# Sufficient conditions for $f$ to be a rotation of $g$

Assume that $f$ and $g$ are analytic and nonvanishing on $B_{2}(0)=$ {$z \in \mathbb{C} :|z|<2$} and that $|f(z)|=|g(z)|$ when $|z|=1$. Can we show that there is an $|a|=1$ such that $f(z)=ag(z)$ for any $z$ in $B_{2}(0)$?

I believe we should look at $f/g$ which would also be analytic. I feel like the $a$ will come out of Schwarz Lemma. Thanks for the help.

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You're right, you want to look at $f/g$ (call this $h$). However, Schwarz Lemma would only help if $f(0) = 0$. Instead, note that $1/\overline{h(1/\overline{z})}$ is analytic in $\{z: |z| > 1/2\}$ and coincides with $h(z)$ for $|z|=1$. What can you conclude from this?