Radius of converge of Laurent series for $\frac{1}{\sin z}$ I want to show that the radius of convergence of Laurent series for $\frac{1}{\sin z}$ is $\pi$
I  showed that:
\begin{align}
\frac 1 {\sin z} & = \frac 1 {z  - \frac{z^3}{3!} + \frac{z^5}{5!} +  \cdots   } =  \frac 1 {z\big(1  - (\frac{z^2}{3!}  - \frac{z^4}{5!}   + \cdots )\big)} \\[10pt]
& = \frac{1}{z} \bigg( 1 + \bigg(\frac{z^2}{3!}  - \frac{z^4}{5!}   + \cdots \bigg) +  \bigg(\frac{z^2}{3!}  - \frac{z^4}{5!}   + \cdots \bigg)^2 + \cdots   \bigg) \\[10pt]
& = \frac{1}{z} +\frac{z}{3!}  + \bigg( \frac{1}{3!}- \frac{1}{5!}\bigg)z^3 + \cdots
\end{align}
This equality holds if and only if:
$$\bigg|\frac{z^2}{3!}  - \frac{z^4}{5!}   + \cdots \bigg| < 1
\iff \bigg|\frac{\sin z}{z}  - 1\bigg| < 1$$
How can I show that the last inequality implies $|z| <  \pi$ ?

After writing my comment, I realize that it is way simpler...apparently:
The function $\;\cfrac1{\sin z}\;$ is analytic at $\;\left\{\,z\in\Bbb C\;/\;0<|z|<\pi\,\right\}\;$ and thus the radius of convergence of (the non principal part of) the Laurent series of the function around zero is at least $\;\pi\;$ . But as already commented, it can't be $\;\ge\pi\;$ as the function has a pole of order one at $\;z=\pi\;$ . Thus , the radius is exactly $\;\pi\;$ .
