Re: residue~~

From: Robin Chapman (rjc_at_ivorynospamtower.freeserve.co.uk)
Date: 06/04/04 

Date: Fri, 04 Jun 2004 11:42:09 +0100

mina_world wrote:

> hello.....sir~
>
>
> evaluate the integral of these function around
>
> the circle |z| = 2 in the positive sense.
>
> f(z) = (z^5) / {1 - (z^3)}

This has simple poles at z = 1, z = exp( 2pi i/3) and
z = exp(4 pi i/3).

> --------------------------------
>
> f(z) = (z^5) / {1 - (z^3)}
>
> = {-(z^5) / (z^3)}*[1 / {1-(1/(z^3))}]
>
> = -(z^2)*{1 + 1/(z^3) + 1/(z^6) +.....}
>
> thus, residue is -1

That's a residue at infinity.

It you want to use residues to evalaute this integral, then
you could add together the residues at the cube roots of 1.
Better though use the substitution w = 1/z. Then
f(z) dz = - f(1/w) w^{-2} dw. This changes the anticlockwise
integration about |z|= 2 to clockwise integration about
|w| = 1/2. We get, reversing the direction of integration,
int_{|w|=1/2} w^{-7}/(1 - w^{-3}) dw
= - int_{|w|=1/2} w^{-4}/(1 - w^3) dw.
The pole at zero has residue -1, so I get -2pi i as the integal
(unless (likely) I have made a silly mistake). 

-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9"
Francis Wheen, _How Mumbo-Jumbo Conquered the World_

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