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Contour Integration: finding residues

  1. Aug 27, 2010 #1
    Hi, I'm stuck on this problem:

    [tex]\int{\frac{1}{z^4+1}}[/tex]

    Writing it as a product of its roots, we get:

    [tex]\frac{1}{(z-e^{\frac{i\pi}{4}})(z-e^{\frac{3i\pi}{4}})(z-e^{\frac{5i\pi}{4}})(z-e^{\frac{7i\pi}{4}})}[/tex]

    Then applying Cauchy's residue theorem for simple poles:
    [tex]\mbox{Res}(f,c)=\lim_{z\rightarrow c}(z-c)f(z)[/tex]

    It's here that I'm stuck - I've got the poles and the function, how do I get the residues in this case?
     
  2. jcsd
  3. Aug 27, 2010 #2

    Office_Shredder

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    For the limit, the (z-c) that you're multiplying by will cancel with the (z-c) in the denominator and you can just plug in the value for the limit. However, you kind of have a mess left to multiply out. It's a lot easier to use the fact that if [tex]f(z)=\frac{g(z)}{h(z)}[/tex] where h(z) has a simple zero at c, the residue at c is [tex]\frac{g(z)}{h'(c)}[/tex] if you've seen this before
     
  4. Sep 1, 2010 #3
    Thanks for that.

    Trying both ways, I come out with the same residue for [tex]c = e^{\frac{i\pi}{4}}[/tex]
    of [tex]\frac{1}{-2\sqrt{2}+2i\sqrt{2}}[/tex]

    But the answer I'm looking for is [tex]\frac{1}{4i\sqrt{i}}[/tex] because that's what you get when you expand the Laurent series.

    So any ideas where I've gone wrong?

    Thanks :)
     
  5. Sep 1, 2010 #4

    HallsofIvy

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    Have you considered expanding [itex]4i\sqrt{i}[/itex] in the form a+ bi? You might find that those are the same.
     
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