TITLE: Finding the residue at $z=0$ for complex function $1/(z^2\sin z$)
QUESTION [8 upvotes]: Find the residue at $z=0$ for complex function $1/(z^2\sin z$)
I know $z=0$ is a pole of order 3 but can't seem to calculate the residue value for it.

REPLY [3 votes]: You can reference the result for a pole of order $3$ at $z=0$:
$$\text{Res}_{z=0} \frac{1}{z^2 \sin{z}} = \frac{1}{2!} \lim_{z \rightarrow 0} \left[\frac{d^2}{dz^2} \frac{z}{\sin{z}}\right]$$
$$\frac{d}{dz} \frac{z}{\sin{z}} = \frac{\sin{z} - z  \cos{z}}{\sin^2{z}} $$
$$\begin{align}\frac{d^2}{dz^2} \frac{z}{\sin{z}} &= z \left(\csc ^3{z}+\cot ^2{z} \csc{z}\right)-2 \cot{z} \csc{z}\\ &=\left (\frac{1}{z^3} + \frac{1}{2}   + \frac{1}{z^3} - \frac{1}{2}\right ) - \frac{2}{z^2} + \frac{1}{3} + O(z)\\ &= \frac{1}{3}\end{align}$$
Therefore the residue is $1/6$.