We can start from the series expansion...

\(\displaystyle \frac{\sin (\pi z)}{\pi z} = 1 - \frac{(\pi z)^{2}}{3!} + \frac{(\pi z)^{4}}{5!} - \dots\) (1)

... and then, setting...

\(\displaystyle f(z) = \frac{\pi z} {\sin (\pi z)} = a_{0} + a_{1} z + a_{2} z^{2} + \dots\) (2)

... find the \(\displaystyle a_{n}\) imposing...

\(\displaystyle f(z) \frac{\sin (\pi z)}{\pi z} = (a_{0} + a_{1} z + a_{2} z^{2} + \dots) \{ 1 - \frac{(\pi z)^{2}}{3!} + \frac{(\pi z)^{4}}{5!} - \dots\} = 1\) (3)

From (3) now we derive directly...

\(\displaystyle a_{0} = 1\)

\(\displaystyle a_{1} =0\)

\(\displaystyle a_{2} - a_{0} \frac{\pi ^{2}}{3!} = 0 \rightarrow a_{2} = \frac{\pi^{2}}{6}\)

\(\displaystyle a_{3} =0\)

\(\displaystyle a_{4} - a_{2} \frac{\pi^{2}}{3!} + a_{0} \frac{\pi^{4}}{5!}=0 \rightarrow a_{4} = \frac{7 \pi^{4}}{360}\) (4)

... and so one, so that is...

\(\displaystyle \frac{\pi z}{\sin (\pi z)} = 1 + \frac {\pi^{2} z^{2}} {6} + \frac{7 \pi ^{4} z^{4}}{360} + \dots\) (5)

... and now, deviding (5) by \(\displaystyle \pi z\), we obtain...

\(\displaystyle \frac{1}{\sin (\pi z)} = \frac{1}{\pi z} + \frac {\pi z} {6} + \frac{7 \pi ^{3} z^{3}}{360} + \dots\) (6)

The Laurent series (6) converges for \(\displaystyle 0 < |z| < 1\) ...

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)