If $|f(z)| \geq |g(z)|$ for $z \in D$ and $E = \{ z \in D : |f(z)| =|g(z)| \}$ has a limit point, then $E=D$. 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. 
 A: I think that you can use the maximum modulus principle to get what you want. 
Suppose first that $f$ has no zeros in the unit disk $D$. Then that means that $\dfrac{g(z)}{f(z)}$ is analytic in $D$. Then your condition that $|f(z)| \geq |g(z)|$ for every $z \in D$ translates to
$$\left | \frac{g(z)}{f(z)} \right | \leq 1 \quad \text{for every $z \in D$}$$
Then since $E$ has a limit point in $D$, there is a sequence of points $z_n \in E$ such that $z_n \to a$ for some $a \in D$ and since for every $z_n \in E$ we have $\left | \dfrac{g(z_n)}{f(z_n)} \right | = 1$
then by continuity also $$\left | \frac{g(z_n)}{f(z_n)} \right | \to \left | \frac{g(a)}{f(a)} \right | $$
so $\left | \dfrac{g(a)}{f(a)} \right | = 1$. Thus we have that 
$$\left | \frac{g(z)}{f(z)} \right | \leq \left | \frac{g(a)}{f(a)} \right | \quad \text{for every $z \in D$}$$
so by the maximum modulus principle we conclude that $g/f$ is constant, say $g/f = c$ and $|c| = 1$. Then this implies that $|f(z)| = |g(z)|$ for all $z \in D$ and this implies that $E = D$ as you wanted to show.

Now, what happens if $f$ has zeros in $D$?
In this case you can use the inequality $|f(z)| \geq |g(z)|$ to conclude that if $f$ has a zero of order $n$ at $a \in D$, then $a$ is also a zero of $g$ of order $m$ and actually $m \geq n$. This implies that the quotient $g/f$ has removable singularities at the zeros of $f$ and then we can redefine the function at those points and again we would have $g/f$ analytic in $D$ and the previous argument applies.
