Laurent series

1. Apr 6, 2006

Warr

Just wondering where to go with this one..

calculate the laurent series of $$\frac{1}{e^z-1}$$

don't even know where to start on it

I know $$e^z={{\sum^{\infty}}_{j=0}}\frac{z^j}{j!}$$

but not much else...

Last edited: Apr 6, 2006
2. Apr 6, 2006

matt grime

The first thing you need to do is figure out where you're taking the Laurent expansion about (presumably zero since that is what your expression for e^z is. Why not put that in to the expression and play around with it?

3. Apr 6, 2006

Warr

Well, I only assumed that I knew that the expansion of e^z was about 0. It only specifies "calculate the laurent expansion of $$\frac{1}{e^z-1}$$ for $$0 < |z| < 2\pi$$"

Last edited: Apr 6, 2006
4. Apr 6, 2006

Warr

I tried that but couldn't really come up with anything..

$$\frac{1}{e^z-1}=\frac{1}{(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+...)-1}=\frac{1}{z+\frac{z^2}{2!}+\frac{z^3}{3!}+...}=\frac{1}{z(1+\frac{z}{2!}+\frac{z^2}{3!}+...)}$$

no idea where to go with this..

I can't see how I could turn the series into a useful series that converges to a algebraic expresion that I could actually rearrange to continue....

Last edited: Apr 6, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook