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Laurent series

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculate the laurent series expansion about he points specified, classify the singularity and sate the region of convergence for.

    [itex]\frac{1}{z^2 - 1}[/itex] at (i) z=1 (ii) z=-1 (iii)z=0

    2. Relevant equations



    3. The attempt at a solution

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

    i observe that f(z) is analytic in {z≠ -1,1} so is anyalytic in -1 < [itex]\left|z\right| < 1[/itex]

    @z=1 [itex]\frac{-1}{2}\frac{1}{1-(-z)} = \frac{-1}{2}\sum (-z)^n[/itex]

    similarly

    @z=-1 [itex]\frac{1}{2}\frac{1}{(z-1)} = \frac{-1}{2}\frac{1}{1-z)} = \frac{-1}{2}\sum (z)^n[/itex]

    and

    @z=0 is it just the sum of both of them? singularities are at 1, -1 and region of convergence -1 < [itex]\left|z\right| < 1[/itex]

    am i anywhere near?
     
  2. jcsd
  3. May 12, 2012 #2
    No. When you expand around a point z0, then the series should be in powers of (z-z0). So for the first one, the first partial term is already in powers of z-1. However for the second term:

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

    that's not. But you can do so by forcing a factor of z-1 in the denominaor by:

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

    which you can then easily express that as a power series in terms of (z-1) right? Same dif with the other ones.
     
  4. May 12, 2012 #3
    thanks a million jackmell
     
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