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Make a function holomorphic

  1. Oct 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Find a constant k such that the function [itex] v(x,y) = y^3-4xy +kx^2y [/itex] can be the imaginary part of a holomorphic function f on C

    2. Relevant equations

    The Cauchy-Riemann equations: [itex]u_x=v_y[/itex] and [itex]u_y=-v_x[/itex]

    3. The attempt at a solution

    So far I have taken the partial derivatives of v w.r.t y and equated it to [itex]u_x[/itex] and then integrated w.r.t x giving:
    [itex]u=3xy^2-2x^2y+(k/3)x^3+f(y)[/itex]

    Then differentiating w.r.t y to give:
    [itex]u_y=6xy-2x^2y+(k/3)x^3y+f'(y)[/itex]

    Next I equate this to [itex]-v_x[/itex] giving:
    [itex]6xy-2x^2y+(k/3)x^3y+f'(y)=4y-2kxy[/itex]

    Now Im not sure which direction to head next or even if this is the correct approach to begin with. Help is greatly appreciated
     
  2. jcsd
  3. Oct 12, 2013 #2

    Dick

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    You are working too hard. Can you show that Cauchy-Riemann implies ##v_{xx}+v_{yy}=0##? You don't really need to find u.
     
    Last edited: Oct 12, 2013
  4. Oct 12, 2013 #3

    LCKurtz

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    Your method is OK and will solve the problem. But you have mistakes in your work. Hard to point out your errors since you omitted some steps. In particular, the term in red is incorrect.
     
  5. Oct 12, 2013 #4
    Dick and LCKurtz thanks both for replying. Dick: following your method I get [itex]v_{xx}+v_{yy}=2ky+6y=0 \rightarrow k=-3[/itex]
    This method is showing that the function is harmonic for k=-3. Is this sufficient for answering the question of which k makes the function holomorphic?
     
  6. Oct 12, 2013 #5

    Dick

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    Yes, if the function is holomorphic then v needs to be harmonic. So the only possible value is k=(-3).
     
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