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Lauarent Series of 1/(1+z^2)

  1. Nov 6, 2007 #1
    Find the Laurent series that converges for [tex]0 < | z - i| < R[/tex] of

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

    I have been given the hint to break it up as

    [tex]\frac {1}{1 + z^2} = (\frac {1}{z - i})(\frac {1}{z + i})[/tex] and then expand [tex]\frac {1}{z + i}[/tex] . I am kind of confused about this, because the series is centered at $i$. I'm not exactly sure how to do it because the center isn't 0.

    The solution is -[tex]\sum_{n = 0}^{\infty}(\frac {i}{2})^{2n + 1}(z - i)^{n - 1}[/tex]
     
  2. jcsd
  3. Nov 6, 2007 #2

    Avodyne

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    You could let z = w+i, and expand in powers of w. Then, at the end, set w = z-i.
     
  4. Nov 6, 2007 #3

    HallsofIvy

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    You learned a long time ago that the general Taylor's series (as opposed to Maclaurin series) is about "x= a" rather than "x= 0". You want to expand [tex]\frac{1}{z+i}[/itex] around z= i: in powers of (z- i). You could do that by taking derivatives and doing an actual Taylor's series expansion, evaluating the derivatives at z= i rather than at z= 0.
     
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