# Let f be entire and |f| >= 1. prove f is constant

1. Jun 7, 2014

### DotKite

1. The problem statement, all variables and given/known data
Let f(z) be entire and let |f(z)| ≥ 1 on the whole complex plane. Prove f is constant.

2. Relevant equations
Theorem 1: Let f be analytic in the domain D. If |f(z)| = k, where k is a constant, then f is constant.

Maximum Modulus Principle: Let f be analytic and non constant in the bounded domain D. If f is continuous on the closed region R that consists of D and all of its boundary points B, then |f(z)| assumes its max value, and does so only at points on the boundary B.

3. The attempt at a solution

Below is my attempt. Let me know if I am even in the right ballpark.

Proof:

Note that from Theorem 1 if |f(z)| = 1 then f is constant and we are done. Therefore we want to show f is constant for
|f(z)| > 1.

To show this we use contradiction. Suppose f(z) is entire, |f(z)| > 1, and f is non constant. Let D = {z: |z| < 1}. Since f is entire it is continuous on the complex plane. Consequently it is continuous on a region R consisting of D and its boundary points B. By the max modulus principle |f(z)| assumes its max value at the boundary points B. Therefore |f(z)| ≤ 1. Which is contradicts our hypothesis. Therefore f is constant.

2. Jun 7, 2014

### Staff: Mentor

What about an f where |f(z)|=1 somewhere, but not everywhere? You do not cover this case.

Okay so far.

Why? All you have is |z|<1.

That approach does not work.

3. Jun 7, 2014

### DotKite

ok can we consider 1/|f(z)|? which is analytic since we were given f is analytic and |f|≥ 1 on the whole complex plane?

Then 1/|f(z)| ≤ 1 for all z. By liouville's theorem 1/f(z) is constant. Am i warmer?

Last edited: Jun 7, 2014
4. Jun 8, 2014

That works.