TITLE: Holomorphic functions: is it true that $f(\bar{z})=\overline {f(z)}$?
QUESTION [7 upvotes]: 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...

REPLY [2 votes]: 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 $\overline{\sum\cdots} = \sum\overline\cdots$ ?)