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Math Help - Complex variable...

  1. #1
    Senior Member Dinkydoe's Avatar
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    Complex variable...

    I have problems solving: \frac{dz}{dt}= \overline{z}e^{it}

    Seperating variables gives:  \frac{dz}{\overline{z}}=e^{it}dt

    But unfortunately, ...I'm not sure if we're able to integrate \overline{z} w.r.t z
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  2. #2
    A Plied Mathematician
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    Just a shot in the dark here, but what about this:

    \dfrac{dz}{\bar{z}}=\dfrac{z\,dz}{z\bar{z}}=\dfrac  {z\,dz}{|z|^{2}}?
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  3. #3
    Senior Member Dinkydoe's Avatar
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    Thanks for the suggestion...but then again: Integrating \frac{z}{|z|^2} w.r.t z is also a problem I wouldn't know how to solve..

    The actual thing I'm trying to show here, is that not all solutions to this equation are bounded..., the question asks to solve the eq. explicitly
    But I've no clue how. However, we may be able to use that |\frac{dz}{dt}|=|z| for all t..., then try to show that all solution except the origin are unbounded.

    (edit: i meant to show that not all solutions to this problem are bounded)
    Last edited by Dinkydoe; November 3rd 2010 at 02:38 PM.
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  4. #4
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    |z|^2 is a constant.

    So you should get

    \displaystyle \frac{1}{|z|^2} \int{z\,dz} = \int{e^{it}\,dt}
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  5. #5
    Senior Member Dinkydoe's Avatar
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    I don't fully understand. I agree if you said:  \int \frac{z}{|z|^2}dz=\int e^{it}dt

    I don't see why you could just treat |z|^2 as a constant. It's a real number depending on z. So why is it you can treat it as such?

    \int \frac{z}{|z|^2}dz = \frac{1}{|z|^2}\int zdz ...is this really true?
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  6. #6
    A Plied Mathematician
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    I'm with dinkydoe on this one. I would agree, from the DE, that

    \displaystyle\left|\frac{1}{\bar{z}}\,\frac{dz}{dt  }\right|=\left|\frac{z}{|z|^{2}}\,\frac{dz}{dt}\ri  ght|=\frac{1}{|z|^{2}}\left|z\,\frac{dz}{dt}\right  |

    is a constant. But I don't see how you can get that |z|^{2} is a constant.
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  7. #7
    Senior Member Dinkydoe's Avatar
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    Ok, so my professor hinted to something like a substitution z= r(t)e^{i\phi}. Then \overline{z}= r(t)e^{-i\phi}

    So we have \frac{d}{dt}[r(t)e^{i\phi}] = e^{it}r(t)e^{-i\phi}

    So... r'(t)= r(t)e^{(t-2\phi)i}\Longrightarrow r(t)= e^{-ie^{(t-2\phi)i}}

    And finally z= e^{[\phi-e^{(t-2\phi)i}]i}..

    But i'm not even sure if this is a valid way of arguing...Furthermore, doesn't this generate only bounded solutions.
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  8. #8
    A Plied Mathematician
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    Yeah, I was thinking of a substitution somewhat like that, but I ran into difficulties. That one is better, I think. Query: does your solution satisfy the original DE?

    I'm not so sure that this generates only bounded solutions. You don't know that the exponent there is purely imaginary, do you?
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