Is there a simpler way to compute the residue of a function at a pole of order 3? The function $$\frac {1}{z^2(e^{i2\pi z}-1)}$$ has a triple pole at z = 0.  To compute the residue of f at z = 0, I can compute the Laurent expansion of f about z = 0, and then read off the coefficient of the 1/z term.  But the fraction requires polynomial long division and I generally make a lot of mistakes during this somewhat heavy computation (by hand).  Is there a better way to compute the residue at 0?
Thanks,

This is a great example that demonstrates the usefulness of asymptotic expansions to facilitate analysis.  We have the function $f$
$$f(z)=\frac{1}{z^2(e^{i2\pi z}-1)}$$
with a pole of order $3$ at $z=0$ and wish to find its residue there.
To that end, we write 
$$\begin{align}
e^{i2\pi z}-1&=i2\pi z+\frac12 (i2\pi z)^2+\frac16 (i2\pi z)^3+O\left(z^4\right)\\\\
&=i2\pi\left(1+\frac12(i2\pi z)+\frac16(i2\pi z)^2+O\left(z^3\right)\right)
\end{align}$$  
and therefore
$$f(z)=\frac{1}{i2\pi z^3}\left(1-\frac12(i2\pi z)+\frac{1}{12}(i2\pi z)^2+O\left(\left(z\right)^3\right)\right)\tag 1$$
where we used $\frac{1}{1+x}=1-x+x^2+O(x^3)$ in arriving at $(1)$.
Finally, the residue is simply the $z^{-1}$ term of $f$ in $(1)$.  This is easily seen to be $i\pi/6$.
