I'm trying to calculate the residue of $g(z)=1/\sin(z)$ at the simple pole of $z=0$.
For some reason, I can't figure out how to do it since when I compute the taylor expansion of $\frac {z}{\sin(z)}$ I do not get a $z^{-1}$ term.
Thanks for your help.
Note that, in a neighbourhood of $z$, $$\frac{1}{\sin(z)}=\frac{1}{z-\frac{z^3}{6}+o(z^3)} =\frac{1}{z}\cdot\frac{1}{1-\frac{z^2}{6}+o(z^2)}\\= \frac{1}{z}\cdot\left(1+\frac{z^2}{6}+o(z^2)\right)= \frac{1}{z}+\frac{z}{6}+o(z).$$ What is the residue at $0$?
Let $g(z)=\sin z$. Since $0$ is a simple pole (the roots of $\sin z$ can be shown to be simple) we have that $$ Res(f,0) = \lim_{z\to 0}\frac{z}{\sin z}=\frac{1}{g'(0)}=\frac{1}{\cos 0}=1. $$
Remember the residue formula is $$Res(f,z_0) = \lim_{z\to z_0}(z-z_0)^k f^{(k-1)}(z),$$ where $k$ is the multiplicity of $z_0.$
