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Residue of an integral

  1. Apr 24, 2017 #1
    1. The problem statement, all variables and given/known data
    use the residue theorem to find the value of the integral,
    integral of [itex] z^3e^{\frac{-1}{z^2}} [/itex] over the contour |z|=5
    3. The attempt at a solution
    When I first look at this I see we have a pole at z=0 , because we cant divide by zero in the exponential term.
    and a pole of order 2, So I multiply the function by the function that causes the singularity , and take the first derivative of that and evaluate it at z=0,
    this give me [itex] 2pi*i[3z^2] [/itex] but this gives me 0,
    my book says the answer should be i*pi , which I can find from the Laurent series , but I cant seem to get it using the residue theorem.
     
  2. jcsd
  3. Apr 24, 2017 #2

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    I'm not sure that I understand. Are you directly calculating Res(f,0) = limz->0(z⋅f(z))?
     
  4. Apr 24, 2017 #3
    so the exponential function, does not have a pole or a removable singularity, since it does not have a pole , I think I just have to expand it in a Laurent series to find the residue, im trying to find the value of that integral over the circle |z|=5, and I am trying to use the residue theorem ,
     
  5. Apr 24, 2017 #4

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    I understand that you are trying to use the residue theorem, but what are the details of how you doing that? Are you directly calculating Res(f,0) = limz->0(z⋅f(z))? Or are you trying some other way to find the coefficient of the 1/z term of the Laurent series?
     
  6. Apr 24, 2017 #5
    I expanded the exponential as a Laurent series , then I multiplied by z^3, and then found the 1/z term , other ways wont work because its not a pole ,
     
  7. Apr 24, 2017 #6

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    I'll buy that. It is an essential singularity.
     
  8. Apr 24, 2017 #7
    thanks for helping me clear it up.
     
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