Exercise · ZW-LIOUVILLE-THEOREM-009 · ★★★ Advanced

Let \(f\) be entire with \(|f(z)|\ge1\) for all \(z\). Which bounded entire function results?

Choose one option.

Show answer
  • C. \(g(z)=1/f(z)\), with \(|g|\le1\)

Explanation

Since \(f\) never vanishes, \(1/f\) is entire, and \(|1/f|\le1\). By Liouville \(1/f\) is constant, hence so is \(f\).

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Review the concept: Liouville's theorem

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

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