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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: $$ \frac{z}{\sin(z)} = \frac{z}{\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}z^{2k+1}} =\frac{1}{\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}z^{2k}} .$$

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    $\begingroup$ For (a), my suggestion for tackling the $k \neq 0$ case is to write $z / \sin(z)$ as $(z - k\pi)/\sin(z - k\pi) + k\pi / \sin(z - k\pi)$. Then use similar logic to the $k = 0$ case. $\endgroup$
    – Kenny Wong
    Commented May 16, 2023 at 17:26

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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.

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    $\begingroup$ It's much simpler than that (having the Mittag Leffer expansion is stronger). The claim follows from the fact that $(z-\pi k)\frac{z}{\sin z}$ tends to a finite nonzero quantity as $z\to\pi k$ $\endgroup$
    – FShrike
    Commented May 16, 2023 at 18:26

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