Rule

Complex Conjugate Root Theorem

Let $p (x)$ be a polynomial function whose coefficients are real. If a complex number $a + b i$ is a root of $p (x),$ then the root's complex conjugate, $a - b i,$ is also a root of $p (x) .$

$p (a + b i) = 0 ⟹ p (a - b i) = 0$

Proof

Let $p (x)$ be a polynomial with real coefficients $c_{k} .$ $p (x) = k = 0 \sum n c_{k} x^{k}$ Using the properties of conjugation, the conjugate of the polynomial evaluated at $a + b i$ can be rewritten.

$\overline{p (a + b i)} = \overline{k = 0 \sum n c_{k} (a + b i)^{k}}$	Conjugate of the polynomial evaluated at $a + b i$
$\overline{p (a + b i)} = k = 0 \sum n \overline{c_{k} (a + b i)^{k}}$	Conjugate of a sum is the sum of the conjugates
$\overline{p (a + b i)} = k = 0 \sum n \overline{c_{k}} \overline{(a + b i)^{k}}$	Conjugate of a product is the product of the conjugates
$\overline{p (a + b i)} = k = 0 \sum n \overline{c_{k}} (\overline{a + b i})^{k}$	Conjugate of a power is the power of the conjugate
$\overline{p (a + b i)} = k = 0 \sum n c_{k} (\overline{a + b i})^{k}$	Conjugate of a real number is itself
$\overline{p (a + b i)} = k = 0 \sum n c_{k} (a - b i)^{k}$	The conjugate of $a + b i$ is $a - b i$
$\overline{p (a + b i)} = p (a - b i)$	The polynomial evaluated at $a - b i$

It is given that $a + b i$ is a root of the polynomial $p (x) .$ $p (a + b i) = 0$ The conjugate of the real number $0$ is $0,$ so the conjugate of the polynomial evaluated at $a + b i$ is $0 .$ $\overline{p (a + b i)} = 0$ Comparing this with the equality in the last line of the table shows that $a - b i,$ the conjugate of $a + b i$ is also a root of the polynomial. $\overline{p (a + b i)} \overline{p (a + b i)} = p (a - b i) = 0 ⟹ p (a - b i) = 0$