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DotKite
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Homework Statement
Let f(z) be entire and let |f(z)| ≥ 1 on the whole complex plane. Prove f is constant.
Homework 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.
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.