Mathematical Methods for Physics and Engineering : A Comprehensive Guide

COMPLEX VARIABLES

C C

C′

(a) (b)

Figure 24.13 The contours used to prove the residue theorem: (a) the original

contour; (b) the contracted contour encircling each of the poles.

to the integral, becausefis analytic betweenCandC′. Now the contribution to

theC′integral from the polygon (a triangle for the case illustrated) joining the

small circles is zero, sincefis also analytic insideC′. Hence the whole value of

the integral comes from the circles and, by result (24.60), each of these contributes

2 πitimes the residue at the pole it encloses. All the circles are traversed in their

positive sense ifCis thus traversed and so the residue theorem follows. Formally,

Cauchy’s theorem (24.40) is a particular case of (24.61) in whichCencloses no

poles.

Finally we prove another important result, for later use. Suppose thatf(z)has

a simple pole atz=z 0 and so may be expanded as the Laurent series

f(z)=φ(z)+a− 1 (z−z 0 )−^1 ,

whereφ(z) is analytic within some neighbourhood surroundingz 0. We wish to

find an expression for the integralIoff(z) along anopencontourC,whichis

the arc of a circle of radiusρcentred onz=z 0 given by

|z−z 0 |=ρ, θ 1 ≤arg(z−z 0 )≤θ 2 , (24.62)

whereρis chosen small enough that no singularity off, other thanz=z 0 , lies

within the circle. ThenIis given by

I=

∫

C

f(z)dz=

∫

C

φ(z)dz+a− 1

∫

C

(z−z 0 )−^1 dz.

If the radius of the arcCis now allowed to tend to zero, then the first integral

tends to zero, since the path becomes of zero length andφis analytic and

therefore continuous along it. OnC,z=ρeiθand hence the required expression

forIis

I= lim

ρ→ 0

∫

C

f(z)dz= lim

ρ→ 0

(

a− 1

∫θ 2

θ 1

1

ρeiθ

iρeiθdθ

)

=ia− 1 (θ 2 −θ 1 ). (24.63)