<s>[INST] Laurent expansion on an annulus problem I have $f(z)= \frac{1}{\sin(z)}$ and am required to show that on the disc {$0<|z|<\pi$} the Laurent expansion is equal to:
$$c_{-1}z^{-1}+\sum_{n=0}^\infty{c_nz^n}$$
My plan is use the expansion of $$\sin(z)= z-\frac{z^3}{3!}+\frac{z^5}{5!}...$$ and rewrite $\frac{1}{\sin(z)}$ as $$\frac{1}{z(1-(\frac{z^2}{3!}-\frac{z^4}{5!}+...))} = \frac{1}{z} \times \frac{1}{(1-(\frac{z^2}{3!}-\frac{z^4}{5!}+...))}.$$
Usually I would ignore the condition about the disc that {$0<|z|<\pi$}
and use the fact for small enough $z$, $$\left|\frac{z^2}{3!}-\frac{z^4}{5!}+...\right|$$ is smaller than $1$ we can use $\frac{1}{1-z}=1+z+z^2...$
so $$\frac{1}{(1-(\frac{z^2}{3!}-\frac{z^4}{5!}+...))}=1+(\frac{z^2}{3!}-\frac{z^4}{5!}+...)+(\frac{z^2}{3!}-\frac{z^4}{5!}+...)^2...$$ so 
\begin{align}
\frac{1}{z}\times\frac{1}{(1-(\frac{z^2}{3!}-\frac{z^4}{5!}+...))}&= \frac{1}{z} \times [1+(\frac{z^2}{3!}-\frac{z^4}{5!}+...)+(\frac{z^2}{3!}-\frac{z^4}{5!}+...)^2...] \\ &=\frac{1}{z}+0+\frac{z}{3!}...
\end{align}
and the rest of the terms when you expand out. But do I need to show that $|\frac{z^2}{3!}-\frac{z^4}{5!}+...|$ is smaller than $1$ using the fact that $z$ lies in the disc first?
 [/INST] $$
             z\frac{1}{\sin(z)}
$$
has no singularities in $0 < |z| < \pi$ and has a limit at $z=0$ of $1$, which means that this function has a removable singularity at $z=0$. So you have a power series expansion of the following in $|z| < \pi$:
$$
                    z\frac{1}{\sin(z)} = 1 + a_1z + a_2 z^2+ a_3z^3+\cdots, \\             \frac{1}{\sin(z)}=\frac{1}{z}+a_1 + a_2 z + a_3 z^2 + \cdots.
$$
</s>