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Math Help - Find an analytic function from its real part

  1. #1
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    Find an analytic function from its real part

    Hi there. I have two doubts on this exercise. The exercise gives me the real part u(x,y), and I have to find from it the imaginary part, such that the function I get accomplishes the given condition, and its an analytic function.

    u(x,y)=x^2-y^2+2x+y,f(i)=0

    The first doubt I got I think its quiet silly. I don't know how I should evaluate f(i). I hoped to have there two coordinates, for x and y. But I have just i. I don't know if I should took that as a complex number expressed in its polar form, does it mean that x=0 and y=i?

    On the other hand I've tried to find v(x,y), the real part it asks me for.

    So I used the Cauchy-Riemann conditions.

    u_x=2x+2=v_y \rightarrow v(x,y)=2xy+2y+\Phi(x)
    -u_y=v_x=2y-1=2y+\Phi'(x)\rightarrow \Phi'(x)=2y-2x-1 \rightarrow \Phi(x)=2xy-x^2-x+K

    So there is something wrong in there, the function I get for Phi depends on y too. Whats wrong with this? I couldn't find the analytic function under this procedure. Did I do something wrong or is it that it doesn't exist an analytic function for the given u(x,y)?

    Bye and thanks.

    PD: Ok, I realized about the first part, it was really silly, it just means f(z)=f(i), z=i.
    Last edited by Ulysses; September 23rd 2011 at 03:00 PM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Find an analytic function from its real part

    Quote Originally Posted by Ulysses View Post
    Hi there. I have two doubts on this exercise. The exercise gives me the real part u(x,y), and I have to find from it the imaginary part, such that the function I get accomplishes the given condition, and its an analytic function.

    u(x,y)=x^2-y^2+2x+y,f(i)=0

    The first doubt I got I think its quiet silly. I don't know how I should evaluate f(i). I hoped to have there two coordinates, for x and y. But I have just i. I don't know if I should took that as a complex number expressed in its polar form, does it mean that x=0 and y=i?

    On the other hand I've tried to find v(x,y), the real part it asks me for.

    So I used the Cauchy-Riemann conditions.

    u_x=2x+2=v_y \rightarrow v(x,y)=2xy+2y+\Phi(x)
    -u_y=v_x=2y-1=2y+\Phi'(x)\rightarrow \Phi'(x)=2y-2x-1 \rightarrow \Phi(x)=2xy-x^2-x+K

    So there is something wrong in there, the function I get for Phi depends on y too. Whats wrong with this? I couldn't find the analytic function under this procedure. Did I do something wrong or is it that it doesn't exist an analytic function for the given u(x,y)?

    Bye and thanks.

    PD: Ok, I realized about the first part, it was really silly, it just means f(z)=f(i), z=i.
    Your porcedure is correct!... from the condition...

    -u_{y}=2y-1=v_{x}= 2y+\phi^{'}(x) (1)

    ... You derive immediately that is...

    \phi^{'}(x)= -1 \implies \phi(x)=-x+c (2)

    ... so that is...

    v=2 x y +2 y -x + c (3)

    Taking into account that f(i)=0 You obtain in (3) c=-2 ...

    Kind regards

    \chi \sigma
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