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Need help with complex integration

  1. Feb 13, 2007 #1
    I am trying to teach myself some complex analysis. I am using Complex Numbers by Churchill & Brown as my reference. I have reached the integration section and I am encountering certain difficulties.
    For e.g. I have this problem:
    [tex]\oint \frac{dz}{z^2 - z -2}, |z|\leq 3[/tex]

    I can split up the integrand using partial fractions but I don't know how to interpret the boundary conditions that they have given.
    Cauchy's integral formula gives:
    [tex]\int_C
    \frac{f(z)dz}{z - a} = 2\pi i f(a)[/tex]
    Where f(z) is analytic and single valued within a closed curve 'C' and 'a' is any point interior to C.

    Cauchy-Goursat theorem states:
    [tex]\int_C f(z)dz = \int_a^b f[z(t)] z'(t) dt , a\leq t \leq b [/tex]

    I don't know whether I should use the Cauchy-integral formula or the Cauchy-Goursat theorem.

    Can someone help me out here?
     
    Last edited: Feb 13, 2007
  2. jcsd
  3. Feb 13, 2007 #2

    Dick

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    It sure looks like a contour integral to me. They have GOT to mean integrate around the circle |z|=3, but they should also specify direction. In which case the Cauchy integral formula looks fine. Besides, you are the teacher! You get to decide what it means. Or are you trying to reconcile with a given answer?
     
    Last edited: Feb 13, 2007
  4. Feb 14, 2007 #3
    Yes, it is a contour integral. This question appeared in one of the past exam papers of my university. I wonder if it is a misprint when it should have been |z|= 3. Cauchy's integral formula doesn't seem to fit in here since I don't have a "z-a" term in the denominator. I will out this problem with the other theorem and post my solution ASAP.

    BTW, I am still a student and don't have any given solutions. :wink:
     
  5. Feb 14, 2007 #4

    Dick

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    I thought you said that you were going to use partial fractions to split up the integral? That would give you something to feed to Cauchy formula. I'm not really sure what I think of as Cauchy-Goursat (which isn't what you wrote!) will help you. Anyway, it's really just a residue problem - but I guess you have'nt gotten there yet.
     
  6. Feb 14, 2007 #5
    Thanks for looking into this. This what I get using partial fractions.

    [tex]\frac{1}{z^2 - z -2} = \frac{1}{(z+1)(z-2)} = \frac{1}{3(z-2)} -\frac{1}{3(z+1)}[/tex]

    This doesn't seem to fit the Cauchy Integral or should I use some manipulation like multiplying and dividing by (z-3)??
     
    Last edited: Feb 14, 2007
  7. Feb 14, 2007 #6

    Dick

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    Why do you say it doesn't fit. You have two terms - each one fits. ??
     
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