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Complex function question

  1. Sep 12, 2006 #1
    Problem: Suppose [tex] \Omega \in \mathbb{C} [/tex] is open and connected, [tex] f [/tex] is differentiable on [tex] \Omega [/tex], and [tex] f(z) \in \mathbb{R} , \ \forall z \in \Omega [/tex]. Prove that [tex] f(z) [/tex] is constant.

    Is this just a matter of solving the Cauchy-Riemann equations? If so, I think the proof is relatively straightforward.

    Since the function is differentiable, it can be written as [tex] f(z) = u(x,y) + iv(x,y) [/tex] where u and v are differentiable real-valued functions.

    Since [tex] f(z) [/tex] is real, [tex] v(x,y) = 0 [/tex].

    Then, the first CR equation tells us that [tex] \frac{ \delta u}{ \delta x} = \frac{ \delta v}{ \delta y} [/tex]. But, since [tex] v(x,y) = 0 [/tex], both of these fractions equal 0.

    Next, the second CR equation tells us that [tex] \frac{ \delta u}{ \delta y} = - \frac{ \delta v}{ \delta x} [/tex]. But, since [tex] v(x,y) = 0 [/tex], both of these fractions equal 0.

    So, if [tex] \frac{ \delta u}{ \delta x} = 0 = \frac{ \delta u}{ \delta y} [/tex], then [tex] u(x,y) [/tex] must be some constant function.

    Is that all I need?
     
  2. jcsd
  3. Sep 12, 2006 #2
    It looks ok yes.
     
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