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Complex analysis / Using analyticity of f to prove f is constant

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm supposed to show that, if [itex]f[/itex] is analytic and [itex]|f|[/itex] is constant on a domain [itex]D \subset \mathbb{C}[/itex], [itex]f[/itex] is constant.


    2. Relevant equations
    The hint is to write [itex]f^* = |f|^2 / f[/itex]. I might also need to use the fact that if [itex]f^*[/itex] is analytic too, then [itex]f[/itex] is constant.


    3. The attempt at a solution
    Well, I followed the hint, and I fail to see how it helps at all. Given the hypotheses of the problem, I guess we know [itex]f^* = A / f[/itex] for some [itex]A > 0[/itex], but this doesn't strike me as particularly useful. Writing [itex]f = u(x,y) + i v(x,y)[/itex] only seems to complicate things, but don't I eventually have to do this? I'm guessing I'm supposed to use the Cauchy-Riemann Equations together, in some way, with the fact (proved in my text) that if [itex]h(x,y)[/itex] is a real-valued function that satisfies [itex]\nabla h = 0[/itex] on a domain, then [itex]h[/itex] is constant on that domain. But taking partial derivatives and trying to use [itex]u_x = v_y[/itex] and [itex]u_y = -v_x[/itex] just makes things messy.
     
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  3. Sep 13, 2009 #2

    Dick

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    You've got all the ingredients in front of you. If f=u+iv is analytic then f*=(u-iv)=A/f is also analytic. Write out the Cauchy-Riemann equations for both f and f*. What do they tell you about the partial derivatives of u and v?
     
  4. Sep 13, 2009 #3
    Right now, I've got a system of equations for [itex]f = u+iv[/itex]: (1) [itex]uu_x = vv_x[/itex] and (2) [itex]uu_y = vv_y[/itex]. These came from [itex]u^2 + v^2 = [/itex] Constant. But I've got to show that [itex]u_x = u_y = 0[/itex] from this and [itex]u_x = v_y[/itex], [itex]u_y = -v_x[/itex]. Any hints? I still can't quite get there...
     
  5. Sep 13, 2009 #4

    Dick

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    I already gave you a hint. f=u+iv and f*=u-iv are analytic. Use Cauchy-Riemann on both of them.
     
  6. Sep 13, 2009 #5
    You did. Sorry. I'll do that :)
     
  7. Sep 13, 2009 #6
    I figured it out. Thanks! That was very helpful.
     
    Last edited: Sep 13, 2009
  8. Jul 26, 2010 #7
    well i figured it out using the hint too,but i didn't have to use the fact that modulus of f is constant on D.i just used the cauchy-riemann eqns from f=u+iv;
    du/dx=dv/dy and from f*=u-iv; du/dx=-dv/dy hence du/dx =0, and found the partial derivatives to be zero and so f has to be constant.so where do we use the modulus of f being constant?is it when we show the analyticity of f*?
    maybe it's something like that if f is analytic then 1/f is also analytic so f*=a.1/f is also analytic where a is modulus of f.but i don't know anything about the analyticity of 1/f and can't we just prove it using u and v?
    -it became kind of long and sort of complicated so sorry if it's not understandable.
     
  9. Jul 26, 2010 #8

    Dick

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    Yes, you use |f|=constant to prove f* is analytic. The problem is that |f| isn't generally an analytic function. A constant IS analytic.
     
  10. Jul 26, 2010 #9

    Office_Shredder

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    You should know something about the analyticity of f! Just from the normal quotient rule that you have from real analysis

    In general: If |f| is not constant, then it's probably not analytic. Also f* is almost never analytic, you just have a very special case here (in fact you can show that the function that takes z to its complex conjugate is not analytic)
     
  11. Jul 26, 2010 #10
    yes,this is really easy with quotient rule.thanks for the help!
     
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