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Laurent Series Expansion coefficient for f(z) = 1/(z-1)^2

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Determine the coefficients [itex]c_n[/itex] of the Laurent series expansion

    [itex]\frac{1}{(z-1)^2} = \sum_{n = -\infty}^{\infty} c_n z^n[/itex]

    that is valid for [itex]|z| > 1[/itex].

    2. Relevant equations

    none

    3. The attempt at a solution

    I found expansions valid for [itex]|z|>1[/itex] and [itex]|z|<1[/itex]:

    [itex]\sum_{n = 0}^{\infty} \left(n-1\right)z^n, |z|>1[/itex] and

    [itex]\sum_{n = 2}^{\infty} \left(n-1\right)z^{-n}, |z|<1[/itex]

    I know that if I negate the n's in the second equation and change the index of the sum to go from -∞ to -2 I can add them together to get the sum from -∞ to ∞, but I don't know what to do about the missing n=1 term. Any suggestions?
     
    Last edited: Dec 3, 2011
  2. jcsd
  3. Dec 3, 2011 #2

    vela

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    There are two different Laurent series for the function, each one being valid in different portions of the complex plane. You can't add them together because, at a particular value of z, only one will converge.

    You might want to show us your work on how you found the series because they're not correct.
     
  4. Dec 3, 2011 #3
    I just noticed that I missed the part in the problem statement that says valid for [itex] |z|>1 [/itex], so I only need

    [itex] \sum_{n=0}^{\infty}\left(n+1\right)z^{-\left(n+2\right)}[/itex].

    I got that by noticing that [itex] \frac{1}{\left(z-1\right)^2} = \frac{1}{z^2\left(1-\frac{1}{z}\right)^2}[/itex]

    Using the geometric series expansion, [itex]= \frac{1}{z^2}\left(1 + \frac{1}{z} + \frac{1}{z^2} + \frac{1}{z^3} + ...\right)^2
    = \frac{1}{z^2}\left(1 + \frac{2}{z} + \frac{3}{z^2} + \frac{4}{z^3} + ...\right)
    = \frac{1}{z^2} + \frac{2}{z^3} + \frac{3}{z^4} + \frac{4}{z^5} + ...
    =\sum_{n=0}^{\infty}\left(n+1\right)z^{-\left(n+2\right)}, for |z|>1[/itex]

    Now that I have that, I'm not sure how I can extend the indices of the summation so that they match what we were given in the problem statement.
     
  5. Dec 3, 2011 #4

    vela

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    You got it. You don't need to extend the summation. The "missing" terms aren't there because cn=0 for those powers of z.
     
  6. Dec 3, 2011 #5
    Ok, thanks so much for commenting!
     
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