entire function with ist imaginary part positive

Question

$f$ 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.

Is not f$(\mathbb{C}) \subseteq {z \in \mathbb{C} \mid \operatorname{Im} z \geq 0}$ enough? — Babai, Jul 26 '17 at 19:51

score 9 · Accepted Answer · answered Apr 20 '13 at 02:50

9

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.

answered Apr 20 '13 at 02:50

Clayton

24,751

thank you,thus the positive imagenary part could be useful. – sopin Apr 20 '13 at 03:24
@Clayton Is not f$(\mathbb{C}) \subseteq {z \in \mathbb{C} \mid \operatorname{Im} z \geq 0}$ enough? – Babai Jul 26 '17 at 19:51
@Babai: Of course it is enough, just see the note that I listed in the answer. – Clayton Jul 28 '17 at 16:29

entire function with ist imaginary part positive

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