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Cauchy Integral

  1. May 8, 2009 #1
    Hi,

    I'm stuck on this question:

    Use the Cauchy's integral to show [tex]\oint[/tex]cos(z) / [ z (z2 +9) ] dz = 2[tex]\pi[/tex]i/9

    C is within the square defined by x= +/- 2 , y = +/- 2

    I can't see how to get the denominator in the form (z-z0) to apply the theorem. Any help would be much appreciated!
    [
     
  2. jcsd
  3. May 8, 2009 #2

    malawi_glenn

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    can you write z^2 + 9 as something of the form (z-z0) ?
     
  4. May 8, 2009 #3

    dx

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    You must specify what the path of integration is. There are singularities at z = 0, and z = ±3i.

    The answer you gave is correct if the contour encloses only z = 0, and is traversed in the counter clockwise direction.

    To find the residue at z = 0, you multiply the integrand by (z - 0) = z and evaluate the result at z = 0. If you do this, you will get 1/9.
     
  5. May 8, 2009 #4
    Not sure, but if you split the fraction up into partial fractions its possible to get an integral of that form. Just not sure how to get the coefficents or indeed if its even the correct method.
     
  6. May 8, 2009 #5
    Sorry, the path is around the square with sides at x= +/- 2 and y= +/-2 .
     
  7. May 8, 2009 #6

    malawi_glenn

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    hint: Yes you can do it, read dx's post - you also need to specify what the integration contour is
     
  8. May 8, 2009 #7
    Sorry, I don't understand. How does that get it in the form f(z)dz / [z-zo] ?
     
  9. May 8, 2009 #8

    dx

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    First of all, what is z0? And why are you trying to get it into that form?
     
  10. May 8, 2009 #9

    malawi_glenn

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    the strategy is to find all points in the complex plane where your integrand is singular, then you find out which one are lying inside your contour of integration.

    Can you please write down the entire Cauchy's integral Theorem for us so we see that you are trying.
     
  11. May 8, 2009 #10
    cauchys integral is : [tex]\int[/tex]f(z)dz / [z-z0] = 2[tex]\pi[/tex]i f(z0) . I am trying to use this result to prove the statement.
     
  12. May 8, 2009 #11

    malawi_glenn

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    you have still not said anything what z0 is ... is it something special about it?
     
  13. May 8, 2009 #12
    z0 is the value of the singularity.
     
  14. May 8, 2009 #13

    malawi_glenn

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    so which possible singularities do your integrand posses (hint: there are three such) and which one is closed within your contour??
     
  15. May 8, 2009 #14

    malawi_glenn

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    One does not need to know the Residue theorem to evaluate this for single poles, in many textbooks Cauchy's formula is introduced earlier than the more general Residue theorem...
     
  16. May 8, 2009 #15

    dx

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    Oh, right. Deleted my prev. post.
     
  17. May 8, 2009 #16
    only the singularity at z=0 is enclosed, with the others being at +/- 3i. But i still cant see how to get it inot the correct form... If i could split up into multiple integrals, one with the singularity in the cauchy form and one without a singularity then i could make use of the fact that the integral of fz is zero if it doesnt enclose a singularity. But I'm not sure how to split it in this way...
     
  18. May 8, 2009 #17

    Cyosis

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    As hinted previously, to find all singularities write [itex]z^2+9=(z-a)(z+a)[/itex]. The value a should be easy to find.
     
  19. May 8, 2009 #18

    malawi_glenn

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    oh my, then you are really in trouble, i don't even know how to explain such simple thing xD

    if you have cos(z)/[ z(z^2 + 9) ] and you want the the form f(z)/(z-0), can you GUESS??
     
  20. May 8, 2009 #19
    No. Has it got anything to do with partial fractions?
     
  21. May 8, 2009 #20

    dx

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    There's already a z in the denominator. Just replace it with (z - 0).
     
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