1.12 Polynomials Over the Complex Field (HL)

1. The Fundamental Theorem of Algebra

A quadratic equation may possess two different real roots, two equal real roots, or two non-real complex roots. It can be stated universally that a quadratic always has exactly two roots in the complex field $\mathbb{C}$ (considering that a real number is also a complex number, and allowing for the repetition of a root). This principle is a specific instance of the Fundamental Theorem of Algebra.

A polynomial of degree $n > 1$ has exactly $n$ roots in $\mathbb{C}$.

Notice: An initial version of this theorem asserts that a polynomial of degree $n > 1$ always has at least one root in $\mathbb{C}$. However, if $f(x)$ is a polynomial of degree $n$ and $x = r_1$ is a complex root, polynomial long division by $(x - r_1)$ yields:

$f(x) = (x - r_1)q(x)$

where $q(x)$ is a polynomial of degree $n-1$. Since $q(x)$ also contains a complex root $r_2$, repeating this division process will ultimately isolate exactly $n$ roots for $f(x)$.

2. Factorization and Roots of a Polynomial

The equation $x^2 - 4x + 13 = 0$ yields two complex roots: $x = 2 \pm 3i$. Thus, a quadratic with a discriminant $\Delta < 0$ always produces two complex roots that are conjugates of each other. This pattern is not accidental.

For any polynomial of any degree with real coefficients, if $z = a + bi$ is a root, then its conjugate $\overline{z} = a - bi$ is also a root.

Lemma and Proof

For the conjugates of complex numbers, the following properties hold:

  • $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$
  • $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$
  • $\overline{z^n} = (\overline{z})^n$

Proof: Consider the polynomial $p(x) = a_n x^n + \dots + a_2 x^2 + a_1 x + a_0$ with real coefficients ($a_0, a_1, \dots, a_n \in \mathbb{R}$). If $z$ is a root, then $\overline{z}$ is also a root.

$z$ is a root $\Rightarrow a_n z^n + \dots + a_2 z^2 + a_1 z + a_0 = 0$
Taking the conjugate of both sides: $\overline{a_n z^n + \dots + a_2 z^2 + a_1 z + a_0} = \overline{0}$
$\overline{a_n z^n} + \dots + \overline{a_2 z^2} + \overline{a_1 z} + \overline{a_0} = 0$
Since coefficients are real, they are their own conjugates: $a_n \overline{z}^n + \dots + a_2 \overline{z}^2 + a_1 \overline{z} + a_0 = 0$
$\Rightarrow \overline{z}$ is a root.

Therefore, a polynomial possesses either real roots or non-real complex roots that always manifest in conjugate pairs.

Summary of Factorization

For a polynomial $p(z)$ of degree $n$:

  • Factorization over $\mathbb{C}$: Consists of $n$ linear factors of the form $p(z) = a_n(z - r_1)(z - r_2)\dots(z - r_n)$, where $r_1, \dots, r_n \in \mathbb{C}$.
  • Factorization over $\mathbb{R}$: Consists of linear factors $(z - r)$ and irreducible quadratic factors $(z^2 + pz + q)$, where $r, p, q \in \mathbb{R}$. This occurs because multiplying conjugate root factors yields $(z - a - bi)(z - a + bi) = (z - a)^2 - (bi)^2 = z^2 - 2az + (a^2 + b^2)$.

3. Examples

EXAMPLE 1

Version 1: Find all three roots of the cubic function $f(z) = z^3 - 5z^2 + 9z - 5$ given that $z=1$ is a real root.
Solution: Dividing $f(z)$ by $(z - 1)$ yields $f(z) = (z - 1)(z^2 - 4z + 5)$. The quadratic factor lacks real roots, making this the finest factorization over $\mathbb{R}$. Extending to $\mathbb{C}$, the quadratic produces two complex roots: $2 + i$ and $2 - i$. The factorization over $\mathbb{C}$ is $f(z) = (z - 1)(z - 2 + i)(z - 2 - i)$.
Version 2: Find all three roots of $f(z) = z^3 - 5z^2 + 9z - 5$ given that $z = 2 + i$ is a complex root.
Solution: The conjugate root $z = 2 - i$ is immediately known. Combining these complex roots forms the quadratic factor $(z - 2 + i)(z - 2 - i) = (z - 2)^2 - i^2 = z^2 - 4z + 5$. Dividing $f(z)$ by this quadratic leaves the factor $(z - 1)$. Hence, $z=1$ is the third root.
Version 3: Consider $f(z) = z^3 + az^2 + bz + c$. Given roots $z=1$ and $z=2+i$, find $a, b, c$.
Solution: The third root must be $z = 2 - i$. Therefore, $f(z) = (z - 1)(z - 2 + i)(z - 2 - i)$. Expanding this gives $(z - 1)(z^2 - 4z + 5) = z^3 - 5z^2 + 9z - 5$. Matching coefficients yields $a = -5$, $b = 9$, and $c = -5$.