Exercise · ZW-LIOUVILLE-THEOREM-003 · ★☆☆ Basic

Since \(\sin z\) is entire and not constant, what does Liouville give?

Choose one option.

Show answer
  • B. That it is unbounded on \(\mathbb{C}\)

Explanation

It is the contrapositive: a nonconstant entire function is unbounded. On the imaginary axis \(|\sin(iy)|=\sinh|y|\) grows without bound. The bound \(1\) holds only on the real axis.

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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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