# Complex function question

1. Sep 12, 2006

### BSMSMSTMSPHD

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?

2. Sep 12, 2006

### ircdan

It looks ok yes.