Problem: Suppose $f$ is a non-vanishing continuous function on $\bar{\mathbb{D}}$(closure of unit disk) that is holomorphic in $\mathbb{D}$. Prove that if $$|f(z)|=1, \mbox{whenever} \hspace{1cm}|z|=1$$ then $f$ is constant.
Hint: Extend $f$ to all of $\mathbb{C}$ by $$F(z)=\frac{1}{\overline{f\left(\frac{1}{\overline{z}}\right)}}$$ for $|z|>1.$
Solution:
This problem has a quick solution using the Maximum Modulus Principle. Doing it as the hint suggests, I'm having a little trouble seeing why is $F(z)$ holomorphic outside of $\mathbb{D}$?
Best Answer
$\frac{1}{\bar z}$ is antiholomorphic, so is $f(\frac{1}{\bar z})$ thus $F(z) = \frac{1}{\overline{f(\frac{1}{\bar z}})}$ is holomorphic. I believe it is something called Schwarz reflection principle. You need also to check that $F$ is holomorphic on the boundary of $D$.