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: Use geometric series to find the Laurent series

  1. May 10, 2012 #1
    Use geometric series to find the Laurent series for f (z) = z / (z - 1)(z - 2) in each annulus
    (a) Ann(1,0,1)
    (b) Ann(1,1,∞)

    a= center, r=smaller radius, R=larger radius

    My attempt:
    f(z)= -1/(z-1) + 2/(z-2)
    geometric series: Σ[n=0 to inf] z^n - 1/2 Σ[n=0 to inf] (z/2)^n

    2/(z-2)=-2(1/(2-z))=-2(1/(1-(z-1)))=-2Σ[n=0 to inf](z-1)^n
    f(z)=-1/(z-1)-2Σ[n=0 to inf](z-1)^n=-(z-1)^(-1)+Σ[n=0 to inf]-2(z-1)^n

    -1/(z-1)=-Σ[n=0 to inf](1/(z-1))^n
    2/(z-2)=2/(z-1-1)=2(1/(z-1))(1/(1-1/(z-1)))=2/(z-1)Σ[n=0 to inf](1/(z-1))^n=2Σ[n=0 to inf](z-1)^-(n+1)
    f(z)=-Σ[n=0 to inf](1/(z-1))^n+Σ[n=-inf to -2]2(z-1)^-(n+1)

    Are my geometric series correct? (I feel good about the second one but am unsure on the first)
    Are my summation bounds in the Laurent series correct? I did something funky in the last one, so I'm not sure. Also, I wasn't sure how to find a Laurent series for -1/(z-1) in part a, so I left it as shown.
    Any help and corrections are appreciated!
  2. jcsd
  3. May 11, 2012 #2
    For ##0<|z-1|<1##, I was able to only do half of it. I can't figure out the other piece unless it is supposed to be just like this.

    \frac{2}{z-2} = -2\frac{1}{1+1-z}=-2\frac{1}{1-(z-1)}=-2\sum_{n=0}^{\infty}(z-1)^n
    f(z) = \underbrace{\frac{-1}{z-1}}_{\text{this one is giving me trouble}} - 2\sum_{n=0}^{\infty}(z-1)^n

    For ##1<|z-1|<\infty##, again I could only figure out one piece of it.

    \frac{2}{z-2} = 2\frac{1}{1-\frac{1}{z-1}}=2\sum_{n=0}^{\infty}\left(\frac{1}{z-1}\right)^n
    f(z) = \frac{-1}{z-1}+2\sum_{n=0}^{\infty}\left(\frac{1}{z-1}\right)^n

    I am not sure how we can contend for ##\frac{-1}{z-1}## in both a and b. Maybe it can be left as is and that is it.
  4. May 11, 2012 #3
    Part a is correct. It's okay to have your series split up like that, you don't have to have a general formula for all of your ##C_n##.

    Okay, part b. ##\frac{-1}{z-1} \neq \displaystyle \sum_{n=0}^{\infty} \left(\frac{1}{z-1}\right)^n##

    Just leave that quantity as it was in part a).

    Now for your second term.

    ##2 \cdot \displaystyle \sum_{n=0}^{\infty} (z-1)^{-(n+1)} \neq 2 \cdot \sum_{-\infty}^{-2} (z-1)^{-(n+1)}##

    I'm not sure how you got that. Remember that when you switch the limits of summation from positive to negative, the sign of only n changes. So,

    ##2 \cdot \displaystyle \sum_{n=0}^{\infty} (z-1)^{-(n+1)} = \sum_{-\infty}^{0} (z-1)^{n-1}##

    Now you can easily change that to ##(z-1)^n##.
    Last edited: May 11, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook