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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?
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?