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Integrate 1/z^2 Over a Closed Curve

  1. Dec 8, 2011 #1
    [tex]\int_\alpha\frac{1}{z^2}dz[/tex]

    I can't figure out how to integrate this over a closed circle which contains the origin on its interior. I'm assuming it is equal to 2πi; is there a way to apply Cauchy's Integral Theorem? If I set f(z)=1/z then that is not analytic on the interior, so I don't see how I can apply the theorem.
     
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  3. Dec 8, 2011 #2

    micromass

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  4. Dec 8, 2011 #3
    oops I mean't cauchy's integral theorem. But that theorem would require that f(z)=1/z by analytic at zero.
     
  5. Dec 8, 2011 #4

    micromass

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    Yeah, you can't use that theorem here...
     
  6. Dec 8, 2011 #5
    oops shoot I meant cauchys integral formula! sorry about that lol. How can I use that? Would f(z) just equal 1?

    Edit: Ah wait so I could use the generalized formula with f(z)=1 and get 2πi, is that right?

    Super Duper Edit: But wait I'm sorry that would give me 0 because the derivative of a constant function is 0, but I'm quite sure it should be 2πi, so something's wrong.
     
    Last edited: Dec 8, 2011
  7. Dec 8, 2011 #6

    Dick

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    It's not 2pi*i. It's 0. Do what you did in the other thread and integrate around the unit circle and put z=exp(it). Change it to an integral dt. Use the cauchy integral theorem on the difference between the unit circle and the contour alpha.
     
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