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Math Help - Cauchy Integral Formula

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

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

    \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: z_{0} is interior to C and z_{0} is exterior to C

    Case 1:

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

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

    and

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

    and obviously:

    \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 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

    \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 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 C (and inside) when 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 z_{0} is outside the countour we can apply the Cauchy-Goursat Theorem and so

    \int_{C} \frac{f'(z)}{z-z_{0}} = \int_{C} \frac{f(z)}{(z-z_{0})^2}<br />
 = 0

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