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Math Help - Complex Analysis: Analytic Function

  1. #1
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    Complex Analysis: Analytic Function

    Here is my question: Let f=u+iv be analytic and suppose that f(i)=1. If u=2x^2+2x+1-2y^2, find v.

    My proof so far follows:

    By differentiating with respect to x and y, we see that u_x=4x+2, u_y=-4y, v_x=4y, and v_y=4x+2. In order to move back to v, I will integrate on v_x and v_y as follows:
    4xy+2y+g(x)+c=v
    4xy+g(y_+c=v
    \rightarrow 4xy+2y+0+c=v

    To solve for c, f=u+iv and f(i)=1, so \cdots

    and now this is the really dumb part... I am not sure if I am supposed to substitute this literally as u + i*v, using the u as given and the v that I have derived, substituting (0,1) for i, or do something else entirely. I feel that I am nearly at the solution, but I am losing it here in something that should be quite simple. I will appreciate your help. Thanks.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Complex Analysis: Analytic Function

    Quote Originally Posted by tarheelborn View Post
    Here is my question: Let f=u+iv be analytic and suppose that f(i)=1. If u=2x^2+2x+1-2y^2, find v...
    The condition f(i)=1 means that u(0,1)=1 and if is u=2x^2+2x+1-2y^2 then u(0,1)=-1... contradiction!...

    Kind regards

    \chi \sigma
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  3. #3
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    Re: Complex Analysis: Analytic Function

    But don't I have to put the v in there somewhere, too, since that's what I've had to find? And I'm trying to solve for c? Since f=u+iv to start with, don't they go together? That's where I am getting confused. I have discussed this problem with classmates and nobody mentioned coming up with a contradiction. I know that we are supposed to solve for c and I really thought that the answer was either 2 or 2i, but for the life of me, I can't get there.
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