How to find the Laurent series about z=0 and therefore the residue at z=0 of f(z) = 1/(z^4 sin(pi z)), where f(z) is a complex valued function?

So I just had a complex analysis exam, went pretty well but I was agonising over this question for quite a few pages of scribble, if someone could put me out of my misery and show me how it should be done that'd be great.

1 Answer
Steve M
Feb 17, 2017

f(z)=1/(piz^5) + (pi)/(6z^3) + (7pi^3)/(360z) + (31pi^5z)/15120 + O(z^3)

Res{f(z)}_(z=0) = (7pi^3)/(360)

Explanation:

Typically in an exam you would need to derive the power series for 1/sin(piz) using the TS for sinx and then use the Binomial Series to expand (sin(pix))^-1. This is quite tedious and it can be looked up in reference books. The series is as follows:

csc x = 1/x + x/6 + (87x^3)/360 + (31x^5)/15120 + ...

And so the Laurent series for f(z)=1/(z^4sin(piz)) is:

f(z)=1/z^4 \ csc(piz)
\ \ \ \ \ \ \ =1/z^4 {1/(piz) + (piz)/6 + (7(piz)^3)/360 + (31(piz)^5)/15120 + O(z^7)}
\ \ \ \ \ \ \ =1/(piz^5) + (pi)/(6z^3) + (7pi^3)/(360z) + (31pi^5z)/15120 + O(z^3)

So then;

Res{f(z)}_(z=0) = (7pi^3)/(360)

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