Although this requires a bit of knowledge on Complex Anlaysis, I recently discovered this new way to prove the Fundamental Theorem of Algebra and I couldn’t help but share it.

First of all, what is the **Fundamental Theorem of Algebra (FTA)**? This very important (hence the name!) result states that:

Every non-constant polynomial with complex coefficients has a complex root.

In order to prove this, we must first be aware of Liouville’s Theorem:

Every bounded, entire function is constant.

**Definitions**

Bounded: a function on a set X is said to be bounded if there exists a real number *M* such that

for all *x* in *X*.

Entire: An entire function is a holomorphic function on the entire complex plane.

Liouville’s theorem is proved using the Cauchy integral formula for a disc, one of the most important results in Complex Analysis. Although I will not describe how to prove it or what it states in this blog post, I encourage you to read about here it as it is truly a remarkable result.

Now armed with Liouville’s Theorem we can prove the FTA.

### Proof

Let P(z) = z^{n} + c_{n-1}z^{n-1} + … + c_{1}z + c_{0} be a polynomial of degree *n* > 0. Then |P(z)| –> ∞ as |z| –> ∞, so there exists R such that |P(z)| > 1 for all z with |z| > R.

Consider f(z) = 1/P(z). If P has no complex zeros then f is entire. So, as f is continuous, f is bounded on {|z| ≤ R}.

As |f(z)| < 1 when |z| > R, f is a bounded entire function, so by **Liouville’s Theorem **f is constant, which is a contradiction.

The only thing we assumed was that P had no complex zeros, and so we contradicted this fact. Hence, P must have at least **one **complex zero. Amazing right!

M x