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If f(z) holomorphic show that f(z) is constant

  1. Mar 7, 2015 #1
    1. The problem statement, all variables and given/known data

    If f(z) = u + iv is a holomorphic function on a domain D satisfying |f(z)| = constant, show that f(z) is constant.

    2. Relevant equations

    f(z) = u(z) + y(z)i = u(x,y) + y(x,y)i

    z = x + yi

    3. The attempt at a solution

    |f(z)| = √u2+v2) = √u(x,y)2+v(x,y)2)

    I know this is a simple problem, but I'm just not seeing the connection. I'm probably missing some obvious relation/equation to consider.
     
  2. jcsd
  3. Mar 7, 2015 #2

    WWGD

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    Maybe you can just quote Liouville? Maybe extending into an entire function with
    constant nodulus.

    If not, use Maximum Modulus http://en.wikipedia.org/wiki/Maximum_modulus_principle

    Maybe let ##c=u^2+v^2 ## , then ##0= 2uu_x+2vv_x ## and ##0 =2uu_y+ 2vv_y ## and use Cauchy -Riemann.
     
    Last edited: Mar 7, 2015
  4. Mar 8, 2015 #3

    SammyS

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    It seems to me that you might want to start out by using the definition of holomorphic .

    You also have a couple of typos in your post.

    I assume you mean: f(z) = u(z) + v(z)i = u(x,y) + v(x,y)i
     
  5. Mar 8, 2015 #4
    Shoot. Yes. Sorry.

    If f is complex differentiable at any point a ∈ D, f is called a holomorphic function on D.

    Then of course f is holomorphic ⇔ Cauchy-Riemann Equations hold.

    If f is a holomorphic function, then f'(z) = ux + ivx
     
    Last edited: Mar 8, 2015
  6. Mar 8, 2015 #5
    WWGD, I had that idea last night. I haven't been able to divine a different "proof," so that's likely what I'll turn in.
     
  7. Mar 8, 2015 #6
    use the cauchy-riemann conditions along with the laplace equation
     
  8. Mar 9, 2015 #7

    WWGD

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    Or, like occh suggested, use the fact that a harmonic function cannot be bounded. Then, if |f| is bounded, both the Real, Complex part must be bounded.
     
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