Residue of $\frac{z^2-z}{1-sinz}$ There is a double pole at $\frac{\pi}{2} + 2n\pi$
However, I am only familiar with the formula
$\mathrm{Res}(f(z))_{z=z_0}=\frac{1}{(m-1)!}\lim_{z\to z_0}\left[\frac{d^{m-1}}{dz^{m-1}}(z-z_0)^m f(z)\right]$
for multiple poles, but I am getting stuck taking the limit (and not sure that this is the equation I'm supposed to be working with) 
 A: In that case, it may be easier to calculate in a direct way. That is, you can calculate Laurant series directly and check the $-1$th coefficient. 
$f(z) = \frac{z^2-z}{1-\sin z}$. If we calculate the residue of $\frac{1}{1-\sin z}$, then the residue of $f(z)$ follows directly (just multiplying $z^2-z$ at the point), so let's calculate the residue of $\frac{1}{1-\sin z}$. To make it much simpler, put $w=\pi/2-z$, and we have $\frac{1}{1-\sin z}=\frac{1+\sin z}{\cos^2 z} = \frac{1+\cos w}{\sin^2 w}$. 
Observe the following:
$1+\cos w = 2 - \frac{w^2}{2!}+\frac{w^4}{4!}-\cdots$
$\sin^2 w = (w-\frac{w^3}{3!}+\frac{w^5}{5!}-\cdots)^2 = w^2 - \frac{2}{3!}w^4 + (\frac{2}{5!}-\frac{1}{3!^2})w^6 + \cdots$
Now we have to divide $1+\cos w$ by $\sin^2 w$ to get the desired Laurant series. The first non-zero term is $2z^{-2}$. The residue is the coefficient for $z^{-1}$, and it is zero. 
In fact, all the terms of $1+\cos w$ and $\sin^2 w$ are of even degree, the Laurant series for $f(z)$ at $\frac{\pi}{2}+2n\pi$ has only even degree terms. Therefore all the residues are 0. 
