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Polar Integrals

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral qywvtv.png

    2. Relevant equations
    I can substitute s4q5pc.png and thus end up with

    3. The attempt at a solution
    I then expand the denominator out and end up with 1/ kd33ap.png
    However I then assume I need to factorise the top line of that fraction as this will be the denominator in my integral, so I want to factorise this and find where it equals zero and thus where the poles are. However I'm unsure how to do this and need help! Thanks
  2. jcsd
  3. Apr 25, 2012 #2
    Just multiply upstairs and downstairs by z, carry z^2 in downstair inside the brackets, and solve the 2nd order equation.
  4. Apr 25, 2012 #3
    sorry I don't really understand what you are saying!
  5. Apr 25, 2012 #4
    [tex]I = -i \oint_{|z|=1} \frac{dz}{z(5-3(\frac{z-z^{-1}}{2i}))^2} = -i \oint_{|z|=1} \frac{ z dz}{\left[5z-3(\frac{z^2-1}{2i})\right]^2} [/tex]
    and then solve
    [tex] 5z-3(\frac{z^2-1}{2i}) = 0 [/tex]
    (and remember how many of each pole there are!)
  6. Apr 26, 2012 #5
    thanks! I'll give it a go
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