$\csc{z} = \frac{1}{\sin{z}}$ 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?
Best Answer
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 $1$.