Calculate residue at the singularity for following function

Question

Calculate residue at the singularity for following function $$f(x)=\frac{\pi\cot(\pi z)}{z^2}.$$ The singularity is of the type pole of order three. But I am not understanding how to calculate the residue. Is polynomial division the only way left here to find residue?

Answer 1

An idea with power series, and $\;z\;$ whose modulus is pretty close to zero (take into account that we're only interested in the coefficient of $\;z^{-1}\;$ , so we don't probably need many summands in the Laurent series...):

$$\frac{\pi\cot\pi z}{z^2}=\frac{\pi\cos\pi z}{z^2}\cdot\frac1{\sin \pi z}\frac{\pi\sin\pi z}{z^2}\cdot\frac1{\pi z-\frac{\pi^3z^3}6+\mathcal O(z^5)}\stackrel{\text{Geom. series}}=$$$${}$$

$$=\frac{\pi\cos\pi z}{z^3}\cdot\left(1+\frac{\pi^2z^2}6+\frac{\pi^4z^4}{36}+\ldots\right)=\frac\pi{z^3}\left(1-\frac{\pi^2z^2}2+\ldots\right)\left(1+\frac{\pi^2z^2}6+\ldots\right)=$$$${}$$

$$=\frac\pi{z^3}\left(1-\frac{\pi^2z^2}{3}+\ldots\right)$$

and thus that's a pole of order $\;3\;$ and its residue is $\;\cfrac{\pi^3}3\;$