How can I define the annulus of convergence for $f(z)=\frac{1}{\sin z}$?

Question

I am thinking about the Laurent series of the function:

$$f(z)=\frac{1}{\sin z}$$

How could I define its annulus of convergence? Since the singularities are $$z=m\pi, m\in\mathbb{Z}$$

Aren't there infinitely many annulus' of convergence for which I can write a Laurent series. For example, one could be:

$$C=\{z\in\mathbb{C}\ |\ \pi<\left|z\right|<2\pi\}$$

So, is it possible to have many different annulus' of convergence? Also, could $$C=\{z\in\mathbb{C}\ |\ 0<\left|z\right|<\pi\}$$ be an annulus of convergence?

Answer 1

We have $$\text{Res}\left(\frac{1}{\sin z},z=k\pi\right) = (-1)^k $$ for any $k\in\mathbb{Z}$, hence $\frac{1}{\sin z}$ has the following Eisenstein-like-series: $$ \frac{1}{\sin z} = \frac{1}{z}+\sum_{m\geq 1}(-1)^m\left(\frac{1}{z-m\pi}+\frac{1}{z+m\pi}\right) $$ with the RHS being a uniformly convergent series over any compact subset of $\mathbb{C}\setminus\pi\mathbb{Z}$.
Such representation can be easily used to derive the Laurent series of $\frac{1}{\sin z}$ over any annulus not enclosing any singularity.

Answer 2

Hint. One may write, for each $m=0,1,2,\cdots,$ $$ f(z)=\frac{1}{\sin z}=\frac{(-1)^m}{(z-m\pi)}\cdot\frac{2i(z-m\pi)\cdot e^{i(z-m\pi)}}{e^{2i(z-m\pi)}-1},\quad 0<|z-m\pi|<\pi, $$ then, by using properties of the Bernoulli numbers, one gets $$ f(z)=\frac{(-1)^m}{(z-m\pi)}+(-1)^m\sum_{n=0}^\infty \frac{2(2^{2n+1}-1)\cdot B_{2n+2}}{(2n+2)!}(z-m\pi)^{2n+1},\quad 0<|z-m\pi|<\pi. $$