Calculating the number of zeroes in a function given analyticity on a set I'm posed with the following problem:
Let $D = \{z : |z| < 1\}$. Let $f$ be analytic and non-constant on $W$.
Suppose that $\overline{D} \subset W$ (the closure of $D$ is contained in $W$), and that $|f|$ is constant on $\partial D$, denoting the boundary of $D$.
I need to prove that $f$ has at least one zero in $D$.
What is a good way to approach this problem?
 A: The Schwarz reflection principle is stated as follows:

Suppose that $F$ is a continuous function on the closed upper half plane $\overline{\Pi^+}=\{z\in\mathbb{C}|\text{Im}(z)\geq 0\}$, analytic on the upper half plane $\Pi^+=\{z\in\mathbb{C}|\text{Im}(z)>0\}$, which takes real values on the real axis. Then $F$ can be extended to a analytic function on the entire plane by letting $\overline{F(\overline{z})}$ for $z\not\in\Pi^+$.

This isn't quite the situation we have here, but we can make it so. Recall the Cayley Transform (under heading conformal map), which is a conformal map $\varphi:\Pi^+\to D$ given by
$$
\varphi(z)=\frac{z-i}{z+i},\hspace{.5 in}\text{having inverse}\hspace{.5 in}\varphi^{-1}(z)=i\frac{1-z}{1+z}.
$$
Since $|f|=r$, for some constant $r$, when $z\in\partial D$, it follows that $f/r:D\to D$. (It's worth mentioning this is not true for the trivial case where $r=0$. In this case, the zeros of $f$ have an accumulation point and hence $f$ is identically zero, which we cannot have by hypothesis.) Define $F=\varphi^{-1}\circ (f/r)\circ\varphi:\Pi^+\to\Pi^+$. It can verified that $F$ maps reals to reals and so the Schwarz reflection principle asserts that $F$ can be extended analytically to an entire function by letting $\overline{F(\overline{z})}$ for $z\not\in\Pi^+$. I'll leave it to you to show that further messing with $\varphi$ and $\varphi^{-1}$ (and noting that $\varphi$ maps to the unit circle when $z$ is on the real axis) shows that $f/r$ can be extended to an entire function $g$ given by
$$
 g(z) =
  \begin{cases}
   f(z)/r & \text{if } |z| \leq 1 \\
   r/\overline{f(1/\overline{z})}       & \text{if } |z| >1
  \end{cases}.
$$
So $g$ is an entire function (which are known to have much more strict behavior than simply analytic functions) which depends wholly on the values of $f$ on $\overline{D}$. 
I believe this to right direction to go, unless it can be shown using more simplistic methods. See if this steers you in the right direction and let me know if you would like clarification on anything.
