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Complex proof

  1. Sep 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose that [tex] f(z) [/tex] and [tex] f(z) - conj(f(z)) [/tex]

    proof [tex] f(z) [/tex] is constant on D

    2. Relevant equations



    3. The attempt at a solution

    If I write the equations as [tex] f(x,y) = u(x,y) + i*v(x,y) [/tex]

    then [tex] f(z) - conj(f(z)) [/tex]

    = [tex] u(x,y) + i*v(x,y) - (u(x,y) - i*v(x,y)) [/tex]

    and the [tex] u(x,y) [/tex] cancel out and we are left with

    [tex] f(z) - conj(f(z)) = 2*i*v(x,y)[/tex]

    and by the Cauchy Riemann eq

    [tex] \frac {du}{dx} = \frac {dv}{dy} [/tex]

    [tex] \frac {du}{dy} = - \frac {dv}{dx} [/tex]

    since [tex] u(x,y) = 0 [/tex] for [tex] f(z) - conj(f(z)) = 2*i*v(x,y)[/tex]

    in order for [tex] \frac {du}{dy} u(x,y) = - 2*i*\frac {dv}{dx}v(x,y) [/tex]

    f(z) has to be constand on D
     
  2. jcsd
  3. Sep 4, 2009 #2

    Mark44

    Staff: Mentor

    I don't understand what you are given here. The things that you are supposing are statements that are assumed to be true. All you have here are two expressions, f(z) and f(z) - conj(f(z)). Is there some relationship between these two expressions, such as an equation?
     
  4. Sep 4, 2009 #3
    Im sorry, the problem is: assume that t [tex] f(z) [/tex] and [tex] f(z) - conj(f(z)) [/tex] are both analytic in a domain D. Prove thet [tex]f(z)[/tex] is constant on D
     
  5. Sep 4, 2009 #4

    Mark44

    Staff: Mentor

    OK, now I follow what you're trying to do.
    I think you're waving your hands here. You have established from the given conditions that u(x, y) = 0, so what you have is (with derivatives changed to partial derivatives):
    [tex] - 2*i*\frac {\partial}{\partial x}v(x,y) = 0[/tex]
    This says that v(x, y) = g(x) + C, where g is a function of x alone.

    From your Cauchy-Riemann equations, what can you say about
    [tex]\frac {\partial}{\partial y}v(x,y)[/tex]?
     
  6. Sep 7, 2009 #5
    thanks for the reply.

    [tex] \frac {\partial}{\partial y} v(x,y) = 0 [/tex] by the riemann eq.

    so this says that

    v(x,y) = g(y) + C

    which is a contradiction to your first statement and hence

    v(x,y) = C

    does this make sense?

    thanks!
     
  7. Sep 7, 2009 #6

    Mark44

    Staff: Mentor

    Yes, makes sense.
     
  8. Sep 7, 2009 #7
    ok but doesn't that show that f(z) - conj(f(z)) is constant on D. how do I tie this into f(z) alone?
     
  9. Sep 8, 2009 #8

    Mark44

    Staff: Mentor

    No, it shows that f(z) is constant on D. Recall that you defined f(z) = f(x,y) = u(x,y) + i*v(x,y) back in your original post.
     
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