What is the residue at z = 0 for this complex function?

[FunkyDwarf]

Hey guys,

Im trying to evaluate the integral round the unit circle to in turn evaluate a real integral, but the bit I am stuck on is finding the residual at zero for the complex function.

Whats on the bottom of the function isn't overly important as that's not what's causing the rukus, but rather on the top i have z^n + 1/z^n (technically i can move the problem bit to the bottom i know) and the denominator of the function is a quadratic in z. I can solve for the residual at the point that the denominator is zero, but how do i get the residual for z = 0 where the 1/z^n blows up? I've tried multiplying by this and differentiating that but it doesn't seem to work =(

Hope that made sense :S
Cheers
-G

 

[Dick]

Shouldn't you do what you suggested and multiply the numerator and denominator by z^n? The pole at zero is going to be a nth order pole, not a simple pole.

 

[FunkyDwarf]

Yeah i know but i found if i do that i get a hideous set of derivatives to (this was an exam question to do in a few minutes) so i figured i was doing something wrong. I'll give it another go though, cheers.

 

[Dick]

Yeah i know but i found if i do that i get a hideous set of derivatives to (this was an exam question to do in a few minutes) so i figured i was doing something wrong. I'll give it another go though, cheers.

I'm going to guess you have something like (z^n+1/z^n)/((z-a)(z-b)). To get the residue around z=0 you can throw out the z^n part. That leaves you with 1/(z^n(z-a)(z-b)). Computing the nth derivative of the inverse quadratic looks complicated. But it's not that bad if you express it using partial fractions.