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 $|z| <  \pi$ ?

The radius of convergence of the Laurent series of $1/\sin z$ about $z=0$ is $\pi$. The reason is that its poles are at $n\pi$, $n\in\Bbb Z$
(because that's where the sine has its zeroes). The nearest poles to zero
are at $\pm \pi$, as distance $\pi$ from the origin.
