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How to show that if $f:\mathbb{C}\to \mathbb{C}$ is an entire function with $f(\mathbb{C})\subset G:=\mathbb{C}\setminus [0, \infty)$, then $f$ is constant.

Approach: I showed that there is a conformal map $g$ that maps $G$ into the unit disc $\mathbb{D}$. So $g\circ f$ is an entire function that maps $\mathbb{C}$ into the unit disc. This function is bounded, so by Liouville's theorem it is constant. How can I conclude from this that $f$ should be constant? Thanks in advance.

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    $\begingroup$ Why not use Little Picard? $\endgroup$ Jun 22, 2017 at 14:13
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    $\begingroup$ Okay thanks, then it's obvious. But, we didn't prove that theorem, so I want to prove it in a different way. $\endgroup$
    – bob
    Jun 22, 2017 at 14:16

2 Answers 2

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Hint : you know that $g$ is an open map since it's conformal, and open maps are stable by composition.

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  • $\begingroup$ Yes, many thanks! So if $f$ is non-constant, it is also an open map. So $f\circ g$ is an open map. But since it is constant, you get a contradiction, right? $\endgroup$
    – bob
    Jun 22, 2017 at 14:20
  • $\begingroup$ Exactly ! ${{{}}}$ (I assume you mean $g \circ f$ ) ? $\endgroup$
    – user171326
    Jun 22, 2017 at 14:20
  • $\begingroup$ Oh yes indeed! Thanks! $\endgroup$
    – bob
    Jun 22, 2017 at 14:23
  • $\begingroup$ @bob glad I could help ! $\endgroup$
    – user171326
    Jun 22, 2017 at 14:25
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We can choose $g$ to be a holomorphic bijection of $G$ onto $\mathbb D.^1$ Since $g\circ f$ is constant, so is $g^{-1}\circ (g\circ f) =f.$

$^1$ For example, $g(z) = \dfrac{\sqrt z -1}{\sqrt z +1}.$

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    $\begingroup$ $g^{-1}$ is not necessarily well-defined. $\endgroup$ Jun 22, 2017 at 14:25
  • $\begingroup$ $g$ does not need to be invertible, e.g $z \mapsto e^z$. $\endgroup$
    – user171326
    Jun 22, 2017 at 14:26
  • $\begingroup$ @ThomasAndrews I was taking $g$ to be a holomorphic bijection of $G$ onto $\mathbb D.$ I'll edit to make sure that is understood. $\endgroup$
    – zhw.
    Jun 22, 2017 at 15:13
  • $\begingroup$ @N.H. I'm not sure why you mention $e^z$ as it takes $G$ onto the plane minus the origin. $\endgroup$
    – zhw.
    Jun 22, 2017 at 15:18
  • $\begingroup$ My point was that conformal maps doesn't need to be invertible, but of course you can pick a bijective conformal map. $\endgroup$
    – user171326
    Jun 22, 2017 at 15:22

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