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Thread: Cauchy's integral formula

  1. #1
    Junior Member piglet's Avatar
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    Cauchy's integral formula

    $\displaystyle \int_{C}\frac{f(z)}{z-z0}dz = 2\pi.i.f(z0) $

    Im using cauchy's integral formula, see above to solve the following integral

    $\displaystyle \int_{C}\frac{z}{z^{2}-1}dz $ where $\displaystyle C $ is a circle of radius 4 centred at 0.

    I have deduced that the function $\displaystyle \frac{z}{z^{2}-1} $ is not analytic at $\displaystyle z = +i $ $\displaystyle z = -i $

    I let $\displaystyle f(z) = \frac{z}{z+i} $ and $\displaystyle z0 = -i $

    so filling into cauchy's integral formula i got

    $\displaystyle \int_{C}\frac{z}{(z+i)(z-i)}dz = \frac{f(z)}{z-i} = 2\pi.i.f(i) $

    Answer = $\displaystyle \pi.i $

    My problem lies in the fact that both $\displaystyle z = +i $ $\displaystyle z = -i $ both lie in $\displaystyle C $. Therefore do i have to take both singularites into account when evaluating the integral using cauchy's integral formula???

    Thanks,
    Piglet
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by piglet View Post
    $\displaystyle \int_{C}\frac{f(z)}{z-z0}dz = 2\pi.i.f(z0) $

    Im using cauchy's integral formula, see above to solve the following integral

    $\displaystyle \int_{C}\frac{z}{z^{2}-1}dz $ where $\displaystyle C $ is a circle of radius 4 centred at 0.

    I have deduced that the function $\displaystyle \frac{z}{z^{2}-1} $ is not analytic at $\displaystyle z = +i $ $\displaystyle z = -i $

    I let $\displaystyle f(z) = \frac{z}{z+i} $ and $\displaystyle z0 = -i $

    so filling into cauchy's integral formula i got

    $\displaystyle \int_{C}\frac{z}{(z+i)(z-i)}dz = \frac{f(z)}{z-i} = 2\pi.i.f(i) $

    Answer = $\displaystyle \pi.i $

    My problem lies in the fact that both $\displaystyle z = +i $ $\displaystyle z = -i $ both lie in $\displaystyle C $. Therefore do i have to take both singularites into account when evaluating the integral using cauchy's integral formula???

    Thanks,
    Piglet
    Yes. See example 30.2 here: http://math.furman.edu/~dcs/courses/...lecture-30.pdf
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  3. #3
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    $\displaystyle
    \frac{z}{z^{2}-1} \: = \:\frac{1}{2(z-1)} \: + \frac{1}{2(z+1)} \:

    $

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