Chapter 6 Residues

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De_nition 6.0.1 Given a Laurent Series for the function f about w,

f(z) =

1 X

􀀀1

ak(z 􀀀 w)k

the value a

􀀀1 is called the residue of f at w.

We write Res[f,w] for the value a

􀀀1.

Example 5.5.2 makes it easy to see that Res[tan z=z2,0] = 1.

There are some nifty tricks for calculating residues. Why we want to calculate

them will appear later, take it that there are good reasons.

For example:

Example 6.0.3 Calculate [Res f,0] for

f(z) =

2

z3 􀀀(i + 1)z2 +iz

Solution:

We can clearly factorise the denominator easily into

z(z 􀀀 1)(z 􀀀 i)

149

150 CHAPTER 6. RESIDUES

so the coe_cient of 1=z is going to come from

2

(z 􀀀 1)(z 􀀀 i)

which when z = 0 is just

2

i

= 􀀀2i

Exercise 6.0.8 Do the last example the long way around and con_rm that

you get the same answer.

And now one reason why we would like to be able to compute residues:

Theorem 6.1 If f has a singularity at w and is otherwise analytic in a

neighbourhood of w, and if c is a simple closed loop going around w once in

an anticlockwise sense, then

Ic

f(z)dz = 2_i Res[f;w]

Proof:

This comes immediately from the de_nition of the Laurent series. 2

And the obvious extension for multiple singularities:

Theorem 6.2 (Cauchy's Residue Theorem)

If c is a simple closed curve in C and the function f is meromorphic on the

region enclosed by c and c itself, with singularities at w1; w2; _ _ _wn in the

region enclosed by c, then

Ic

f(z)dz = 2_i

n

X1

Res[f;wk]

where the integration is taken in the positive sense.

Proof:

The usual argument which replaces the given curve by circuits around each

singularity will do the job. 2

151

There is a clever way to compute residues for poles of order greater than one:

Theorem 6.3 If f has a pole of order k at w,

Res[f;w] =

1

(k􀀀1)!

lim

z!w

[(z 􀀀 w)kf(z)][k􀀀1]

where the exponent [q] refers to the q-fold derivative.

Proof:

We have

f(z) =

a

􀀀k

(z􀀀w)k+

a

􀀀k+1

(z 􀀀 w)k􀀀1+_ _ _+

a

􀀀1

(z 􀀀w)

+ a0 + a1(x􀀀w)+a2(x􀀀w)2+_ _ _

since f has a pole of order k. Then the function g(z) = (z 􀀀 w)kf(z) is

analytic at w and has derivatives of all orders, and they go:

g(z) = (z 􀀀 w)kf(z)

= a

􀀀k + a

􀀀k+1(z 􀀀 w) + a

􀀀k+2(z 􀀀 w)2 + _ _ _+a

􀀀1(z 􀀀w)k􀀀1 +a0(z 􀀀w)k +_ _ _

g0(z) = a

􀀀k+1 + 2a

􀀀k+2(z 􀀀 w) + _ _ _ + (k 􀀀1)a

􀀀1(z 􀀀 w)k􀀀2 + _ _ _ ...

g[k􀀀1](z) = (k 􀀀 1)! a

􀀀1 + k! a1(z 􀀀 w) + _ _ _

And as z ! w, the higher terms all vanish to give the result 2

Example 6.0.4 Calculate

Ic

dz

z4 + (1􀀀i)z3 􀀀iz2

Where c is the circle centred on the origin of radius 2.

Solution The long way around is fairly long. So we use residues and the

last theorem.

First we rewrite the function f:

f(z) =

1

(z2)(z + 1)(z 􀀀 i)

152 CHAPTER 6. RESIDUES

and observe that it has poles at zero, 􀀀1 and i. All are within the circle c.

The (z+1) pole has coe_cient

1

(z2)(z 􀀀 i)

which at z = 􀀀1 is 􀀀1=(1 + i) = (i􀀀1)=2.

The z 􀀀 i pole has coe_cient

1

(z2)(z + 1)

which at z = i is 􀀀1=(1 + i) = (i􀀀1)=2.

And _nally we compute the residue at 0 which is

lim

z!0

d

dz _ 1

(z + 1)(z 􀀀 i)_

= lim

z!0􀀀[

2z + (1􀀀i)

((z + 1)(z 􀀀 i))2

] = 1􀀀i

The integral is therefore

2_i[(1 􀀀 i) + (i􀀀1)=2 + (i􀀀1)=2)] = 0

Exercise 6.0.9 Do it the long way around to convince yourself that I haven't

blundered.

The quick way saves some messing around with partial fractions. In fact we

can use the above results to calculate the partial fractions; it is a neat trick

which you will _nd in Mathews and Howell, pp 249-250. If you ever need to

compute a lot of partial fractions, look it up1.

1Some people love knowing little smart tricks like this. It used to be thought the best

thing about Mathematics: you can use sneaky little tricks for impressing the peasantry.

Some schoolteachers use them to impress teenagers. This tells you a lot about such folk.

6.1. TRIGONOMETRIC INTEGRALS 153

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