Prove that no such $f$ exists

Question

I have come upon this question and could not find a solution for it.

Prove that there is no entire function $f$ such that $\forall z\in \mathbb{C}$ $\quad|f(z)|>|z|$.

Any ideas on how to solve this question?

Answer 1

Let $g(z)=\frac z {f(z)}$ for $z \neq 0$. (Note that $f(z) \neq 0$ for $z \neq 0$). Then $g$ is analytic in $\mathbb C \setminus \{0\}$. Since it is bounded it extends to an entire function $G$. This entire function is bounded. By Liouville's Theorem it must be a constant. This means $f(z)=cz$ for some constant, but then the given inequality does not hold at $0$.

Answer 2

Clearly there is no $z$ for which $f(z) = 0$. Therefore, $g(z) := \frac{z}{f(z)}$ is entire. Since $|z| < |f(z)|$, $|g(z)| < 1$ for all $z \in \mathbb{C}$. Liouville implies $g(z) \equiv c$ is constant, i.e. $z \equiv cf(z)$. Clearly $c \not = 0$, so $f(0) = 0$, contradiction.

Answer 3

We have that $f(z) \ne 0$ for all $z$, hence the function $h(z):= \frac{z}{f(z)}$ is entire and we have $|h(z)| \le 1$ for all $z$.

Liouville tells us, that $h$ is constant. Thus, for some complex $c$ we have

$$\frac{z}{f(z)}=c$$

for all $z$. With $z=0$ we get $ c=0$, a contradiction.