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Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be a non-constant entire function.

  1. by Liouville's Theorem, $f(\mathbb{C})$ is dence in $\mathbb{C}$.
  2. by the Open Mapping Theorem $f(\mathbb{C})$ is open in $\mathbb{C}$.
  3. by a simple argument from general topology $f(\mathbb{C})$ is connected.

What more can we say, about $f(\mathbb{C})$?

Fly by Night
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Fin8ish
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    Have you heard of Picard's theorem yet? – Daniel Fischer Aug 18 '14 at 21:37
  • I saw it in wiki, but it seems not to be elementry like liouville and open mapping theorem. Does it have simple proof like liouville and open mapping theorem? – Fin8ish Aug 18 '14 at 21:49
  • No, not as simple. To construct the modular function $\lambda$, you need the reflection principle and some tedious work checking properties of a subgroup of automorphisms of the upper half-plane. But the statement of the theorem is simple and beautiful enough that it is worth the work. – Daniel Fischer Aug 18 '14 at 21:53
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    See also http://math.stackexchange.com/questions/820315/is-there-a-generalization-of-the-fundamental-theorem-of-algebra-for-power-series. – lhf Aug 18 '14 at 21:58

1 Answers1

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By Picard's little theorem, the image of a non-constant entire function is either $\mathbb{C}$ or $\mathbb{C}\setminus \{a\}$ for some $a\in\mathbb{C}$.

Ultimately, that is because the modular function $\lambda$ is a covering $\lambda \colon \mathbb{H} \to \mathbb{C}\setminus \{0,1\}$ where $\mathbb{H}$ is the upper half-plane, so by general properties of coverings, since $\mathbb{C}$ is simply connected, every entire function with values in $\mathbb{C}\setminus \{0,1\}$ lifts to an entire functions with values in $\mathbb{H}$ and is therefore constant. If the values of an entire function lie in $\mathbb{C}\setminus \{a,b\}$ with $a\neq b$, a linear transformation puts us in the situation $a = 0,\, b = 1$.

Daniel Fischer
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