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Integral using cauchy theorem

  1. Feb 9, 2014 #1
    Hello,
    I don't get why using the fact that ∫dz/z = 2*pi*i accros the circle
    This integral gives:

    1/3∫(1/(z-2)-1/(z-1/2))dz = 1/3(-2*pi*i) accross the circle

    Thanks !
     
  2. jcsd
  3. Feb 9, 2014 #2

    Dick

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    Why do you think 1/z has anything to do with (1/3)*(1/(z-2)-1/(z-1/2))??
     
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