I am studying elementary complex variable topic. An exercise in Brown-Churchill book asks us to show that if $f$ is analytic in a domain $D$ and real-valued at all points then $f$ must be constant throughout $D$.
This should be doable by following the proof of showing zero derivative implies constant. However, can I use the fact (which is already proved in the book): If a function and its conjugate are both analytic in $D$, then $f$ is constant there. ???
I am thinking that since in the exercise, $f$ and its conjugate are equal because $f$ is real-valued, then they are both analytic as $f$ is, and the conclusion follows. Is this all right?
Any help is appreciated! Thanks for the guidance!
Here is an application of Cauchy Riemann equations.
Suppose $f = u + iv$ be real valued , analytic function then $v = 0$. So by Cauchy Riemann equation we must have $u_x = u_y = 0$ implying that $u$ is a constant and consequently $f$ is constant.