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...

Solution 1:

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 $\overline{f(z)} = \overline{g(z)} = f(\bar z)$.

Solution 2:

Use that $f$ is locally a power series: $$f(z) = \sum_{n=0}^\infty a_n(z-c)^n,$$ $${f(\bar{z})} = \sum_{n=0}^\infty a_n(\bar{z}-c)^n,$$ $$\overline{f(z)} = \sum_{n=0}^\infty \bar{a_n}(\bar{z}-\bar{c})^n.$$ (Why $\overline{\sum\cdots} = \sum\overline\cdots$ ?)