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 $n \ge 1$ has exactly $n$ roots in $\mathbb{C}$.

Notice: An initial version of this theorem asserts that a polynomial of degree $n \ge 1$ 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 strictly 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 ($a_0, a_1, \dots, a_n \in \mathbb{R}$). We want to prove that if $z$ is a root, then $\overline{z}$ is also a root.

$$\begin{aligned} p(z) &= 0 \implies a_n z^n + \dots + a_2 z^2 + a_1 z + a_0 = 0 \\ &\implies \overline{a_n z^n + \dots + a_2 z^2 + a_1 z + a_0} = \overline{0} \quad \text{(taking conjugate of both sides)} \\ &\implies \overline{a_n z^n} + \dots + \overline{a_2 z^2} + \overline{a_1 z} + \overline{a_0} = 0 \quad \text{(splitting addition)} \\ &\implies \overline{a_n}(\overline{z})^n + \dots + \overline{a_2}(\overline{z})^2 + \overline{a_1}(\overline{z}) + \overline{a_0} = 0 \quad \text{(splitting powers/multiplication)} \end{aligned}$$

Since the coefficients are purely real, they are their own conjugates (i.e., $\overline{a_k} = a_k$). Therefore:

$$\begin{aligned} a_n (\overline{z})^n + \dots + a_2 (\overline{z})^2 + a_1 (\overline{z}) + a_0 &= 0 \\ p(\overline{z}) &= 0 \implies \overline{z} \text{ is a root.} \end{aligned}$$

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

Re (Real)
Im (Imaginary)
$z = a + bi$
$\overline{z} = a - bi$
Real Root

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 the complex conjugate root factors inherently removes all imaginary numbers:
    $$\begin{aligned} (z - (a + bi))(z - (a - bi)) &= ((z - a) - bi)((z - a) + bi) \\ &= (z - a)^2 - (bi)^2 \\ &= (z - a)^2 - b^2 i^2 \\ &= z^2 - 2az + a^2 + b^2 \end{aligned}$$

3. Examples

EXAMPLE 1: Factoring from a Real Root

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 the known linear factor $(z - 1)$ yields: $$f(z) = (z - 1)(z^2 - 4z + 5)$$
The quadratic factor $(z^2 - 4z + 5)$ lacks real roots ($\Delta = 16 - 20 = -4 < 0$), making this the finest factorization over $\mathbb{R}$.
Extending to $\mathbb{C}$, the quadratic produces two complex roots via the quadratic formula: $$z = \dfrac{4 \pm \sqrt{-4}}{2} = \dfrac{4 \pm 2i}{2} \implies z = 2 + i \text{ and } z = 2 - i$$
The absolute factorization over $\mathbb{C}$ is: $$f(z) = (z - 1)(z - 2 + i)(z - 2 - i)$$

EXAMPLE 2: Factoring from a Complex Root

Find all three roots of $f(z) = z^3 - 5z^2 + 9z - 5$ given that $z = 2 + i$ is a complex root.

Solution:

By the Conjugate Root Theorem, the conjugate $z = 2 - i$ is immediately known to be a root as well.
Combining these complex roots forms an irreducible quadratic factor: $$\begin{aligned} (z - (2 + i))(z - (2 - i)) &= ((z - 2) - i)((z - 2) + i) \\ &= (z - 2)^2 - i^2 \\ &= (z^2 - 4z + 4) - (-1) \\ &= z^2 - 4z + 5 \end{aligned}$$
Dividing the original cubic $f(z)$ by this quadratic factor leaves the linear factor $(z - 1)$.
Hence, $z=1$ is the third and final root.

EXAMPLE 3: Building a Polynomial from its Roots

Consider $f(z) = z^3 + az^2 + bz + c$. Given roots $z=1$ and $z=2+i$, find the real coefficients $a, b, c$.

Solution:

Since the coefficients are real, the third root must be the conjugate $z = 2 - i$.
Therefore, we can construct the polynomial by multiplying its factors: $$f(z) = (z - 1)(z - (2 + i))(z - (2 - i))$$
Expanding this (utilizing the quadratic expansion from Example 2) gives: $$\begin{aligned} f(z) &= (z - 1)(z^2 - 4z + 5) \\ &= z(z^2 - 4z + 5) - 1(z^2 - 4z + 5) \\ &= z^3 - 4z^2 + 5z - z^2 + 4z - 5 \\ &= z^3 - 5z^2 + 9z - 5 \end{aligned}$$
Matching coefficients with $z^3 + az^2 + bz + c$ yields: $$a = -5, \quad b = 9, \quad c = -5$$