1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook