MATH F112 (MATHEMATICS-II)

Gaurav Dwivedi
Department of Mathematics
BITS Pilani, Pilani Campus.
 Module 12
Series and Residues
Sequence
Definition (Sequence)
A sequence is function f : N → C. A sequence is
denoted by {z1 , z2 , . . . , zn , . . .} or {zn }.

Definition (Limit)
We say that limn→∞ zn = z0 , if for every ϵ > 0, there
exists n0 ∈ N such that

|zn − z0 | < ϵ whenever n ≥ n0 .

If limn→∞ zn = z0 , then we say that {zn } converges to
z0 .
Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 3 / 78
Series
Definition (Series)
P∞
For a given sequence {zn }, n=0 zn is called an infinite
series.

Definition (Convergence of Series)

∞
X
A series zn is convergent if the sequence
n=0
N
X
SN = zn
n=1
P∞
converges to S; we then write n=0 zn = S.
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Taylor Series

Taylor Series
Let f (z) be analytic throughout a disk |z − z0 | < R.
Then f (z) has the power series representation
∞
X
f (z) = an (z − z0 )n ,
n=0

f (n) (z0 )
in |z − z0 | < R, where an = n! , n = 0, 1, 2 . . ..

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Maclaurin Series

Maclaurin Series
Taylor Series about the point z0 = 0 is called Maclaurin
series, i.e.,
X∞
f (z) = an z n ,
n=0
f (n) (0)
in |z| < R, where an = n! , n = 0, 1, 2 . . ..

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Exercises

You can obtain the following expansions:

∞
z
X zn
1. e = (|z| < ∞).
n=0
n!
∞
z 2n+1
X
n
2. sin z = (−1) (|z| < ∞).
n=0
(2n + 1)!
∞
X z 2n
3. cos z = (−1)n (|z| < ∞).
n=0
(2n)!

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 7 / 78
Exercises
∞
X z 2n+1
4. sinh z = (|z| < ∞).
n=0
(2n + 1)!
∞
X z 2n
5. cosh z = (|z| < ∞).
n=0
(2n)!
∞
1 X
6. = z n (|z| < 1).
1−z n=0
∞
1 X
7. = (−1)n z n (|z| < 1).
1+z n=0
∞
1 X
8. = (−1)n (z − 1)n (|z − 1| < 1).
z n=0
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Laurent’s Theorem

Laurent’s Theorem
Suppose that a function f (z) is analytic throughout an
annular domain D : R1 < |z − z0 | < R2 centered at z0
and let C denote any positively oriented simple closed
contour around z0 and lying in D. Then, at each point in
D, f (z) has the series representation
∞ ∞
X
n
X bn
f (z) = an (z − z0 ) + ,
n=0 n=1
(z − z0 )n

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 9 / 78
where
Z
1 f (z)
an = dz, n = 0, 1, 2, . . .
2πi C (z − z0 )n+1
and
Z
1 f (z)
bn = dz, n = 1, 2, . . .
2πi C (z − z0 )−n+1

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 10 / 78
Question

Find the Laurent series representation of

1
f (z) =
(z − 2)(z − 3)
in the following domains:
1 D : 1 < |z| < 2.
2 D : 2 < |z| < 3.
3 D : 3 < |z|.

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Solution 1
1 1
We have f (z) = − z−2 + z−3 . Now in the given domain,
we have |z| |z|
2 < 1 and so 3 < 1. Thus, we write

1 1
f (z) = −
2 1 − z2 3 1 − z3

1 z −1 1  z −1
= 1− − 1−
2 2 3 3
∞   ∞   ∞ ∞
1 X z n 1 X z n X zn X zn
= − = n+1
− n+1
.
2 n=0 2 3 n=0 3 n=0
2 n=0
3

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 12 / 78
Solution 2
1 1
We have f (z) = − z−2 + z−3 . Now in the given domain,
2
we have |z| < 1 and |z|
3 < 1. Thus, we write

1 1
f (z) = − 2 −
3 1 − z3



z 1− z
 −1
1 2 1 z −1
=− 1− − 1−
z z 3 3
∞ n ∞ ∞ ∞
2n zn
 
1X 2 1 X  z n X X
=− − =− n+1
− n+1
.
z n=0 z 3 n=0 3 n=0
z n=0
3

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 13 / 78
Solution 3
1 1
We have f (z) = − z−2 + z−3 . Now in the given domain,
3 2
we have |z| < 1 and so |z| < 1. Thus, we write
1 1
f (z) = − +
z 1 − z2 z 1 − z3



 −1  −1
1 2 1 3
=− 1− + 1−
z z z z
∞   n ∞   n
1X 2 1X 3
=− +
z n=0 z z n=0 z
∞ ∞
X 2n X 3n
=− n+1
+ n+1
.
n=0
z n=0
z
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Question
Show that, when 0 < |z − 1| < 2, the Laurent series
z
representation of (z−1)(z−3) is
∞
X (z − 1)n 1
−3 − .
n=0
2n+2 2(z − 1)

Sol. Note that we need to express f (z) in powers of

z − 1. So write f (z) in terms of z − 1 as
1 3
f (z) = − + .
2(z − 1) 2(z − 1 − 2)
|z−1|
Now in the given domain, we have 2 < 1.
Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 15 / 78
Thus, we write
1 3
f (z) = − −
2(z − 1) 4 1 − z−1


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

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Exercises
1. Show 
that ∞
2 1 X (−1)n
z sin 2 = 1 + 4n
, (0 < |z| < ∞)
z n=1
(2n + 1)!z
2. Show that
∞
" #
z n
e 1 X (z + 1) 1 1
= + + ,
(z + 1)2 e n=0 (n + 2)! z + 1 (z + 1)2

(0 < |z + 1| < ∞).

X (−1)n+1 ∞
1
3. Show that = n
, (1 < |z| < ∞).
1+z n=1
z
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Exercises

4. Find the Taylor or Laurent series expansion of

1
f (z) = 2 in the following domains:
(z + 1)(z 2 + 2)
a |z| < 1, √
b 1 < |z|√< 2
c |z| > 2.

Ans. (a) ∞ 1
P n

2n
n=0 (−1) 1 − n+1 z
P∞ n+1 z 2n
P2∞ 1
(b) n=0 (−1) 2n+1 + n=0 (−1)n z 2n+2
n
(c) ∞ n (1−2 )
P
n=0 (−1) z 2n+2

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Exercises

5. Find the Laurent series expansion of

1
f (z) = 2 in powers of (z − 1) indicating the
z − 4z
region where the expansion is valid.

n

Ans. f (z) = − 14 ∞n=0(−1)n(z − 1)n + 31 ∞n=0 z−1
P P
3 (|z − 1| < 1)
P 
∞ (−1)n P∞ z−1 n
f (z) = − 14 + 13


n=0 (z−1)n n=0 3 (1 < |z − 1| < 3)
P n n

1 ∞ (−1) −3
f (z) = − 4 n=0 (z−1)n+1 (|z − 1| > 3)

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Exercises
6. Find the Laurent series expansion of
( 1
e−e z−i z ̸= i
f (z) =
0 z=i
in the powers of z − i in the domain
D = {z ∈ C : |z| > 2}.
Ans.
  
1 1 1
e−1 1 − + + ...
z − i 2! (z − i)2
 2 #
1 1 1 1
+ + + ... + ...
2! z − i 2! (z − i)2
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Exercises

7. Find the Laurent series expansion of e−1/z and use it

to show that
Z 1
e−t 
−1
p  π
√ 2
cos 2 cos (t) + 1 − t dt = ·
−1 1 − t2 2

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 Chapter 6
Residues and Poles

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Laurent’s Theorem
Suppose that a function f (z) is analytic throughout
D : R1 < |z − z0 | < R2 and let C be a positively
oriented simple closed contour around z0 and lying in D.
Then, at each point in D
∞ ∞
X
n
X bn
f (z) = an (z − z0 ) + n
.
n=0 n=1
(z − z0 )
Z
1 f (z)
an = dz, n = 0, 1, 2, . . .
2πi C (z − z0 )n+1
and
Z
1 f (z)
bn = dz, n = 1, 2, . . .
2πi C (z − z0 )−n+1
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Residue

Residue
1
The coefficient of z−z 0
(that is b1 ) in the Laurent series
is called the residue of f (z) at z0 and we write it as

b1 = Res f (z).
z=z0

On putting n = 1 in the formula for bn ’s, we get

Z
f (z) dz = 2πi Res f (z).
C z=z0

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Singular Point
Let a function f (z) is not analytic at a point z0 , but
analytic at some points in every nbd of z0 . Then z0 is
called a singular point or a singularity of f (z).

Isolated Singularity
A singularity z0 of f (z) is called isolated, if f (z) is
analytic in some deleted nbd of z0 .

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Example
The function
z+1
f (z) = ,
z(z 2 + 1)
has three isolated singular points z = 0 and z = ±i.

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Example
z = 0 is a singular point of principal branch

Log(z) = ln r + iΘ (r > 0, −π < Θ < π)

of the logarithm function. It is not, however, an isolated

singular point since every deleted ϵ neighborhood of it
contains points on the negative real axis and the branch
is not analytic there.

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Example
The function
1
f (z) =
sin(π/z)
has singular points z = 0 and z = n1 (n = ±1, ±2, . . .).
Each singular point except z = 0 is isolated. The singular
point z = 0 is not isolated because every ϵ neighborhood
of z = 0 contains other singular points of f.(Why?)

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Laurent Series

If z0 is an isolated singularity of f (z), then f (z) is

analytic in the deleted neighborhood 0 < |z − z0 | < R
for some R. Consequently, f (z) has a Laurent series
representation in 0 < |z − z0 | < R of the form
∞ ∞
X
n
X bn
f (z) = an (z − z0 ) + .
n=0 n=1
(z − z0 )n

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Cauchy’s Residue Theorem

Cauchy’s Residue Theorem

Let C be a positively oriented simple closed contour.
Suppose that f (z) is analytic inside and on C except for
a finite number of singular points zk , (k = 1, 2, . . . , n)
inside C. Then
Z Xn
f (z) dz = 2πi Res f (z).
C z=zk
k=1

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Theorem(Section 77)
If a function f is analytic in the finite plane except for a
finite number of singular points interior to a positively
oriented simple closed contour C, then
Z   
1 1
f (z) dz = 2πi Res 2 f .
C z=0 z z

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Classification
of isolated singularities

Removable Singularity
If there is no term in PP, i.e., bn = 0, for all n, then z0 is
called a removable singularity of f (z).

Examples
1. We have
sin z z2 z4
= 1 − + + ···
z 3! 5!
Since there is no term in PP so z = 0 is a removable
singularity.
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2. We have
1 − cos z 1 z2 z4
= − + + ···
z2 2! 4! 6!
Since there is no term in PP so z = 0 is a removable
singularity.

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Classification
of isolated singularities

Pole
If there are finite number of terms in PP i.e., bm ̸= 0, for
some m and bn = 0 for all n > m, then z0 is called a
pole of order m. Thus if
b1 b2 bm
PP = + + · · · + ,
z − z0 (z − z0 )2 (z − z0 )m
then z0 is a pole of order m.
If m = 1, then z0 is a pole of order 1 and is called a
simple pole.
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Example

Examples
z+1
1. For f (z) = z 2 −2z , we have

z z2
 
3 1
f (z) = − 1 + + + ··· − ,
4 2 4 2z

where 0 < |z| < 2. Since there is only one term in PP so

z = 0 is a simple pole.

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Examples
1
2. For f (z) = z 2 (1+z) , we have

1 1
f (z) = 1 − z + z 2 + · · · − + 2
z z
where 0 < |z| < 1. Since there are two terms in PP so
z = 0 is a pole of order 2.

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Classification
of isolated singularities

Essential Singularity
If there are infinitely many terms in PP i.e., bn ̸= 0, for
infinitely many values of n, then z0 is called an essential
singularity of f (z). Thus if
b1 b2 bn
PP = + + · · · + + ··· ,
z − z0 (z − z0 )2 (z − z0 )n

then z0 is an essential singularity of f (z).

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Example
Examples
1. We have
 
1 1 1 1 1 1
sin = − . 3 + . 5 + ··· ,
z z 3! z 5! z

where 0 < |z| < ∞. Since there are infinitely many terms
in PP so z = 0 is an essential singularity.
2.
1 1 1 1 1
e1/z = 1 + + . 2 + . 3 + · · · ,
z 2! z 3! z
where 0 < |z| < ∞. Since there are infinitely many terms
in PP, z = 0 is an essential singularity.
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Question
1
Find the residue of f (z) = z+z 2 at z = 0 .
1
Sol. Clearly f (z) = z(1+z) has two isolated singularities,
z = 0 and z = −1. Since we want to find residue at
z = 0, we expand f (z) as a Laurent series around z = 0.
As the function f (z) is analytic in the annular region
0 < |z| < 1, we can write
1 1 1 1 −1 1
2
= − = −(1+z) = −(1−z+z 2 −· · · )
z+z z z+1 z z
Hence Res f (z) = 1.
z=0

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Exercises

1 In the solution to the problem on previous slide, can

we use 0 < |z| < 1/2 as our domain for Laurent
series? What about 0 < |z| < 1/3? If yes, then
would the residue change? Justify.
2 In the solution to the problem on previous slide, can
we use 0 < |z| < 2 as our domain for Laurent series?
1
Find the residue of f (z) = z+z 2 at z = −1.
3

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Question

Find and classify the singularities of ze1/z .

Sol. z = 0 is an isolated singular point. We have
1 1 1 1
ze1/z = z + 1 + . + . 2 + ··· ,
2! z 3! z
where 0 < |z| < ∞. Since there are infinitely many terms
in PP so z = 0 is an essential singularity of ze1/z .

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Question

2z
e
Show that the singular point of f (z) = (z−1) 2 is a pole.

Determine the order of the pole and the corresponding

residue.
Sol. Clearly z = 1 is an isolated point for f (z). To find
the type of the singularity z = 1, we expand f (z) in
terms of z − 1 in domain 0 < |z − 1| < ∞.

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We have
e2z e2z−2+2
=
(z − 1)2 (z − 1)2
2(z−1)
2 e
=e
(z − 1)2
e2
 
4
= 2
1 + 2(z − 1) + (z − 1)2 + (z − 1)3 + · · ·
(z − 1) 3
2 2
e 2e 4
= 2
+ + e2 + e2 (z − 1) + · · ·
(z − 1) z−1 3

Thus z = 1 is a pole of order 2 and Resf (z) = 2e2 .

z=1

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Question
1
R 2


Evaluate C z sin z dz, where C : |z| = 1.

Sol. Clearly, z =
0 is an isolated singularity of

f (z) = z 2 sin z1 . Since f (z) is analytic in the annular
domain 0 < |z| < ∞, we can represent f (z) by a
Laurent series
 
1 1 1 1 1
z 2 sin = z − . + . 3 + ···
z 3! z 5! z
 
1 1
Thus, Res z 2 sin =− .
z=0 z 3!
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Hence
Z    
1 1 πi
z 2 sin dz = 2πi − =− .
|z|=1 z 3! 3

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Question

Use Cauchy’s residue theorem to evaluate

Z
f (z) dz,
C

e−z
where f (z) = and C : |z| = 3.
(z − 1)2
Sol. Clearly f (z) has an isolated singularity z = 1 within
C. So to find the residue at z = 1, we expand f (z) in
0 < |z − 1| < ∞ in terms of z − 1.

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 46 / 78
 e−z
f (z) =
(z − 1)2
e−(z−1+1)
=
(z − 1)2
e−1
= e−(z−1)
(z − 1)2
e−1 (z − 1)2 (z − 1)3
 
= 1 − (z − 1) + − + ···
(z − 1)2 2! 3!
Thus Res f (z) = −e−1 . By Cauchy’s residue theorem
z=1
Z
f (z) dz = −2πie−1 .
C
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Question

Use Cauchy’s residue theorem to evaluate

Z
f (z) dz,
C

z+1
where f (z) = z 2 −2z and C : |z| = 3.

Sol. Clearly f (z) has two isolated singularities z = 0 and

z = 2 within C. So, we’ll find the residues at both of
these singularities. To find the residue at z = 0, we
expand f (z) in 0 < |z| < 2 in terms of z.

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We have
z+1
f (z) =
z 2 − 2z
3 1
= −
2(z − 2) 2z
3 z −1 1
=− 1− −
4 2 2z 
2
3 z z 1
=− 1 + + + ··· − .
4 2 4 2z

Therefore Res f (z) = −1/2.

z=0

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Now to find the residue at z = 2, we expand f (z) in
0 < |z − 2| < 2 in term of z − 2. We have
z+1 3 1
f (z) = 2
= −
z − 2z 2(z − 2) 2(z − 2 + 2)
3 1
= −
2(z − 2) 4 1 + z−2


2
 −1
3 1 z−2
= − 1+
2(z − 2) 4 2
z − 2 (z − 2)2
 
3 1
= − 1− + + ··· .
2(z − 2) 4 2 4

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Therefore Res f (z) = 3/2.
z=2

Hence, by Cauchy’s residue theorem

Z
f (z) dz = 2πi(−1/2 + 3/2) = 2πi.
C

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QuestionEvaluate Z
f (z) dz,
C
z+1
where f (z) = z 2 −2z and C : |z| = 3.

z+1
Sol. We have 1/z 2 f (1/z) = z(1−2z) . To find the residue
at z = 0, we expand f (z) in 0 < |z| < 1/2 in terms of z.
We have

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 52 / 78
 z+1
f (z) =
z(1 − 2z)
1 3
= +
z 1 − 2z
1
= + 3(1 + 2z + 4z 2 + · · · ).
z
Therefore Res f (z) = 1. Hence by above theorem
z=0
Z
f (z) dz = 2πi(1) = 2πi.
C

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Question

f (z)
Suppose f (z) is analytic at z0 , and write g(z) = z−z 0
.
Show that
(a) if f (z0 ) ̸= 0, then z0 is a simple pole of g, with
residue f (z0 ).
(b) if f (z0 ) = 0, then z0 is a removable singular point
of g.

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Residue at Poles
Let z0 be an isolated singular point a function f (z). The
following two statements are equivalent:
1 z0 is a pole of order m(m = 1, 2, . . .) of f ;
2 f (z) can be written in the form
ϕ(z)
f (z) = ,
(z − z0 )m
where ϕ(z) is analytic and non-zero at z0 .
Moreover, if the statements (1) and (2) are true,

ϕ(z0 ) if m = 1,
Res f (z) = 1
h m−1
d
i
z=z0  (m−1)! dz m−1 ϕ(z) if m ≥ 2.
z=z0

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 55 / 78
Theorem

Theorem
If f (z) has a pole of order m at z0 , then
 m−1 
1 d
Res f (z) = lim m−1
(z − z0 )m f (z) .
z=z0 (m − 1)! z→z 0 dz

In particular, if z0 is a pole of order one(i.e. Simple

Pole), then

Res f (z) = lim [(z − z0 )f (z)] .

z=z0 z→z0

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 56 / 78
Question

2
+2
Show that the singular point of f (z) = zz−1 is a pole.
Determine the order m of each pole, and find the
corresponding residue.
2 ϕ(z)
Sol. Here f (z) = zz−1
+2
is of the form (z−z0)
m where

ϕ(z) = z 2 + 2 is analytic and nonzero at z = 1.

Therefore z = 1 is a pole of order 1. Also,
Resz=1 f (z) = ϕ(1) = 3.

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 57 / 78
Question

Show that
Log z π + 2i
Res 2 2
=
z=i (z + 1) 8
2
Sol. Here f (z) = Log(z−i)
z/(z+i)
2 is of the form (z−zϕ(z)
0)
m where

ϕ(z) = Log z/(z + i)2 is analytic and nonzero at z = i.

Therefore z = i is a pole of order 2. Also
d  z + i − 2z Log z  π + 2i
Res f (z) = ϕ(z) = = .

dz z(z + i)3 8

z=i z=i z=i

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 58 / 78
Question
Find the value of the integral
3z 3 + 2
Z
2
dz,
C (z − 1)(z + 9)
taken counterclockwise around the circle |z − 1| = 2.
3
ϕ(z) 3
3z +2 3z +2
Sol. Here f (z) = (z−1)(z 2 +9) = z−1 where ϕ(z) = (z 2 +9)

is nonzero and analytic at z = 1. So f (z) has a simple

pole at z = 1 and so Resf (z) = ϕ(1) = 1/2. Hence, by
z=1
residue theorem
3z 3 + 2
Z
2
dz = πi.
C (z − 1)(z + 9)
Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 59 / 78
Question
Find the value of the integral

3z 3 + 2
Z
2
dz,
C (z − 1)(z + 9)

taken counterclockwise around the circle |z| = 4.

Sol. Here
3z 3 + 2 3z 3 + 2
f (z) = = ,
(z − 1)(z 2 + 9) (z − 1)(z + 3i)(z − 3i)
has three singularities z = 1, 3i, −3i within C. Note that
all these singularities are simple pole. So
Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 60 / 78
 3z 3 + 2
 
1
Res f (z) = =
z=1 (z + 3i)(z − 3i) z=1 2
3z 3 + 2
 
75 + 245i
Res f (z) = =
z=3i (z − 1)(z + 3i) z=3i 60
3z 3 + 2
 
75 − 245i
Res f (z) = = .
z=−3i (z − 1)(z − 3i) z=−3i 60

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 61 / 78
Hence, by residue theorem

3z 3 + 2
 
1 75 + 245i 75 − 245i
Z
2
dz = 2πi + +
C (z − 1)(z + 9) 2 60 60
= 6πi.

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 62 / 78
Alternate Solution
Here
3 + 2z 2
 
1 1
f = 2 ,
z2 z z (1 − z)(1 + 9z 2 )
has a pole of order two at z = 0. So

3 + 2z 2
    
1 1 d
Res 2 f = = 3.
z=0 z z dz (1 − z)(1 + 9z 2 ) z=0

Hence,
3z 3 + 2
Z
dz = 2πi(3) = 6πi.
C (z − 1)(z 2 + 9)

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 63 / 78
Question
Find the value of the integral
Z
1
3
dz,
C z (z + 4)
taken counterclockwise around the circle |z + 2| = 3.

Sol. Here z = 0 is a pole of order 3 and z = −4 is a

pole of order 1 (both lie inside C).
1 d2
 
1 1
Res f (z) = =
z=0 2! dz 2 z + 4 z=0 64
 
1 1
Res f (z) = 3 =− .
z=−4 z z=−4 64
Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 64 / 78
Hence,
Z  
1 1 1
dz = 2πi − = 0.
C z 3 (z + 4) 64 64

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 65 / 78
Zeros of an analytic function
Let f (z) be analytic in a domain D. If f (z0 ) = 0 for
some z0 ∈ D, then z0 is called a zero of f (z).
If f (z0 ) = f ′ (z0 ) = f ′′ (z0 ) = · · · = f (m−1) (z0 ) = 0 and
f (m) (z0 ) ̸= 0, then z0 is a zero of order m.
A zero of order 1 is called a simple zero.
Theorem
Let f (z) be analytic at a point z0 . Then z0 is a zero of f
of order m if and only if f (z) can be written in the form

f (z) = (z − z0 )m ϕ(z),

where ϕ is analytic at z0 and ϕ(z0 ) ̸= 0.

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 66 / 78
Example
The entire function f (z) = z(ez − 1) has a zero of order
m = 2 at the point z0 = 0 since
f (0) = f ′ (0) = 0 and f ′′ (0) = 2 ̸= 0.
So, we can write
f (z) = (z − 0)2 g(z),
where g is analytic and g(0) ̸= 0. Note that here g is
defined by means of the equations
( z
e −1
z z ̸= 0
g(z) =
1 z=0
Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 67 / 78
Zeros and Poles

Theorem
Suppose that
(a) two functions p and q are analytic at a point z0 ,
(b) p(z0 ) ̸= 0 and q has a zero of order m at z0 .
Then the quotient p(z)/q(z) has a pole of order m at z0 .

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 68 / 78
Theorem
Suppose that
(a) two functions p and q are analytic at a point z0 ,
(b) p(z0 ) ̸= 0, q(z0 ) = 0 and q ′ (z0 ) ̸= 0.
Then z0 is a simple pole of the quotient p(z)/q(z) and

p(z) p(z0 )
Res = ′ .
z=z0 q(z) q (z0 )

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 69 / 78
Question

Show that the point z = 0 is a simple pole of the

function
1
f (z) = .
sin z

Sol. Here both p(z) = 1 and q(z) = sin z are analytic at

z = 0. Also, p(0) ̸= 0 and q(z) has a simple zero at
z = 0 (as q(0) = sin 0 = 0 and q ′ (0) = cos 0 = 1 ̸= 0).
Therefore, by above theorem z = 0 is a simple pole of
f (z).

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 70 / 78
Question
Evaluate the integral
Z
tan z dz,
C

where C denote the positively oriented circle |z| = 2.

sin z
Sol. Here f (z) = tan z = cos z . Clearly the only
singularities of f (z) are the zeros of cos z which are
given by nπ + π2 , n = 0, ±1, ±2, . . . (see p-107). Only
z = −π/2 and z = π/2 lie in C. It is easy to see that
these two are simple poles of f (z).

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 71 / 78
Now
Res f (z) = lim (z + π/2)f (z)
z=−π/2 z→−π/2
sin z
= lim (z + π/2) = −1,
z→−π/2 cos z
Res f (z) = lim (z − π/2)f (z)
z=π/2 z→π/2
sin z
= lim (z − π/2) = −1.
z→π/2 cos z
Hence,
Z
tan z dz = 2πi(−1 − 1) = −4πi.
C
Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 72 / 78
Example
Show that z = 0 is simple pole of
cos z
cot z =
sin z
and find the residue. both p(z) = cos z and q(z) = sin z
are analytic at z0 = 0. Also,
p(0) = cos 0 = 1, q(0) = sin 0 = 0 and
q ′ (0) = cos 0 = 1. Therefore, z0 = 0 is a simple pole of
cot z. Also,
p(0)
Res cot z = ′ = 1.
z=0 q (0)

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 73 / 78
Exercises I
Z
dz π
1 Show that 2 2
= √ , where C is the
C (z − 2) + 3 2 2
positively oriented boundary of the rectangle whose
sides lie along the lines x = ±2, y = 0, and y = 1.
2 Show that f (z) = csc2 z has a pole of order 2 at
z = 0 and Resf (z) = 0.
Zz=0
1
3 Show that dz = −πi, where C is
C sinh 2z
positively oriented circle |z| = 2.
(3z + 2)2
Z
4 Show that = 9πi, where C is
C z(z − 1)(2z + 5)
the positively oriented circle |z| = 3.
Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 74 / 78
Exercises II

z 3 e1/z
Z
5 Show that 3
= 2πi, where C is the
C 1 + z
positively oriented circle |z| = 3.
6 Show that f (z) = cot z
z 4 has a pole of order 5 at
1
z = 0, and Resf (z) = − 45 .
z=0
7 Determine the nature of isolated singularity
(Removable, pole, essential) for the functions:
sin2 z
a (Ans. Simple pole at z = 0.)
z3
(ez − 1)2
b (Ans. z = 0 is a removable singular point.)
1 − cos z

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 75 / 78
Exercises III

8 Compute the residues of f (z) = cot2 z at all

isolated singular points.
Ans. zn = nπ, n ∈ Z are poles of order 2 and Res f (z) = 0.
z=zn
3
(log z)
9 Consider f (z) = , where
z2 + 1
log z = ln r + iθ (r > 0, 0 < θ < 2π). Find and classify all the
singularities of f (z) in finite complex plane C with justification
and hence, find the residues of f (z) it isolated singularities.
3 3
Ans. z = i, −i are simple poles Res = 27π 16
, Res = − π16 ,
z=−i z=i
{z ∈ C : Re(z) ≥ 0, Im(z) = 0} are non isolated singular
points.

Gaurav Dwivedi (BITS Pilani) MATH F112 (MATHEMATICS-II) June 26, 2023 76 / 78
Exercises IV
cosh z 2 − 1
10 Consider the function f (z) = . Classify the
z9
singularity z = 0 of the function in terms of removable
singularity, pole or essential singularity. In case, z = 0, is a
pole, specify the order of the pole andZ hence compute residue
of f (z) at z = 0. Further, evaluate f (z)dz, where
C
C : |2z − 2 − 2i| = 3 is traversed in positive direction.
1 πi
Ans. Pole of order 5, residue is 24 and integral is 12 .
f (z)
11 Suppose f (z) is analytic at z0 , and g(z) = z−z 0
. Show that
(a) if f (z0 ) ̸= 0, then z0 is a simple pole of g, with residue
f (z0 ).
(b) if f (z0 ) = 0, then z0 is a removable singular point of g.

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