This is my problem:
Let $D := \{ z \in \mathbb{C} : |z| <1 \}$. Let $f$ and $g$ be analytic functions on $D$. Suppose $|f(z)| \geq |g(z)|$ for all $z \in D$. Define $E = \{ z \in D : |f(z)| =|g(z)| \}$. Show that if $E$ has a limit point in $D$, then $E=D$.
This looks like it should be related to the identity theorem for analytic functions, but I can't seem to get it to work out.
Going for a proof by contradiction, I assume that $E \;$ has a limit point in $D\;$ and that $E \neq D\;$. Then there is a sequence $z_n$ of points in $E$ which converges to a limit point $z_0$ in $D$. By continuity, $z_0 \in E$; that is, $f(z) = re^{i\theta}$ and $g(z) = re^{i\phi}$ for some $r$, $\theta$, and $\phi$.
And here I stop. Any advice on how to proceed?
Thanks.