Q:

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

Since the isolated singularities are $z=k\pi, k \in \mathbb Z$, so we divided the complex plane into the disjoint annulus, i.e. $\{z: n\pi <|z|<(n+1)\pi\}, n \in \mathbb N \cup \{0\}$. On these annuli, $f$ is analytic, so has a unique Laurent series.
First, let's consider $\{z: 0 < |z|<\pi\}$, then the solution says
$$\begin{aligned}
f(z)& =\frac{1}{\sin z}\\
& =\frac{1}{z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots}\\
& =\frac{1}{z}\cdot \frac{1}{1-\left(\frac{z^2}{3!}-\frac{z^4}{5!}+\cdots\right)}\\
& =\frac{1}{z}\left[1+\left(\frac{z^2}{3!}-\frac{z^4}{5!}+\cdots\right)+\left(\frac{z^2}{3!}-\frac{z^4}{5!}+\cdots\right)^2+\cdots\right]
\end{aligned}$$
My questions are:
Do we need $\left|\frac{z^2}{3!}-\frac{z^4}{5!}+\cdots\right|<1$ to get the last step? If we do, then why it is less than 1?
Thanks for any help!

A:

We have that
$$g(z) = 1-\left(\frac{z^2}{3!}-\frac{z^4}{5!}+\cdots\right)$$
equals $\sin(z)/z$ for all $z\neq 0$.
Since $\sin(\pi/2) = 1$, we have $g(\pi/2) = 2/\pi$, that is, $|g(\pi/2)|<1$.
Now, $g$ is entire (why?), so in particular it is continuous.
It follows that $|g|<1$ in some open set $U\subset \mathbb C$ containing $\pi/2$, so the last step is valid for $z\in U$.
At this point, it suffices to invoke the uniqueness of the Laurent series.

The uniqueness of Laurent series' coefficients is a very powerful tool for calculating them.
It implies that if your 'chain of equalities' (in your calculations) holds true in an open subset of your annulus, then the end result holds true in all of the annulus.

