Is there a special property that unites all functions for which $f(x^*) = f^*(x)$ holds?
Naively I can show that it is true for any function that has a series expansion with only real coefficients, and does not hold for a function that has a series expansion with complex coefficients. But this statement is not very rigorous, at least because a series expansion might not be valid for the entire domain. Is there a more formal way to write this statement?
Just curious, no coursework involved
First of all, your condition assumes that, if $D$ is the domain of $f$, then$$z\in D\implies\overline z\in D.\tag1$$So, I will deal with a function whose domain is an open set that satisfies that condition and which is connected too. I will also assume that $f$ if analytic.
A connected set for which the condition $(1)$ holds must have a non-empty intersection with $\mathbb R$. Then the conditions
are equivalent. In fact, if the first condition holds and if $z\in D\cap\mathbb{R}$, then $\overline z=z$ and therefore $f(z)=f\left(\overline z\right)=\overline{f(z)}$, which implies that $f(z)\in\mathbb R$. And if the second condition holds, then let $g(z)=\overline{f\left(\overline z\right)}$. The function $g$ is analytic too (this follows from the Cauchy-Riemann equations, for instance) and $(\forall z\in D\cap\mathbb{R}):f(z)=g(z)$. Therefore, $f=g$, by the identity theorem. But then$$z\in D\implies f\left(\overline z\right)=g\left(\overline z\right)=\overline{f(z)}.$$