What is the Laurent series for \frac{1}{e^z - 1} in the given domain?

[Combinatus]

Homework Statement

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.

Homework Equations

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?

 

[micromass]

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

 

[Combinatus]

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

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)}}?