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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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2 Answers

up vote 1 down vote accepted

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
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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
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