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,
On
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 $i\pi/6$.
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.