2
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ Why not use Little Picard? $\endgroup$ – Sahiba Arora Jun 22 '17 at 14:13
  • 1
    $\begingroup$ 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 '17 at 14:16
4
$\begingroup$

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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 '17 at 14:20
  • $\begingroup$ Exactly ! ${{{}}}$ (I assume you mean $g \circ f$ ) ? $\endgroup$ – user171326 Jun 22 '17 at 14:20
  • $\begingroup$ Oh yes indeed! Thanks! $\endgroup$ – bob Jun 22 '17 at 14:23
  • $\begingroup$ @bob glad I could help ! $\endgroup$ – user171326 Jun 22 '17 at 14:25
3
$\begingroup$

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

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ $g^{-1}$ is not necessarily well-defined. $\endgroup$ – Thomas Andrews Jun 22 '17 at 14:25
  • $\begingroup$ $g$ does not need to be invertible, e.g $z \mapsto e^z$. $\endgroup$ – user171326 Jun 22 '17 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 '17 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 '17 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 '17 at 15:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.