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Thread: How To Do Integral Of Complex Conjugate?

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    How To Do Integral Of Complex Conjugate?

    $\displaystyle \int\limits_{1+i}^{2+i}s\, dx$

    Where s is the complex conjugate of x

    I have the answer but I don't have the foggiest idea how it is obtained.

    answer: [(2^2 -1 )/2] -i
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    Quote Originally Posted by soma View Post
    $\displaystyle \int\limits_{1+i}^{2+i}s\, dx$

    Where s is the complex conjugate of x

    I have the answer but I don't have the foggiest idea how it is obtained.

    answer: [(2^2 -1 )/2] -i
    Define $\displaystyle x = e^{i\theta}$.

    Then $\displaystyle \overline{x} = e^{-i\theta}$

    and $\displaystyle dx = i\,e^{i\theta}$.

    Also note that $\displaystyle \theta = \frac{\log{x}}{i}$.


    So the integral becomes

    $\displaystyle \int{\overline{x}\,dx} = \int{e^{-i\theta}\,i\,e^{i\theta}\,d\theta}$

    $\displaystyle = \int{i\,d\theta}$

    $\displaystyle = i\theta + C$

    $\displaystyle = i\left(\frac{\log{x}}{i}\right) + C$

    $\displaystyle = \log{x} + C$.


    Now you should be able to substitute the limits of integration.

    Remember that $\displaystyle \log{x} = \ln{|x|} + i\,\textrm{arg}\,{x}$.
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