Chapter 6.

Types of Singularities & residue theorem

Dr.Rehab Alsultan

Umm AlQura University

rasultan@uqu.edu.sa

July 13, 2024

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 1 / 21
Overview

1 Introduction

2 Classification of isolated singularities

3 Cauchy’s Residue Theorem

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 2 / 21
Introduction

The Cauchy–Goursat theorem states that if a function is analytic at all

points interior to and on a simple closed contour C, then the value of the
integral of the function around that contour is zero. If, however, the
function fails to be analytic at a finite number of points interior to C,
there is, as we shall see in this chapter, a specific number, called a residue,
which each of those points contributes to the value of the integral.

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 3 / 21
Singularity

In mathematics, a singularity is in general a point at which a given

mathematical object is not defined, or a point of an exceptional set where
it fails to be well-behaved in some particular way, such as differentiability.

For example, the function

1
f (x) =
x
on the real line has a singularity at x = 0, where it seems to ”explode” to
x = ±∞ and is not defined.

The function
g (x) = |x|
also has a singularity at x=0, since it is not differentiable there.

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 4 / 21
Isolated Singularity

A point z0 is called an Isolated Singularity of a function f (z) if f (z) has a

singularity at z0 but single valued and analytic in the annular region :
0 < |z − z0 | < R i.e., analytic in neighborhood of z0 .
Otherwise it is non-isolated singular points.
Example:

z2 − 1
f (z) =
z(z 2 + 1)(z − 2)
Here z = 0, z = ±i and z = 2 are isolated singularities.

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 5 / 21
Types of isolated singularities

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 6 / 21
Removable Singularity

If in the Laurent’s series expansion, the principal part is zero, i.e.,

∞
X
f (z) = an (z − z0 )n + 0
n=0

z0 is removable singularity if

lim f (z) = a0
z→z0

exists and is equal to a0 ∈ C

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 7 / 21
Example:
z
f (z) =
z
solution:
z0 = 0 is isolated singularity. In particular
z
lim =1
z→0 z

z0 = 0 is removable singularity.

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 8 / 21
Example:

sin(z)
f (z) =
z
Solution:
z0 = 0 is isolated singularity f (z0 ) = 00
need to check if the limz→0 sin(z)
z exsit, useing the rule of l’Hospital

sin(z) cos(z)
lim = lim =1
z→0 z z→0 1
Hence, z0 = 0 is removable singularity.

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 9 / 21
Poles
If the principal part of Laurent’s series has finite number of terms, i.e.,
∞
X b1 b2 bn
fz = an (z − z0 )n + + 2
+ ··· +
(z − z0 ) (z − z0 ) (z − z0 )n
n=0

Then the singularity z = z0 is said to be the pole of order n, if b1 6= 0

and b2 = b3 = · · · = 0, then
∞
X b1
fz = an (z − z0 )n +
(z − z0 )
n=0
Then singularity z = z0 is said to be pole of order 1 or a Simple Pole.

z0 is pole of order m: ∃m ∈ N, m ≥ 1 ∃c ∈ C, c 6= s.t

lim (z − z0 )m f (z) = c
z→z0

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 10 / 21
Example :
1
f (z) = , m ∈ N, m ≥ 1
(z − z0 )m
z = z0 is a pole of order m, because
1
lim (z − z0 )m f (z) = lim (z − z0 )m = 1 6= 0/0
z→z0 z→z0 (z − z0 )m

Example :
1
f (z) = ,
(z − 1)2 (z − 2)
solution:
z0 = 1 is pole of order 2
z1 = 2 is pole of order 1

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 11 / 21
Example
1
f (z) =
sin(z)
solution: z = 0 is ploe of order 1 because

z 1
lim zf (z) = lim = useing l 0 Hopital = lim = 1 6= 0
z→0 z→0 sin(z) z→0 cos(z)

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Essential Singularity

If in the Laurent’s series expansion, the principal part contains an

infinite number of terms, then the singularity z = z0 is said to be an
Essential Singularity.

z0 is Essential Singularity: z0 is neither removable nor a pole of order

m.

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 13 / 21
Example:
1
f (z) = e z
Solution:
z0 = 0 is an Essential Singularity
z0 is not removable
z0 is not a pole of order m since limz→0 z m f (z) doesn’t converge
also,
1 1 1
f (z) = e z = 1 + + + ···
z 2!(z)2
Since the number of negative power terms of (z-0) is infinite, z = 0 is an
essential singularity.

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 14 / 21
Cauchy’s Residue Theorem

Theorem: Cauchy’s Residue Theorem if C is a simple closed positively

oriented contour and f is analytic on and inside C except at some point
z0 , z1 , · · · , zn inside C , then
Z n
X
f (z) = 2πi Res(Zj )
C j=1

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 15 / 21
 What is a residue?
The residue of a function f at an isolated singularity z0 , denoted
Res(z0 ) is the coefficient a−1 in the Laurent Series around z0
a−2 a−1
f (z) = · · · + 2
+ + a0 + a1 (z − z0 ) + a2 (z − z0 )2 + · · ·
(z − z0 ) (z − z0 )

How to determine a residue at z0 ?

1. write out the laurant series around z0

2. If z0 is a simple pole → Res(z0 ) = limz→z0 (z − z0 )f (z)

3. If z0 is a pole of order
1 d m−1
m → Res(z0 ) = limz→z0 (m−1) dz m−1
((z − z0 )m f (z))

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 16 / 21
Example:determine all integral by using residus
Z
cos z
dz
C z
solution:
Res(0) = limz→0 (z − 0) cosz z = limz→0 cos z = cos 0 = 1
Z
cos z
dz = 2πiRes(0) = 2πi1 = 2πi
C z

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 17 / 21
Example:determine all integral by using residus
Z
cos z
dz
C z −1
Solution:
Z
cos z
dz = 2πiRes(1)
C z −1
cos z
Res(1) = limz→1 (z − 1) (z−1) = limz→1 cos z = cos 1

Z
cos z
dz = 2πiRes(1) = 2πi cos 1
C z −1

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 18 / 21
Example:determine all integral by using residus

1 − 2z
Z
dz
C z(z − 1)
Solution:

1 − 2z
Z
dz = 2πi(Res(0) + Res(1))
C z(z − 1)
1−2z 1−2z 1−2(0)
Res(0) = limz→0 (z − 0) z(z−1) = limz→0 (z−1) = 0−1 = −1
1−2z 1−2(1)
Res(1) = limz→1 (z − 1) z(z−1) = limz→1 1−2z
(z) = 1 = −1

1 − 2z
Z
dz = 2πi(Res(0) + Res(1)) = 2πi(−1 + (−1)) = −4πi
C z(z − 1)

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 19 / 21
Example:determine all integral by using residus
Z
2z + 1
2
dz
C z(z − 1)
Solution:
Z
2z + 1
dz = 2πi(Res(0) + Res(1))
C z(z − 1)2
2z+1 2z+1 2(0)+1
Res(0) = limz→0 (z) z(z−1)2 = limz→0 (z−1)2
= (0−1)2
=1

1 d 2−1 2 2z+1
Res(1) = limz→1 (2−1)! dz 2−1 ((z − 1) . z(z−1)2 )
1 d
= limz→1 1! . dz (2 + z1 ) = limz→1 − z12 = − 112 = −1
Z
2z + 1
dz = 2πi(Res(0) + Res(1)) = 2πi(1 + (−1)) = 0
C z(z − 1)2

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 20 / 21
 The End

Dr.Rehab Alsultan (UQU) Chapter 6.Singularities & residue theorem July 13, 2024 21 / 21