# Let $f(z) = \frac{z^{-2}}{\sin( \pi z )}$. What is the residue for $z \neq 0$?

Let $f(z) = \frac{z^{-2}}{\sin( \pi z )}$. What is the residue for $z \neq 0$?

In a pdf online, it states we may calculate the residue using the "derivative trick" to get: $$\mathrm{Res}(f,n) = \frac{n^{-2}}{\pi \cos(\pi n)} = \frac{(-1)^n}{\pi n^2}.$$ What is the "derivative trick" that is being referred to? I know from the definition of a simple pole that $$\mathrm{Res}(f,n) = \lim_{z \to n} (z-n) f(z).$$ So I suppose this could be the "derivative trick" since $$\lim_{z \to n} \frac{\sin(\pi z)}{(z-n)} = \frac{d}{dz} \sin(\pi n) = \pi \cos(\pi z).$$ But this is just the definition of a simple pole. Or is there another "derivative trick" for calculating residues that I'm not aware of? Perhaps I'm reading into the text too much. It can be found here at the end of page 4.

• You understood the text correctly and the "derivative trick" is what you have explained. It will not work though for the residue at $z=0$. – Fabian Jan 1 '16 at 6:21

In situations like this I prefer to use the definition of the residue, which is simply the coefficient of $(z-z_n)^{-1}$ in the Laurent expansion about a pole at $z=z_n$. In this case, the poles are at integers $z=n$, so we expand about $\zeta=z-n$:
$$f(z) = \frac{(n+\zeta)^{-2}}{\sin{\pi(n+\zeta)}} = \frac{(-1)^n}{n^2 \sin{\pi \zeta} } \left (1+\frac{\zeta}{n} \right )^{-2}$$
Near a pole $z=n$, i.e., for small $\zeta$, the coefficient of $\zeta^{-1}$ is $(-1)^n/(\pi n^2)$. This is the residue of $f$ about the pole $z=n$.
• @mathjacks: $\sin{\pi \zeta} = \pi \zeta [1+O(\zeta^2)]$ so that $$\frac1{\sin{\pi \zeta}} = \frac1{\pi \zeta} + O(\zeta)$$ – Ron Gordon Jan 3 '16 at 1:06