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

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.

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

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

Here is a generalization.

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

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.

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.

• the Picard theorem is way too complicated for this question (proving the Liouville theorem takes 3 lines from the Cauchy integral formula) – reuns Oct 18 '16 at 15:20
• @user1952009 Agreed. I just wanted to give the OP a second approach to the problem for the sake of completeness. – glpsx Nov 16 '16 at 16:48