Question

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.

Answer 1

`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 `sin`x 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)`