I'm struggling a bit to arrive at the conclusion that $f(z)$ is a constant.
Suppose $f(z)$ is holomorphic on and inside the unit disk, and that it has no zeroes on the interior. Also, assume that $|f(e^{i\theta}|$ = 1, for $0<\theta< 2\pi$.
Then by the maximum principle, the maximum of $f(z)$ is attained on the boundary -- that is, the unit circle. But since $g:=\large \frac {1}{f(z)}$ is holomorphic on and inside the unit disk, too, then it also attains its maximum on the boundary, and so this forces $f(z)$ to be minimal on the circle as well.
Knowing now that $f(z)$ attains both its max and min on the circle, how can I conclude that it is constant? This is tricky of course since $f(z)$ could assume infinitely many different values that have modulus = 1.
So, I hope to make use of the open mapping theorem.
Is this too ambitious? If I can say that the open set, namely the interior of the unit disk, maps onto the unit circle in the w-plane, then we know that the image is compact, and by Heine-Borel's theorem, the image is closed and bounded. In particular, the image is closed. By the open mapping theorem, the holomorphic function $f(z)$ must be constant.
...unfortunately I don't see why $f(z)$ has to be an onto mapping carrying the interior of the unit disk onto the unit circle in the w-plane.
Any hints or comments are welcome.
Thanks,