Chapter 11: laurents Transform

Question 1

Definition 11

isolated singularity

Answer 1

A function f which is analytic on the punctured disc
D’ = {z ∈ C : 0 < |z − α| < R} but not on {z ∈ C : |z − α| < R} is said to have an
isolated singularity at α

“isolated singularity” = “bad point”

Question 2

EXAMPLE

1/sin z
has singularities

Answer 2

1/sin z
has singularities at z = nπ for n ∈ Z and is analytic on {z : 0 < |z| < π} for
example.

Question 3

EXAMPLE

1/ (z-1)(z+2) has singularities

Answer 3

1/ (z-1)(z+2) has singularities at 1 and at −2, and is analytic on the punctured discs
{z : 0 < |z − 1| < 3} and {z : 0 < |z + 2| < 3}.

Question 4

Theorem 11.2 (Laurent’s Theorem)

Answer 4

Suppose that f has an isolated singularity at α
(so f is analytic on some punctured disc D0 = {z : 0 < |z − α| < R}). Then f can be
represented on D0
by a Laurent series about z = α, i.e.,

f(z) = Σ from n =-∞ to ∞ of
a_n (z-α)^n
for z in D’.

The laurent coefficients are given by

a_n
= (1/2πi)∫_Cr f(w)/(w − α)^(n+1) . dw,

where Cr : w = α + reit (0 ≤ t ≤ 2π) for any 0 < r < R.

Question 5

for two sets analytic a_n….

Answer 5

∫_Cr1 f(w)/(w − α)^(n+1) . dw

The an are independent of r: if 0 < r1 < r2 < R, we know that the function
f(w)/(w−α)^(n+1) is analytic on the D0 and is therefore analytic between Cr1 and Cr2

By Theorem
9.1, we see that

∫_Cr2 f(w)/(w − α)^(n+1) . dw,

and so is independent of r

Question 6

DEF 11.3 residue and laurents

Answer 6

The Laurent coefficient a−1 is called the residue of f at α. We write
a−1 = Res {f; α}

f f has a singularity at α and Cr is a positively oriented circle centred on α of radius r,
then

∫_Cr f(z)dz = 2πi Res {f; α}.

Question 7

EXAMPLE:

Find the residue of sinh z/z^4 at the origin.

Answer 7

Let f(z)= sinh z/z^4

Then f has a singularity at 0 and that it is analytic on C \ {0}

for z ≠ 0

f(z) = sinh z/z^4
= [z + (z^3 /6) + (z^5/120) + … ]/z^4
=
1/z^3 + ((1/6)/z) + (z/ 120) +…

and so Res {f : 0} =1/6

Question 8

EXAMPLE:
Find the Laurent series of 1/ z(z−1) about z = 1 giving an expression for the general
term. Where is this expansion valid?

diagram D* and point

Answer 8

Let g(z) = 1/z(z-1)

g is analytic in C \ {0, 1}. It has isolated singularities at
the points 0,1. It is analytic on the punctured disc D∗ = {z ∈ C : 0 < |z − 1| < 1}
about 1. Hence g(z) has a Laurent series expansion, in powers of z − 1 valid in D∗
.

Let w=z-1

For 0 < |z − 1| < 1, i.e. 0 < |w| < 1,

g(z)= 1/ z(z−1)
= (1/(z-1)) - 1/z
= (1/w) - (1/(1+w))
= (1/w) - Σ from n =0 to ∞ of (-1)^n w^n

= (1/(z-1)) - Σ from n =0 to ∞ of (-1)^n (z-1)^n

is the Laurent series expansion for g(z) valid in D∗

Question 9

if isolated singularity

Answer 9

If the function f has an isolated singularity at α, then f(z) has a Laurent Series
expansion
f(z) = Σ∞ n=−∞ of a_n(z − α)^n valid for z in some punctured disc D0
centred on α.

Question 10

if analytic

Answer 10

If f is analytic at z0, then f(z) has a Taylor Series expansion f(z) = Σ∞ n=0 of a_n(z −z_0)^n
valid for z in some disc ∆ centred on z0.

Question 11

A REMOVABLE SINGULARITY.

Definition 11.4

Answer 11

If a_n = 0 for all n < 0,
(so that b_n = 0 for all positive integers n,

i.e. all the negative powers of (z − α) in the Laurent expansion have coefficient 0) then α
is called a removable singularity.

Question 12

REMOVABLE SINGULARITY FORM

Answer 12

f(z) = Σ∞ n=-∞ of a_n(z −α)^n

Suppose that f has an isolated singularity at α. Then f(z) has a Laurent series expansion

Σ from n=1 to ∞of b_n/(z −α)^n +
Σ from n=0 to ∞of a_n(z−α)^n

valid in some punctured disc about α, where bn = a−n for all positive integers n

Question 13

If the function f has a removable singularity at α, then f(α) is undefined

Answer 13

If we define f(α) by (or change the value of f(α) so that) f(α) = a0,
then we get a function which is defined on a disc centred at α given by a Taylor series, i.e.
an analytic function. We have, then, removed the “removable singularity” by defining (or
redefining) f(α).

Question 14

EXAMPLE

sinz/z
has a singularity at 0 as it is not defined there.

Answer 14

But the function

f(z) =
{sinz/z if z≠0
{ 1 if z=0

is analytic on C and

f(z) = 1 - (z^2/6) + (z^4/120) - (z^6/7!) +…

is its Taylor series about 0, valid for all z ∈ C.

Question 15

B ISOLATED ESSENTIAL SINGULARITY.

Definition 11.5

Answer 15

If an ≠ 0 for infinitely many n < 0, (so that infinitely many of the
coefficients bn are non-zero

i.e. there are infinitely many negative powers of (z −α) with
non-zero coefficient in the Laurent expansion) then f has an isolated essential singularity
at α.

Question 16

e^(1/z) and sin(1/z)

are both analytic on the punctured disc
D_0 = {z ∈ C : 0 < |z|},

Answer 16

so z = 0 is an isolated singularity of e^(1/z) and sin(1/z)

for z ≠ 0

e^(1/z) = 1 + (1/(1!z) + (1/2!z^2) + (1/3!z^3) + …

sin(1/z) = (1/z) - (1/3!z^3) + (1/5!z^5) -….

thus both e^(1/z) and sin(1/z) have isolated essential singularities at the origin

Note The even negative powers of z in the Laurent expansion of sin(1/z)
about the origin
all have coefficient zero. However, there are still an infinite number of negative powers
with non-zero coefficient

Question 17

C. POLE
def 11.6

Answer 17

If we can find k ∈ N such that a_{−k} ≠ 0 and a_n = 0 for n < −k, then
f is said to have a pole of order k at α

b_k ≠ 0 and b_n =0 for ALL n bigger than k and Laurent series only contains finite # of neg powers of (z− α) with non-zero coefficients

The Laurent series looks like

[a_{-k} / (z − α)^k]
\+
[a_{-{k-1}} / (z − α)^{k-1}] + 
....+
a_-1 / (z − α)  + a_0  + a_1(z − α) + .....

Question 18

The Laurent series

of
sinhz /z^4
looks like:

Answer 18

(POLE)
for z≠ 0

sinhz /z^4 = (1/z^3) + (1/6)/z + z/120 +…

so sinhz /z^4 has a pole of order 3 @ the origin
. Note that 0 is an isolated singularity as
the function is analytic on the punctured disc D0 = {z ∈ C : 0 < |z|}

Note that sinh z has a zero to order 1 at the origin and z
4 has a zero to order
sinhz /z^4
therefore has a zero of order −3 in some sense

Question 19

THM 11.7

f pole order k when

Answer 19

The function f has a pole of order k at α if and only if f(z) can be
expressed in the form
f(z) = g(z)/(z − α)^k
in some punctured disc D’ = {z ∈ C : 0 < |z − α| < R}, (R > 0) where, g is analytic and
non-zero in the disc D = D’ ∪ {α}

Question 20

THM 11.8 zero order k and pole

Answer 20

If the function f has a zero of order k at α, then 1/f

has a pole of order k at α.

Question 21

COROLLARY 11.9

sero order and pole quotient functs

Answer 21

If the function f has a zero of order m at α and the function g has a zero of order n at α, then

(i) f/ g
has a pole of order (n − m) at α if n > m;
(i) f/g
has a removable singularity at α if n ≤ m

Question 22

SIMPLE POLE

DEF 11.10

Answer 22

If k = 1, we say that α is a simple pole.

Question 23

Find all the singularities in the complex plane of

1/(1+z^2)

Answer 23

1/(1+z^2) is analytic in C \ {± i} and so has isolated singularities at ± i.

1+z^2 = (z-i)(z+i)

and so (1+z^2) has has zeros of order 1 at the points ± i.

thus 1/(1+z^2) has simple poles at ± i.

Question 24

Find all the singularities in the complex plane of

sinz/z^2

Answer 24

sinz/z^2 is analytic on C \ {0} and so 0 is an isolated singularity. Now
sin z has a zero of order 1 at the origin and z^2 has a zero of order 2 at the origin.

Hence by Corollary 11.9
sinz/z^2
has a simple pole at the origin.

Question 25

Find all the singularities in the complex plane of

1/(e^z -1)

Answer 25

e^z =1 if and only if z = 2nπi, where n is an integer.

let f(z) =(e^z -1)
then 

f(2nπi) = 0
f'(2nπi) = [e^z]_z=2nπi = e^(2nπi) =1 ≠ 0

Thus f has zeros of order 1 at the points 2nπi(n ∈ Z).

Hence 1/f
has isolated singularities at the points 2nπi(n ∈ Z) and these are simple
poles by Theorem 11.8

Question 26

Determine the singularity of cotz/z

at the origin.

Answer 26

Let h(z) = cotz/z = cosz/zsinz = (f(z)/g(z)) where f(z) = cos z , g(z) = z sin z .

function h is analytic on the punctured disc {z ∈ C : 0 < |z| < π} and
so h has a isolated singularity at the origin

sin z has a zero of order 1 at the origin and, therefore, g has a zero of order 2 at the origin

By theorem 11.8, 1/g
has a pole of order 2 at the origin. Hence h =f/g
has a pole of order 2 i.e. a double pole at the origin since f is analytic in C and f(0) = cos 0 ≠ 0

Question 27

Explain how Laurent expansions are used to classify isolated singularities.
Find all the singularities in the complex plane of funct and classify them

find the residue at each of the singularities

cos(1/(z-1))

Answer 27

Let g(z) =  cos(1/(z-1)) 
g is analytic in C\{1}. For z ≠ 1,

g(z) = cos(1/(z-1))
= 1 - (1/2!)( 1/ (z-1))^2 + (1/4!)( 1/ (z-1))^4 - …

is the Laurent expansion of g(z) about the isolated singularity at
1. valid in C{1} and

Res{g;1} =0

Question 28

Explain how Laurent expansions are used to classify isolated singularities.
Find all the singularities in the complex plane of funct and classify them

find the residue at each of the singularities

zcos(1/(z-1))

Answer 28

Let h(z) = zcos(1/(z-1))
h is analytic in C\{1}. For z ≠ 1,

h(z) = [(z-1) +1]cos(1/(z-1))
= [(z-1) +1] [1 - (1/2!)( 1/ (z-1))^2 + (1/4!)( 1/ (z-1))^4 - …]

=
1 + (z-1) - (1/2!)(1/(z-1))^2 - (1/2!)(1/(z-1)) + - (1/4!)(1/(z-1))^4 + - (1/4!)(1/(z-1))^3-…

is the Laurent expansion of h(z) about the isolated singularity at
1. valid in C{1} and
Res{h;1} = -1/2! = -0.5

Question 29

THM 11.1

Quick ways of calculating residues at poles

Answer 29

1. Suppose that f has a pole of order k at α. Then
   Res {f ; α} =
   [1/(k − 1)!] limz→α of [d^{k−1}/dz^{k−1}] [(z − α)^k f(z)]
2. If f =g/h
   where g and h are analytic at α and g(α) ≠ 0, h(α) = 0, h’(α) ≠ 0 (so h has a simple zero at α), then f has a simple pole at α and

This shortcut will work when f(z) is given by a formula involving a fraction with an obvious factor (z − α)^n
in the denominator.
In other cases you will need to find a Laurent expansion and pick out the
coefficient a_{−1}

Question 30

Find the singularities in the complex plane of the following and calculate the residues at
each of them:

sinz/(z^10)

Answer 30

Let q(z) = sinz/z^10

the q is analytic in C except for an isolated singularity. Now
sin z has a zero of order 1 at the origin and z
10 has a zero of order 10 at the origin. From
Corollary 11.9, we see that q has a pole of order 9 at the origin

(no shortcut and no 11.1, 1) )
laurent expansion for z≠0

sinz/(z^10) = (1/z^9) - 1/3!z^7) + (1/5!z^5) - (1/7!z^3)) + 1/(9!z) - z/11! + ….

Res(q;0} = 1/9!

Question 31

Find the singularities in the complex plane of the following and calculate the residues at
each of them:

1/(1+z^2)

Answer 31

Let f(z) = 1/(1+z^2)
 We know that f is analytic in C \ {± i} and has simple poles at ± i

let
g(z) = 1 , h(z) = 1 + z^2

h’(z) = 2z
g(i) =1
h(i)=0
h’(i) =2i

thus

Res{f;i}= g(i)/h’(i) = 1/2i

Res{f;-i}= g(-i)/h’(-i) = -1/2i

by Theorem 11.11 part (2).

Question 32

Find the singularities in the complex plane of the following and calculate the residues at
each of them:
1/(e^z - 1)

Answer 32

Let k(z) = 1/(e^z - 1)

k is analytic in C except for simple poles at the
points 2nπi (n ∈ Z)

Theorem 11.11 part (2):

Res(k;  2nπi} =
[1/(d/dz) (e^z − 1)]_z=2nπi
=
[1/e^z]_z=2nπi
= 1.

Question 33

Find the singularities in the complex plane of the following and calculate the residues at
each of them:
e^z/(z-1)^2

Answer 33

Let m(z) = e^z/(z-1)^2.

Then m is analytic on C except for a singularity at the point 1.

(z-1)^2 e^-z has a zero of order 2 at z = 1 and so its reciprocal m has a double pole
at 1.

thm 11.11 part 1 with k=2 and α = 1 gives,

Res{m;1} = (1/(2-1)!) lim(z→1) of (d/dz)[(z-1)^2 [e^z/(z-1)^2]]

= limz→1
of d(e^z)/dz
= e.

Question 34

Find the singularities in the complex plane of the following and calculate the residues at
each of them:

1/(1+z^2)^9

Answer 34

Let p(z) = 
1/(1+z^2)^9

function p is analytic on C except for isolated
singularities at ± i.
(1 + z^2)^9 = (z + i)^9(z − i)^9 has zeros of order 9 at ± i, we see that 1/(1+z^2)^9 has poles
of order 9 at ± i, i.e. m has poles of order 9 at ± i . Using Theorem 11.11 part (1) with
k = 9 gives,

Res{p;i}
= 16!/ ((8!)^2 2^17 i)

Res{p;-i}
= 16!/ ((8!)^2 2^17 i)