Calculating the Laurent Series of $\frac{1}{e^z-1}$

[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...

 

[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 into the expression and play around with it?

 

[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"

 

[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...