Show that if \(\Re(f(z)) \geq 0\) for all \(z\in \CC\), then \(f\) is constant.
Solution
Define \(g(z) \da e^{-f(z)}\), then \[ \abs{g(z)} = e^{-\Re(f(z))} \leq e^0 = 1 .\] Since \(g\) is entire and bounded, \(g\) is constant and thus so is \(f\).
Entire functions with nonnegative real part are constant
E-ZDVLE
- Area
- Complex Analysis
- Topics
- Entire Functions, Harmonic Functions, Liouville's Theorem
- Status
- draft