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.
A: Yes, you do need that sum, which is the same as $(\sin{z})/z-1$, to be smaller than $1$ to make the expansion. This actually occurs on a smaller region than that where the Laurent series turns out to converge:

(there is no nice expression for this set since it is specified using exponentials and powers.)
However, as Fimpellizieri notes in the their answer, that this region contains an open set is enough to carry out the calculation, then the principle of analytic continuation implies that the series you find as a result of the calculation converges to the function on a larger set: it being a power series, this is the disk that has radius the distance from the expansion point to the nearest non-removable singularities, namely $\pi$.

As for the Laurent series for $\pi<\lvert z \rvert < 2\pi$, the function
$$ f_i(z) = \frac{1}{\sin{z}} - \left( \frac{1}{z} - \frac{1}{z-\pi} - \frac{1}{z+\pi} \right) $$
is analytic inside $\lvert z \rvert < 2\pi$, while
$$ f_o(z) = \frac{1}{z} - \frac{1}{z-\pi} - \frac{1}{z+\pi} $$
is analytic in $ \pi > \lvert z \rvert$. These therefore provide an appropriate decomposition $f=f_o+f_i$ for Laurent's theorem to apply. We then expand $f_i(z)$ in a power series about $z=0$, and $f_o(z)$ as a power series in $1/z$, i.e. about $z=\infty$.
