# Residue of pole of order $3$

I was trying to find out the residue of the function: $$f(z)=\frac{\pi\cot(\pi z)}{z^2}$$

It is evident that we have $$z=0$$ as a pole of order $$3$$.

So we have: $$\operatorname*{Res}_{z = 0}f(z)=\frac{1}{2}\lim_{z \to 0}\frac{d^2}{dz^2}\left(z^3\times \frac{\pi\cot(\pi z)}{z^2}\right)$$ So we get:

$$\operatorname*{Res}_{z=0}f(z)=\frac{\pi}{2}\lim_{z \to 0}\frac{d}{dz}\left(-\pi z \csc^2 \pi z+\cot(\pi z)\right)$$

But it’s tedious to continue from here. Is there any alternate way?

## 2 Answers

The formula you’re applying is perhaps best viewed as a bookkeping device to extract the desired coefficient of $$z^{-1}$$ of the Laurent series of $$f$$ at $$z=0$$. In the present case, it seems easier to work directly with the series. From Laurent series for $\cot (z)$, we have

$$\cot z=\frac1z-\frac z3+O\left(z^3\right)\;,$$

so

$$\frac{\pi\cot(\pi z)}{z^2}=\frac1{z^3}-\frac{\pi^2}3\cdot\frac1z+O(z)\;,$$

from which you can read off that the residue at $$z=0$$ is $$-\frac{\pi^2}3$$.

$$Res(f,0)=\frac{1}{2}\lim_{z \to 0}\frac{d^2}{dz^2}\left(z^3\times \frac{\pi\:\cot(\pi z)}{z^2}\right)$$

$$=lim_{z\rightarrow 0}\frac{d^2}{dz^2}(πz\times \frac{\pi cos(π z)}{sin(πz)})$$

$$= -π^2/2$$ as $$\lim_{z\rightarrow 0}\frac{sin(πz)}{πz}=1$$