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Liouville's theorem (complex analysis)
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===Fundamental theorem of algebra=== There is a short [[Fundamental theorem of algebra#Complex-analytic proofs|proof of the fundamental theorem of algebra]] using Liouville's theorem.<ref>{{cite book|author1=Benjamin Fine|author2=Gerhard Rosenberger|title=The Fundamental Theorem of Algebra|url=https://books.google.com/books?id=g0KHD7EIl4cC&pg=PA70|year=1997|publisher=Springer Science & Business Media|isbn=978-0-387-94657-3|pages=70β71}}</ref> {{Math proof|title=Proof ([[Fundamental theorem of algebra]])|drop=hidden|proof= [[Proof by contradiction|Suppose for the sake of contradiction]] that there is a nonconstant polynomial <math>p</math> with no complex root. Note that <math>|p(z)| \to \infty</math> as <math>z \to \infty</math>. Take a sufficiently large [[ball (mathematics)|ball]] <math>B(0, R)</math>; for some constant <math>M</math> there exists a [[sufficiently large]] <math>R</math> such that <math>1/|p(z)| < 1</math> for all <math>z \not\in B(0, R)</math>. Because <math>p</math> has no roots, the function <math>q(z) = 1/p(z)</math> is [[entire function|entire]] and [[holomorphic function|holomorphic]] inside <math>B(0, R)</math>, and thus it is also [[continuous function|continuous]] on its [[closed ball|closure]] <math>\overline B(0, R)</math>. By the [[extreme value theorem]], a continuous function on a closed and bounded set obtains its extreme values, implying that <math>1/|p(z)| \le C</math> for some constant <math>C</math> and <math>z \in \overline B(0, R)</math>. Thus, the function <math>q(z)</math> is bounded in <math>\mathbb C</math>, and by Liouville's theorem, is [[constant function|constant]], which contradicts our assumption that <math>p</math> is nonconstant. }}
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