Laurent Series for $\cot(\pi z)$ Find the Laurent Series for $f(z) = \cot(\pi z)$ at $z= 0$.
I just need hints; I am not sure how to do this at all. I am very new to this.
 A: Summon the Maclaurin series (which is the Taylor series centered at $z=0$) of both $\cos(\pi z)$ and $\sin(\pi z)$, to obtain $$\cot(\pi z) = \frac{\cos(\pi z)}{\sin(\pi z)}=\frac{1-\frac 1{2!}(\pi z)^2+\frac 1{4!}(\pi z)^4-\frac 1{6!}(\pi z)^6+\cdots}{\pi z-\frac 1{3!}(\pi z)^3+\frac 1{5!}(\pi z)^5-\frac 1{7!}(\pi z)^7+\cdots}.$$
Can you do long division of the following?
$$\pi z-\frac 1{3!}(\pi z)^3+\frac 1{5!}(\pi z)^5-\frac 1{7!}(\pi z)^7+\cdots\overline{\left){1-\frac 1{2!}(\pi z)^2+\frac 1{4!}(\pi z)^4-\frac 1{6!}(\pi z)^6+\cdots} \right. }$$
Your quotient will be the first few terms of your Laurent series of $\cot(\pi z)$ centered at $z=0$.
A: \begin{align}
\pi \cot(\pi z)
&=\frac{d}{dz}\ln(\sin(\pi z))\\
&=\frac{d}{dz}\left[\ln\pi z\prod^\infty_{n=1}\left(1-\frac{z^2}{n^2}\right)\right]\\
&=\frac{1}{z}-2\sum^\infty_{n=1}\frac{1}{n^2}\frac{z}{1-\frac{z^2}{n^2}}\\
&=\frac{1}{z}-2\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{z^{2k-1}}{n^{2k}}\\
&=\frac{1}{z}-2\sum^\infty_{k=1}\zeta(2k)z^{2k-1}
\end{align}
