I want to determine the singularities for:
$f(z) = \frac{z}{\sin(z)}$ for $z_0 = 2\pi k $, where $ k\in \mathbb{Z} $.
If the singularity is removable, I have to compute the limit.
If there is a pole, I have to specify the principal part of the Laurent series.
My thoughts:
if $ k = 0 $ the singularity is removable, and the limit is $1$.
For $ k \neq 0 $ we have a pole of order $1$, but here's is my question: how do I determine the principal part of the Laurent series?
Let me quickly give you my notation: Consider the Laurent series $$ \sum_{k = -\infty}^{\infty} a_k(z-z_0)^k = \sum_{k = 1}^{\infty} a_{-k}(z-z_0)^{-k} + \sum_{k = 0}^{\infty} a_k(z-z_0)^k .$$
The series $\sum_{k = 1}^{\infty} a_{-k}(z-z_0)^{-k}$ is called the principal part.
How do I get the principal part for this function? All I know is that:
$$ \frac{z}{\sin(z)} = \frac{z}{\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}z^{2k+1}} =\frac{1}{\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}z^{2k}} .$$
Best Answer
Use pole expansion,
$$\csc z=\sum_{n=-\infty}^\infty \frac{(-1)^n}{z-n\pi}$$
hence,
$$\frac{z}{\sin z}=\sum_{n=-\infty}^\infty (-1)^n\frac{z}{z-n\pi}$$
So all $z=n\pi, n\neq 0$ are simple poles.