If $f$ is analytic in a disk $|z|<R$ then so is $g(z)=\overline{f(\bar z)}$ in the disk

Question

How to prove that if $f$ is analytic in a disk $|z|<R$ then $g(z)=\overline{f(\overline z)}$ is also analytic in the disk and also $f=g$ iff $f$ is real valued in $(-R,R)$

Answer 1

As $f(z)$ is analytic there is a power series for $f$ $$f(z)=\sum_{k=0}^\infty a_k z^k $$ We know $$g(z)=\overline{\sum_{k=0}^\infty a_k \overline{z}^k}=\sum_{k=0}^\infty \overline{a_k} z^k$$ When $f$ is real valued on $(-R,R)$ all $a_k$ must be real and hence $\overline{a_k}=a_k$.