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,

Suppose $f$ has a pole of order $n$ at $z_0$, i.e.
$$
f(z)=\frac{a_{-n}}{(z-z_0)^n}+\frac{a_{-n+1}}{(z-z_0)^{n-1}}+\dots+\frac{a_{-1}}{z-z_0}+g(z)
$$
in a neighborhood $U\ni z_0$, where $g(z)$ is holomorphic. Thus
$$
(z-z_0)^nf(z)=a_{-n}+a_{-n+1}(z-z_0)+\dots+a_{-1}(z-z_0)^{n-1}+g(z)(z-z_0)^n
$$
$$
\left(\frac{d}{dz}\right)^{n-1}\left[(z-z_0)^nf(z)\right] = (n-1)!a_{-1}+n!g(z)(z-z_0)+\dots+g^{(n-1)}(z)(z-z_0)^n
$$
$$
\text{Res}_{z_0}f(z) = a_{-1}=\frac{1}{(n-1)!}
\lim_{z\to z_0}\left(\frac{d}{dz}\right)^{n-1}\left[(z-z_0)^nf(z)\right]
$$
This formula requires some differentiation, which can get a bit ugly when $(z-z_0)^nf(z)$ is a rational function such as $\dfrac{z}{e^{2\pi iz}-1}$ in your case. You might find this easier than computing the Laurent series in many cases, though.
