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Thread: Cauchy Integral Formula

  1. #1
    Member Haven's Avatar
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    Cauchy Integral Formula

    Show that if $\displaystyle f$ is analytic within an on a simple closed contour C and $\displaystyle z_{0}$ is not on C, then

    $\displaystyle \int_{C} \frac{f'(z)}{z-z_{0}} = \int_{C} \frac{f(z)}{(z-z_{0})^2}$

    I've broken this down into two cases: $\displaystyle z_{0}$ is interior to C and $\displaystyle z_{0}$ is exterior to C

    Case 1:

    Since $\displaystyle z_{0}$ is interior to C we may apply the Cauchy Integral Formula so

    $\displaystyle f'(z_{0})*2\pi\\i = \int_{C} \frac{f'(z)}{z-z_{0}} $

    and

    $\displaystyle f'(z_{0})*2\pi\\i = \int_{C} \frac{f(z)}{(z-z_{0})^2}$

    and obviously:

    $\displaystyle \int_{C} \frac{f'(z)}{z-z_{0}} = \int_{C} \frac{f(z)}{(z-z_{0})^2}$

    However I have no idea what to do when $\displaystyle z_{0}$ is exterior to C. We're only given that f is analytic in the closure of C. I was hoping to use contour deformation or the Cauchy-Goursat Theorem but i don't know how.
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  2. #2
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    Quote Originally Posted by Haven View Post

    $\displaystyle \int_{C} \frac{f'(z)}{z-z_{0}} = \int_{C} \frac{f(z)}{(z-z_{0})^2}$

    However I have no idea what to do when $\displaystyle z_{0}$ is exterior to C. We're only given that f is analytic in the closure of C. I was hoping to use contour deformation or the Cauchy-Goursat Theorem but i don't know how.
    Arent' the integrands both analytic on $\displaystyle C$ (and inside) when $\displaystyle z_0$ is outside the contour? Then what?
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  3. #3
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    So if I'm getting this right. Since both integrands are analytic when $\displaystyle z_{0}$ is outside the countour we can apply the Cauchy-Goursat Theorem and so

    $\displaystyle \int_{C} \frac{f'(z)}{z-z_{0}} = \int_{C} \frac{f(z)}{(z-z_{0})^2}
    = 0$

    Right?
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  4. #4
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    That's what I'd say.
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