Why does csc(z) only have 1st order poles?

Question

$\csc{z} = \frac{1}{\sin{z}}$ is said (in my text book) to have only simple (1st order) poles.

I can see that this is justified since the Laurent series expansion is:

$$ \csc{z} = \frac{1}{z} + \frac{z}{6} + \frac{7 z^3}{360} + ... $$

However, I do not understand how to derive this expansion and why there are not more $z^{-n}$ terms since the Taylor series expansion of $\sin{z}$ has an infinite number of positive exponent terms, so my intuition is that $1/\sin{x}$ should have an infinite number of negative exponent terms.

Where does this Laurent series come from? and why are the poles first order and not infinite order?

Answer 1

You know that $z=0$ is an isolated singularity and it is a pole ($\csc z \rightarrow \infty$ as $z\rightarrow 0$). Then, $\lim_{z\rightarrow 0} z\cdot \csc z =1$ implies that this pole is of order $1$.

Answer 2

The (finite) poles of $\csc z$ are the zeros of $\sin z$, with the same order. All zeros of the sine are simple, because $$ \lim_{z\to k\pi}\frac{\sin z}{z-k\pi}=\lim_{w\to0}\frac{\sin(w+k\pi)}{w}= \lim_{w\to0}(-1)^k\frac{\sin w}{w}=(-1)^k $$