3Laurent series and singularities

IB Complex Methods

3.4 Residues

So far, we’ve mostly been making lots of definitions. We haven’t actually used

them to do much useful things. We are almost there. It turns out we can

easily evaluate integrals of analytic functions by looking at their Laurent series.

Moreover, we don’t need the whole Laurent series. We just need one of the

coefficients.

Definition

(Residue)

.

The residue of a function

f

at an isolated singularity

z

0

is the coefficient

a

−1

in its Laurent expansion about

z

0

. There is no standard

notation, but shall denote the residue by res

z=z

0

f(z).

Proposition. At a simple pole, the residue is given by

res

z=z

0

f(z) = lim

z→z

0

(z − z

0

)f(z).

Proof. We can simply expand the right hand side to obtain

lim

z→z

0

(z − z

0

)

a

−1

z − z

0

+ a

0

+ a

1

(z − z

0

) + ···

= lim

z→z

0

(a

−1

+ a

0

(z − z

0

) + ···)

= a

−1

,

as required.

How about for more complicated poles? More generally, at a pole of order

N, the formula is a bit messier.

Proposition. At a pole of order N , the residue is given by

lim

z→z

0

1

(N − 1)!

d

N−1

dz

N−1

(z − z

0

)

N

f(z).

This can be proved in a similar manner (see example sheet 2).

In practice, a variety of techniques can be used to evaluate residues — no

single technique is optimal for all situations.

Example.

Consider

f

(

z

) =

e

z

z

3

. We can find the residue by directly computing

the Laurent series about z = 0:

e

z

z

3

= z

−3

+ z

−2

+

1

2

z

−1

+

1

3!

+ ··· .

Hence the residue is

1

2

.

Alternatively, we can use the fact that

f

has a pole of order 3 at

z

= 0. So

we can use the formula to obtain

res

z=0

f(z) = lim

z→0

1

2!

d

2

dz

2

(z

3

f(z)) = lim

z→0

1

2

d

2

dz

2

e

z

=

1

2

.

Example. Consider

g(z) =

e

z

z

2

− 1

.

This has a simple pole at

z

= 1. Recall we have found its Laurent series at

z

= 1

to be

e

z

z

2

− 1

=

e

2

1

z − 1

+

1

2

+ ···

.

So the residue is

e

2

.

Alternatively, we can use our magic formula to obtain

res

z=1

g(z) = lim

z→1

(z − 1)e

z

z

2

− 1

= lim

z→1

e

z

z + 1

=

e

2

.

Example.

Consider

h

(

z

) = (

z

8

−w

8

)

−1

, for any complex constant

w

. We know

this has 8 simple poles at

z

=

we

nπi/4

for

n

= 0

, ··· ,

7. What is the residue at

z = w?

We can try to compute this directly by

res

z=w

h(z) = lim

z→w

z − w

(z − w)(z − we

iπ/4

) ···(z − we

7πi/4

)

=

1

(w −we

iπ/4

) ···(w − we

7πi/4

)

=

1

w

7

1

(1 − e

iπ/4

) ···(1 − e

7iπ/4

)

Now we are quite stuck. We don’t know what to do with this. We can think

really hard about complex numbers and figure out what it should be, but this is

difficult. What we should do is to apply L’Hˆopital’s rule and obtain

res

z=w

h(z) = lim

z→w

z − w

z

8

− w

8

=

1

8z

7

=

1

8w

7

.

Example.

Consider the function (

sinh πz

)

−1

. This has a simple pole at

z

=

ni

for all integers

n

(because the zeros of

sinh z

are at

nπi

and are simple). Again,

we can compute this by finding the Laurent expansion. However, it turns out it

is easier to use our magic formula together with L’Hˆopital’s rule. We have

lim

z→ni

z − ni

sinh πz

= lim

z→ni

1

π cosh πz

=

1

π cosh nπi

=

1

π cos nπ

=

(−1)

n

π

.

Example.

Consider the function (

sinh

3

z

)

−1

. This time, we find the residue by

looking at the Laurent series. We first look at

sinh

3

z

. This has a zero of order

3 at z = πi. Its Taylor series is

sinh

3

z = −(z − πi)

3

−

1

2

(z − πi)

5

.

Therefore

1

sinh

3

z

= −(z − πi)

−3

1 +

1

2

(z − πi)

2

+ ···

−1

= −(z − πi)

−3

1 −

1

2

(z − πi)

2

+ ···

= −(z − πi)

−3

+

1

2

(z − πi)

−1

+ ···

Therefore the residue is

1

2

.

So we can compute residues. But this seems a bit odd — why are we

interested in the coefficient of

a

−1

? Out of the doubly infinite set of coefficients,

why a

−1

?

The purpose of this definition is to aid in evaluating integrals

H

f

(

z

) d

z

,

where

f

is a function analytic within the anticlockwise simple closed contour

γ

,

except for an isolated singularity

z

0

. We let

γ

r

be a circle of radius

r

centered

on z

0

, lying within γ.

z

0

γ

r

Now

f

has some Laurent series expansion

P

∞

n=−∞

a

n

(

z − z

0

)

n

about

z

0

. Recall

that we are allowed to deform contours along regions where f is analytic. So

I

γ

f(z) dz =

I

γ

r

f(z) dz

=

I

γ

r

∞

X

n=−∞

a

n

(z − z

0

)

n

dz

By uniform convergence, we can swap the integral and sum to obtain

=

∞

X

n=−∞

a

n

I

γ

r

(z − z

0

)

n

dz

We know how to integrate around the circle:

I

γ

r

(z − z

0

)

n

dz =

Z

2π

0

r

n

e

inθ

ire

iθ

dθ

= ir

n+1

Z

2π

0

e

i(n+1)θ

dθ

=

(

2πi n = −1

r

n+1

n+1

e

i(n+1)θ

= 0, n 6= −1

.

Hence we have

I

γ

f(z) dz = 2πia

−1

= 2πi res

z=z

0

f(z).

Theorem. γ

be an anticlockwise simple closed contour, and let

f

be analytic

within γ except for an isolated singularity z

0

. Then

I

γ

f(z) dz = 2πia

−1

= 2πi res

z=z

0

f(z).