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Method of residues

  1. Aug 18, 2012 #1
    1. The problem statement, all variables and given/known data

    evaluate using the method of residues [itex]\int^{\pi}_{-\pi}[/itex] [itex]\frac{d\theta}{1+sin^2\theta}[/itex] (=[itex]\pi\sqrt{2})[/itex]

    2. Relevant equations



    3. The attempt at a solution [itex] sin^2 \theta = \frac{1}{2i}(z^2 - \frac{1}{z^2}) [/itex] so our integral becomes [itex]\int \frac{2iz^2}{z^4+2iz^2-1}\frac{dz}{iz} [/itex]

    but before i continue on from here im not sure what the story with the limits. Normally i've just seen limits of 0 to 2[itex]\pi[/itex] so should iI put a [itex]\frac{1}{2}[/itex] in from of the integral?
     
  2. jcsd
  3. Aug 18, 2012 #2
    sorry got that wrong [itex] sin^2 \theta = (\frac{1}{2i}(z - \frac{1}{z}))^2 [/itex] should be [itex] \int \frac{-4z^2}{z^4-2z^2+1}\frac{dz}{iz}[/itex] but still unsure about my limits....
     
  4. Aug 18, 2012 #3
    As theta varies from -pi to pi, how much of the unit circle does it sweep out?
     
  5. Aug 18, 2012 #4
    I think its Half a sweep, thats why i think i put a half in front of the integral
     
  6. Aug 18, 2012 #5
    Are you sure?
     
  7. Aug 18, 2012 #6
    a whole sweep is 2[itex]\pi[/itex], half a sweep is o to [itex]\pi[/itex] so is [itex]\pi[/itex] to -[itex]\pi[/itex] half a sweep...backwards? so i put -[itex]\frac{1}{2}[/itex] in front of it?
     
  8. Aug 18, 2012 #7
    not sure...
     
  9. Aug 18, 2012 #8
    What is pi - (-pi)?
     
  10. Aug 18, 2012 #9

    uart

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    Science Advisor

    ^ Also, double check your denominator there, you've made a mistake in the algebra.

    When you sort that out you'll find it comes out fairly easy, four real roots with two of them inside your contour.
     
  11. Aug 18, 2012 #10
    2[itex]\pi[/itex] thanks :)
     
  12. Aug 18, 2012 #11
    [itex]\int \frac{-4z^2}{z^4-2z^2+2}\frac{dz}{iz} [/itex]

    thanks a million
     
  13. Aug 18, 2012 #12
    **** thats not right either
     
  14. Aug 18, 2012 #13
    Now i'm getting [itex] \int \frac{-4z^2}{z^4-6z^2+2}\frac{dz}{iz} [/itex] which doesnt work out that easy. So im guessing I got it wrong again. im going to use formula and see what it throws up anywho.thanks guys. Do either of you have a clue what my other post (jordans lemma) is about? i havent a clue...
     
  15. Aug 18, 2012 #14
    I think you need to double-check your algebra, one more time.
     
  16. Aug 18, 2012 #15
    yes your right, it should be [itex] \int \frac{-4z^2}{z^4-6z^2+1}\frac{dz}{iz} [/itex]
     
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