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Math Help - [SOLVED] holomorphic function

  1. #1
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    [SOLVED] holomorphic function

    Hello,

    I have to find all holomorphic functions F(z)=F(x+iy)

    F(x+iy)=P(x,y) + i Q(x,y)

    with the condition (1) that P(x,y)/Q(x,y) is a function of x only.

    Combining (1) with cauchy relations I reach that the module of F depends only on y. But I can't get more precise, though I think it must be possible.

    Thanks for any help.

    civodul
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  2. #2
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    I'm no expert with this but I don't think there are any non-constant functions which satisfy this relation:

    Let f(z)=u(x,y)+iv(x,y):

    If \frac{u}{v}=G(x) then \frac{u}{v}=\frac{g(x)h(y)}{r(x)h(y)}=\frac{g}{r}=  G(x). Now if f is holomorphic, then u(x,y)=g(x)h(y) satisfies Laplace's equation: g''h+h''g=0 or \frac{g''}{g}=-\frac{h''}{h}. The only way that's true is if g(x)=ax+b and h(y)=cy+d. Letting u(x,y)=(ax+b)(cy+d), then the harmonic conjugate of this is found to be v(x,y)=\frac{acy^2}{2}+ady-\frac{cax^2}{2}-cbx+k and this is not a product solution of the form v(x,y)=r(x)h(y).

    I hope someone here can confirm or prove this wrong.
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  3. #3
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    Write F(z) = re^{i\theta}, where r is a function of y only, and \theta is a function of x and y.

    The condition that P/Q is a function of x alone tells us (implicitly) that F(z) is nonzero. Therefore (locally, at least) we can form \ln(F(z)) = \ln r + i\theta, which will be an analytic function. Apply the Cauchy–Riemann equations to this function. They tell you that \theta is a function of x only, and in fact \theta(x) = \alpha x+\beta (where \alpha and \beta are real), and that r'(y)/r(y) = -\alpha. That leads to the conclusion that F(z) = \lambda e^{i\alpha z}, where \lambda\in\mathbb{C} and \alpha\in\mathbb{R} are constants.
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  4. #4
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    Quote Originally Posted by Opalg View Post
    The condition that P/Q is a function of x alone tells us (implicitly) that F(z) is nonzero.
    Thanks for your very clear answer.
    There is still one point that is not so obvious to me:
    Why would the fact that P/Q is a function of x alone entail that F(z) is non zero?

    civodul
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  5. #5
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    Quote Originally Posted by civodul View Post
    Why would the fact that P/Q is a function of x alone entail that F(z) is non zero?
    What I meant was that the question, by referring to the fraction P/Q, implicitly assumes that Q (and therefore F) is nonzero.
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  6. #6
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    Quote Originally Posted by Opalg View Post
    What I meant was that the question, by referring to the fraction P/Q, implicitly assumes that Q (and therefore F) is nonzero.
    That is what I assumed you meant but I wasn't completely sure.

    Civodul
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