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Laurent series

  1. May 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Calculate a Laurent series about z = 0 for 1/(z2(z+i)) in the region D = {z: |z| < 1}

    2. Relevant equations



    3. The attempt at a solution
    I used partial fractions to get 1/(z2(z+i)) = 1/z -1/z2 - 1/(z+i) but where do I go from here.
     
  2. jcsd
  3. May 29, 2010 #2

    HallsofIvy

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    Expand [tex]\frac{1}{z+i}[/tex] in a McLaurin series (Taylor's series about z= 0) and multiply each term by [itex]z^{-2}[/itex].

    You can get the McLaurin series most easily by writing [tex]\frac{1}{z+i}= -i\frac{1}{1- (-z/i)}[/tex][tex]=-i\frac{1}{1- (iz)}[/tex] and writing it as a geometric series.
     
  4. May 29, 2010 #3
    I let t = z
    So 1/(t+i) = 1/t(1+i/t) which is a geometric series
    = 1/t(1-(i/t)+(-1/t)^2-...
    = 1/t - i/t2 + i2/t3 + ...
    = 1/z - i/z2 - 1/z3 + ...
     
    Last edited: May 30, 2010
  5. May 30, 2010 #4
    How do I find the geometric series for 1/z and -i/z2
     
  6. May 30, 2010 #5

    vela

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    That series doesn't converge for |z|<1.

    You don't. The 1/z and 1/z2 terms are already in the correct form.
     
  7. May 30, 2010 #6
    Cheers.
     
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