Pairs of Complex Roots for Polynomials with Real Coefficients

# Pairs of Complex Roots for Polynomials with Real Coefficients

Suppose that $p(x) = a_0 + a_1x + ... + a_mx^m$ is a polynomial with real coefficients $a_0, a_1, ..., a_m \in \mathbb{R}$. Recall that the complex conjugate of a complex number $z = a + bi = \Re (z) + \Im (z) i$ is denoted by $\bar{z}$ and $\bar{z} = a - bi = \Re(z) - \Im (z) i$.

The following theorem will tell us that the complex roots of $p(x)$ come in pairs, that is if $\lambda \in \mathbb{C}$ is a complex root of $p(x)$ then so is its complex conjugate, $\bar{\lambda}$. Note that all real numbers are complex numbers, and so it should be relatively obvious that if $\lambda = a + 0i = a$ is a strictly real root of $p(x)$, then $\bar{\lambda} a - 0i = a$ is trivially also a root of $p(x)$. The important part of the following theorem lends itself to strictly complex roots.

 Theorem 1: Let $p(x) \in \wp (\mathbb{R})$. If $\lambda \in \mathbb{C}$ is a complex root of $p(x)$ then the complex conjugate $\lambda$, $\bar{\lambda}$ is also a root of $p(x)$.
• Proof: Let $p(x) \in \wp (\mathbb{R})$ such that $p(x) = a_0 + a_1x + ... + a_mx^m$ where $a_0, a_1, ..., a_m \in \mathbb{R}$ and suppose that $\lambda \in \mathbb{C}$ is a complex root of $p(x)$. Then $p(\lambda) = 0$ and so:
(1)
\begin{align} \quad p(\lambda) = a_0 + a_1 \lambda + ... + a_m \lambda ^m \\ \quad 0 = a_0 + a_1 \lambda + ... + a_m \lambda ^m \\ \end{align}
• Now take the complex conjugate of both sides of the equation above, and thus:
(2)
\begin{align} \quad \bar{0} = \overline{a_0 + a_1 \lambda + ... + a_m \lambda ^m} \\ \end{align}
• Recall that the complex conjugate of a real number is itself. Also recall that for $y, z \in \mathbb{C}$ the additive property of a complex conjugate is $\overline{y + z} = \bar{y} + \bar{z}$, and multiplicative property of a complex conjugate is $\overline{y \cdot z} = \bar{y} \cdot \bar{z}$. Applying these three properties to the righthand side of the equation above we get that:
(3)
\begin{align} \quad 0 = a_0 + a_1 \bar{\lambda} + ... + a_m \bar{\lambda}^m \end{align}
• Therefore $p(\bar{\lambda}) = 0$ and so $\bar{\lambda}$ is a root of $p(x)$. $\blacksquare$

The following corollary is also extremely useful and tells us that polynomials with real coefficients of odd degree must have at least one real root.

 Corollary 1: Let $p(x) \in \wp ( \mathbb{R} )$. If $\mathrm{deg} (p)$ is odd, then $p(x)$ contains at least one real root.
• Proof: Suppose that $p(x) \in \wp ( \mathbb{R} )$ and $\mathrm{deg} (p) = n$ is odd. By Theorem 1, since the complex roots of $p(x)$ come in pairs, this implies that the number of complex roots $\lambda = a + bi$ where $b \neq 0$ is even, and so there must be at least one root in the form $a + 0i$, and so $p(x)$ has at least one real root. $\blacksquare$

## Example 1

Verify Theorem 1 by finding all roots to the polynomial $p(x) = x^2 + 4x + 5$.

Using the quadratic formula we have that:

(4)
\begin{align} \quad x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ \quad x = \frac{-4 \pm \sqrt{16 - 20}}{2} \\ \quad x = -2 \pm \sqrt{-4}{2} \\ \quad x = -2 \pm i \end{align}

So the roots of $p(x)$ are $x_1 = -2 + i$ and $x_1 = -2 - i$ which are the complex conjugates of one another.