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Math Help - Complex function C-R stuff

  1. #1
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    Complex function C-R stuff

    Hey,

    I've been struggling with this tute question,



    I really don't understand where im supposed to be going,
    I've been trying to show the u derivatives are zero because the hint says to do that,
    So far ive just been messing around with

    f(z) = u + iy, |f(z)|=u^2+y^2 =k

    then i took d/dx partial derivative

    2u(du/dx) + 2v(dv/dx) = 0 all partial derivatives

    Now im stuck plugging in C-R equations trying to find a way to show du/dx and du/dy are zero,
    I don't really understand what to do after that also,

    Thanks in advanced,
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Complex function C-R stuff

    Quote Originally Posted by Daniiel View Post
    Hey,

    I've been struggling with this tute question,



    I really don't understand where im supposed to be going,
    I've been trying to show the u derivatives are zero because the hint says to do that,
    So far ive just been messing around with

    f(z) = u + iy, |f(z)|=u^2+y^2 =k

    then i took d/dx partial derivative

    2u(du/dx) + 2v(dv/dx) = 0 all partial derivatives

    Now im stuck plugging in C-R equations trying to find a way to show du/dx and du/dy are zero,
    I don't really understand what to do after that also,

    Thanks in advanced,
    If |f(z)|=k= \text{constant}, then is...

    f(z)= k\ e^{i\ \theta(x,y)} (1)

    Now if we write the C-R equations in 'nonstandard' form...

    u_{x}-v_{y}=0

    u_{y}+v_{x}=0 (2)

    ... from (1) and (2) we obtain...

    \sin \theta\ \frac{d \theta}{dx} + \cos \theta\ \frac{d \theta}{d y}=0

    \cos \theta\ \frac{d \theta}{dx} - \sin \theta\ \frac{d \theta}{d y}=0 (3)

    Now the linear equations system (3) is nonsingular because its determinant is \delta=-1 \ne 0, so that the (3) has the only solution \frac{d \theta}{d x} = \frac{d \theta}{d y} =0...

    Kind regards

    \chi \sigma
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  3. #3
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    Re: Complex function C-R stuff

    Thanks alot,

    So you let u = cos(theta) v=sin(theta) and f(z) was = k(u+iv)

    im a little confused though,

    If |f(z)|=k=|u+iv|

    how can f(z) = k(u+iv) ?
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Complex function C-R stuff

    Quote Originally Posted by Daniiel View Post
    Thanks alot,

    So you let u = cos(theta) v=sin(theta) and f(z) was = k(u+iv)...
    Not exactly!... the correct procedure is...

    |f(z)|=k \implies f(z)= k\ (\cos \theta + i\ \sin \theta) \implies u= k\ \cos \theta\ ,\ v=k\ \sin \theta

    Kind regards

    \chi \sigma
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  5. #5
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    Re: Complex function C-R stuff

    Sorry i'm still a little confused,

    I don't understand how f(z)=k(u+iv) and how its not just f(z) = k

    because k = u^2 + v^2 or 1 if using cos and sin

    k=|f(z)|=|u+iv|

    so f(z) = k(u+iv) = k only if u = 1 and v = 0
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  6. #6
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    Re: Complex function C-R stuff

    Ok that parts all sweet now,

    But how do I apply this in answering the question about |z|<100,
    do i just say as it is true for |f(z)| and that = k
    it is also true for f(z) since it also = k?
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