Proving an entire, complex function with below bounded modulus is constant.

Question

How would I show that an entire function $f$ with the property $|f(z)| \geq 1$ must be constant?

I'm aware of Liouville's theorem, just not sure how to apply it here.

Answer 1

The map $z\mapsto 1/f(z)$ is

• well defined;
• entire;
• bounded on the complex plane.

Here is a generalization.

Answer 2

If $|f| \geq 1$, then $|\frac{1}{f}| \leq 1$. Apply Liouville.

Answer 3

As a stronger alternative to Liouville's theorem, it is worth mentioning that we can apply Picard's little theorem to this particular example.

Little Picard Theorem. If a function $f : \mathbb{C} → \mathbb{C}$ is entire and non-constant, then the set of values that $f(z)$ assumes is either the whole complex plane or the plane minus a single point.

In layman's terms, Picard's little theorem says that any entire analytic function whose range omits two points must be a constant function.

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Note that the image of $f$ is missing infinitely many points, namely anything in the unit disk. Thus, $f$ must be constant by Picard's little theorem.