Proving zeros inside a disk using Maximum Principle?

Question

Let $f$ be non-constant analytic in a neighborhood of the closed unit disk such that $|f|=1$ on the unit circle. Show that $f$ has a zero inside the unit disk. It has been suggested to argue this by contradiction and using the maximum principle. However, is it possible to argue this question using properties of conformal maps since doesn't $f$ map the unit circle to itself? Regardless, I don't understand what information the Maximum Principle would give me to argue by contradiction. Why couldn't $0<|f|<1$ inside the disk?

Answer 1

If $f$ does not have a zero inside the unit disk, then $g=1/f$ is also analytic in a neighborhood of the closed unit disk with $|g|=1$ on the unit circle. Using the maximum principle implies that $|g(0)|\leq 1$, or equivalently $|f(0)|\geq1$, so $f$ attains its maximum on the closed unit disk at a point inside the disk. By the maximum principle this implies that $f$ is constant which is absurd. This contradiction proves that $f$ must have a zero inside the unit disk.