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Math Help - Cauchy's integral formula

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

     \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

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

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

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

    so filling into cauchy's integral formula i got

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

    Answer =  \pi.i

    My problem lies in the fact that both  z = +i  z = -i both lie in  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
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    mr fantastic's Avatar
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    Quote Originally Posted by piglet View Post
     \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

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

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

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

    so filling into cauchy's integral formula i got

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

    Answer =  \pi.i

    My problem lies in the fact that both  z = +i  z = -i both lie in  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
    Senior Member
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    Mar 2010
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    <br />
\frac{z}{z^{2}-1} \: = \:\frac{1}{2(z-1)} \: + \frac{1}{2(z+1)} \:<br /> <br />

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