Calculating residue of $f(z)=\frac{z^2+z^3}{({\sin z})^3}$ at its singularities So the singularities would be at $n\pi$, $n\in \mathbb{Z}$.
The residue in $z=0$, I have already calculated (through shifting the series) and it equals $1$ however for the others $n\pi$ I am stuck. Help would be appreciated

Hint:-
$z^{2}=(z-n\pi)^{2}+2n\pi(z-n\pi)-2n^{2}\pi^{2}$ and
$z^{3}=(z-n\pi)^{3}+3n\pi(z-n\pi)^{2}+3n^{2}\pi^{2}(z-n\pi)+n^{3}\pi^{3}$
Use these and write $\frac{z^{2}+z^{3}}{\sin(z)}=(-1)^{n}\frac{(z-n\pi)^{2}+2n\pi(z-n\pi)-2n^{2}\pi^{2}+(z-n\pi)^{3}+3n\pi(z-n\pi)^{2}+3n^{2}\pi^{2}(z-n\pi)+n^{3}\pi^{3}}{(\sin(z-n\pi))^{3}}$.
And now calculate the residue as you would by computing the coefficient of $\frac{1}{z-n\pi}$ in the Laurent Series expansion.
Another hint:- This whole thing can be done by calculating the residue at $0$ of
$$(-1)^{n}\frac{z^{2}+2n\pi z-2n^{2}\pi^{2}+z^{3}+3n\pi z^{2}+3n\pi^{2} z+n^{3}\pi^{3}}{\sin^{3}(z)}$$
Caution:- I often make errors in calculation so the expressions might not be entirely accurate but I think the hint is enough to provide you with a method to proceed by!!!.
I wish you a succesful dig for the residue.
