TITLE: Characterizing non-constant entire functions with modulus $1$ on the unit circle
QUESTION [44 upvotes]: Is there a characterization of the nonconstant entire functions $f$ that satisfy $|f(z)|=1$ for all $|z|=1$?

Clearly, $f(z)=z^n$ works for all $n$. Also, it's not difficult to show that if $f$ is such an entire function, then $f$ must vanish somewhere inside the unit disk. What else can be said about those functions?
Thank you.

REPLY [4 votes]: For your $f$ we know there are no zeros on the boundary of the circle, and only a finite number of zeros inside. So you can form the Blaschke product $b$ for the non-zero zeros of $f$ inside the circle, and $f(z)/(z^{n}b(z))$ is non-vanishing and holomorphic on an open neighborhood of the closed unit disk for some natural number $n$, and has modulus $1$ on the unit circle. Of course, it may be that $f$ has no non-zero zeros inside the disk, in which case, define $b\equiv 1$.
So $\ln|f(z)/(z^{n}b(z))|$ is harmonic on an open neighborhood of the closed unit disk, and it vanishes on the unit circle. So this harmonic function is identically $0$, which means $|f(z)/(z^{n}b(z))|\equiv 1$ inside the disk. Now, by the maximum modulus principle for holomorphic functions, $f(z)=Cz^{n}b(z)$ for some unimodular constant $C$, and for all $|z| \le 1$. By the identity theorem, this must be true for all $z$ where $b(z)$ is finite, which means that the Blaschke factor $b$ must not be present in the factoring because $f$ is entire. Therefore, $f=Cz^{n}$ for some constant $C$ with $|C|=1$ and for some $n \ge 0$.