determine type of singularities and compute residue of a function Determine the type of singularities and residue of $$\frac{1}{\sin^2(z)}$$
For this problem, this is the way I approach this: 
we have : 
$$\sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots$$
$$= z\left(1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots\right)$$
Let $h(z) = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots$. Then, we have: 
$$\sin(z) = z\cdot h(z)$$
$$\sin^2(z) = z^2 \cdot h^2(z)$$
Therefore, 
$$\frac{1}{\sin^2(z)} = \frac{1}{z^2} \cdot \frac{1}{h^2(z)}$$
Thus, it has a simple pole at $z_0 = 0$.
But, I don't know how to calculate the residue of this function. Can someone please how me how to compute its residue. 

You are on the right track.  For the pole at $z=0$, let's write
$$\begin{align}
\frac{1}{\sin^2 z}&=\frac{1}{\left(z-\frac16 z^3++O(z^5)\right)^2}\\\\
&=\frac{1}{z^2\left(1-\frac16 z^2++O(z^4)\right)^2}\\\\
&=\frac{\left(1+\frac16 z^2+O(z^4)\right)^2}{z^2}\\\\
&=\frac{1}{z^2}+\frac13+O(z^2)
\end{align}$$
Thus, we find the singularity is a pole of order $2$.  
We can perform a similar expansion for around any of the zeros of the sine function and see that there are singularities of $\csc^2 z$ that are poles of order $2$ at $z=n\pi$ for all integer values of $n$.
Another way to see this is to recall the infinite product representation of the sine function.  Then,
$$\sin z=z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2\pi^2}\right)$$
implies that
$$\csc^2 z=\frac{1}{z^2 \prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2\pi^2}\right)^2}$$
which clearly shows the second order poles at $z=n\pi$.
