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$f$ is an entire function, and it satisfies $f(\mathbb{C}) \subseteq \{z \in \mathbb{C} \mid \operatorname{Im} z > 0\}$. Show that $f$ is constant.

I want to take advantage of the Liouville's Theorem, but I just can't figure out the relationship between its image part with its module.

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  • $\begingroup$ Apply Liouville to 1/(f+ic), for a real positive c. $\endgroup$
    – D M
    Commented Apr 20, 2013 at 2:46
  • $\begingroup$ Is not f$(\mathbb{C}) \subseteq \{z \in \mathbb{C} \mid \operatorname{Im} z \geq 0\}$ enough? $\endgroup$
    – Babai
    Commented Jul 26, 2017 at 19:51

1 Answer 1

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Consider $$g(z)=\frac{1}{f(z)+i}.$$ Clearly it is entire, and $$\left|g(z)\right|=\frac{1}{|f(z)+i|}\leq\frac{1}{1}=1.$$ Thus, $g(z)$ is constant. Applying Liouville's theorem, $g(z)$ is constant, implying $f(z)$ is constant.

NOTE: This argument can be used to show that $f(\Bbb C)$ is dense in $\Bbb C$ by using a proof by contradiction.

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  • $\begingroup$ thank you,thus the positive imagenary part could be useful. $\endgroup$
    – sopin
    Commented Apr 20, 2013 at 3:24
  • $\begingroup$ @Clayton Is not f$(\mathbb{C}) \subseteq \{z \in \mathbb{C} \mid \operatorname{Im} z \geq 0\}$ enough? $\endgroup$
    – Babai
    Commented Jul 26, 2017 at 19:51
  • $\begingroup$ @Babai: Of course it is enough, just see the note that I listed in the answer. $\endgroup$
    – Clayton
    Commented Jul 28, 2017 at 16:29

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