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Math Help - Cauchy Integral #2

  1. #1
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    Cauchy Integral #2

    Evaluate the following closed contour integral

    1)
    \oint\frac{zdz}{z-1}
    with  |z|=2

    Since  z =1 is inside  |z| =2

    Let f(z) = z

    And by the Cauchy Integral formula
     \oint\frac{f(z)}{z-z_0} = 2\pi i f(z_0)

    So
     \oint\frac{z}{z-z_0} = 2\pi i f(z_0)
     z_0 = 1

    Thus
     2\pi i * f(z_0) = 2 \pi i

    Is this right?

    But in this following question, I have a problem
     \oint\frac{zdz}{z-2}
    with |z|=1

    As z=2 is not inside |z|=1

    How should I go upon this question?
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    The solution to 1) is correct. For the second problem, the function z/(z–2) is analytic everywhere inside the circle |z|=1, so by Cauchy's theorem the integral is 0.
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