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[parent] proof of the fundamental theorem of algebra (Liouville's theorem) (Proof)

Let $ f \colon \mathbb{C}\rightarrow\mathbb{C}$ be a polynomial, and suppose $ f$ has no root in $ \mathbb{C}$. We will show $ f$ is constant.

Let $ g=\frac{1}{f}$. Since $ f$ is never zero, $ g$ is defined and holomorphic on $ \mathbb{C}$ (ie. it is entire). Moreover, since $ f$ is a polynomial, $ \vert f(z)\vert\rightarrow\infty$ as $ \vert z\vert\rightarrow\infty$, and so $ \vert g(z)\vert\rightarrow 0$ as $ \vert z\vert\rightarrow\infty$. Then there is some $ M>0$ such that $ \vert g(z)\vert<1$ whenever $ \vert z\vert>M$, and $ g$ is continuous and so bounded on the compact set $ \{ z\in\mathbb{C}:\vert z\vert\leq M\}$.

So $ g$ is bounded and entire, and therefore by Liouville's theorem $ g$ is constant. So $ f$ is constant as required.$ \square$



"proof of the fundamental theorem of algebra (Liouville's theorem)" is owned by Evandar.
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Cross-references: Liouville's theorem, compact set, bounded, continuous, entire, holomorphic, root, polynomial

This is version 3 of proof of the fundamental theorem of algebra (Liouville's theorem), born on 2002-02-13, modified 2004-09-16.
Object id is 1928, canonical name is ProofOfTheFundamentalTheoremOfAlgebra.
Accessed 6882 times total.

Classification:
AMS MSC: 12D99 (Field theory and polynomials :: Real and complex fields :: Miscellaneous)
 30A99 (Functions of a complex variable :: General properties :: Miscellaneous)
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