Can somebody explain to me why $\frac{1}{\sin(z)}$ having a residue of $1$ at $z=0$ means that $\frac{1}{\sin(z)}$ has a Laurent series around $z=0$ of the form $\frac{a_0}{z}+a_1 +a_2z+a_3z^2+...$ (where $a_0=1$) according to the webpage:
How do you know it doesn't start from $\frac{a_0}{z^6}$?
EDIT: How do you know the coefficients of all terms $\frac{1}{z^m}$ where $m>1$ is $0$?
Multiply your function by z. You get $\frac{z}{\sin(z)}$
You know the limit of this function at zero, and therefore, you know that the series for this function, if it exists, also has the same limit.
But which terms contribute to this limit? Well, any term that isn't $\frac{1}{z^n}$ for n>0 will not contribute, and any term that has $n \neq 1$ will be infinite. So you can therefore realize that there is a $\frac{1}{z}$ term, and nothing with a higher negative power.
