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Cauchy Integral Formula

  1. Apr 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluating using CIF.

    |z| = 4

    Integral [itex]{\frac {{{\rm e}^{2\,iz}}{\it dz}}{ \left( 3\,z-1 \right) ^{2}}}[/itex]



    3. The attempt at a solution

    So the singularity here is z = 1/3 which is inside the circle.

    Therefore using the formula [itex]2\,i\pi \,f[/itex] and substituting in the z = 1/3

    We get f(1/3) = exp(2/3i)

    So I get the answer of 2*Pi*i(exp(2/3i))

    However the answer given is (-4Pi/9)(exp(2/3i))

    I'm thinking it has something to do with the repeated root, but I'm not sure.
     
  2. jcsd
  3. Apr 28, 2012 #2
    Found out that when dealing with powers, a generalised CIF has to be used. Now however I get -4Pi(exp(2/3i))

    Where did that 1/9 come from?
     
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