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Laurent Series

  1. Nov 4, 2008 #1
    1. The problem statement, all variables and given/known data
    We have
    [tex]f(z) = \frac{1}{z^2 - 2z - 3}[/tex]
    For this function, we want to find the Laurent Series around z=0, that converges when z=2 and we want to find the area of convergence.

    2. Relevant equations
    [tex]\frac{1}{z-3} = -\frac{1}{3}\bigg( \frac{1}{1 - \frac{z}{3}}\bigg) = -\frac{1}{3}\sum_{n=0}^{\infty}\frac{z^n}{3^n} = -\sum_{n=0}^{\infty}\frac{z^n}{3^{n+1}}[/tex]

    3. The attempt at a solution
    There are two things I'm having trouble with.
    I know that the function is analytic (holomorph) at the disk [itex]D_1 = \{z \in C : |z| < 1 \}[/itex], the annulus [itex]D_2 = \{z \in C : 1 < |z| < 3 \}[/itex] and in the area [itex]D_3 = \{z \in C: 3 < |z|\}[/itex].

    Am I correct when I focus my attention to the area [itex]D_2[/itex]? Since we are interested in the area where f is convergent for [itex]z=2[/itex]?

    I will try to find the Laurent series, and do so by first rewriting the function.
    [tex]f(z) = \frac{1}{4}\bigg(\frac{1}{z-3} - \frac{1}{z+1}\bigg)[/tex]

    My first problem is that for [itex]|z| > 1[/itex] I know that
    [tex]\frac{1}{1+z} = \sum_{n=0}^{\infty}\frac{(-1)^n}{z^{n+1}}[/tex]
    but I don't know why! I can only find information when [itex]|z| < 1[/itex]. Can someone please explain why?

    Regardless, I end up with the Laurent series:
    [tex]f(z) = \frac{1}{4}\bigg( -\sum_{n=0}^{\infty}\frac{z^n}{3^{n+1}} -\sum_{n=0}^{\infty}\frac{(-1)^n}{z^n+1}\bigg)[/tex]

    The second problem is to find the area of convergence. I usually do the ratio test, but I don't really know where to start with this one.

    Any help will be very appreciated!
     
    Last edited: Nov 4, 2008
  2. jcsd
  3. Nov 4, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    The series in your first problem comes from writing 1/(1+z) as (1/z)*(1/(1+1/z)). Now do a geometric series expansion on the second factor. As for determining the region of convergence, a Laurent series generally converges on some sort of an annulus, r<|x|<R. You get the inner radius from testing the negative powers in your expansion and the outer radius from testing the positive powers.
     
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