Module 3: Laurent Series

Dr. T. Phaneendra

January 8, 2020

1 Singularities

Definition 1.1 (Singularity). A singularity of a function w = f (z) is a point z0

at which f is not analytic, but is analytic at some point in every neighborhood of
z0 .
Definition 1.2 (Isolated singularity). A singularity z0 such that f is analytic at
all points of some deleted neighborhood N (z0 ) − {z0 } of z0 .
Example 1.1. Let f (z) = 1/z for all z ∈ C. Then f is analytic at all points of
the complex plane except at z = 0, and hence z = 0 is an singularity of f .
Example 1.2. The origin z = 0 is an isolated singularity of f (z) = sin z/z 4 .
Example 1.3. The function f (z) = 1/(z 2 + 1) is analytic at all points of the
complex plane except at z = ±i, and hence z = ± i are isolated singularities.
Example 1.4. For f (z) = 1/ sin(π/z), the singularities are z = 0, ±1, ±1/2,
±1/3, ... which lie on the real axis from z = −1 to z = 1. Each z = 1/n is an
isolated singularity. But z = 0 is not an isolated singularity, since every deleted
neighborhood of it contains infinitely many singularities 1/n other than itself.
Example 1.5. The functions f (z) = |z|2 and g(z) = |z| are nowhere analytic
and hence the criterion of singularity is not satisfied at any point of the complex
plane. Thus they do not have singularities.

1
CVPDE (MAT3003) Module 3

2 Laurent Series about an Isolated Singularity

Let f (z) be analytic in the annulus A : 0 < r < |z − z0 | < R, where z0 is an

isolated singularity of f . Then the Laurent series of f about z0 is given by
∞ ∞ ∞
an (z − z0 )n = a−n (z − z0 )−n + an (z − z0 )n
P P P
f (z) = (2.1)
n=−∞ n=1 n=0
| {z } | {z }
Principal Part Analytic Part

where a0n s are called its coefficients.

1
Remark 2.1. The coefficient of the first negative power z−z0
in the expansion
(2.1) is known as the residue of f at z = z0 , and is denoted by Res (f ; z0 ). That is
Res (f ; z0 ) = a−1
1
Example 2.1. Consider f (z) = z 2 (1−z)
for all z. Then are . We wish to find a
Laurent’s series of f about its isolated singularities z = 0 and 1. z = 0 in the
regions R1 : 0 < |z| < 1 and R2 : |z| > 1.

On R1 : 0 < |z| < 1, we see that

∞ ∞
1 1
zn = z n−2 = 1 1
+ 1 + z + z2 + · · ·
P P
z 2 (1−z)
= z2 z2
+ z
n=0 n=0

Note that Res (f ;0) = the coefficient of 1/z in the Laurent series = 1.

While, on R2 : |z| > 1, we see that  z1  < 1. So,

 

∞
1 −1
1
= − z13 1 − = − z13 zn

P
z 2 (1−z) z
n=0
∞
z n−3 = − z13 − 1 1
− 1 − z − z2 − · · ·
P
=− z2
− z
n=0

1
Example 2.2. Consider f (z) = z 2 (1−z)
for all z of Example 2.1. Find a Laurent’s
series of f about z = 1, which is convergent on an appropriate domain R.

Dr. T. Phaneendra 2 511, A10, SJT

CVPDE (MAT3003) Module 3

Let z − 1 = u. Then

1 1 1 −2
z 2 (1−z)
= − (u+1)2 u = − u (1 + u)

∞ ∞
= − u1 (−1)n (n + 1)un = (−1)n+1 (n + 1)un−1 ,
P P
n=0 n=0

which converges on 0 < |u| < 1.

∞
(n + 1)(−1)n+1 (z − 1)n−1 .
P
Thus the Laurent series of f about z = 1 is f (z) =
n=0
1
Example 2.3. Find the Laurent series of f (z) = (z+1)(z−2)
about its isolated
singularities z = −1 and z = 2, and find the residues there at.
Let z + 1 = u. Then,
∞ ∞
1 1
(1 − u/3)−1 = − 3u
1
(u/3)n = − un−1
P P
f (z) = u(u−3)
= − 3u 3n+1
,
n=0 n=0

which converges on 0 < |u/3| < 1, that is on R1 : 0 < |z + 1| < 3. Thus the
∞
P (z+1)n−1
Laurent series of f about z = −1 is f (z) = − 3n+1
.
n=0

Also, Res (f; −1) = a1 = −1/3.

While, on the domain R2 : |z + 1| > 3, we see that  z+1

 3   
 < 1 so that  3  < 1.
u
Therefore,
∞ ∞ ∞
3 −1 3n 3n
1 1 1
(3/u)n =

P P P
f (z) = u(u−3)
= u2
1− u
= u2 un+2
= (z+1)n+2
.
n=0 n=0 n=0

1
Exercise 2.1. Find the Laurent series of f (z) = z(z+5)
about its isolated singu-
larities z = 0 and z = −5, and find the residues of f there at.
The Laurent series of f about z = 0 converge on the domains R1 : 0 < |z| < 5
and R2 : |z| > 5. Then identify the coefficient of 1/z in the expansion in the
punctured region R1 , which will be the residue of f at z = 0. Similarly, the

Dr. T. Phaneendra 3 511, A10, SJT

CVPDE (MAT3003) Module 3

Laurent series of f about z = −5 converge on the domains R3 : 0 < |z + 5| < 1

and R2 : |z + 5| > 1. Then identify the coefficient of 1/(z + 5) in the expansion
in R3 , which will be the residue of f at z = −5.
1
Example 2.4. Consider f (z) = z(1−z)
for all z. We find a Laurent’s series of f
about z = 1, which is convergent on the domain R : |z − 1| < 1. Let z − 1 = u.
Then

1 1 1 −2
z 2 (1−z)
= − (u+1)2 u = − u (1 + u)

∞ ∞
= − u1 (−1)n (n + 1)un = (−1)n+1 (n + 1)un−1 ,
P P
n=0 n=0

which converges on R : |u| < 1, that is on R : |z − 1| < 1. Thus the Laurent

series of f about z = 1 is
∞
(n + 1)(−1)n+1 (z − 1)n−1 .
P
f (z) =
n=0

Exercise 2.2. Find a Laurent’s series of each of the following functions f (z)
about its isolated singularities, which is convergent on an appropriate domain
of validity, and find the residues, where ever possible:
1
(a) f (z) = z(z+2)
; z = 0, −2
1
(b) f (z) = z 2 +4
; z = 2i
z+4
(c) f (z) = z 2 (z 2 +3z+2)
; z = −1, −2
z
(d) f (z) = (z−1)(z−3)
; z = 1, 3
7z 2 −9z−18
(e) f (z) = z 3 −9z
; z = 0, ±3
1
(f) f (z) = (z−1)2 (z+3)
; z = 1, −3
8z+1
(g) f (z) = z(1−z)
; z = 0, 1

Dr. T. Phaneendra 4 511, A10, SJT

CVPDE (MAT3003) Module 3

3 Laurent Series about a Point of Analyticity in an Annulus

1
Example 3.1. Suppose we need the Laurent series of f (z) = z(1−z)
about z = 2
in the region R : 1 < |z − 2| < 2, which does not contain these singularities.
Also, f is analytic at z = 2 ∈ R. Still, we can develop a Laurent series as follows:

1 1 1
Let w = z − 2. Then f (z) = − (w+2)(w+1) = 2+w
− w+1
.

Now, in the region R : 1 < |w| < 2, we see that  w1  < 1 and  w2  < 1.
   

Therefore, the Laurent series about z = 2 is given by:

1 1

−1 1 1 −1
= 12 1 − w2


f (z) = 2−w
− w+1 −w 1+ w
∞ ∞
1 w n 1 n 1 n
P
P

= 2 2
− w
(−1) w
n=0 n=0
∞ ∞
wn n+1 1 n+1
P P

= 2 n+1 + (−1) w
n=0 n=0
∞ ∞
P (z−2)n P (−1)n+1
= 2n+1
+ (z−2)n+1
n=0 n=0

Exercise 3.1. Find the Laurent’s series of each of the following functions f (z)
about an indicated point of analyticity, which converges in the given annulus:
1
(a) f (z) = z(z+2)
about z = −1 in 1 < |z − 1| < 3
z2
(b) f (z) = z 2 −3z+2
about z = −1 in 1 < |z| < 2 and 1 < |z − 3| < 2
z
(c) f (z) = (z−1)(z−3)
about the origin in 1 < |z| < 3
z−1
(d) f (z) = z2
about z = 2 in the annulus 1 < |z − 2| < 2
1
(e) f (z) = z(z−1)
about z = −1 in 1 < |z − 2| < 2

Dr. T. Phaneendra 5 511, A10, SJT