Laurent series of $\frac{z}{\sin({\frac{\pi}{z+1}})}$ in the roots of the denominator

Question

I'd like to compute the first few terms of the Laurent series of $\frac{z}{\sin(\frac{\pi}{z+1})}$ at $z=\frac{1}{k}-1, k\in\mathbb{Z}$. I assume I know the espansion of $\frac{1}{\sin{z}}$ in $z=0$, so i'm trying to express $\frac{\pi}{z+1}$ in terms of $[z-(\frac{1}{k}-1)], k\in\mathbb{Z}$, but I can't get anything useful to be put in the expansion of $\frac{1}{\sin{z}}$.

Answer 1

Hint:

Let $z=w+\left(\frac1k-1\right)$ then $$ f(z)=F(w)=\frac{w+\left(\frac1k-1\right)}{\sin\left(\frac{\pi}{w+\frac1k}\right)}=\frac{w+\left(\frac1k-1\right)}{\sin\left(\frac{k\pi}{1+kw}\right)} $$ Now recall that the expansion of $k\pi(1+kw)^{-1}$ at $0$: for $|w|<\frac1k$, $$ \frac{k\pi}{1+kw}=k\pi\sum_{n=0}^{\infty} (-1)^n k^nw^n=\pi\sum_{n=0}^{\infty} (-1)^n k^{n+1}w^n $$ and, for $v \in \mathbb{C}$, $$ \sin(v)=v-\frac{v^3}{3!}+\frac{v^5}{5!}-\dots\tag{1} $$ Can you take it from here?