proving $h(1/z)= \overline{1/h(\overline{z})}$

Question

Suppose $h$ is a holomorphic function on the disk $B_2(0)$ such that $|h(z)=1$ if $|z|=1$.

I want to prove that $h(1/z)= \overline{1/h(\overline{z})}$ when $1/2<|z|<2$.

I wanted to use schwarz Lemma but I don't know if the image of the disk is a disk or if $h(0)=0$. I tried constructing an other function so that I could apply Schwarz Lemma to the comoposition but I couldn't.

Answer 1

$$ g(z)=h(1/z)·\overline{h(\bar z)}-1 $$ is a holomorphic function on $\frac12<|z|<2$ with $g(z)=0$ for $|z|=1$, as on the unit circle $1/z=\bar z$. This implies $g(z)=0$ everywhere on that annulus as otherwise the roots of $g$ would have to be isolated

Answer 2

A hint:

Use the reflection principle. Its basic version is the following: If $f$ is analytic in a disc $D_r$ around the origin, and if $f(x)\in{\mathbb R}$ for $-r<x<r$ then $$f(\bar z)=\overline{f(z)}\quad\forall \ z\in D_r\ .$$