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  1. Let $\Omega \subset \mathbb{C}$ open. How do I prove that if $f(\Omega) \subseteq \text{ a line }$ then $f$ is constant?
  2. How do I prove that if $f$ is holomorphic in $\mathbb{C}$ and there exists $r>0$ such that $f(\mathbb{C})\subset \mathbb{C}-B(0,r)$, then $f$ is constant?

2 Answers 2

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  1. Since $f$ is holomorphic it maps open sets to open sets,since $f(\Omega)\subset \text{line}\implies f \text{is constant}$
  2. Use Picards Theorem ,If the range of $f$ excludes more than two points of $\Bbb C$ then $f$ is constant.Here it excludes uncountably many points.
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    The first line seems bizzare: take $f:\Bbb C \to \Bbb C$, $\forall z\,f(z)=0$. It is a holomorphic function, yet it maps an open set $\Bbb C$ to a non-opet set $\{0\}$, – 2017-01-09
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    @TZakrevskiy It is actually if $f$ is holomorphic and not constant it is an open map (see https://en.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis)). – 2017-01-09
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    @TZakrevskiy; I think the comment is useless here because that is clear from the answer – 2017-01-09
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    @learnmore Also you could use Picards Theorem on the first case and the open mapping theorem on the second case (assuming your disc is open). But still nice to see both in action. – 2017-01-09
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    Yes that's true @ctst – 2017-01-09
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    @ctst thank you for this clarification. – 2017-01-09
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    @learnmore I would not say so, because this is not clear from your answer. A slight reformulation "If $f$ is not constant and holomorphic, then it maps..." eliminates all the confusion once and for all. – 2017-01-09
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    Thanks a lot. That's a very clear and straightforward reasoning. One bonus question: can you think of a more "hands-on" way to prove the statements without using these powerful theorems? – 2017-01-09
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    Ummm.. Open mapping Theorem is a very elementary theorem in Complex Analysis;If you want to neglect Picard's Theorem you can try the second one using Open Mapping Theorem also – 2017-01-09
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    In case you are satisfied with the answer ,you can accept it – 2017-01-10
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    I'm wondering this: the open map theorem holds if $\Omega$ is connected. Does 1 hold if $\Omega$ is just open? Or is there a counterexample? – 2017-01-14
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    No the result holds if $\Omega$ is open only ;you will only need the whole space to be open & connected and here the whole domain is $\Bbb C$ – 2017-01-14
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A proof of 2. without Picard:

We have $|f(z)| \ge r$ for all $z \in \mathbb C$. Let $g=1/f$. Then $g$ is a bounded entire function. By Liouville, $g$ is constant, hence $f$ is constant.