation of the circle 𝐶0 is |𝑧 − 𝑧0 | = 𝑟
Or 𝑧 − 𝑧0 = 𝑟𝑒 𝑖𝜃 ⇒ 𝑑𝑧 = 𝑟𝑖𝑒 𝑖𝜃 𝑑𝜃
0
𝑓(𝑧)
𝑓(𝑧0 + 𝑟𝑒 𝑖𝜃 )
𝑑𝑧 = − ∫
𝑟𝑖𝑒 𝑖𝜃 𝑑𝜃
𝑖𝜃
𝑧
−
𝑧
𝑟𝑒
0
𝐶
2𝜋

∴∮

2𝜋

= 𝑖 ∫0 𝑓(𝑧0 + 𝑟𝑒 𝑖𝜃 ) 𝑑𝜃
2𝜋

= 𝑖 ∫0 𝑓(𝑧0 ) 𝑑𝜃

[as 𝑟 → 0]

= 2𝜋𝑖
...
Evaluate ∮𝐶

𝑧 2 −4𝑧+4
𝑑𝑧
𝑧+𝑖

where 𝐶 is the circle |𝑧| = 2
...
Let 𝑓(𝑧) = 𝑧 2 − 4𝑧 + 4
which is analytic and 𝑧0 = −𝑖 is within 𝐶
...
𝑓(−𝑖)
= 2𝜋𝑖
...
[−1 + 4𝑖 + 4]
= 2𝜋𝑖(3 + 4𝑖) = 2𝜋(−4 + 3𝑖)
8|Page

𝑧

Example
...

Solution
...
Thus
∮𝐶

𝑧
(𝑧+3𝑖)

(𝑧−3𝑖)

1

𝑑𝑧 = 2𝜋𝑖
...
2 = 𝜋𝑖

𝑧

∴ ∮𝐶 𝑧2 +9 𝑑𝑧 = 𝝅𝒊
...
Evaluate ∮𝐶 𝑧2 +2𝑧−3 𝑑𝑧 where 𝐶 is the circle |𝑧 − 2| = 2
...
Here,
5𝑧+7

5𝑧+7

∮𝐶 𝑧2 +2𝑧−3 𝑑𝑧 = ∮𝐶 (𝑧+3)(𝑧−1) 𝑑𝑧
1

1

= 3∮𝐶 𝑧−1 𝑑𝑧 + 2∮𝐶 𝑧+3 𝑑𝑧
∵ 𝑧 = 1 is interior to 𝐶 and 𝑧 = −3 is exterior to C
...


Example
...
Let 𝐼 = ∮𝐶

sin 𝜋𝑧 2 +cos 𝜋𝑧 2
𝑑𝑧
(𝑧−1)(𝑧−2)

where 𝐶 is the circle |𝑧| = 3
...
Then
sin 𝜋𝑧 2 +cos 𝜋𝑧 2
𝑑𝑧
(𝑧−1)(𝑧−2)
1

𝐼 = ∮𝐶

sin 𝜋𝑧 2 +cos 𝜋𝑧 2
𝑑𝑧
(𝑧−1)(𝑧−2)
2

+ ∮𝐶

Where, 𝐶1 & 𝐶2 are two circles of very small radii with centres at 𝑧 = 1 & 𝑧 = 2
respectively
...
[

𝑑𝑧 + ∮𝐶

2

sin 𝜋𝑧 2 +cos 𝜋𝑧 2
]
𝑧−2
𝑧=1

0−1

0+1
)
1

= 2𝜋𝑖 ( −1 ) + 2𝜋𝑖 (

[

sin 𝜋𝑧2 +cos 𝜋𝑧2
]
𝑧−1

𝑧−2

+ 2𝜋𝑖
...
Use Cauchy’s integral formula to evaluate ∮𝐶 (𝑧−1)(𝑧−2) 𝑑𝑧 where 𝐶 is the circle |𝑧| = 3
...
Let 𝐼 = ∮𝐶 (𝑧−1)(𝑧−2) 𝑑𝑧
Here, 𝑧 = 1 𝑎𝑛𝑑 𝑧 = 2 are two points within 𝐶, at which the integrand is not analytic
...

𝑒2𝑧

𝐼 = ∮𝐶

1

[𝑧−2]
𝑧−1

𝑒2𝑧

𝑑𝑧 + ∮𝐶

[𝑧−1]
𝑧−2

2

𝑒2

𝑒4

−1

1

𝑒 2𝑧

𝑒 2𝑧

𝑑𝑧 = 2𝜋𝑖
...
[𝑧−1]

𝑧=2

= 2𝜋𝑖 ( ) + 2𝜋𝑖 ( ) = 𝟐𝝅𝒊𝒆𝟐 (𝒆𝟐 − 𝟏)

Cauchy’s integral formula for the derivative of an analytic function

Statement: If a function 𝑓(𝑧) is analytic in the region 𝐷,then its derivative at any point 𝑧 = 𝑧0 of 𝐷 is also analytic in
𝐷 and is given by
1

𝑓(𝑧)
2 𝑑𝑧
0)

𝑓 ′ (𝑧0 ) = 2𝜋𝑖 ∮𝐶 (𝑧−𝑧

𝑓(𝑧)
2 𝑑𝑧
0)

OR ∮𝐶 (𝑧−𝑧

= 2𝜋𝑖
...

Proof
...

Thus, the derivative of an analytic function is analytic
...
𝑓 𝑛 (𝑧0 )

Where, 𝐶 is any closed contour in 𝐷 surrounding the point 𝑧 = 𝑧0
...
We shall prove this theorem by Mathematical Induction
...
−1] 𝑓(𝑧)𝑑𝑧

+ ⋯ …
...
] 𝑓(𝑧)𝑑𝑧

Taking 𝑙𝑖𝑚𝑖𝑡 as ℎ → 0
lim

𝑓𝑘 (𝑧0 +ℎ)−𝑓𝑘 (𝑧0 )
ℎ
ℎ→0

=

𝑓𝑘 (𝑧0 +ℎ)−𝑓𝑘 (𝑧0 )
ℎ
ℎ→0

=

lim

⇒ 𝑓 𝑘+1 (𝑧0 ) =

(𝑘+1) !
2𝜋𝑖

(𝑘+1) !
2𝜋𝑖
(𝑘+1) !
2𝜋𝑖

1

∮𝐶 (𝑧−𝑧

0)

𝑘+2

𝑓(𝑧)𝑑𝑧

𝑘+2

𝑓(𝑧)𝑑𝑧

1

∮𝐶 (𝑧−𝑧

0)

∮𝐶 (𝑧−𝑧

1

0)

(𝑘+1)+1

𝑓(𝑧)𝑑𝑧

Hence, the theorem is true for 𝑛 = 𝑘 + 1
𝑛!

𝑓(𝑧)
𝑛+1
0)

∴ 𝑓 𝑛 (𝑧0 ) = 2𝜋𝑖 ∮𝐶 (𝑧−𝑧

----------------------- (1)

Since, 𝑧0 is a point of 𝐷, so by eqn (1), 𝑓 𝑛 (𝑧0 ) is analytic in 𝐷
...


Liouville’s Theorem
Statement: If a function 𝑓(𝑧) is analytic for all finite values of 𝑧 and is bounded, then it is a constant
...
Since 𝑓(𝑧) is bounded, so |𝑓(𝑧)| ≤ 𝑀 where 𝑀 is a positive constant
...
Take the contour 𝐶 to be a large circle with its centre at the origin and
radius 𝑅 enclosing the points 𝑧1 𝑎𝑛𝑑 𝑧2 so that, 𝑅 > |𝑧1 | and 𝑅 > |𝑧2 |
...

∫ 𝑅𝑑𝜃
2𝜋 (𝑅−|𝑧2 |)(𝑅−|𝑧1 |) 0
|𝑧1 −𝑧2 |2𝜋𝑅
1

...

Note: A function which is analytic in the whole of the 𝑧 − 𝑝𝑙𝑎𝑛𝑒 is called an entire function
...
Evaluate ∮𝐶 (𝑧+1)5 𝑑𝑧 where 𝐶 is the circle |𝑧| = 2
...
Let 𝐼 = ∮𝐶 (𝑧+1)5 𝑑𝑧
Here, 𝑧 = −1 is a point within 𝐶, at which the integrand is not analytic
...
Evaluate ∮𝐶 𝑧2 (𝑧2−4) 𝑑𝑧 where 𝐶 is the circle |𝑧| = 1
...
Let 𝐼 = ∮𝐶 𝑧2 (𝑧2 −4) 𝑑𝑧 = ∮𝐶 𝑧2(𝑧+2)(𝑧−2) 𝑑𝑧
Here, 𝑧 = 0 is only point within 𝐶, at which the
integrand is not analytic
...
2𝑧
𝑑
𝑒 −𝑧
= 2𝜋𝑖
...
[
]
(𝑧 2 − 4)2
𝑑𝑧 𝑧 − 4 𝑧=0
𝑧=0
(0 − 4)(−𝑒 −0 ) − 0
= 2𝜋𝑖
...
[16] =

𝝅𝒊

...
Show that ∮𝐶 (𝑧2 +4)2 = 16; 𝐶 ≡ |𝑧 − 𝑖| = 2
...
Let 𝐼 = ∮𝐶 (𝑧2 +4)2 = ∮𝐶 (𝑧+2𝑖)2(𝑧−2𝑖)2
Here, 𝑧 = 2𝑖 is only point within 𝐶, at which the
integrand is not analytic
...
[𝑑𝑧 {(𝑧+2𝑖)2 }]

−2

= 2𝜋𝑖
...
[−64𝑖] = 𝟏𝟔
...
Evaluate ∮𝐶 (𝑧+1)2(𝑧2 +4) 𝑑𝑧, where 𝐶 is the circle |𝑧| = 3
...
Let 𝐼 = ∮𝐶 (𝑧+1)2(𝑧2+4) 𝑑𝑧 = ∮𝐶 (𝑧+1)2(𝑧+2𝑖)(𝑧−2𝑖)
𝑧2 −2𝑧

𝐼=

[ 2 ]
𝑧 +4
∮𝐶 (𝑧+1)
2 𝑑𝑧
1

𝑧2 −2𝑧

+ ∮𝐶

[(𝑧+1)2 (𝑧−2𝑖)]
𝑧+2𝑖

2

𝑧2 −2𝑧

𝑑𝑧 + ∮𝐶

[(𝑧+1)2(𝑧+2𝑖)]

3

𝑧−2𝑖

𝑑𝑧

Where, 𝐶1 , 𝐶2 & 𝐶3 are circles of very small radius with
centres at 𝑧 = −1, 𝑧 = −2𝑖 & 𝑧 = 2𝑖 respectively
...

where ak are complex constants, is called a power series in (𝑧 − 𝑧0 )
...


Circle of Convergence
𝑘
Every complex power series ∑∞
𝑘=0 𝑎𝑘 (𝑧 − 𝑧0 ) has radius of convergence R and has a circle of convergence defined
by |z – z0| = R, 0 < R < 
...

(ii) a finite number (converges at all interior points of the circle (|z − z0| = R)
...


A power series may converge at some, all, or none of the points on the circle of
convergence
...
Consider the series ∑∞
𝑘=1
lim |

𝑛→∞

𝑧𝑛+2
𝑛+1
𝑧𝑛+1
𝑛

𝑛+1
|𝑧|
𝑛→∞ 𝑛

| = lim

𝑧 𝑘+1
,
𝑘

by ratio test
...
O
...
using ratio test for the power series ∑∞
𝑘=0 𝑎𝑘 (𝑧 − 𝑧0 )

(i)
(ii)
(iii)

𝑎𝑛+1
|
𝑛→∞ 𝑎𝑛
𝑎
lim | 𝑛+1 |
𝑛→∞ 𝑎𝑛
𝑎
lim | 𝑎𝑛+1 |
𝑛→∞
𝑛

lim |

1

= 𝐿 ≠ 0, the R
...
C
...

= 0 the R
...
C
...

= ∞ the R
...
C
...


To find R
...
C
...
O
...
is 
...
O
...
using root test
6𝑘+1 𝑘

𝑘
Consider the power series ∑∞
𝑘=1 (2𝑘+5) (𝑧 − 2𝑖)

𝑎𝑛 = (

6𝑛 + 1 𝑛
6𝑛 + 1
𝑛
) , lim √|𝑎𝑛 | = lim
=3
𝑛→∞
𝑛→∞ 2𝑛 + 5
2𝑛 + 5
1

1

This root test shows the R
...
C
...
The circle of convergence is |𝑧 − 2𝑖| = 3 ; the series converges absolutely for
1

|𝑧 − 2𝑖| <
...
𝑓 𝑛 (𝑧0 ) = 𝑛! 𝑎𝑛
𝑎𝑛 =

𝑓 𝑛 (𝑧0 )
 ,  𝑛 ≥ 0
𝑛!

When 𝑛 = 0, We interpret the zeroth derivative as 𝑓(𝑧0 )
Now, we have
𝑓(𝑧) = ∑∞
𝑘=0

𝑓𝑘 (𝑧0 )
(𝑧
𝑘!

− 𝑧0 )𝑘

This series is called the Taylor series for f centered at 𝑧0
...


Laurent’s Series
If 𝑓(𝑧) is analytic inside and on the boundary of the annular region 𝑅 bounded by two concentric circles 𝐶1 𝑎𝑛𝑑 𝐶2 of
radii 𝑟1 𝑎𝑛𝑑 𝑟2 (𝑟1 > 𝑟2 ) respectively having centre at 𝑧0 , ∀ 𝑧 𝑖𝑛 𝑅
∞
𝑛
−𝑛
𝑓(𝑧) = ∑∞
𝑛=0 𝑎𝑛 (𝑧 − 𝑧0 ) + ∑𝑛=1 𝑏𝑛 (𝑧 − 𝑧0 )

Where,
𝑎𝑛 =

1
𝑓(𝑤)
∮
𝑑𝑧 ; 𝑛 = 0,1,2,3, … …
...


∞
n
−n
we have, f(z) = ∑∞
-------------------- (1)
n=0 a n (z − z0 ) + ∑n=1 b(z − z0 )
The part with negative powers is called the principal part of (1) and will converge for |𝑧 − 𝑧0 | > 𝑟1
...

Hence the sum of these parts converges when𝑟1 < |𝑧 − 𝑧0 | < 𝑟2
...
The function 𝑓(𝑧) =

sin 𝑧
𝑧3

is not analytic at 𝑧 = 0 and hence can not be expanded in a Maclaurin series
...
Thus
sin 𝑧 = 𝑧 −
𝑓(𝑧) =

sin 𝑧
𝑧3

𝑧3
3!

+
1

𝑧5
5!

−
1

𝑧7
7!

= 𝑧2 − 3! +

+ ⋯ … …
...


This series converges for all 𝑧 except 𝑧 = 0, 0 < |𝑧|
...
Expand 𝑓(𝑧) = 𝑧(𝑧−1) in a Laurent series valid 1 < |𝑧 − 2| < 2
Solution
...
]

− ⋯ …
...
]
1

1

1

= 𝑧−2 − (𝑧−2)2 + (𝑧−2)3 − ⋯ …
...
Expand 𝑓(𝑧) = 𝑧(1−𝑧) in a Laurent series valid for 0 < |𝑧| < 1
...
We can write
8𝑧+1

𝑓(𝑧) = 𝑧(1−𝑧) =

8𝑧+1 1
[1−𝑧]
𝑧

1

= (8 + 𝑧) [1 + 𝑧 + 𝑧 2 + ⋯ …
...

𝑧3

Example
...

𝑧3

Solution
...
Expand 𝑓(𝑧) = 𝑧2 −1 about 𝑧 = 𝑖
...
𝑓(𝑧) = 𝑧2 +1 = 2𝑖 (𝑧−𝑖 − 𝑧+𝑖) = 2𝑖 (𝑧−𝑖 − 2𝑖+𝑧−𝑖)
∞

1
1
1
1
1 1
1
1 𝑛
= (
− ⋅
)=
−
∑
(−
) (𝑧 − 𝑖)𝑛
2𝑖 𝑧 − 𝑖 2𝑖 1 + 𝑧 − 𝑖
2𝑖 𝑧 − 𝑖 (2𝑖)2
2𝑖
𝑛=0
2𝑖
𝑖 1
1 𝑖
=−
+ + (𝑧 − 𝑖) … …
...

An analytic function 𝑓(𝑧) is said to have a zero of order 𝑚 if 𝑓(𝑧) is expressible as
𝑓(𝑧) = (𝑧 − 𝑎)𝑚 𝜙(𝑧), where 𝜙(𝑧) is analytic and 𝜙(𝑎) ≠ 0
Singularity:
A point 𝑧0 is said to be a singular point or singularity of 𝑓(𝑧) if 𝑓(𝑧) fails to be analytic at 𝑧0 but is analytic at some
point in the neighbourhood of 𝑧0
...

𝑧+1

Example
...
Thus 𝑧 = 0 and 𝑧 = 2 are the only singularities of this function
...
Hence 𝑧 = 0 and 𝑧 = 2 are the
isolated singularities of this function
...
at the points where
⇒𝑧=

1
𝑛

𝜋
𝑧

= 𝑛𝜋

(1, 2, 3, … … … … )
1 1
2 3

Thus 𝑧 = 1, , , … … …
...
Therefore, 𝑧 = 0
is an non-isolated singularity of the given function
...
+
+ ⋯…
(𝑧 − 𝑧0 )𝑛
𝑧 − 𝑧0 (𝑧 − 𝑧0 )2 (𝑧 − 𝑧0 )3

The first series on the right side of Laurent expansion is the regular part ( Taylor series) while series with negative
powers of (𝑧 − 𝑧0 ) is called Principal Part
...
P
...
Moreover, the nomenclature is justified
as it can be removed by appropriately defining the function at 𝑧 = 𝑧0
...


Note that it has a removable singularity at 𝑧 = 0 due to the absence of the Principal Part in the Laurent expansion
...

Essential Singularity: If the Principal Part in the Laurent series expansion of a function 𝑓(𝑧)in some deleted
neighbourhood of 𝑧0 contains infinitely many terms then 𝑧0 is called an essential singularity of 𝑓(𝑧)
...

Suppose 𝑏𝑚 ≠ 0, 𝑏𝑚+1 = 𝑏𝑚+2 = 𝑏𝑚+3 = ⋯ … … … … = 0, then the point 𝑧0 is termed as a Pole of order 𝑚
...

𝑒𝑧

The function 𝑓(𝑧) = (𝑧−1)3 has a pole of order 3 since
𝑒𝑧

𝑓(𝑧) = (𝑧−1)3 =

𝑒
...

(2) The limit point of the poles of a function 𝑓(𝑧) is a non-isolated essential singularity
...
Find out the zero and discuss the nature of the singularity of 𝑓(𝑧) =

𝑧−2
𝑧2

1

sin 𝑧−1

Solution
...

1

1

sin 𝑧−1 = 0 ⇒ 𝑧−1 = 𝑛𝜋,
1

⇒ 𝑧 = 1 + 𝑛𝜋 ,

𝑛∈ℤ

𝑛∈ℤ

which are also zeros of order 1
...

Again, to obtain the poles of 𝑓(𝑧) equating to zero its denominator we get
𝑧 2 = 0 ⇒ 𝑧 = 0, 0
Hence 𝑧 = 0 is a pole of order 2
...
Discuss the nature of the singularity of 𝑓(𝑧) =
Solution
...


1

= 𝑧3 (𝑧 − sin 𝑧)

− (𝑧 −

𝑧2
5!

𝑧−sin 𝑧
𝑧3

+

𝑧4
7!

𝑧3
3!

+

𝑧5
5!

−

𝑧7
7!

+ ⋯ … … … )] =

1 𝑧3
(
𝑧 3 3!

−

𝑧5
5!

+

𝑧7
7!

+ ⋯………)

+ ⋯……

Since there is no term in the principal part of the given function hence 𝑧 = 0 is a removable singularity
...
Discuss singularity of 𝑓(𝑧) = (𝑧−𝑎)2 at 𝑧 = 𝑎 and 𝑧 = ∞
...
We have 𝑓(𝑧) = (𝑧−𝑎)2 = (𝑧−𝑎)2 sin 𝜋𝑧
Poles of 𝑓(𝑧) are given by (𝑧 − 𝑎)2 sin 𝜋𝑧 = 0
⇒ z = 𝑎, 𝑛, 𝑛 ∈ ℤ
z = 𝑎 is a pole of order 2 and since z = ∞ is a limit point of these poles, therefore z = ∞ is a non-isolated essential
singularity
...
⋯
0

0

0

0

1

The residue of a function is the coefficient of the term 𝑧−𝑧 (𝑖
...

0

−2𝑧+3

1

1

1

𝑧

Example
...
1−𝑧 = 1 + 𝑧 + 𝑧 2 + 𝑧 3 + ⋯ … … …
...

There are two methods:
1
...
By Formula (Use formula directly)

Formula for finding the residue for a simple pole:
If 𝑓(𝑧) has a simple pole at 𝑧0 , then the Laurent series is:
𝑏

𝑓(𝑧) = 𝑎0 + 𝑎1 (𝑧 − 𝑧0 ) + ⋯ + 𝑧−𝑧1 (0 < |𝑧 − 𝑧0 | < 𝑅)Or (𝑧 − 𝑧0 )𝑓(𝑧) = 𝑎0 (𝑧 − 𝑧0 ) + 𝑎1 (𝑧 − 𝑧0 )2 + ⋯ + 𝑏1
0

19 | P a g e

Taking 𝑙𝑖𝑚𝑖𝑡 as 𝑧 → 𝑧0
lim ( 𝑧 − 𝑧0 )𝑓(𝑧) = 𝑏1

𝑧→𝑧0

OR Res 𝑓(𝑧) = 𝑏1
𝑧=𝑧0

= lim ( 𝑧 − 𝑧0 )𝑓(𝑧)
𝑧→𝑧0

∴ Res 𝑓(𝑧) = lim ( 𝑧 − 𝑧0 )𝑓(𝑧)
𝑧→𝑧0

𝑧=𝑧0

2𝑧−𝑖

Example
...

Solution
...
[

1
{1+

𝑧−𝑖 2
}
2𝑖

]

(𝑧−𝑖)
𝑖
(𝑧−𝑖)2
(𝑧−𝑖)3

...
]
]
2
4(𝑧−𝑖)
2𝑖
(2𝑖)
(2𝑖)3
𝑖 1
1
5
1
− 4 𝑧−𝑖 − 4 − 16 (𝑧 − 𝑖) + 2 (𝑧 − 𝑖)2 ⋯ … (0 < |𝑧 − 𝑖| < 2)
1
2

= [− −
=

𝑧+1

Example
...
Res𝑓(𝑧) = lim 𝑧 𝑧(𝑧−2) = lim 𝑧−2 = − 2
𝑧→0

𝑧=0

𝑧→0

𝑧+1

𝑧+1
𝑧→2 𝑧

Res𝑓(𝑧) = lim( 𝑧 − 2) 𝑧(𝑧−2) = lim
𝑧→2

𝑧=2

3

=2

(Residues by definition) Laurent series are:
𝑧+1

𝑓(𝑧) = 𝑧(𝑧−2) = −
=−
𝑓(𝑧) =

𝑧+1
𝑧
{1 + 2 +
2𝑧

𝑧+1
𝑧(𝑧−2)

=

𝑧+1
1
{
}
2𝑧 1−𝑧/2

=

𝑧2
22

1

(𝑧−2)+3
1
[
]
(𝑧−2) 2+(𝑧−2)

=

(𝑧−2)+3
𝑧−2
(𝑧−2)2
−
+ 2
[1
2(𝑧−2)
2
2
3

1

1

= 2
...
]

(𝑧−2)2
8

+ ⋯ …
...
Find the residues of 𝑓(𝑧) = 𝑧2 +9
𝑧+1

𝑧+1

Solution
...

𝟔

Example
...
Here 𝑓(𝑧) = cot 𝑧 or 𝑓(𝑧) =

cos 𝑧
sin 𝑧

Poles of 𝑓(𝑧) are given by
sin 𝑧 = 0 ⇒ 𝑧 = 𝑛𝜋
cos 𝑧

[𝑅𝑒𝑠
...

Differentiate
𝑑
[(𝑧
𝑑𝑧

− 𝑧0 )2 𝑓(𝑧)] = 2𝑎0 (𝑧 − 𝑧0 ) + 3𝑎1 (𝑧 − 𝑧0 )2 + ⋯ … ⋯ + 𝑏1

Taking 𝑙𝑖𝑚𝑖𝑡 as 𝑧 → 𝑧0
lim

𝑑

𝑧→𝑧0 𝑑𝑧

[(𝑧 − 𝑧0 )2 𝑓(𝑧)] = 𝑏1

∴ Res 𝑓(𝑧) = lim

𝑑

𝑧→𝑧0 𝑑𝑧

𝑧=𝑧0

[(𝑧 − 𝑧0 )2 𝑓(𝑧)]
1

Example
...

Solution
...
Find the residue of 𝑓(𝑧) =
Solution
...
⋯}

21 | P a g e

=
=

(𝑧−1)
(𝑧−1)+1
1
1
1
{(𝑧−1)2 − 3(𝑧−1) + 32 − 33 … …
...
(𝑧−1) − 27 + 81 − ⋯ … … ⋯ (0
3(𝑧−1)2

< |𝑧 − 1| < 3)

Formula for finding the residue for a pole of any order:
If 𝑓(𝑧) has a pole of order 𝑚 at 𝑧0 , then the Laurent series is
𝑓(𝑧) = 𝑎0 + 𝑎1 (𝑧 − 𝑧0 ) + ⋯ …
...
⋯ + 𝑏1 (𝑧 − 𝑧0 )𝑚−1
+𝑏2 (𝑧 − 𝑧0 )𝑚−2 + ⋯ … … ⋯ + 𝑏𝑚
Now, differentiate (𝑚 − 1) times and let 𝑧 → 𝑧0 to get
𝑑 𝑚−1
[(𝑧
𝑧→𝑧0 𝑑𝑧 𝑚−1

∴ lim

− 𝑧0 )𝑚 𝑓(𝑧)] = (𝑚 − 1)! 𝑏1
𝑑 𝑚−1

1

Res 𝑓(𝑧) = (𝑚−1)! lim [𝑑𝑧𝑚−1 {(𝑧 − 𝑧0 )𝑚 𝑓(𝑧)}]
𝑧→𝑧0

𝑧=𝑧0

𝑑 𝑚−1

1

OR Res 𝑓(𝑧) = (𝑚−1)! [𝑑𝑧𝑚−1 {(𝑧 − 𝑧0 )𝑚 𝑓(𝑧)}]
𝑧=𝑧0

Example
...
Here 𝑓(𝑧) =

𝑧=𝑧0

sinh 𝑧

...
𝑜𝑓 𝑓(𝑧)]𝑧=∞ = 𝑙𝑖𝑚[−𝑧𝑓(𝑧)]
𝑧→0

OR
1
𝑖𝑛 𝑡ℎ𝑒 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑓(𝑧)]
𝑧

[𝑅𝑒𝑠
...
Find the residue of 𝑓(𝑧) = 𝑧2 −1 at 𝑧 = ∞
...
We have, 𝑓(𝑧) = 𝑧2 −1 =

𝑧3
1
𝑧

𝑧 2 (1− 2 )

1 −1

= 𝑧 (1 − 𝑧2 )
1

1

1

= 𝑧 + 𝑧 + 𝑧3 + 𝑧 5 + ⋯ ……
1

[𝑅𝑒𝑠
...
Evaluate ∫𝑐 𝑧 sin 𝑧 , 𝑤ℎ𝑒𝑟𝑒 𝐶: |𝑧| = 1
1

Solution
...
𝑜𝑓 𝑓(𝑧)]𝑧=0 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 = 0
𝑧
∴ ∫𝑐

𝑑𝑧
𝑧 sin 𝑧

= 2𝜋𝑖 × 0 = 0
...
Evaluate ∳𝐶 (𝑧−1)2(𝑧−3) 𝑑𝑧 , where 𝐶 is the circle |𝑧| = 2
...
Since only the pole 𝑧 = 1 lies within the circle, then there is only one singular point 𝑧 = 1 within 𝐶, then
1

∳𝐶 (𝑧−1)2 (𝑧−3) 𝑑𝑧 = 2𝜋𝑖
...
1! lim [𝑑𝑧 (𝑧 − 1)2 (𝑧−1)2 (𝑧−3)]
𝑧→1

= 2𝜋𝑖
...
If a function 𝑓 is analytic
on and within 𝐶, except at a finite number of singular points 𝑧1 , 𝑧2 , 𝑧3 , … … … …
...
∑𝑛𝑘=1[𝑅𝑒𝑠𝑖𝑑𝑢𝑒 𝑜𝑓 𝑓(𝑧)]𝑧𝑘
Proof
...
+∮𝐶 𝑓(𝑧)𝑑𝑧
1

2

𝑛

= ∑𝑛𝑘=1 [∮𝐶 𝑓(𝑧)𝑑𝑧]
𝑘

= ∑𝑛𝑘=1[2𝜋𝑖
...
Evaluate ∮𝐶 (𝑧−1)2(2𝑧+3) 𝑑𝑧, where 𝐶 ≡ |𝑧| = 2
...
Let 𝑓(𝑧) = (𝑧−1)2(2𝑧+3)
Here, 𝑧 = 1 & 𝑧 =

3
−2

are two poles and both lie within 𝐶
...
𝑓(𝑧)]𝑧=1 = lim [ (𝑧 − 1)2
...
𝑓(𝑧)]𝑧=−3 = lim3 [(𝑧 + )
...
[𝑠𝑢𝑚 𝑜𝑓 𝑟𝑒𝑠𝑖𝑑𝑢𝑒𝑠 𝑎𝑡 𝑝𝑜𝑙𝑒𝑠 𝑤𝑖𝑡ℎ𝑖𝑛 𝐶]
86

86

= 2𝜋𝑖
...

2+𝑧

Example
...

2+𝑧

2+𝑧

2+𝑧

Solution
...
{lim 𝑧−1 + lim
𝑧→0

2+𝑧

∴ ∮𝐶 𝑧2 −𝑧 𝑑𝑧 = 2𝜋𝑖
...
Evaluate ∫𝑐 𝑧(𝑧−2)4

Solution
...
𝑜𝑓 𝑧(𝑧−2)4 ]

=

𝑧=2

𝑑𝑧

1
𝑑3
lim [ 3 (𝑧
3 ! 𝑧→2 𝑑𝑧

1

∴ ∫𝑐 𝑧(𝑧−2)2 = 2𝜋𝑖 (− 16) = −

− 2)4

1
]
𝑧(𝑧−2)4

=−

C

1
16

𝝅𝒊

...
show ∫𝑐𝑒 𝑧2 𝑑𝑧 = 0 , where 𝐶: |𝑧| = 1
1

Solution
...
𝑜𝑓 𝑒 𝑧2 ]

+

1
3!

=0
𝑧=0

1
𝑧6

+ ⋯
...
Evaluate ∫𝑐 𝑧(𝑧−1) 𝑑𝑧

where 𝐶: |𝑧| = 2

Solution
...
𝑓(𝑧)]𝑧=0 = lim [𝑧
...
𝑜𝑓 𝑓(𝑧)]𝑧=1 = lim [(𝑧 − 1) 𝑧(𝑧−1)] = 3
𝑧→1

∴

5𝑧−2
∫𝑐 𝑧(𝑧−1)

0

2

𝑑𝑧 = 2𝜋 𝑖 (2 + 3) = 𝟏𝟎𝝅𝒊
...
Poles are 𝑧 = 0, 𝑧 = 1
(a) when 0 < |𝑧| < 1
5𝑧−2

𝑓(𝑧) = 𝑧(𝑧−1) =

5𝑧−2
𝑧


...
)
𝑏1 = 𝐵1 = [𝑅𝑒𝑠
...
1+(𝑧−1)
𝑧−1
3
= (5 + 𝑧−1) [1 − (𝑧

=

− 1) + (𝑧 − 1)2
...

𝑧

1

Example
...

𝑧

𝑧

Solution
...

𝑧

[𝑅𝑒𝑠
...

]
(𝑧−2)(𝑧−1)

|𝑧 − 2| =

1
2

𝑧→2

𝑧

= [𝑧−1]

=2

𝑧=2

By Cauchy’s Residue Theorem
𝑧

∮𝐶 𝑧2 −3𝑧+2 𝑑𝑧 = 2𝜋𝑖
...
[2] = 𝟒𝝅𝒊
...
Evaluate ∮𝐶 𝑧2 +2𝑧+5 𝑑𝑧, where 𝐶 is the circle |𝑧 + 1 − 𝑖| = 2
...
Let 𝑓(𝑧) = 𝑧2 +2𝑧+5 = (𝑧−𝛼)(𝑧−𝛽), where 𝛼 = −1 + 2𝑖, 𝛽 = −1 − 2𝑖
Here, poles are 𝑧 = 𝛼 & 𝑧 = 𝛽, but only 𝑧 = 𝛼 lies within 𝐶
...
𝑜𝑓 𝑓(𝑧)]𝑧=𝛼 = lim [(𝑧 − 𝛼)
...
[𝑠𝑢𝑚 𝑜𝑓 𝑟𝑒𝑠𝑖𝑑𝑢𝑒𝑠 𝑎𝑡 𝑝𝑜𝑙𝑒𝑠 𝑤𝑖𝑡ℎ𝑖𝑛 𝐶]
= 2𝜋𝑖
...


25 | P a g e

Contour Integration
Contour Integral: The contour integral of a complex function 𝑓: 𝐶 → 𝐶 is a generalization of the integral for realvalued functions
...
Contour integration is closely related to the calculus of residues
...

Consider a circle C: |𝑧| = 1 as contour |𝑧| = 1 ⇒ 𝑧 = 𝑒 𝑖𝜃
1

1

1

1

1

1

cos 𝜃 = 2 (𝑒 𝑖𝜃 + 𝑒 −𝑖𝜃 ) = 2 (𝑧 + 𝑧), sin 𝜃 = 2𝑖 (𝑒 𝑖𝜃 − 𝑒 −𝑖𝜃 ) = 2𝑖 (𝑧 − 𝑧)
and 𝑑𝜃 =

𝑑𝑧
𝑖𝑧

; 0 ≤ 𝜃 ≤ 2𝜋
𝑧+𝑧 −1 𝑧−𝑧 −1 𝑑𝑧
,
)
2
2𝑖
𝑖𝑧

2𝜋

∴ ∫0 𝐹 (cos 𝜃 , sin 𝜃)𝑑𝜃 = ∳𝐶 𝐹 (

2𝜋

Example
...
Let I = ∫0

1
𝑑𝜃
(2+cos 𝜃)2

1
𝑑𝜃
(2+cos 𝜃)2
1

1

put cos 𝜃 = [𝑧 + ] , 𝑑𝜃 =
2
𝑧
4
𝑖

∴ 𝐼 = ∳𝐶
4
𝑖

= ∳𝐶

𝑧
 𝑑𝑧
(𝑧 2 +4𝑧+1)2

𝑑𝑧
𝑖𝑧

, where 𝐶: |𝑧| = 1

|𝒛|
C= 𝟏

𝑧
 𝑑𝑧
(𝑧−𝑧1 )2 (𝑧−𝑧2 )2

where, 𝑧1 = −2 + √3 , 𝑧2 = −2 − √3
Since only 𝑧1 is inside the unit circle 𝐶
...
2πi
...
of f(z)]z=z1
Now, [Res
...

√

26 | P a g e

2𝜋

Example
...
Let I = ∫0

1

1

put cos 𝜃 = 2 [𝑧 + 𝑧] , 𝑑𝜃 =
2𝑧

𝑑𝑧
𝑖𝑧

𝑑𝑧

∴ 𝐼 = ∳𝐶 [𝑏𝑧2 +2𝑎𝑧+𝑏] ( 𝑖𝑧 ) , where 𝐶: |𝑧| = 1
2

= 𝑖𝑏 ∳𝐶

|𝒛| = 𝟏
C

𝑑𝑧
2𝑎
[𝑧 2 + 𝑧+1]
𝑏

2

= 𝑖𝑏 ∳𝐶 (𝑧−𝑧

𝑑𝑧

1 )(𝑧−𝑧2 )

𝑎

where 𝑧1 = − 𝑏 +
Let 𝑓(𝑧) = (𝑧−𝑧

√𝑎 2 −𝑏2
𝑏

𝑎

and 𝑧2 = − 𝑏 −

1

√𝑎 2 −𝑏2
𝑏

2

∴ 𝐼 = 𝑖𝑏 ∳𝐶 𝑓(𝑧)𝑑𝑧

1 )(𝑧−𝑧2 )

----------(1)

Poles of 𝑓(𝑧) are given by
(𝑧 − 𝑧1 )(𝑧 − 𝑧2 ) = 0 ⇒ 𝑧 = 𝑧1 , 𝑧2 (both are simple)
it is given that 𝑎 > |𝑏| ⇒ |𝑧2 | > 1
∵ 𝑧1
...
𝑧2 | = 1
⇒ |𝑧1 |
...
𝑜𝑓 𝑓(𝑧)]𝑧=𝑧1 = lim (𝑧 − 𝑧1 )𝑓(𝑧)
𝑧→𝑧1

= lim

1

𝑧→𝑧1 (𝑧−𝑧2 )

=

1
𝑧1 −𝑧2

∴ ∳𝐶 𝑓(𝑧)𝑑𝑧 = 2𝜋𝑖 (
2

From (1), 𝐼 = 𝑖𝑏 [

=

𝑏
2√𝑎 2 −𝑏2

𝑏
2√𝑎2 − 𝑏 2

𝑏𝜋𝑖
√𝑎2 −𝑏2

2𝜋

Solution
...
Evaluate ∫0

)=

√𝑎 2 −𝑏2

𝑑𝜃
𝑎+𝑏 sin 𝜃

, where 𝑎 > |𝑏|

dθ
𝑎+𝑏 sin 𝜃
1

1

put sin 𝜃 = 2𝑖 [𝑧 − 𝑧] , 𝑑𝜃 =
∴ 𝐼 = ∳𝐶 [𝑏𝑧2

2𝑖𝑧
𝑑𝑧
( ),
+2𝑖𝑎𝑧−𝑏] 𝑖𝑧

2

= 𝑏 ∳𝐶

𝑑𝑧
𝑖𝑧

where 𝐶: |𝑧| = 1

|𝒛| = 𝟏
C

𝑑𝑧
[𝑧 2 +

2

2𝑖𝑎
𝑧−1]
𝑏

𝑑𝑧
1 )(𝑧−𝑧2 )

= 𝑏 ∳𝐶 (𝑧−𝑧
where 𝑧1 = −

𝑖𝑎
𝑏

+

𝑖√𝑎 2 −𝑏2
𝑏

1
)(𝑧−𝑧
1
2)

Let 𝑓(𝑧) = (𝑧−𝑧

and 𝑧2 = −

𝑖𝑎
𝑏

−

𝑖√𝑎 2 −𝑏2
𝑏

2

∴ 𝐼 = 𝑏 ∳𝐶 𝑓(𝑧)𝑑𝑧 ---------- (1)

Poles of 𝑓(𝑧) are given by
27 | P a g e

(𝑧 − 𝑧1 )(𝑧 − 𝑧2 ) = 0 ⇒ 𝑧 = 𝑧1 , 𝑧2 (both are simple)
it is given that 𝑎 > |𝑏| ⇒ |𝑧2 | > 1
∵ 𝑧1
...
𝑧2 | = 1
⇒ |𝑧1 |
...
𝑜𝑓 𝑓(𝑧)]𝑧=𝑧1 = lim (𝑧 − 𝑧1 )𝑓(𝑧)
𝑧→𝑧1

1

= lim

𝑧→𝑧1 (𝑧−𝑧2 )

1

=𝑧

1 −𝑧2

𝑏

∴ ∳𝐶 𝑓(𝑧)𝑑𝑧 = 2𝜋𝑖 (

2𝑖√𝑎 2 −𝑏2

2

From (1), 𝐼 = 𝑏 [

𝑏𝜋
√𝑎 2 −𝑏2

=

)=

𝑏
2𝑖√𝑎 2 −𝑏2
𝑏𝜋

√𝑎 2 −𝑏2

2𝜋

]=

√𝑎 2 −𝑏2

2𝜋 cos 3𝜃
𝑑𝜃
5−4 cos 𝜃

Example
...
Let I = ∫0

2𝜋

𝐼 = 𝑅𝑒𝑎𝑙 𝑝𝑎𝑟𝑡 𝑜𝑓 ∫0
put 𝑒 𝑖𝜃 = 𝑧 , 𝑑𝜃 =

𝑒 3𝑖𝜃
dθ
5−4 cos 𝜃

2𝜋

= R
...
[∫0

𝑒 3𝑖𝜃
dθ]
5−2(𝑒 𝑖𝜃 +𝑒 −𝑖𝜃 )

𝑑𝑧
𝑖𝑧

𝑧3

𝑑𝑧

∴ I = R
...
[∳𝐶 5−2(𝑧+𝑧−1 ) ( 𝑖𝑧 )], where 𝐶: |𝑧| = 1
𝑧3

1

1

= R
...
[− 𝑖 ∳𝐶 2𝑧2 −5𝑧−2 𝑑𝑧] = R
...
[− 2𝑖 ∳𝐶
𝑧3
,
1 )(𝑧−𝑧2 )

Let 𝑓(𝑧) = (𝑧−𝑧
∴ 𝐼 = R
...
[−

where 𝑧1 =

1
2

𝑧3
5
𝑧 2 − 𝑧−1
2

𝑑𝑧]

𝑎𝑛𝑑 𝑧2 = 2

1
∳ 𝑓(𝑧)𝑑𝑧]
2𝑖 𝐶

---------------- (1)

Poles of 𝑓(𝑧) are given by (𝑧 − 𝑧1 )(𝑧 − 𝑧2 ) = 0 ⇒ 𝑧 = 𝑧1 , 𝑧2
(both are simple) it is very clear that only z1 lies inside the unit circle 𝐶
...
𝑜𝑓 𝑓(𝑧)]𝑧=𝑧1 = lim (𝑧 − 𝑧1 )𝑓(𝑧) = lim
𝑧→𝑧1

∴ ∳𝐶 𝑓(𝑧)𝑑𝑧 = 2𝜋𝑖 (−

𝑧1 3
1 −𝑧2

=𝑧

1

= − 12

1
𝑖𝜋
)=−
12
6
1

𝑖𝜋

𝝅

From (1), 𝐼 = R
...
[− 2𝑖 (− 6 )] = 𝟏𝟐
...
Evaluate ∫0

2+cos 𝜃

2𝜋 𝑠𝑖𝑛2 𝜃−2 cos 𝜃

Sol
...
P
...
P
...
P
...

Now, [𝑅𝑒𝑠
...
P
...

1−𝑎 sin 𝜃
(𝑧−1/𝑧)/2𝑖 𝑑𝑧
1
𝑧 2 −1
1
∮𝐶 1−𝑎(𝑧−1/𝑧)/2𝑖 𝑖𝑧 = 𝑖 ∮𝐶 𝑧 −𝑎𝑧2 +2𝑖𝑧+𝑎 𝑑𝑧

Example
...
𝐼 =

1−𝑧 2

1

= 𝑖𝑎 ∮𝐶 𝑧(𝑧2 −2𝑖𝑧/𝑎−1) 𝑑𝑧
We have 3 simple poles, 𝑧0 = 0 and 𝑧 2 − 2𝑖𝑧/𝑎 − 1 = 0
⇒𝑧=

𝑖
(1 ± √1 − 𝑎2 )
...
Res 𝑓(0)
1 − 𝑧2
1
= lim [𝑧
]=

...
Res
1 (𝑧1 −𝑧2 )

=𝑧

C

𝑖
−2𝑖
(1+√1−𝑎2 )⋅( 𝑎 √1−𝑎 2 )
𝑎
2
𝑎

(1+√1−𝑎2 )√1−𝑎2

𝑎2
(1+√1−𝑎2 )√1−𝑎2

=

=

2𝜋𝑎

(1+√1−𝑎2 )√1−𝑎2

2𝜋𝑎
(1+√1−𝑎 2 )√1−𝑎 2

29 | P a g e

Semi Circular Contour
∞ 𝑓(𝑥)
𝑑𝑥 ,
𝑔(𝑥)

Integral of the type ∫−∞

where 𝑓(𝑥) 𝑎𝑛𝑑 𝑔(𝑥) are polynomials in 𝑥, such that [𝑥

𝑓(𝑥)
]
𝑔(𝑥)

→ 0 as 𝑥 → ∞ and

(i) 𝑔(𝑥) has no zeros on the real axis
(ii) the degree of 𝑔(𝑥) is greater than that of 𝑓(𝑥) by at least 2
...

𝑓(𝑧)

Let 𝐹(𝑧) = 𝑔(𝑧)
By Residue theorem, we have
∮𝐶 𝐹(𝑧)𝑑𝑧 = 2𝜋𝑖
...
𝑜𝑓 𝐹(𝑧)]𝑧𝑘
𝑅

∫−𝑅 𝐹(𝑥)𝑑𝑥 = −∫𝐶𝑅 𝐹(𝑧)𝑑𝑧 + 2𝜋𝑖 ∑𝑛𝑘=1[𝑅𝑒𝑠
...
𝑜𝑓 𝐹(𝑧)]𝑧𝑘 ----------(1)
𝑅

𝑅→∞

𝑅→∞

𝜋

Now, lim ∫𝐶 𝐹(𝑧)𝑑𝑧 = ∫0 𝐹(𝑅𝑒 𝑖𝜃 )𝑖𝑅𝑒 𝑖𝜃 𝑑𝜃 = 0 [𝑤ℎ𝑒𝑛 𝑅 → ∞]
𝑅→∞

𝑅

∞

From (1), We have ∫−∞ 𝐹(𝑥)𝑑𝑥 = 2𝜋𝑖 ∑𝑛𝑘=1[𝑅𝑒𝑠
...
Evaluate the Cauchy principal value of ∫−∞ (𝑥 2+1)(𝑥2+9) 𝑑𝑥
1

1

Solution
...

R

1

1

1
2
2 +9)  dz
(z
+1)(z
R

∮C (z2 +1)(z2 +9) 𝑑𝑧 = ∫−R (x2 +1)(x2 +9)  dx + ∫C
= I1 + I2

𝐼1 = 2𝜋𝑖
...

Now let R →  and note that on CR:
|(𝑧 2 + 1)(𝑧 2 + 9)| = |(𝑧 2 + 1)||(𝑧 2 + 9)|
≥ ||𝑧|2 − 1| ||𝑧|2 − 9| = (𝑅 2 − 1)(𝑅 2 − 9)
From the ML-inequality
|𝐼2 | = |∫
𝐶𝑅
𝑅

1

(𝑧 2

1
𝜋𝑅
|𝐼 | → 0 as 𝑅 → ∞
𝑑𝑧| ≤ 2
2
+ 1)(𝑧 + 9)
(𝑅 − 1)(𝑅 2 − 9) 2

𝜋

Thus, lim ∫−𝑅 (𝑥 2 +1)(𝑥2 +9) 𝑑𝑥 = 12
𝑅→∞

Behavior of Integral as R → 
𝑓(𝑧)

Suppose 𝐹(𝑧) = 𝑔(𝑧), where the degree of 𝑓(𝑧) is 𝑛 and the degree of 𝑔(𝑧) is 𝑚 ≥ 𝑛 + 2
...
Evaluate the Cauchy principal value of ∫−∞ 𝑥 4+1 𝑑𝑥
Solution
...


We also knew that

𝑖

1

P
...
∫−∞ 𝑥 4 +1 𝑑𝑥 = 2𝜋𝑖
...
Evaluate the Cauchy principal value of ∫0

𝑥 2 +9

𝑑𝑥

Solution
...
Firstly, we check that the integrand is even, so we have
∞ 𝑥 sin 𝑥

∫0

𝑥 2 +9

∞ 𝑥 sin 𝑥
𝑑𝑥
𝑥 2 +9

1

𝑑𝑥 = 2 ∫−∞

𝑧

𝑧
𝑒 𝑖𝑧 𝑑𝑧
2
𝑧
+9
𝑅

∫𝐶 𝑧2 +9 𝑒 𝑖𝑧 𝑑𝑧 = ∫𝐶

𝑅

𝑥

+ ∫−𝑅 𝑥 2 +9 𝑒 𝑖𝑥 𝑑𝑥

𝑪𝑹

𝑥𝑒 𝑖𝑥

𝑅

= 0 + ∫−𝑅 𝑥 2 +9 𝑑𝑥
𝑧𝑒 𝑖𝑧

= 2𝜋𝑖
...
[(𝑧 − 3𝑖) (𝑧−3𝑖)(𝑧+3𝑖)]
𝑧𝑒 𝑖𝑧

𝑧

∫𝐶 𝑧2 +9 𝑒 𝑖𝑧 𝑑𝑧 = 2𝜋𝑖
...
[
Or

∞ 𝑥 cos 𝑥
𝑑𝑥
𝑥 2 +9

∫−∞

∞ 𝑥 sin 𝑥
𝑥 2 +9

3𝑖𝑒 −3
6𝑖

∞ 𝑥 sin 𝑥
𝑑𝑥
𝑥 2 +9

+ 𝑖 ∫−∞

∞ 𝑥 cos 𝑥
𝑑𝑥
𝑥 2 +9

⇒ ∫−∞
∴ ∫0

𝑹

𝑧=3𝑖

𝜋𝑖

] = 𝑒3

=

𝜋𝑖
𝑒3

∞ 𝑥 sin 𝑥
𝑑𝑥
𝑥 2 +9

= 0 and ∫−∞

𝑧=3𝑖

=

𝜋
𝑒3

𝝅

𝑑𝑥 = 𝟐𝒆𝟑
...
When

𝑓(𝑧) has a pole at 𝑧 = 𝑐, where 𝑐 is a real number, we must use the indented contour as in figure
...

If 𝐶𝑟 is the contour defined by 𝑧 = 𝑐 + 𝑟𝑒 𝑖𝜃 , 0 ≤ 𝜃 ≤ 𝜋 then
lim

∫
𝑟→0 𝐶𝑟

𝑓(𝑧) 𝑑𝑧 = 𝜋𝑖
...
𝑓(𝑧) and 𝑔 is analytic at 𝑐
...

we have
31 | P a g e

𝜋 𝑖𝑟𝑒 𝑖𝜃

∫𝐶 𝑓(𝑧) 𝑑𝑧 = 𝑏1 ∫0
𝑟

𝑟𝑒 𝑖𝜃

𝜋

 𝑑𝜃 + 𝑖𝑟 ∫0 𝑔(𝑐 + 𝑟𝑒 𝑖𝜃 )  𝑒 𝑖𝜃 𝑑𝜃

= 𝐼1 + 𝐼2
First, we see
𝜋 𝑖𝑟𝑒 𝑖𝜃
𝑑𝜃
𝑟𝑒 𝑖𝜃 +9

𝐼1 = 𝑏1 ∫0

𝜋

= 𝑏1 ∫0 𝑖𝑑𝜃 = 𝜋𝑖𝑏1 = 𝜋𝑖
...

Hence, It follows that lim |𝐼2 | = 0 & lim𝐼2 = 0
...
Evaluate the Cauchy principal value of ∫−∞ 𝑥(𝑥 2−2𝑥+2) 𝑑𝑥
...
We consider the contour integral ∳𝐶 𝑧(𝑧2 −2𝑧+2)
1

Let 𝑓(𝑧) = 𝑧(𝑧2 −2𝑧+2)
𝑓(𝑧) has simple poles at 𝑧 = 0 and 𝑧 = 1 + 𝑖 in the upper half-plane
...
[ Res 𝑓(𝑧)𝑒 𝑖𝑧 ]
𝑅

𝑧=1+𝑖

𝑟

Taking the limits as R →  and r → 0, we have
𝑒 𝑖𝑥

∞

P
...
∫−∞ 𝑥(𝑥 2 −2𝑥+2) 𝑑𝑥 − 𝜋𝑖
...
[ Res 𝑓(𝑧)𝑒 𝑖𝑧 ]
𝑧=0

𝑧=1+𝑖

1

Now,[Res𝑓(𝑧)𝑒 𝑖𝑧 ] = 2
𝑧=0

1
[ Res 𝑓(𝑧)𝑒 𝑖𝑧 ] = 𝑒 −1+𝑖 (1 + 𝑖)
𝑧=1+𝑖
4
∞

𝑒 𝑖𝑥

1

1

P
...
∫−∞ 𝑥(𝑥 2 −2𝑥+2) 𝑑𝑥 = 𝜋𝑖
...
[− 4 𝑒 −1+𝑖 (1 + 𝑖)]
Using 𝑒 −1+𝑖 = 𝑒 −1 [cos 1 + sin 1], then
∞

P
...
∫
−∞
∞

𝑥(𝑥 2

cos 𝑥
𝜋
𝑑𝑥 = 𝑒 −1 [cos 1 + sin 1]
− 2𝑥 + 2)
2

sin 𝑥

𝜋

P
...
∫−∞ 𝑥(𝑥 2 −2𝑥+2) 𝑑𝑥 = 2 [1 + 𝑒 −1 (sin 1 − cos 1)]
...
∫𝟎
2
...

4
...

6
...


𝒅𝜽

; 𝒂 > |𝒃|

𝒂+𝒃 𝐜𝐨𝐬 𝜽
𝟐𝝅
𝒅𝜽
∫𝟎 √𝟐−𝐜𝐨𝐬 𝜽
𝟐𝝅
𝒅𝜽
∫𝟎 𝟏−𝟐𝒂 𝐬𝐢𝐧 𝜽+𝒂𝟐 ; 𝟎
𝟐𝝅
𝒅𝜽
∫𝟎 𝟓+𝟒 𝐬𝐢𝐧 𝜽
𝝅 𝐜𝐨𝐬 𝟑𝜽𝒅𝜽
∫𝟎 𝟓−𝟒 𝐜𝐨𝐬 𝜽;
𝟐𝝅 𝐜𝐨𝐬 𝟑𝜽𝒅𝜽
∫𝟎 𝟓+𝟒 𝐜𝐨𝐬 𝜽;
𝝅 𝐜𝐨𝐬 𝟑𝜽𝒅𝜽
∫𝟎 𝟓+𝟒 𝐜𝐨𝐬 𝜽;

Ans:

𝝅
√𝒂𝟐 −𝒃𝟐

Ans: 𝟐𝝅
<𝒂<𝟏

Ans:
Ans:
Ans:
Ans:
Ans:

𝟐𝝅
𝟏−𝒂𝟐
𝟐𝝅
𝟑
𝝅
𝟐𝟒
−𝝅
𝟏𝟐
−𝝅
𝟐𝟒

E-resources:
https://video
...
edu
...
php?vid=a88c1806e
https://video
...
edu
...
php?vid=60af682a7
https://video
...
edu
...
php?vid=ae7d90e1c
https://video
...
edu
...
php?vid=f671ef694
https://video
...
edu
...
php?vid=bd359002b
https://video
...
edu
...
php?vid=edca44590
https://video
...
edu
...
php?vid=808b126d6

33 | P a g e



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