We are asked to solve the following integral using residue theorem
$\int_{|z|=1} \frac{1}{z\sin^2z} dz$.
I was able to show that it has one pole of order 3 inside $|z|=1$ given by $z=0$. We know that $$\int_{|z|=1} \frac{1}{z\sin^2z} dz = 2\pi i Res\left(\frac{1}{z\sin^2z},0\right).$$ where, $Res\left(\frac{1}{z\sin^2z},0\right) = \displaystyle\lim_{z\rightarrow 0}\left(\frac{1}{2!}\frac{d^2}{dz^2}\left(\frac{z^2}{\sin^2z}\right)\right)$.
The problem is that this limit seem to not exist with my computation. Is my approach right?
By Taylor expansion when $z\to 0$ \begin{equation} \frac{1}{z\sin^2z} = \frac{1}{z^3}\left(1 -\frac{z^2}{6} + o(z^2)\right)^{-2} =\frac{1}{z^3}+\frac{1}{3z} + o\left(\frac{1}{z}\right) \end{equation}