Entire functions, such that $g(\mathbb{C})\setminus h(\mathbb{C})=\{ z\in \mathbb{C}: |z|\geq 1\}$

Question

Proof, there are no holomorphic functions $g, h:\mathbb{C}\to\mathbb{C}$, such that $g(\mathbb{C})\setminus h(\mathbb{C})=\{ z\in \mathbb{C}: |z|\geq 1\}$ Can I assume, that $h(\mathbb{C})$ must be the unit disc, so by that h must be constant, therefore there are no such functions? Or that it must be bounded somehow? Any hints would be helpfull.

Answer 1

Note that $h(\mathbb{C})\subseteq \mathbb{D}=\{z\in\mathbb{C}: |z|<1\}$.

Now, $h$ is entire and it's bounded given the above. Then, by Liouville's Theorem, $h$ is constant.

As we have that $g(\mathbb{C})\setminus h(\mathbb{C})=\{ z\in \mathbb{C}: |z|\geq 1\}$ and $h(\mathbb{C})=\{z_0\}$ for some $z_0\in \mathbb{D}$, we deduce that $g(\mathbb{C})$ is at most $\{ z\in \mathbb{C}: |z|\geq 1\}\cup \{z_0\}$.

But it is well known that the image of an non-constant entire function is dense in the complex plane, so $g$ must be constant too. Then, it is obvious that $g(\mathbb{C})\setminus h(\mathbb{C})\neq\{ z\in \mathbb{C}: |z|\geq 1\}$.