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Find all the holomorphic functions $f$ holomorphic on an open set $G$ containing the closed unit ball $\bar{\mathbb{D}}$ such that $|f(z)|=1$ for every $z$ with $|z|=1$.

I think that the functions are of the form $f(z)=cz^n$ for $n\geq 0$ and $|c|=1$. I am also looking, if possible, for a solution that does not make any use of the Schwarz's Reflection Principle.

user163644
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  • There are more such functions. For example every automorphism of the disk has that property. The product of finitely many functions with that property has that property. – Daniel Fischer Aug 02 '14 at 14:15
  • Yes you are right. So we are dealing with Blasche products? – user163644 Aug 02 '14 at 14:52
  • With a "k", Blaschke. Can you prove it? – Daniel Fischer Aug 02 '14 at 15:00
  • as a simple explicit example you can look at $z \to \frac{1+az}{a+z}$ where $a$ is a real number of absolute value $\gt 1$. for $|z|=1$ we have $\left| \frac{1+az}{a+z} \right|^2 =\frac{(1+az)(1+az^)}{(a+z)(a+z^)} =\frac{1+a^2+2a\mathfrak{Re}(z)}{a^2+1+2a\mathfrak{Re}(z)}=1$ – David Holden Aug 02 '14 at 15:23

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As you guessed in a comment, such functions are exactly (finite) Blaschke products. Indeed, the assumptions imply that the function $f$ has finitely many zeros in the unit disk. Construct a Blaschke product $B$ that has the same zeros with same multiplicities. Argue that $f/B$ is a constant.