How to determine Laurent series associated to $f(z)$

Question

The function is

$$f(z)= \frac{1}{(e^z -1)},$$

$z$ belong to $\mathbb{C}$ and $0<|z|<1$.

I need a general expression in term of a sum from 0 to infinity

Answer 1

The Laurent series at $0$ is defined with the help of the Bernoulli numbers.

Answer 2

$$\frac{1}{z}-\frac{1}{2}+\frac{z}{12}-\frac{z^3}{720}+\frac{z^5}{30240}-\frac{z^7}{1209600}+\frac{z^9}{4790016}-\frac{691z^{11}}{1307674368000}+O(z^{13})$$ I think it is no simple expression,so this is the result used mathematica.

Answer 3

Use that the power series of the exponential function converges for any $\;z\in\Bbb C\;$ :

$$\frac1{e^z-1}=\frac1{1+z+\frac{z^2}2+\ldots-1}=\frac1{z\left(1+\frac z2+\mathcal O(z^2)+\ldots\right)}=$$

$$=\frac1{z}\left(1-\frac z2+\frac{z^2}4-\ldots\right)=\frac1z-\frac12+\ldots$$

The above is already enough to know, for example, the residue at $\;z=0\;$ of the function.