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Need help finding a Laurent Series

  1. Apr 6, 2014 #1
    1. The problem statement, all variables and given/known data

    Let [itex]f(z) = \frac{1}{z^2-1}[/itex]. Find Laurent Series valid for the following regions.

    • 0<|z−1|<2
    • 2<|z−1|<∞
    • 0<|z|<1

    2. Relevant equations

    [tex]\frac{1}{1-z}=\sum^{\infty}_{n=0}z^n,\: |z|<1[/tex]
    [tex]f(z)=\sum^{\infty}_{n=0}a_n(z-z_0)^n+\sum^{\infty}_{n=1}b_n(z-z_0)^{-n}[/tex]

    3. The attempt at a solution

    I really have no idea what to do, especially for the first two regions. I have written the function as
    [tex]f(z)=\frac{1}{(z+1)(z-1)}[/tex] and then attempted to find the laurent series for[itex]\frac{1}{z-1}[/itex] & [itex]\frac{1}{z-1}[/itex], then shifted it to 1<|z|<3 and multiply the two sums together, but I think this the wrong way to do it. It's been a couple days and I still can't figure this out!
     
  2. jcsd
  3. Apr 6, 2014 #2

    micromass

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    Let's start with the first one. You need to find the Laurent series in the region ##0<|z-1|<2##. This means that you need to find a series of the form

    [tex]\sum_{k=-\infty}^{+\infty} a_k (z-1)^k[/tex]

    which converges on the right region.

    Can you first try to express

    [tex]\frac{1}{z+1}[/tex]

    as such a series? What do you get? Hint:

    [tex]\frac{1}{z+1} = \frac{1}{2 + (z-1)} = \frac{1}{2} \frac{1}{1 + (z-1)/2}[/tex]

    Can't you just multiply the series with the factor ##\frac{1}{z-1}## to get the final answer?
     
  4. Apr 6, 2014 #3
    i'm still quite confused. I don't really remember from calculus how to get series expansions that aren't around z=0.

    I tried [tex]\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z-1}{2}\right)^n=\sum_{n=0}^{\infty}\frac{(-1)^n(z-1)^n}{2^{n+1}}[/tex]
     
  5. Apr 6, 2014 #4

    micromass

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    Yep, that is exactly right. But you should always mention for which ##z## this series expansion is valid.

    Basically, you have the series

    [tex]\frac{1}{1-a} = \sum a^n[/tex]

    but this is only valid for ##|a| < 1##. You applied this series expansion to ##a = -(z-1)/2##, which is fine, but that only works for

    [tex]|(z-1)/2|<1[/tex]

    or ##|z-1|<2##.

    Ok, now what about

    [tex]\frac{1}{(z-1)(z+1)}[/tex]

    Just multiply the series you have with the factor ##1/(z-1)## and you're finished!
     
  6. Apr 6, 2014 #5
    so just [tex]f(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n}(z-1)^{n-1}}{2^{n+1}},\: |z-1|<2[/tex]
     
  7. Apr 6, 2014 #6
    is this correct?

    for 2 < |x-1|< ∞ I got
    [tex]\sum^\infty_{n=0}\frac{2^n(-1)^n}{(z-1)^{n+2}}[/tex]
    and for 0 <|z| < 1 I got
    [tex]\sum^\infty_{n=0}-z^{2n}[/tex]
     
  8. Apr 6, 2014 #7

    micromass

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    Fine, but here you also want ##0<|z-1|## since it isn't defined in ##1##.

    So, now about ##2<|z-1|<+\infty##. You can't apply the previous series decomposition anymore. But this time you do know that

    [tex]\left|\frac{2}{z-1}\right|<1[/tex]

    So that means that this time you can use that

    [tex]\frac{1}{1 - 2/(z-1)}= \sum \left(\frac{2}{z-1}\right)^n[/tex]

    so you must use this somehow.
     
  9. Apr 6, 2014 #8

    micromass

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    That seems ok.
     
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