A question related to Liouville's Theorem

Question

Suppose f is entire and $\lvert f(z)\rvert \geq 1 $ on the whole complex plane. Then, f must be a constant function.

I know that if f is bounded on C then it is constant. But I couldn't relate this fact to this question. Unfortunately, I have any other ideas to prove this.

Actually, it looks very counter-intuitive to me, I don't even see why this is true. Because I know that if f is entire and $lim_{z \rightarrow \infty}f(z) = \infty$, then f is a polynomial. In particular, isn't this f (as in the limit) satisfies the condition in the question?

Any help is appreciated.

polynomials tend to have zeroes – Angina Seng Dec 04 '19 at 13:08 — Angina Seng, Dec 04 '19 at 13:08
@LordSharktheUnkown Not tend to but always. – trisct Dec 04 '19 at 13:11 — trisct, Dec 04 '19 at 13:11

score 2 · Accepted Answer · answered Dec 04 '19 at 13:08

2

Note that $\frac{1}{f(z)}$ is entire since $f(z) \neq 0,\forall z \in \Bbb{C}$

Also $|\frac{1}{f(z)}|\leq 1,\forall z \in \Bbb{C}$

Now apply Liouville's theorem $g(z)=\frac{1}{f(z)}$

answered Dec 04 '19 at 13:08

Marios Gretsas

21,906

A question related to Liouville's Theorem

1 Answers1