<s>[INST] Computing with Cauchy Residue theorem how do I calculate $$\operatorname{Res}\left(\frac{1}{z^2 \cdot \sin(z))}, 0\right)$$ What is the order of the pole? $3$?
 [/INST] Yes, the pole is of order 3.  Thus
$$\begin{align}\mathrm{Res}_{z=0} \frac{1}{z^2 \sin{z}} &= \frac{1}{2!} \lim_{z \rightarrow 0} \frac{d^2}{d z^2} \left [ z^3 \frac{1}{z^2 \sin{z}} \right ] \\ &= \frac{1}{2}\lim_{z \rightarrow 0} \frac{d^2}{d z^2} \left ( \frac{z}{\sin{z}} \right )\\ &= \frac{1}{2} \lim_{z \rightarrow 0} [z (\csc^3{z} + \cot^2{z} \csc{z}) - 2 \cot{z} \csc{z}]\\ &= \lim_{z \rightarrow 0}\frac{1}{2} \left (\frac{2}{z^2}+\frac{1}{3}-\frac{2}{z^2} \right )\\ &= \frac{1}{6} \end{align}$$
</s>