surjective holomorphic function in a special domain

Question

Let $ D\subset \mathbb{C}$ be open, bounded, connected and with smooth boundary. Let $f$ be a nonconstant holomorphic function in a neighborhood of the closure of $D$ , such that $|f(z)|=c \forall z\in \partial D$, show that $f$ takes on each value $a$, such that $|a| < |c| $ at least once in $D$.

Answer 1

The underlying principle in this problem is the open mapping property for holomorphic functions. However, this problem can be cleaned up by using some more specialized results.

Claim 1. If $f(z)$ must vanish somewhere on $D$.

Proof: As $f$ is nonconstant, then by maximum modulus principle, $|f(z)| < c$ on $D$. However, if $f(z)$ doesn't vanish on $D$, then by the minimum modulus principle, $|f(z)| > c$, a contradiction.

Claim 2. For every $a$ such that $|a| < c$, $f(z) - a$ has a zero in $D$.

Proof: Notice that for all $z \in \partial D$, $|2f(z) - (f(z) - a)| = |f(z) + a| \le c + |a| < 2c = |f(z)|$. Therefore, by Rouche's theorem, the function $2f(z)$ and the function $f(z) - a$ must share the same number of zeros in $D$. By Claim 1, $f(z)$ vanishes somewhere in $D$, and hence $f(z) - a$ vanishes somewhere in $D$.

Answer 2

I should preface this by saying there must be a better solution than this.

Let $B$ be the open disk $B = \{z\in \mathbb{C} : |z|<c\}$. We want to show $B\subset f(D)$. Define $S = \{z\in B : z\notin f(D)\}$. First note that $S$ is closed in $B$, since by the open mapping theorem $f(D)$ is open, and $S = B\smallsetminus f(D)$.

We want to show next that $S$ is open in $B$. If $w\in S$, then, from the assumption that $|f(z)| = c$ for all $z\in \partial D$, we have that $w\notin f(\overline{D})$. Thus $S = B\smallsetminus f(\overline{D})$. Since $f(\overline{D})$ is compact and hence closed, we conclude $S$ is open in $B$.

Since $B$ is connected, it follows that $S = \varnothing$ or $S = B$. Note that the latter case cannot happen. Indeed, if $S = B$, then the maximum modulus principle implies that $f(D)\subset \{|z| = c\}$, which is not possible by the open mapping theorem. Thus $S = \varnothing$, completing the proof.

I don't see how the assumption of smoothness on the boundary comes in, so maybe there's a mistake in here somewhere.