I'm facing the following function:
$$f(z) = \frac{z}{\sin z}$$
I know there are singularities at $z_k=k\pi$ with $k \in \mathbb{Z}$. Using the basical properties of the $\sin(z)$ function, I found: $$\sin(z-k\pi) = (-1)^k \sin(z)$$
And as $\sin(z)$ can be developped with a Taylor series: $$\sin(z) = \sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)!} z^{2n+1}$$ I turn my initial equation into: $$\frac{z}{\sin z} = \frac{(z-k\pi) + k\pi}{(-1)^k \sin (z-k\pi)} = \sum_{n=0}^{+\infty}\frac{(2n+1)!}{(-1)^n(-1)^k}(z-k\pi)^{-2n} + \dots$$ I conclude thus that all those poles are essential, meaning the Laurent development of this series has infinite terms different from zero.
Yet, my book claims that they are all... simple poles !
- Where is my mistake ?
- How to prove these are single poles ?