Is $\overline{f(z)}=f(\overline z)$ always?

Question

Given that $w=f(z)$ is a regular function of $z$ such that $f'(z)\neq0$. Is $\overline{f(z)}=f(\overline z)$ always?

This suggests that it is true for basic operations but is it always true when $f$ is an elementary or non elementary function? I am unable to come up with a counterexample.

I encountered this step in class while proving $\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)\log|f'(z)|=0$

Answer 1

If $\omega\in\mathbb C\setminus\mathbb R$ and if $f(z)=\omega z$, then we don't have$$(\forall z\in\mathbb C):\overline{f(z)}=f\left(\overline z\right),$$since, for instance, $\overline{f(1)}=\overline{\omega}$, whereas $f\left(\overline 1\right)=f(1)=\omega\neq\overline\omega$.

Answer 2

This is not true in general. If we choose any regular function $f$ for which there is an $a \in \Bbb R$ such that $f(a) \not\in \Bbb R$, then $$f(\bar a) = f(a) \neq \overline{f(a)} .$$ There are many such functions: If a chosen regular function $f$ whose domain intersects $\Bbb R$ admits no such $a$ then $i f$ does.

The statement is true, however, in an important case: The Schwarz Reflection Principle (sometimes called the Riemann-Schwarz Principle) implies that the identity holds when an analytic function $f$ (with connected domain) maps real values to real values, that is, when $f(\Bbb R \cap \operatorname{dom} f) \subseteq \Bbb R$.