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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
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