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Math Help - Is this correct? - showing analytic function is constant.

  1. #1
    Junior Member platinumpimp68plus1's Avatar
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    Is this correct? - showing analytic function is constant.

    for analytic function f:A-->C, where A is an open connected set. If f(A) is completely contained in the line {x+iy | x=2y}, show that f is constant.

    My answer: i said that the analytic function f(x,y)=u(x,y)+iv(x,y) has to be in the form u(x,y)=x, v(x,y)=2x. Then Cauchy-Riemann relations:

    du/dx=1 =/= 0=dv/dy
    du/dy=0 =/= -2=-dv/dx

    So the only way they can hold is if x=c, c some constant, so that all the partial derivatives are equal to 0.

    Is this correct? (I feel like its a little too simple...)
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Are you sure the function must be of that form? What about u(x,y)=xg(x,y) and v(x,y)=2xg(x,y) where g: \mathbb{R}^2 \rightarrow \mathbb{R} is an arbitrary function?
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  3. #3
    Junior Member platinumpimp68plus1's Avatar
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    Oh hm, well that would work too I suppose. Do I use the same process (ie. show that their partials don't satisfy C-R)?
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Give it a try!
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  5. #5
    Junior Member platinumpimp68plus1's Avatar
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    okay, so i plugged it in, equated terms for g(x,y), and found dg/dy must be equal to 0 (because the dg/dx's cancelled out). consequently, du/dy and dv/dy are also equal to 0, hence for C-R to hold, du/dx and dv/dx are 0 as well --> f is constant.

    yay?
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