Residue of $\frac{z^3+z^2}{\sin(z)^3}$

Question

I want to compute the residue for $f(z)=\frac{z^3+z^2}{\sin(z)^3}$ in all isolated singularities. For $0$ this is easy, since $0$ is a pole or order 1.

But for $k \pi, k \in \mathbb{Z}: k \neq 0$ I am stuck,these are poles of order 3, but using the formula with the derivative doesnt really seem to work out for me.

I would be thankful for some help,

Greetings

Answer 1

Yes we do have a pole of order $3$.

I would make a Laurent series expansion of the $1/\sin^3z$ factor about each pole $z=k\pi$. Plug in $w=z-k\pi$ and note that we will then have, from trigonometry:

$\sin z=(-1)^k\sin w$

So

$(\sin z)^{-3}=(-1)^kw^{-3}[1-(w^2/6)+O(z^4)]^{-3}$

$=(-1)^k[w^{-3}+w^{-1}/2+O(w)]$

The $O(w)$ component cannot contribute to the $w^{-1}$ term when we multiply by the numerator

$z^3+z^2=(w+k\pi)^3+(w+k\pi)^2.$

Thus we multiply this polynomial by the terms shown from the Laurent series given above and then read off the $w^{-1}$ coefficient of the product.