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

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Homework Help Overview

The discussion revolves around calculating the Laurent series of the function \(\frac{1}{e^z-1}\), particularly focusing on the region where \(0 < |z| < 2\pi\). Participants are exploring the series expansion of \(e^z\) and its implications for the given function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are considering the expansion of \(e^z\) around zero and questioning how to manipulate this series to find the Laurent series for \(\frac{1}{e^z-1}\). There is uncertainty about the starting point and how to proceed with the series once substituted into the expression.

Discussion Status

The discussion is ongoing, with participants attempting to substitute the series for \(e^z\) into the function and exploring the resulting expression. Some guidance has been offered regarding the location of the Laurent expansion, but there is still a lack of clarity on how to effectively manipulate the series for further progress.

Contextual Notes

There is a specific constraint mentioned regarding the region for the Laurent expansion, which is \(0 < |z| < 2\pi\). Participants are also grappling with the implications of this constraint on their calculations.

Warr
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Just wondering where to go with this one..

calculate the laurent series of [tex]\frac{1}{e^z-1}[/tex]

don't even know where to start on it

I know [tex]e^z={{\sum^{\infty}}_{j=0}}\frac{z^j}{j!}[/tex]

but not much else...
 
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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?
 
Well, I only assumed that I knew that the expansion of e^z was about 0. It only specifies "calculate the laurent expansion of [tex]\frac{1}{e^z-1}[/tex] for [tex]0 < |z| < 2\pi[/tex]"
 
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I tried that but couldn't really come up with anything..

[tex]\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!}+...)}[/tex]

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...
 
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