# Complex Conjugate Root Theorem

In algebra, the Complex Conjugate Root Theorem states that if $P(x)$ is a polynomial with real coefficients, then a complex number is a root of $P(x)$ if and only if its complex conjugate is also a root.

A common intermediate step in intermediate competitions is to recognize that when given a complex root of a real polynomial, its conjugate is also a root.

## Proof

Let $P(x)$ have the form $a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$, where constants $a_n, a_{n-1}, \ldots, a_1, a_0$ are real numbers, and let $z$ be a complex root of $P(x)$. We then wish to show that $\overline{z}$, the complex conjugate of $z$, is also a root of $P(x)$. Because $z$ is a root of $P(x)$, $$P(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0 = 0.$$ Then by the properties of complex conjugation, \begin{align*} \overline{a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0} = \overline{0} \\ \overline{a_n z^n} + \overline{a_{n-1} z^{n-1}} + \cdots + \overline{a_1 z} + \overline{a_0} = 0 \\ a_n \overline{z^n} + a_{n-1} \overline{z^{n-1}} + \cdots + a_1 \overline{z} + a_0 = 0 \\ a_n \overline{z}^n + a_{n-1} \overline{z}^{n-1} + \cdots + a_1 \overline{z} + a_0 = 0 \\ P(\overline{z}) = 0, \end{align*} which entails that $\overline{z}$ is also a root of $P(x)$, as required. $\square$