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Thread: Cauchy Integral #2

  1. #1
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    Cauchy Integral #2

    Evaluate the following closed contour integral

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

    Since $\displaystyle z =1 $ is inside $\displaystyle |z| =2 $

    Let $\displaystyle f(z) = z$

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

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

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

    Is this right?

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

    As $\displaystyle z=2$ is not inside $\displaystyle |z|=1$

    How should I go upon this question?
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  2. #2
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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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