Having trouble calculating the residue at 0 for this integral within the unit circle I understand that its a pole of order 3 because both the z^2 and the sinz have singularities at 0. Is there an easier way to find this residue without resorting to finding the laurent series around z of $\frac{cscz}{z^2}$?
$\int \frac{1}{z^2sinz}~dz$
You may want to use the following formula :
If $f(z)$ has pole of order $n$ at $c$ then residue at $c$ is given by
$$\mathrm{Res}(f,c) = \frac{1}{(n-1)!} \lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}}\left( (z-c)^{n}f(z) \right)$$