Show that if \(f\) is entire and \(\abs{f(z)} > 1\) for all \(z\), then \(f\) is constant.
Solution
The inequality forces \(f\neq 0\) anywhere, so \(1/f\) is entire and bounded by 1. By Liouville, \(1/f\) is constant, and thus so is \(f\).
Entire functions with $|f(z)|>1$ are constant
E-U2X4C
- Area
- Complex Analysis
- Topics
- Entire Functions, Liouville's Theorem
- Status
- draft