Exercise · ZW-LIOUVILLE-THEOREM-010 · ★★★ Advanced

Let \(f\) be entire with \(f(z+1)=f(z)\) and \(f(z+i)=f(z)\) for all \(z\). What follows?

Choose one option.

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  • A. \(f\) is constant

Explanation

All values of \(f\) repeat on the square \([0,1]\times[0,1]\), which is compact, so \(f\) is bounded. By Liouville it is constant. \(e^{2\pi iz}\) does not satisfy \(f(z+i)=f(z)\).

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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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