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Remind me how to do this complex integral

  1. Sep 7, 2014 #1
    integrate e^z/(1-cosz) dz over circle of radius, say 2

    i can't seem to recall how it is done.

    singularity at z=0

    2*pi*i * res (at z=0) would be the solution

    any shortcut to find this residue?
     
    Last edited: Sep 7, 2014
  2. jcsd
  3. Sep 7, 2014 #2
    btw, is there any kind of theorem saying integral of 1/(1-sinz) dz over circle of radius 2 would be zero? this one has 2 singularities at equal distance apart. i feel like there's a theorem somewhere. anymore care to refresh my memory?
     
  4. Sep 8, 2014 #3
  5. Sep 8, 2014 #4

    WWGD

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    The only one I know is the one where you use the sum of the residues at each of the singularities, so it would integrate to 0 only if the residue at one singularity is the negative
    of the other one.
     
  6. Sep 8, 2014 #5
    nvm what i said before. the only singularity inside our contour is z = 0. rest are outside and don't even matter.

    and yes

    integral = 2*pi*i * [sum of residues at each singularities]
     
  7. Sep 8, 2014 #6
    i've tried several methods (that i can think of) for this problem and none of them seem to work.

    also tried expanding as series and didn't get anywhere.
     
  8. Sep 8, 2014 #7

    WWGD

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    If you have done the expansion, the residue is the coefficient of 1/z , since all other terms integrate to 0.
     
  9. Sep 8, 2014 #8
    why are you integrating?

    this is what i was doing

    e^z = 1+z+z^2/2 + O(z^3)
    1-cosz = z^2/2-z^4/4! + O(z^6)

    i couldn't really single out 1/z part
     
  10. Sep 8, 2014 #9

    WWGD

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    No I'm not integrating, just stating that if you did term-by-term integration of the series, the only term that would be nonzero is the term containing 1/z. Let me think it thru a bit more.
     
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