Proof of the Fundamental Theorem of Algebra 📂Complex Anaylsis

Proof of the Fundamental Theorem of Algebra

Theorem ¹

An $n$th-degree polynomial function $P(x) = a_{0} + a_{1} x + a_{2} x^2 + \cdots + a_{n} x^{n}$ has exactly $n$ roots, counting multiplicities.

Explanation

In fact, when we solve polynomials we naturally proceed as if a solution exists, but there is no guarantee that this is always so. For example, the quadratic polynomial $x^2+1 = 0$ has no real root. However, if we allow complex numbers here, we can see that there exist two solutions, $\pm i$.

To put it as a fact, when solving a polynomial, if we allow imaginary roots then a solution always exists no matter what, and there are exactly as many as the degree. As with every fundamental theorem, its importance goes without saying. The core idea is Liouville’s theorem, and mathematical induction is used to generalize it for a natural number $n$.

Proof

First, suppose that no solution satisfying $P(z) = 0$ exists. Then $\displaystyle {{1} \over {P(z)}}$ is an entire function, and since $\displaystyle \lim_{|z| \to \infty} \left| {{1} \over {P(z)}} \right| = 0$, it is bounded.

Liouville’s theorem: If $f$ is an entire function and bounded, then $f$ is a constant function.

By Liouville’s theorem, $P$ must be a constant function, but since this contradicts the assumption, $P(z) = 0$ has at least one solution.

Now let us generalize to a natural number. We said that $P(z) = 0$ has at least one solution; letting that solution be $z = \alpha$, we have $$ P(z) = (z-\alpha) Q(z) $$ where $Q(z) = b_{0} + b_{1} x + b_{2} x^2 + \cdots + b_{n-1} x^{n-1} = 0$ also has at least one solution. Repeating this process, by mathematical induction the $n$th-degree polynomial $P(z) = 0$ has exactly $n$ solutions.

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