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Math Help - Integrating a complex function

  1. #1
    Junior Member piglet's Avatar
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    Integrating a complex function

    Dont know if this is thr correct place to put this but im looking for a couple of pointers on a recent maths quiz i got.

    I have to integrate Ze^(z^2) around simple closed curve C C: from 1 along axis from i to i.

    I understand that you must check to see if the function is analytic before deciding which method of integration to use but i seem to becoming unstuck with my partial derivatives. Here's what i've done


    f(z) = Ze^(z^2)

    as z = x+iy

    => f(z) = (x + iy)e^((x+iy)^2)

    f(z) = xe^((x+iy)^2) + iye^((x+iy)^2)

    let u(x,y) = xe^((x+iy)^2) let v(x,y) = iye^((x+iy)^2)

    do i need to use the product rule to find Ux and Vy?
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  2. #2
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    Hi Piglet. Ain't that zero. Cauchy's Theorem and all. The function f(z)=ze^{z^2} is entire right. Why is the integral of that over a closed contour zero then? Something about the Cauchy-Riemann equations, and Green's Theorem I think. Try and go over that proof then use Cauchy's Theorem to conclude the integral is zero or well, just integrate it the regular way and show the antiderivative is analytic then F(a)-F(a)=0 since it's a closed contour.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Because f(z) is analytic in the whole complex plane, its line integral depends only from the starting and final point, so that we can integrate along the imaginary axis setting z=i\omega. The integral becomes...

    \int_{-i}^{i} f(z)\cdot dz = - \int_{-1}^{1} \omega\cdot e^{-\omega^{2}}\cdot d\omega (1)

    ... and because the integrand function is odd the integral (1) is equal to 0...

    Kind regards

    \chi \sigma
    Last edited by chisigma; February 23rd 2010 at 10:07 AM. Reason: Trivial error corrected...
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  4. #4
    Junior Member piglet's Avatar
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    Quote Originally Posted by chisigma View Post
    Because f(z) is analytic in the whole complex plane, its line integral depends only from the starting and final point, so that we can integrate along the imaginary axis setting z=i\omega. The integral becomes...

    \int_{-i}^{i} f(z)\cdot dz = - \int_{-1}^{1} \omega\cdot e^{-\omega^{2}}\cdot d\omega (1)

    ... and because the integrand function is odd the integral (1) is equal to 0...

    Kind regards

    \chi \sigma
    Thanks guys for the help an quick responses. To say im a bit fuzzy on this topic is an understatement and im finding your input really helpful.

    I had it in my hand that f(z) was not analytic at z = 0 and therefore i couldn't use cauchy's integral theorem to state that the integral of f(z) is = 0... Hence i would have to use the 2nd method of integration (which im unsure of) to solve the integral of f(z)

    Is my whole train of thought wrong on this matter?
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  5. #5
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    Quote Originally Posted by piglet View Post
    Dont know if this is thr correct place to put this but im looking for a couple of pointers on a recent maths quiz i got.

    I have to integrate Ze^(z^2) around simple closed curve C C: from 1 along axis from i to i.


    What does this mean, anyway?? Please be accurate.


    I understand that you must check to see if the function is analytic before deciding which method of integration to use but i seem to becoming unstuck with my partial derivatives. Here's what i've done


    f(z) = Ze^(z^2)

    as z = x+iy

    => f(z) = (x + iy)e^((x+iy)^2)

    f(z) = xe^((x+iy)^2) + iye^((x+iy)^2)

    let u(x,y) = xe^((x+iy)^2) let v(x,y) = iye^((x+iy)^2)

    do i need to use the product rule to find Ux and Vy?

    Check that \frac{1}{2}e^{z^2} is a primitive for your function.

    Tonio
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  6. #6
    Junior Member piglet's Avatar
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    Sorry im only after spotting my error in my question

    The function should be integrated along the axis from 1 to i
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  7. #7
    MHF Contributor chisigma's Avatar
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    All right!... in that case the integral can be perfomed first along the real axis from 1 to 0 and after along the imaginary axis from 0 to i...

    \int_{1}^{i} z\cdot e^{z^{2}}\cdot dz = - \int_{0}^{1} \sigma\cdot e^{\sigma^{2}}\cdot d\sigma - \int_{0}^{1} \omega\cdot e^{-\omega^{2}}\cdot d\omega=

    -\frac{1}{2}\cdot |e^{\sigma^{2}}|_{0}^{1} + \frac{1}{2}\cdot |e^{-\omega^{2}}|_{0}^{1}= -\frac{1}{2}\cdot (e-\frac{1}{e})

    Kind regards

    \chi \sigma
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