An entire function Satisfying $| f ( z ) | \geqslant | z | ,~~~ \forall z \in \mathbb C$

Question

Let $f$ be an entire function Satisfying $| f ( z ) | \geqslant | z | ,~~~ \forall z \in \mathbb C$ Then cleary $f ( z )$ is a polynomial since $\lim _ { | z | \rightarrow \infty } \left|f ( z ) \right| = \infty$

But can we say that $f ( z ) = a z$ for some $a \in \mathbb C$ with $| a | \geqslant 1 ?$

Answer 1

Note that $|f(0)| >0$ for $z \neq 0$. If $f$ has a zero of order $n$ at $0$ then there exists an analytic function $g$ such that $f(z)=z^{n}g(z)$ and $g(0)\neq 0$. Now $|z^{n}g(z)| \geq |z|$ so $|z^{n-1}g(z)| \geq 1$ for $z \neq 0$. Letting $z \to 0$ we get the contradiction $0 \geq 1$ unless $n=1$. Hence the order of zero at $z=0$ is either $0$ or $1$. This implies that $\frac z {f(z)}$ is an entire function. Since $|\frac x {f(z)}| \leq 1$ it follows (by Louiville's Theorem) that $\frac z {f(z)}$ is a constant $c$ (which cannot be $0$). Hence $f(z)=\frac 1 c z$.

Answer 2

In general, if you have two entire function $f, g$ with $\lvert f \rvert \leq \lvert g \rvert$ you can show that $f = a g$ for some constant $a$ using that $f / g$ is bounded and applying Liouville.