If $f$ is entire and omits one value from $\Bbb{C}$ such that $|f(z)|\leq 2|z|^3$, $\forall |z|\geq 2$, then $f$ is a constant function.

Question

Let $f$ be an entire and omits one value from $\mathbb{C}$ such that $|f(z)|\leq 2|z|^3,~\forall |z|\geq 2$. Then show that $f$ is a constant function.

I really don't have any idea to prove this. Kindly don't down vote it. Any hints will be appreciated. Thanks beforehand!

Answer 1

Picard's Little Theorem If $f\colon \mathbb C \to \mathbb C$ is entire, $f(\mathbb C)$ is either a singleton or contains the complement of singleton.

1. It suffices to show that all non-constant entire functions $f$ with $|f(z)|<=2|z|^3$ for $|z|\geq 2$ satisfy $f(\mathbb C)=\mathbb C$.

2. By the generalized Liouville Theorem (see the first answer here), $f$ is a polynomial.

3. It is equivalent to the Fundamental Theorem of Algebra that every non-constant polynomial $p\colon \mathbb C\to\mathbb C$ is surjective.

4. By 3. we have that $f(\mathbb C)=\mathbb C$, and the conclusion follows from 1.