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Complex Analysis - Proving an analytic function f(z) is constant

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f(z) be an analytic function in the complex plane ℂ, and let [itex]\phi[/itex] be amonotonic function of a real variable.

    Assume that U(x,y) = [itex]\phi[/itex](V(x,y)) where U(x,y) is the real part of f(z) and V(x,y) is the imaginary part of f(z). Prove that f is constant.

    2. Relevant equations

    The analytic function f(z) is constant if f'(z)= 0 everywhere.

    The Cauchy Riemann equation...

    ∂u/∂x=∂v/∂y,∂v/∂x=−∂u/∂y


    3. The attempt at a solution

    I'm honestly a bit lost on where to start. I know that if f(z) is analytic then it is differentiable, so I thought that using the Cauchy Riemann equations for the partial derivatives might be helpful where U[itex]_{x}[/itex]=V[itex]_{y}[/itex] and U[itex]_{y}[/itex]=-V[itex]_{x}[/itex], but I don't know how to work with these when the function [itex]\phi[/itex] is there.

    Just any information on what theory is best to look at would be helpful. Thank you.
     
    Last edited: Mar 18, 2012
  2. jcsd
  3. Mar 18, 2012 #2
    simply substitute into the C-R equation and make use of the chain rule of taking derivatives. Also note that phi is monotonic (amonotonic=a monotonic, as I assume), so that phi'≠0
     
  4. Mar 19, 2012 #3
    Actually managed to figure it out just about

    Assume that f(z) = u(x,y) + i v(x,y).

    We know that u(x, y) = ϕ(v(x, y)).

    By Cauchy-Riemann, we have
    ∂u/∂x = ∂v/∂y ==> ϕ'(v(x,y)) * ∂v/∂x = ∂v/∂y, and
    ∂u/∂y = -∂v/∂x ==> ϕ'(v(x,y)) * ∂v/∂y = -∂v/∂x.

    Substituting the first equation into the second equation yields
    ϕ'(v(x,y)) * [ϕ'(v(x,y)) * ∂v/∂x] = -∂v/∂x
    ==> [(ϕ'(v(x,y)))^2 + 1] ∂v/∂x = 0.
    ==> ∂v/∂x = 0, since (ϕ'(v(x,y)))^2 + 1 is nonzero.

    Substituting this into the first equation yields ∂v/∂y = 0.

    Since ∂v/∂x = ∂v/∂y = 0, we conclude by Cauchy-Riemann that ∂u/∂x = ∂u/∂y = 0 as well.
    Hence, both u and v are constant.
    Therefore, f = u + iv is also constant.

    Thanks though.
     
  5. Mar 19, 2012 #4
    good work, better than what I did.
     
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