Fundamental Theorem of Algebra/Proof 2

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Every non-constant polynomial with coefficients in $\C$ has a root in $\C$.


Let $\map P z = a_n z^n + \dots + a_1 z + a_0, \ a_n \ne 0$.

Aiming for a contradiction, suppose that $\map P z$ is not zero for any $z \in \C$.

It follows that $1 / \map P z$ must be entire; and is also bounded in the complex plane.

In order to see that it is indeed bounded, we recall that $\exists R \in \R_{>0}$ such that:

   $\cmod {\dfrac 1 {\map P z} } < \dfrac 2 {\cmod {a_n} R^n}, \text{whenever} \ \cmod z > R$.

Hence, $1 / \map P z$ is bounded in the region outside the disk $\cmod z \le R$.

However, $1 / \map P z$ is continuous on that closed disk, and thus it is bounded there as well.

Furthermore, we observe that $1 / \map P x$ must be bounded in the whole plane.

Through Liouville's Theorem, $1 / \map P x$, and thus $\map P x$, is constant.

This is a contradiction.