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
 A: 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.
A: The computation is simpler if you set $z=n\pi+w$.
$$f(n\pi+w) = (-1)^n [(n\pi+w)^2 + (n\pi+w)^3]\sin^{-3}(w).$$
You develop the numerator and you use twice Taylor-Young formula, first for $\sin(w)$, next for $(1+h)^{-3}$ as $h \to 0$, to get an asymptotic expansion
\begin{eqnarray*}
f(n\pi+w) 
&=& (-1)^n \big(a+bw+cw^2+o(w^2)\big) \Big(w-\frac{w^3}{3}+o(w^3)\Big)^{-3} \\
&=& (-1)^n \big(a+bw+cw^2+o(w^2)\big) w^{-3}(1-w^2+o(w^2)) \\
&=& (-1)^n w^{-3}\big(a+bw+(c-a)w^2+o(w^2)\big) \\
&=& (-1)^n \big(aw^{-3}+bw^{-2}+(c-a)w^{-1}+o(w^{-1})\big).
\end{eqnarray*}
The residue of $f$ at $n\pi$ is $(-1)^n(c-a)$.
