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Homework Help: Laurent Series Expansion Centered on z=1

  1. Jul 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Find a Laurent Series of f(z)=[itex]\frac{1}{(2z-1)(z-3)}[/itex] about the point z=1 in the annular domain [itex]\frac{1}{2}<|z-1|<2[/itex].

    2. Relevant equations

    3. The attempt at a solution
    By partial fraction decomposition, [itex]f(z)=\frac{1}{(2z-1)(z-3)}=\frac{1}{5}\frac{1}{z-3}-\frac{2}{5}\frac{1}{2z-1}[/itex].

    When [itex]\frac{1}{2}<|z-1|<2, \frac{(z-1)}{2}<1[/itex].

    Hence: [itex]\frac{1}{z-3}=-\frac{1}{3-z}=-\frac{1}{3-z+1-1}=-\frac{1}{2-(z-1)}=-\frac{1}{2}\frac{1}{1-\frac{(z-1)}{2}}[/itex]

    Therefore: [itex]\frac{1}{z-3}=-\sum_{n=0}^{\infty}\frac{(z-1)^{n}}{2^{n+1}}[/itex]

    And: [itex]f(z)=-\frac{1}{5}\sum_{n=0}^{\infty}\frac{(z-1)^{n}}{2^{n+1}}-\frac{2}{5(2z-1)}.[/itex]

    Is this complete? Is there something I could similarly do with the other term? Ive tried but havent been able to find anything that works.
     
    Last edited: Jul 16, 2012
  2. jcsd
  3. Jul 16, 2012 #2
    I would try explicitly calculating a few terms of the taylor series of the second term and try seeing a pattern.
     
  4. Jul 16, 2012 #3
    It's this, right? [itex]\frac{1}{2z-1}=\sum^{\infty}_{n=0}(-1)^{n+1}(z-1)^{n}*2^{n}[/itex]

    So the expansion would become: [itex]f(z)=-\frac{1}{5}\sum^{\infty}_{n=0}(z-1)^{n}[(-2)^{n+1}-2^{-n-1}][/itex]
     
  5. Jul 16, 2012 #4
    I'm not too sure about the n+1 on top of the -1. I'm also not too sure whether there should be a factorial there somewhere.
     
  6. Jul 16, 2012 #5
    You're right about the -1. It should be raise to n, not n+1. I mistakingly thought the index started at 1 instead of 0.
    I'm pretty sure the coefficient of the (z-1) term is a multiple of 2 in this case.
     
  7. Jul 16, 2012 #6
    You're correct, sorry for that. I was forgetting to divide by n! in the formula for a_n of a taylor series.
     
  8. Jul 16, 2012 #7
    No biggie. So is the last expression the answer then?
     
  9. Jul 16, 2012 #8
    I still think there are a couple of sign and index errors. Try going through your algebra again to find them (or to prove me wrong).
     
  10. Jul 16, 2012 #9
    I think I figured the other fraction out:

    [itex]\frac{1}{2}<|z−1|<2,\frac{1}{2(z-1)}<1.[/itex]

    Hence: [itex]\frac{1}{2z-1}=\frac{1}{2z-1-1+1}=\frac{1}{2(z-1)+1}=\frac{1}{2(z-1)}\frac{1}{1+\frac{1}{2(z-1)}}=\frac{1}{2(z-1)}\frac{1}{1-(-\frac{1}{2(z-1)})}=\frac{1}{2(z-1)}\sum^{\infty}_{n=0}(-1)^{n}[2(z-1)]^{-n}=\sum^{\infty}_{n=0}(-1)^{n}[2(z-1)]^{-n-1}[/itex]

    That looks much better. I hope.
     
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