Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra was first proved by Carl Friedrich Gauss in 1799. It is the core of algebra because it states that every nonconstant polynomial equation has at least one complex root. Thus, this theorem is indispensable to understanding the behaviour of polynomials and their solutions in complex number theory.

1.0Insight into the Fundamental Theorem of Algebra 

Theorem Statement

"Every non-constant polynomial equation with complex coefficients has at least one complex root."

That means no matter how complicated the polynomial may look, it must have at least one solution (root) in the system of complex numbers. 

A polynomial in x with complex coefficients can be written as: 

$P(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+....+a_{1}x+a_0$

Here, a_n, a_n-1, …., a₁, a₀ are complex numbers and a_n 0, then there exists a complex number z₀ such that f(z₀) = 0. 

Proof

Step 1: Consider the polynomial f(z) at large values of z

Let the polynomial be: 

$f(z)=a_n{z}^n+a_{n-1}{z}^{n-1}+...+a_{1}z+a_0$

where $a_n\ne 0$

• As $\left|z\right|\to \infty$, the highest-degree term a_nzⁿ dominates the polynomial because it grows the fastest.
• For very large $\left|z\right|$, the value of $\left|f(z)\right|$ behaves approximately like: $\left|f(z)\right|\approx \left|a_n\right|{\left|z\right|}^n$

Thus, as $\left|z\right|$ becomes large, f(z) grows without bound (tends to infinity).

Step 2: Use the Extreme Value Theorem for continuous functions

• The function f(z) is a continuous function in the complex plane (because it is a polynomial).
• As $\left|z\right|\to \infty$, we know that $\left|f(z)\right|\to \infty$, meaning the polynomial grows large as z moves farther away from the origin.

According to the Extreme Value Theorem, since the function f(z) is continuous in the complex plane, f(z) must attain at least one minimum value, and this minimum value should not be $\infty \left|f(z)\right|\to \infty$ as $\left|z\right|\to \infty$

Thus, some point z₀ exists in the complex plane where f(z₀) reaches its minimum.

Step 3: Show that f(z₀) = 0

• If f(z₀) were non-zero, we could be able to find a point where f(z) is smaller than $\left|f(z_0)\right|$, which contradicts the idea that f(z₀) is the minimum. 
• Therefore, the only way f(z₀) can achieve a minimum value is if f(z₀) = 0, because the polynomial function can’t get any smaller than 0. 

Thus, by the observation we can say that there exists at least one point z₀ in the complex plane where f(z₀) = 0, which means the polynomial has a root. 

2.0Solved Examples

Problem: Find the roots of the quadratic equation x² + 2x + 5 = 0.

Solution: By using the quadratic formula: 

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

In the given equation: a = 1, b = 2, c = 5

$x=\frac{-2\pm \sqrt{2^2-4\times 1\times 5}}{2\times 1}$

$x=\frac{-2\pm \sqrt{4-20}}{2}$

$x=\frac{-2\pm \sqrt{-16}}{2}$

$x=\frac{-2\pm 4i}{2}$

$x=-1\pm 2i$

The roots of quadratic equation are

$x=-1+2iand-1-2i$

Problem: Solve the quartic equation:

x⁴ + 4x² + 5 = 0

Solution: Let y = x² hence the given equation becomes 

y² + 4y + 5 = 0

By using a quadratic formula: 

$y=\frac{-4\pm \sqrt{4^2-4(1)(5)}}{2(1)}$

$y=\frac{-4\pm \sqrt{16-20}}{2}$

$y=\frac{-4\pm \sqrt{-4}}{2}$

$y=\frac{-4\pm 2i}{2}$

$y=-2\pm i$

Roots y = – 2 + i, –2 – i. But y = x² hence, 

x² = – 2 + i, –2 – i

The roots of the equation are: 

$x=\pm \sqrt{-2+i}$, $\pm \sqrt{-2-i}$

Problem: Solve the equation x⁴ − 5x² + 6 = 0. 

Solution: Let x² = y

y² – 5y + 6 = 0

y² – 3y – 2y +6 = 0

y(y–3) – 2(y–3) = 0 

(y–2)(y–3) = 0 

y–2=0, y–3=0

y = 2, 3 

y = x²

x² = 2, 3 

$x=\pm \sqrt{2}$, $\pm \sqrt{3}$