Proof of the Fundamental Theorem of Algebra 📂Complex Anaylsis

Proof of the Fundamental Theorem of Algebra

Theorem ¹

$n$th degree polynomial $P(x) = a_{0} + a_{1} x + a_{2} x^2 + \cdots + a_{n} x^{n}$ has exactly $n$ roots, including multiple roots.

Explanation

In fact, when we solve a polynomial, we usually assume that there exists a solution, but there’s no guarantee that this is always the case. For example, the quadratic polynomial $x^2+1 = 0$ does not have real roots. However, if complex numbers are allowed, then there are two solutions, which are $\pm i$.

To state a fact, if complex roots are allowed when solving a polynomial, then there will always be a solution, and exactly as many as its degree. The importance of all fundamental theorems is evident without further ado. The core idea is Liouville’s theorem, and mathematical induction is used to generalize it for natural numbers $n$.

Proof

Let’s assume first that there is no solution satisfying $P(z) = 0$, 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, which contradicts our assumption; therefore, $P(z) = 0$ has at least one root.

Now, let’s generalize for the natural numbers. Assuming that $P(z) = 0$ has at least one root, let’s say $z = \alpha$, then $$ P(z) = (z-\alpha) Q(z) $$ 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 root. Repeating this process, by mathematical induction, $n$th degree polynomial $P(z) = 0$ has exactly $n$ roots.

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