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Find v such that f(z) = u+iv is analytic

  1. Mar 20, 2017 #1
    1. The problem statement, all variables and given/known data
    Find v such that f(z) = u+iv is analytic.

    2. Relevant equations
    du/dx = dv/dy and dv/dx=-du/dy

    3. The attempt at a solution
    I'm not sure what I'm supposed to do. I think I need to find U in order to find V because if a function is analytic it satisfies the Cauchy Riemann equations. I tried to play around with cauchy riemann equations to get dv in terms of everything else but that's not helping. Can I also use the Laplace equation to solve this? This is so general that I know it's probably simple, but at the same time it makes it hard to understand what my answer is supposed to look like. What should my approach be?
     
  2. jcsd
  3. Mar 21, 2017 #2

    mfb

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    Staff: Mentor

    You can find v as function of u.
    It is a bit confusing to use the same symbols for the complex components of the argument (first equation) and the complex components of the function ("relevant equations").
     
  4. Mar 21, 2017 #3
    Suppose you are given [itex]u:=u(x,y)[/itex] and you know [itex]f[/itex] is differentiable so it satisfies C-R, in particular,[itex]\frac{du}{dx} = \frac{dv}{dy}[/itex].
    Therefore [itex]v = \int u_x dy + g(x)[/itex], where [itex]u_x[/itex] is the partial derivative w.r.t [itex]x[/itex]. You add a function of [itex]x[/itex] because when you differentiate w.r.t [itex]y[/itex], the [itex]g(x)[/itex] vanishes.
     
    Last edited: Mar 21, 2017
  5. Mar 21, 2017 #4

    mfb

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    That's not how I would interpret the u and v given in the first post.

    @Vitani11: Is that the full and exact problem statement? "Find v such that f(z) = u+iv is analytic." Nothing else given?
     
  6. Mar 21, 2017 #5
    The problem has to start from somewhere. We must either have [itex]u[/itex] or [itex]v[/itex]. I agree that the problem is ambiguous.
     
  7. Mar 21, 2017 #6
    That is 100% all that was given, promise. Anyway thank you I now think I can solve this.
     
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