Proving Constant Function from Cauchy-Riemann Equations

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SUMMARY

The discussion focuses on proving that a differentiable function \( f \) defined on an open and connected set \( \Omega \subset \mathbb{C} \) is constant if \( f(z) \in \mathbb{R} \) for all \( z \in \Omega \). By applying the Cauchy-Riemann equations, it is established that since the imaginary part \( v(x,y) = 0 \), both partial derivatives of the real part \( u(x,y) \) equal zero. Consequently, \( u(x,y) \) must be a constant function, confirming that \( f(z) \) is constant throughout \( \Omega \).

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  • Understanding of Cauchy-Riemann equations
  • Complex function theory
  • Knowledge of differentiability in the context of complex analysis
  • Familiarity with real-valued functions of complex variables
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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?
 
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It looks ok yes.
 

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