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Thread: Cauchy Integral

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    Evaluate $\displaystyle \oint_{C} \left(\frac{1}{(z+i)^{3}}- \frac{5}{z+i}+8 \right) \ dz $ where $\displaystyle C $ is the circle $\displaystyle |z+i| = 1 $ with positive orientation.

    So we have $\displaystyle \int_{C} \frac{1}{(z+i)^{3}} \ dz + -5 \int_{C} \frac{1}{z+i} \ dz + \int_{C} 8 \ dz $. The last integral is $\displaystyle 0 $ by Cauchy-Goursat. For the two other integrals why can't we say that they are both $\displaystyle 0 $ by Cauchy-Goursat? Because they both have singularities at $\displaystyle z = -i $.

    Only the first one is $\displaystyle 0 $ and the second one is $\displaystyle -5(2 \pi i) = -10 \pi i $ by Cauchy Integral Formula. But thet are both analytic within and on $\displaystyle C $. So why are they not both $\displaystyle 0 $?

    yeah you had to use Cauchy Integral formula for derivatives because the function was "too singular."
    Last edited by mr fantastic; Mar 11th 2009 at 05:14 AM. Reason: Merged posts
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    Quote Originally Posted by manjohn12 View Post
    Evaluate $\displaystyle \oint_{C} \left(\frac{1}{(z+i)^{3}}- \frac{5}{z+i}+8 \right) \ dz $ where $\displaystyle C $ is the circle $\displaystyle |z+i| = 1 $ with positive orientation.

    So we have $\displaystyle \int_{C} \frac{1}{(z+i)^{3}} \ dz + -5 \int_{C} \frac{1}{z+i} \ dz + \int_{C} 8 \ dz $. The last integral is $\displaystyle 0 $ by Cauchy-Goursat. For the two other integrals why can't we say that they are both $\displaystyle 0 $ by Cauchy-Goursat? Because they both have singularities at $\displaystyle z = -i $.

    Only the first one is $\displaystyle 0 $ and the second one is $\displaystyle -5(2 \pi i) = -10 \pi i $ by Cauchy Integral Formula. But thet are both analytic within and on $\displaystyle C $. So why are they not both $\displaystyle 0 $?
    The integrands of the other two integrals are not analytic inside and on $\displaystyle C$ (there is a singularity at $\displaystyle z = -i$) so the Cauchy-Goursat Theorem cannot be applied.


    You know that $\displaystyle f^{(n)}(\alpha) = \frac{n!}{2 \pi i} \oint \frac{f(z)}{(z - \alpha)^{n+1}} \, dz$. Therefore:


    1. $\displaystyle \oint_{C} \frac{1}{(z+i)^3} \ dz = 0$ since $\displaystyle n = 2$ and $\displaystyle f(z) = 1$.


    2. $\displaystyle \oint_{C} \frac{1}{(z+i)} \ dz = 2 \pi i$ since $\displaystyle n = 0$ and $\displaystyle f(z) = 1$.
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