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Laurent series

  1. Apr 6, 2006 #1
    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...
     
    Last edited: Apr 6, 2006
  2. jcsd
  3. Apr 6, 2006 #2

    matt grime

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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 in to the expression and play around with it?
     
  4. Apr 6, 2006 #3
    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]"
     
    Last edited: Apr 6, 2006
  5. Apr 6, 2006 #4
    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....
     
    Last edited: Apr 6, 2006
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