Q:

Holomorphic functions: is it true that $f(\bar{z})=\overline {f(z)}$?

Is it true that  $ f(\bar{z})=\overline {f(z)}$,
Where z is complex?
I think it holds when $f(z)$ is holomorphic since we have $f(z)=p(x,y)+iq(x,y)=p(z,0)+iq(z,0)$
Any help...

A:

It is true if (and only if) $f$ is real-valued on the real axis.
For simplicity, assume that $f$ is holomorphic on the whole plane. Let $g(z) = \overline{f(\bar z)}$. It follows from Cauchy-Riemann's equations that $g$ is also holomorphic everywhere, see for example this question
Hence, if $f$ is real on the real axis, it follows that $f(z) = g(z)$ for $z \in \mathbb{R}$. By the identity principle, $f(z) = g(z)$ for all $z \in \mathbb{C}$, i.e. that $\overline{f(z)} = \overline{g(z)} = f(\bar z)$.

