If $f$ is a non constant entire function which is real on the real axis then $\arg f(\overline{z})=-\arg f(z) $

Question

If $f$ is a non constant entire function which is real on the real axis then prove that $$\arg f(\overline{z})=-\arg f(z) $$ where $z=x+iY$, $0<x<1$ and $f$ is non zero on the horizontal line segment $z=x+iY$, $0<x<1$.

By Schwarz reflection principle we have $$\arg f(\overline{z})=\arg \overline{f(z)} $$ Since $\arg$ is a multivalued function, so after this I am stuck. Please solve the question.

Answer 1

If $f$ is real on the real axis then the $a_n $ are real in the power series expansion.

Hence $\operatorname{im} f(\overline{z}) = \sum_n a_n \operatorname{im} \overline{z}^n = \sum_n a_n \operatorname{im} \overline{z^n} = \sum_n a_n (-\operatorname{im} z^n ) = - \operatorname{im} f(z) $ and similarly $\operatorname{re} f(\overline{z}) = \operatorname{re} f(z) $.

Answer 2

Following the comment &zhoraster fis real on real line so $\bar {f(\bar z)}=f(\bar z)$ Let $f(z)=u+iv, u,v \in \Re \implies Arg(f( z())= Arg(u+iv)=\tan^{-1} v/u$. Then $Arg(f(\bar z))=Arg(u-iv)=\tan^{-1} (-v/u)=-\tan^{-1} (v/u)=-Arg(f(z))$