Sum of all Residues at the poles of the function $f(z)=\frac{\cot(\pi z)}{(z+a)^2}$, where $a \in \mathbb{Z}-{0}$

Question

Sum of all Residues at the poles of the function $f(z)=\frac{\cot(\pi z)}{(z+a)^2}$, where $a \in \mathbb{Z}-\{0\}$

(a)$\frac{1}{\pi} \sum_{n=-\infty}^{\infty}\frac{1}{(n+a)^2}-\pi cosec^2 \pi a$

(b)$-\frac{1}{\pi} \sum_{n=-\infty}^{\infty}\frac{1}{(n+a)^2}-\pi cosec^2 \pi a$

(c)$\frac{1}{\pi} \sum_{n=-\infty}^{\infty}\frac{1}{(n+a)^2}+\pi cosec^2 \pi a$

(d) $-\frac{1}{\pi} \sum_{n=-\infty}^{\infty}\frac{1}{(n+a)^2}+\pi cosec^2 \pi a$

We want $$\sum_{a \in \mathbb{Z}-\{0\}} Res_{z=-a}\frac{\cot(\pi z)}{(z+a)^2}$$

$$Res_{z=a}\frac{\cot(\pi z)}{(z-a)^2}$$

$$\lim_{z\rightarrow a} (z-a)\frac{\cot(\pi z)}{(z-a)^2}$$ $$\lim_{z\rightarrow a} \frac{\cot(\pi z)}{(z-a)}=cot(\pi a)$$ Am I correct?. I don't get any series given the option. Please help me.

Answer 1

Form $\cot\pi z=\dfrac{\cos\pi z}{\sin\pi z}$ we know the points $\pi z=kz$ or $z=k\in\mathbb{Z}$ are poles. If $-a\neq k$ then we have a simple pole in $z=k$ so \begin{align} \operatorname{Res}_{z=k}\frac{\cot\pi z}{(z+a)^2} &= \lim_{z\to k} (z-k)\frac{\cot\pi z}{(z+a)^2} \\ &= \lim_{w\to0} w\frac{\cos\pi(w+k)}{(w+k+a)^2\sin\pi(w+k)} \\ &= \lim_{w\to0} w\frac{\cos(k\pi+w\pi)}{(w+k+a)^2\sin(k\pi+w\pi)} \\ &= \dfrac{1}{\pi(k+a)^2} \end{align} If $-a=k$ then we have a pole of order $3$ in $z=k$ so \begin{align} \operatorname{Res}_{z=-a}\frac{\cot\pi z}{(z+a)^2} &= \lim_{z\to-a}\frac{1}{2!}\dfrac{d^2}{dz^2}\left((z+a)^3\frac{\cot\pi z}{(z+a)^2}\right) \\ &= \frac{1}{2!}\times-2\pi(1+\cot^2\pi(-a)) \\ &= -\pi\csc^2\pi a \end{align} then the sum of residues is $$\color{blue}{\sum_{k=-\infty,k\neq -a}^\infty\dfrac{1}{\pi(k+a)^2}-\pi\csc^2\pi a}$$