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Cauchy Integral Formula

  1. Apr 28, 2012 #1
    Cauchy Integral Formula - Multiple Repeated Poles

    1. The problem statement, all variables and given/known data

    C.I.F is doing my head in.

    Evaluate [itex]∫ {\frac {{z}^{2}+1}{ \left( z-3 \right) \left( {z}^{2}-1 \right) }}[/itex]

    For the closed path |z| = 2



    3. The attempt at a solution

    This is a circle of radius 2, with singularities [itex]z = 3, z = -1, z = 1[/itex]

    Since z = 3 falls outside the circle, we focus on the two singularities inside.

    Since the C.I.F is [itex]2\,i\pi \,f \left( z_{{0}} \right) [/itex]

    f(z) here would be [itex]{\frac {{z}^{2}+1}{z-3}}[/itex]

    Substituting z = 1, and z = -1

    We get -1 and -1/2 from f(z)

    So that is -2Pi*i -Pi*i = -3Pi*i

    However the answer given is -i*Pi/2

    I have been stuck on this for a while and I can't figure out what I'm doing wrong.
     
    Last edited: Apr 28, 2012
  2. jcsd
  3. Apr 28, 2012 #2
    answer is -i*pi/2.why are you leaving the factor z^2-1 in denominator.at z=1 residue is -1/2 and at -1 it is 1/4 and so the answer.
     
  4. Apr 28, 2012 #3
    I am sorry, I did not quite understand what you meant. I am not leaving the factor (z^2 - 1) in the denominator.

    How did you get those two values (-1/2 and 1/4)?

    When z = 1, isn't it (1^2 + 1)/(1 - 3) = 2/-2 = -1 ?

    z = -1; (-1)^2 + 1/(-1 -3) = 2/-4 = -1/2
     
  5. Apr 28, 2012 #4
    I think I see what you mean, I must also include the factor (z -1) and (z + 1) with (z - 3) every time I evaluate at one of the singularity. Now I get that correct answer.

    Thank you.
     
  6. Apr 28, 2012 #5
    congratulations.
     
  7. Apr 28, 2012 #6
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