# surjective holomorphic function in a special domain

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 $D$.

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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 $D$.

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Very elegant solution! – Daniel Oct 15 '12 at 17:38

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.

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I did not understood , how you concluded that S is open – Daniel Oct 15 '12 at 1:11
@Daniel: The first step is to show $S = B\smallsetminus f(\overline{D})$. By definition, $S = B\smallsetminus f(D)$. But by assumption $f(\partial D)\cap B = \varnothing$, so it follows that $S = B\smallsetminus f(\overline{D})$. Since $f(\overline{D})$ is closed, $S$ is open. – froggie Oct 15 '12 at 1:16
Sorry for being so stupid , I'm still not understanding why $w\notin \overline{f(D)}$ – Daniel Oct 15 '12 at 1:41
@Daniel: Not a problem. We've taken $w\in S$. By the definition of $S$, this means that $|w|<c$ and $w\notin f(D)$. If $z\in \partial D$, then we know $|f(z)| = c$. Since $|w|<c$, this implies that $f(z)\neq w$. From this we can conclude that $w\notin f(\partial D)$. On the other hand, by assumption $w\notin f(D)$. Combining these gives $w\notin f(\overline{D}) = f(D)\cup f(\partial D)$. – froggie Oct 15 '12 at 2:34