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Cauchy Integral Formula

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Using the Cauchy Integral Formula compute the following integrals,where C is a circle of radius 2a centered at z=o, where 2a<pi


    2. Relevant equations

    [tex]\oint\frac{(z-a)e^{z}}{(z+a)sinz}[/tex]

    3. The attempt at a solution
     
  2. jcsd
  3. Feb 10, 2010 #2

    Dick

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    Integrated around |z|=2a? If you write your integrand as f(z)/z (you figure out what f(z) would need to be) the Cauchy integral formula would tell you what the integral is in terms of f(0). Which would be very nice but the Cauchy integral formula doesn't apply because f(z)/z isn't holomorphic inside |z|=2a. Aside from a removable singularity at z=0 there's a pole at z=a. Are you sure you wrote the problem down correctly?
     
  4. Feb 10, 2010 #3
    yes.
    I have done the form like this:
    [tex]\oint\frac{(z-a)e^{z}}{(z+a)}\frac{dz}{sinz}[/tex] + [tex]\oint\frac{(z-a)e^{z}}{sinz}\frac{dz}{(z+a)}[/tex]
    however the first one is not the standard Cauchy Integral Formula
     
    Last edited: Feb 10, 2010
  5. Feb 10, 2010 #4

    Dick

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    I really don't understand what you are saying there. Did you alter the original problem statement? If so, what was the original?
     
  6. Feb 10, 2010 #5
    The original problem is this:
    [tex]\oint\frac{(z-a)e^{z}}{(z+a)sinz}[/tex]dz c=2a centered at z=0 2a<pi

    we can express the integral around the contour as the sum of the integral around z1 and z2 where the contour is a small circle around each pole. Call these contours C1 around z1 and C2 around z2.

    So I am tying to express this original problem like:
    [tex]\oint\frac{(z-a)e^{z}}{(z+a)}\frac{dz}{sinz}[/tex] + [tex]\oint\frac{(z-a)e^{z}}{sinz}\frac{dz}{(z+a)}[/tex]
     
  7. Feb 10, 2010 #6

    Dick

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    Ok, so you are really doing as a residue theorem problem, not just a Cauchy integral problem. In one of those integrals you should multiply by (z-0) and let z approach 0 and in the other one multiply by (z+a) and let z approach -a, right?
     
  8. Feb 10, 2010 #7
    Well,thanks. Could you please give a quick explanation of Cauchy Residue Theorem?
     
  9. Feb 10, 2010 #8

    Dick

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    You should probably look it up. I don't necessarily explain things that well. The Cauchy Integral Theorem just says f(a)=(1/(2*pi*i)) times the contour integral f(z)/(z-a) over a circle where f(z) is holomorphic. The residue theorem is the obvious generalization of that to the case where you have multiple poles in a single domain and you cut out a circle around each one and add them up. Which is what you are doing.
     
  10. Feb 10, 2010 #9
    Thanks, I have worked it out
     
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