If $f:\mathbb{C}\longrightarrow\mathbb{C}$ is an entire function such that $f(z)\neq w$ for all $z\in \mathbb{C}$ and for all $w\in [0,1]\subset \mathbb{R}$, how to prove that $f$ is constant (without using Picard's little theorem).

Any hint would be appreciated.


closed as off-topic by user223391, B. Mehta, Isaac Browne, Xander Henderson, GNUSupporter 8964民主女神 地下教會 May 2 '18 at 11:54

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    $\begingroup$ What have you tried? As a weak hint, whenever a problem begins "Let $f$ be an entire function..." and ends with "...prove $f$ is constant", a particular theorem is begging to be used (or at least thought of). $\endgroup$ – Dan Z Aug 31 '14 at 3:09
  • $\begingroup$ Hint: Think about the codomain of the entire function $\frac{1}{f(z)}$. $\endgroup$ – Chris Janjigian Aug 31 '14 at 3:30

$f(\Bbb C)\subset \Bbb C\setminus [0,1]$, take $g=1-\dfrac{1}{f}$, $g$ is an entire function and $g(\Bbb C)\subset \Bbb C\setminus \Bbb R_-$, now we can composite with $\log$, and we obtain an entire function $h=\log \circ g$.

$h$ is an entire function and $h(\Bbb C)\subset \Bbb R\times ]-\pi, \pi [$.

To finish remark that the function $z\mapsto \dfrac{1}{h(z)-3\pi i}$ is an entire function and it is bounded, by the Liouville's theorem it is constant, hence $f$ is constant.


First show that there is an (injective) holomorphic mapping from $\Bbb C \setminus [0, 1]$ into the unit disk:

  • Start with $z \to 1 - \frac 1z$, this maps $\Bbb C \setminus [0, 1]$ into $\Bbb C \setminus (-\infty, 0]$.
  • Continue with the principal branch of the square root which maps $\Bbb C \setminus (-\infty, 0]$ to the right half-plane.
  • Finally, $z \to \frac{z-1}{z+1}$ maps the right half-plane to the unit disk.

Denoting the composition of these mappings with $\phi$ , $\varphi \circ f$ is a bounded holomorphic function.

Liouville's theorem implies thtat $\varphi \circ f$ is constant. Now conclude that $f$ is constant as well.


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