# Laurent series of 1/(e^z - 1)

1. Feb 7, 2012

### Combinatus

1. The problem statement, all variables and given/known data

Obtain the first few terms of the Laurent series for the following function in the specified domain:

$\frac{1}{e^z-1}$ for $0 < |z| < 2\pi.$

2. Relevant equations

3. The attempt at a solution

I've attempted a few approaches, but haven't really gotten anywhere. For instance, using a Maclaurin series for $e^z$ yields $\frac{1}{e^z-1} = \frac{1}{z} \cdot \frac{1}{1+\frac{z}{2!}+\frac{z^2}{3!}+\cdots}$

Of course, $\frac{1}{z}$ can be written as $\frac{1}{2\pi - (-z + 2\pi)} =\frac{1}{2\pi} \cdot \frac{1}{1-\frac{-z+2\pi}{2\pi}}$, and since the latter can be written as a geometric series in the given annulus, we have $\frac{1}{z} = \sum_{k = 0}^{\infty} (-1)^k (\frac{z-2\pi}{2\pi})^k.$

Actually, after further contemplation, I think I could set $g(z) = \frac{1}{1+\frac{z}{2!}+\frac{z^2}{3!}+\cdots}$ and differentiate that a few (infinitely many) times to find its Maclaurin series. I think termwise differentiation should be allowed in the annulus in question, since g should be analytic there. Then multiplication by 1/z yields the Laurent series?

2. Feb 7, 2012

### micromass

Do you know the Laurent series of $\frac{1}{z-1}$??

3. Feb 7, 2012

### Combinatus

Regular Maclaurin (Taylor at z=0) series for $|z|<1$, i.e. $-1-z-z^2-z^3-\cdots$, and $\sum_{k = 0}^{\infty} \frac{1}{z^{k+1}}$ for $|z|>1$?

So, $\frac{1}{e^z - 1} = \sum_{k = 0}^{\infty} \frac{1}{e^{z(k+1)}}$?