Residue of $\frac{z}{\sin^2 z}$ at $\pi$ I determined that $\frac{1}{\sin^2 z}$ has a pole of order $2$ at $\pi$ by examining the Taylor series of $\sin^2 z$ about $\pi$. However, I have found the residue of $f(z)=\frac{z}{\sin^2 z}$ at $\pi$ rather difficult to compute.

Is there an intelligent way to compute the residue (in an exam environment) other than by 
  $$Res\left(f,\pi\right)=\lim_{z\to \pi}\frac{d}{dz}(z-\pi)^2f(z)?$$

 A: Think of the residue as being the coefficient of $(z-1)^{-1}$ in the Laurent expansion 
The residue of $\frac{1}{sin^2 z}$ must be zero because of it's even parity about $z=\pi$ so its Laurent series can only contain even powers of $(z-\pi)$
therefore the residue of $ \frac{z}{sin^2 z}$ equals the residue of  $ \frac{z-\pi}{sin^2 (z-\pi)}$ which has only a simple pole so its residue can be calculated without taking a derivative.
A: It does look messy using the residue limit formula, since (after some simplification):
$$
\frac{d}{dz}\left[(z-\pi)^2 \frac{z}{\sin^2z}\right] = \frac{\left(z-\pi\right)\left[3z - \pi - 2z(z-\pi)\cot z\right]}{\sin^2z}
$$
The limit of the numerator and denominator are both zero as $z\rightarrow \pi$, so you could use L'Hôpital's rule. You have to apply it a couple of times to get limits that don't vanish for both, but eventually it gives a residue of $1$. 
If you really want to take a different approach you could use the Cauchy integral formula directly:
$$
\text{Res}\, f(z) = \frac{1}{2\pi i} \oint \frac{z}{\sin^2 z} dz \ .
$$
