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Thread: Evaluate Complex Integral

  1. #1
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    Evaluate Complex Integral

    Use Cauchy's Integral Formula to evaluate the integral $\displaystyle \oint_{C}f(z)dz$ over a contour C, where C is the boundary of a square with diagonal opposite corners at z = −(1 + i )R and z = (1 + i )R, where R > a > 0, and where $\displaystyle f(z)=\frac{e^z}{z-\frac{i\pi}{4} a}$

    I know f(z) would not be analytic at $\displaystyle \frac{i\pi}{4} a$.

    Then by Cauchy's Integral Formula:

    $\displaystyle f(a) = \frac{1}{2i\pi} \oint_{C}\frac{f(z)}{z-a} $

    $\displaystyle f(\frac{i\pi}{4} a) = \frac{1}{2i\pi} \oint_{C}\frac{e^z}{z-\frac{i\pi}{4} a} $

    Am I going in the right direction so far?
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  2. #2
    Senior Member
    Joined
    Feb 2008
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    Re: Evaluate Complex Integral

    Quote Originally Posted by ivinew View Post
    Use Cauchy's Integral Formula to evaluate the integral $\displaystyle \oint_{C}f(z)dz$ over a contour C, where C is the boundary of a square with diagonal opposite corners at z = −(1 + i )R and z = (1 + i )R, where R > a > 0, and where $\displaystyle f(z)=\frac{e^z}{z-\frac{i\pi}{4} a}$

    I know f(z) would not be analytic at $\displaystyle \frac{i\pi}{4} a$.

    Then by Cauchy's Integral Formula:

    $\displaystyle f(a) = \frac{1}{2i\pi} \oint_{C}\frac{f(z)}{z-a} $

    $\displaystyle f(\frac{i\pi}{4} a) = \frac{1}{2i\pi} \oint_{C}\frac{e^z}{z-\frac{i\pi}{4} a} $

    Am I going in the right direction so far?
    Well the symbol $\displaystyle f$ is already "taken," so basically you want to let $\displaystyle g(z)=e^z$ and notice that

    $\displaystyle e^{i\pi a/4}=g(i\pi a/4)=\frac{1}{2\pi i}\int_C\frac{g(z)}{z-(i\pi a/4)} dz$,

    and you're done.
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