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 $D$.
surjective holomorphic function in a special domain
2 Answers
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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0Very elegant solution! – 2012-10-15
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|
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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0@Daniel: Not a problem. We've taken $w\in S$. By the definition of $S$, this means that |w|
and $w\notin f(D)$. If $z\in \partial D$, then we know $|f(z)| = c$. Since |w| , 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 $w\notin f(\overline{D}) = f(D)\cup f(\partial D)$. – 2012-10-15