Exercise · ZW-LIOUVILLE-THEOREM-008 · ★★★ Advanced
Let \(f\) be entire with \(\operatorname{Re} f(z)\le0\) for all \(z\). To which bounded function would you apply Liouville to show \(f\) is constant?
Choose one option.
A \(g(z)=f(z)^2\) B \(g(z)=e^{f(z)}\) C \(g(z)=1/f(z)\) D \(g(z)=\overline{f(z)}\) Show answer ← Back to the question
Explanation \(|e^{f}|=e^{\operatorname{Re}f}\le1\), and \(e^{f}\) is entire. It is constant, and hence so is \(f\). \(f^2\) and \(\bar f\) need not be bounded; \(1/f\) may have poles .
Liouville's theorem says that every entire function \(f\), that is, holomorphic on the whole plane, bounded by a constant \(M\) is constant.
\[|f(z)|\le M\ \ \forall z\in\mathbb{C}\ \Longrightarrow\ f\ \text{constant}\]
More practice problems: liouville's theorem