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Probably easy Laurent expansion question

  1. May 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the Laurent expansion of f(z)= 1/(z^3 - 6z^2 + 9z) in the annulus |z-3|>3.

    2. Relevant equations

    none

    3. The attempt at a solution

    I've been spending way too long on this problem.. I can't seem to think of a way to manipulate f to use the geometric series, other than getting something like 1/(1-(-3/z-3)) and arguing |z-3|>3>-3. Unless there's an other, better way to do this problem, could someone just let me know if I'm on the right track?
     
  2. jcsd
  3. May 17, 2012 #2

    micromass

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    Start with a partial fraction decomposition.
     
  4. May 17, 2012 #3
    Okay.. so now I have:

    ((1/9z) - (1/9(z-3)) + 1/(3(z-3)^2))(1/z)

    I'm probably missing something completely obvious, but I'm still stuck at where I was before..
     
    Last edited: May 17, 2012
  5. May 18, 2012 #4

    HallsofIvy

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    The only thing you are missing is that you almost have the answer! A Laurent expansion of f(x) "in the annulus |z-3|>3" is a power series in terms of powers of x- 3. You will still need to expand 1/z= 1/((z-3)+ 3).
     
  6. May 18, 2012 #5
    Okay, so I expand 1/z = 1/((z-3)+3) = (1/(z-3)) * (1/(1-(-3/(z-3))) =

    1 - 3/(z-3) + 9/(z-3)^2 - 27/(z-3)^3 + ...

    Is this right? My issue keeps coming up with a way to use the geometric series while keeping |c|<1 (if 1/(1-c) = 1+c+c^2+...)

    I think the answer requires that c= (z-3)/3 ..how does that make sense?
     
    Last edited: May 18, 2012
  7. May 18, 2012 #6
    Maybe there is no Laurent expansion since the series doesn't converge?
     
  8. May 19, 2012 #7

    vela

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    You forgot the factor of 1/(z-3) you pulled out front.

    No, you have it backwards. You're in the region where |z-3|>3, right? If you divide that inequality by 3 on both sides, you get |(z-3)/3| > 1, so with your choice for c, you'd have |c|>1, which is what you don't want. You want z-3 in the denominator so the terms get smaller.
     
  9. May 19, 2012 #8
    Ah, that's what I thought, I was just doubting myself.. thanks so much for the help!
     
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