Suppose $R(z)$ be a rational function such that $|R(z)|=1$ for $|z|=1$. Show that $\alpha$ is a zero or a pole or order $m$, if and only if $1/\overline{\alpha}$ is a pole or zero or order $m$ respectively.
I am wondering that if I can first show that $M(z)=R(z)$ the complex conjugate of $R(1/z)$ is a rational function such that $M(z)=1$ on $|z|=1$. Can I assume $z=1$ and then substitute all $z$ with $1$? Thank you very much.
Hint. Note that $F(z):= R(z)\cdot \overline{R(1/\overline{z})}$ is a the rational function such that for $|z|=1$: $$F(z)=R(z)\cdot \overline{R(1/\overline{z})}=R(z)\cdot \overline{R(z)}=|R(z)|^2=1.$$ Since a non-constant rational function attains a value only finitely often, it follows that $F$ is identically equal to $1$. Hence $$R(1/\overline{z}) = \frac{1}{\overline{R(z)}}.$$