Exercise · ZW-FUNDAMENTAL-THEOREM-OF-ALGEBRA-009 · ★★★ Advanced

In the proof via Liouville, if \(p\) had no roots, which function would be entire and bounded?

Choose one option.

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  • C. \(1/p\)

Explanation

\(1/p\) would be entire (no zeros to produce poles) and \(|1/p(z)|\to0\) as \(|z|\to\infty\), hence bounded; by Liouville it would be constant, and so would \(p\): a contradiction.

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Review the concept: Fundamental theorem of algebra

Every non-constant polynomial \(p\) with complex coefficients, of degree \(n\), has at least one root in \(\mathbb{C}\); hence it splits into \(n\) linear factors \(z-z_k\), with leading coefficient \(a_n\).

\[p(z)=a_n\prod_{k=1}^{n}(z-z_k)\]

More practice problems: fundamental theorem of algebra

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