TITLE: Laurent series expansion of 1/sin(z) - why is everything before residue 0?
QUESTION [6 upvotes]: I seem to have taken some bad lecture notes about this...
I have written (about $z = 0$)
$$\frac{1}{\sin(z)} = \ldots +\frac{a_{-3}}{z^3}+\frac{a_{-1}}{z^{-1}}+{a_{1}}{z^1}+{a_{3}}{z^3}+\ldots   $$
Mentioned $\sin(z)$ being an odd function, and everything including and before the term $a_{-3}$ is $0$. I don't understand that bit. Also, why isn't $a_{-1}=0$ as well?
The next part is to evaluate $$\oint_{|z| = \frac{\pi}{2}} \frac{1}{\sin(z)}\ dz$$
Thanks

REPLY [4 votes]: Another approach: define
$$f(z):=\frac1{\sin z}\implies \lim_{z\to 0}\,zf(z)=\lim_{z\to 0}\frac z{\sin z}=1$$
Which means $\,z=0\,$ is a pole of order $\;1\;$ of $\,f(z)\,$ , and this means that in the Laurent series around zero for this function we get
$$a_n=0\;\;\forall\,n\in\Bbb Z\;,\;n<-1\;\;\wedge\;\;f(z)= \frac 1z+a_0+a_1z+\ldots$$