1. ## [SOLVED] Laurent series

Hello, I want to calculate laurent series and its convergence radius of:

$f(z)=\frac{1}{sin({\pi z})}$

Here it is what i've done, I'm not sure if we can do it in this way:

$sin z =\sum \frac{(-1)^{n}}{(2n+1)!}z^{2n+1}=z-\frac{z^3}{3!}+\frac{z^5}{5!}-O(z^7)$ for $-1

so:

$sin (\pi z) =\sum \frac{(-1)^{n}}{(2n+1)!}(\pi z)^{2n+1}=\pi z-\frac{(\pi z)^3}{3!}+\frac{(\pi z)^5}{5!}-O((\pi z)^7)$ for $-1

so:

$\frac{1}{sin (\pi z)} =\sum \frac{(2n+1)!}{(-1)^{n}}\frac{1}{(\pi z)^{2n+1}}=\frac{1}{\pi z}-\frac{3!}{(\pi z)^3}+\frac{5!}{(\pi z)^5}-\frac{1}{O((\pi z)^7)}$ for $-1

Is it right?

Thank you.

2. warning
$\sum \frac{A}{B} \neq (\sum \frac{B}{A})^(-1)$

3. We can start from the series expansion...

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

... and then, setting...

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

... find the $a_{n}$ imposing...

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

From (3) now we derive directly...

$a_{0} = 1$

$a_{1} =0$

$a_{2} - a_{0} \frac{\pi ^{2}}{3!} = 0 \rightarrow a_{2} = \frac{\pi^{2}}{6}$

$a_{3} =0$

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

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

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

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

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

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

Kind regards

$\chi$ $\sigma$