TITLE: Laurent expansion of a function
QUESTION [5 upvotes]: Consider the function 
$$f(z) = \frac{e^z+1}{e^z-1}$$
This function has a Laurent expansion about $0$ of the form
$$f(z) = \frac{a}{z} + \sum_{n=0}^\infty b_nz^n$$
for constants $a, b_1,\ldots, b_n$. Show that $b_n=0$ for all even $n$.

What is the best way to do this? All I can think to do is find the laurent expansion and then show that a few are zero but obviously this doesn't prove it for all even $n$, just a few.

If we split $f(z)$ into even and odd parts we have
$$f(z) = \frac az +\sum_{k=0}^\infty b_{2k}z^{2k}+\sum_{k=0}^\infty b_{2k+1}z^{2k+1}$$
and how the problem turns into showing the first summation is zero.

REPLY [5 votes]: Check that $f(-z) = -f(z).$ Thus
$$\frac{a}{-z} + \sum_{n=0}^\infty b_n(-z)^n = -\left (\frac{a}{z} + \sum_{n=0}^\infty b_nz^n\right ) = -\frac{a}{z} + \sum_{n=0}^\infty (-b_n)z^n.$$
Equating coefficients gives the desired result.