- #1
BSMSMSTMSPHD
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Problem: Suppose \Omega \in \mathbb{C} is open and connected, f is differentiable on \Omega, and f(z) \in \mathbb{R} , \ \forall z \in \Omega. Prove that f(z) 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 f(z) = u(x,y) + iv(x,y) where u and v are differentiable real-valued functions.
Since f(z) is real, v(x,y) = 0.
Then, the first CR equation tells us that \frac{ \delta u}{ \delta x} = \frac{ \delta v}{ \delta y}. But, since v(x,y) = 0, both of these fractions equal 0.
Next, the second CR equation tells us that \frac{ \delta u}{ \delta y} = - \frac{ \delta v}{ \delta x}. But, since v(x,y) = 0, both of these fractions equal 0.
So, if \frac{ \delta u}{ \delta x} = 0 = \frac{ \delta u}{ \delta y}, then u(x,y) 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 f(z) = u(x,y) + iv(x,y) where u and v are differentiable real-valued functions.
Since f(z) is real, v(x,y) = 0.
Then, the first CR equation tells us that \frac{ \delta u}{ \delta x} = \frac{ \delta v}{ \delta y}. But, since v(x,y) = 0, both of these fractions equal 0.
Next, the second CR equation tells us that \frac{ \delta u}{ \delta y} = - \frac{ \delta v}{ \delta x}. But, since v(x,y) = 0, both of these fractions equal 0.
So, if \frac{ \delta u}{ \delta x} = 0 = \frac{ \delta u}{ \delta y}, then u(x,y) must be some constant function.
Is that all I need?
