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Math Help - Cauchy Integral Formula

  1. #1
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    Cauchy Integral Formula

    The question
    Evaluate the following integral using Cauchy Integral Formula,
    \oint_{|z| = 4}{\frac{e^{2iz} \ dz}{(3z - 1)^2}}

    My attempt
    Not analytic at:
    (3z - 1)^2 = 0
    z = \frac{1}{3}

    This is inside the contour, so we use CIF as follows:
    2\pi i e^{\frac{2i}{3}}

    However, the solution is \frac{-4\pi}{9}e^{\frac{2i}{3}}

    Where have I gone wrong? Thanks.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Cauchy Integral Formula

    ... may be that the reason is that in z=\frac{1}{3} the function has a pole with multeplicity two... in that case the formula for the residue is different respect to the case of a pole with multeplicity one...

    Kind regards

    \chi \sigma
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  3. #3
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    Re: Cauchy Integral Formula

    Ahh, I just discovered the generalised CIF equation.

    Still not sure how to get the correct solution though, I did:
    \frac{2\pi i}{!1} 2ie^{2iz}
    = -4\pi e^{\frac{2i}{3}}

    Not sure where the divide by 9 is coming from. :/
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Cauchy Integral Formula

    For a pole of order n is...

    r= \lim_{z \rightarrow z_{0}} \frac{d^{n-1}}{d z^{n-1}} \{(z-z_{0})^{n}\ f(z) \} (1)

    In our case is...

    f(z)= \frac{1}{9}\ \frac{e^{2 i z}}{(z-\frac{1}{3})^{2}} (2)

    ... so that...

    r= \lim_{z \rightarrow \frac{1}{3}} \frac{d}{d z} \frac{e^{2 i z}}{9} = \frac{2i}{9}\ e^{\frac{2}{3} i} (3)

    ... and for any closed path c containing the point z=\frac{1}{3} is...

    \int_{c} f(z)\ dz = 2 \pi i r = - \frac{4 \pi}{9}\ e^{\frac{2}{3} i} (4)

    Kind regards

    \chi \sigma
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  5. #5
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    Re: Cauchy Integral Formula

    Ahh, I just had to pull the 3 out of the denominator.

    Thanks.
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