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Complex contour integration problem.

  1. Nov 27, 2011 #1
    1. The problem statement, all variables and given/known data

    i have the closed ∫ (cos(z))/(3z-π)) dz

    2. Relevant equations

    i presume this works closed ∫ f(z)/(z-w) dz = f(w)2πi

    3. The attempt at a solution

    i think the answer is cos(pi)*2πi
     
  2. jcsd
  3. Nov 27, 2011 #2

    Dick

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    The answer is going to depend on whether your contour encloses the pole or not. But even if it does your answer is wrong. You don't have z-pi in the denominator, you've got 3*z-pi. That makes a difference. Write 3*z-pi=3*(z-pi/3). Now what do you get?
     
  4. Nov 27, 2011 #3
    i can't believe i didn't factor. now i get 1/3*pi*i
     
  5. Nov 27, 2011 #4

    Dick

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    That's better. But the answer still depends on the closed contour. I'd feel better if you described what it was.
     
  6. Nov 27, 2011 #5
    1. evaluate each integral around the counter-clockwise circle lzl =1 repeat for lzl=2.

    (a) ∮ cos(z)/(3z-pi) dz, (b) ∮ z /((9-z^2)*(2z-i)) dz

    thats all it says. i think that the pole may be not be significant
     
  7. Nov 27, 2011 #6

    Dick

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    I think it's very significant. The pole in (a) is at z=pi/3. That's outside of the counter-clockwise circle |z|=1. What's the answer in that case?
     
  8. Nov 27, 2011 #7
    so that would mean that the integrand is analytic in the interior and on the contour, so it would equal 0?
     
  9. Nov 27, 2011 #8

    Dick

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    Yes! Now continue with the other parts and pay attention to which poles are inside of the contour.
     
  10. Nov 27, 2011 #9
    kk thanks alot
     
  11. Nov 27, 2011 #10
    so being aware of the poles, in the case of (b) we have poles at z=-3,3,i/2. but the only one that matters is the i/2 because it is the only one inside the contour line. my only problem is now trying to get ∮ z/((9-z^2)(2z-i)) dz into the form f(w) ∮ 1/(z-w) dz
     
  12. Nov 27, 2011 #11

    Dick

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    Split the expression 1/((9-z^2)(2z-i)) up using partial fractions. Remember that trick? Or even simpler, just put f(z)=z/(9-z^2) and make the expression f(z)/(2z-i).
     
  13. Nov 27, 2011 #12
    i did both, and the partial fractions is a little more ugly. but where i became stuck, is that i began to wonder if i had to do it twice the other one being where f(z) = z/(2z-i)
     
  14. Nov 27, 2011 #13

    Dick

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    I'm not sure where the stuckness comes in. f(z)=z/(9-z^2) is analytic inside of both |z|=1 and |z|=2, isn't it? Just do f(z)/(2z-i). Just apply your theorem to that.
     
  15. Nov 28, 2011 #14
    i guess my confusion is whether we can set f(z) = z/(9-z^2) OR f(z)= z/(2z-i). but i'd much rather just do z/(2z-i). so i suppose my question is just inquisitive. but i could see it expanded out to something like ∮ 1/(z^2-z+2) dz for lzl=1. in this case it isn't so easy to find an f(x) to set up the nice form of 2*pi*i f(w) from f(w) ∮ 1/(z-w) dz
     
  16. Nov 28, 2011 #15
    actually upon further thinking of my question, it would have to have bounds ∮ 1/(z^2-z+2) dz for lzl=1 would be 0 but if i were to extend the radius out to lzl =2 then it gets more interesting
     
  17. Nov 28, 2011 #16

    Dick

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    How did you get from z/((9-z^2)*(2z-i)) to 1/(z^2-z+2)??
     
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