# Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs

## Theorem

Let $\map f z = a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0$ be a polynomial over complex numbers where $a_0, \ldots, a_n$ are real numbers.

Let $\alpha \in \C$ be a root of $f$.

Then $\overline \alpha$ is also a root of $f$, where $\overline \alpha$ denotes the complex conjugate of $\alpha$.

That is, all complex roots of $f$ appear as conjugate pairs.

## Proof 1

Let $\alpha \in \C$ be a root of $f$.

Then $f \left({\alpha}\right) = 0$ by definition.

Suppose $\alpha$ is wholly real.

$\alpha = \overline \alpha$

and so $\overline \alpha$ is a root of $f$ a priori.

Now let $\alpha \in \C$ not be wholly real.

By definition of complex conjugate, we have that:

$\overline 0 = 0$

and so:

$f \left({\alpha}\right) = \overline{f \left({\alpha}\right)}$
$\overline{f \left({\alpha}\right)} = f \left({\overline \alpha}\right)$

from which it follows that:

$f \left({\overline \alpha}\right) = 0$

That is, $\overline \alpha$ is also a root of $f$.

$\blacksquare$

## Proof 2

Let $\alpha = p + q i$.

Let $p + q i$ be expressed in exponential form as $\alpha = r e^{i \theta}$.

As $\alpha = r e^{i \theta}$ satisfies $\map f \alpha = 0$, it follows that:

$a_n r^n e^{n i \theta} + a_{n - 1} r^{n - 1} e^{\paren {n - 1} i \theta} + \dotsb + a_1 r e^{i \theta} + a_0 = 0$

Taking the conjugate of both sides:

$a_n r^n e^{-n i \theta} + a_{n - 1} r^{n - 1} e^{-\paren {n - 1} i \theta} + \dotsb + a_1 r e^{-i \theta} + a_0 = 0$

it follows that $\overline \alpha = p - q i$ is also a root of $f$.

If any of the $a_k$ are complex, then the conjugate of $f$ is not the same polynomial as $f$.

Therefore the result holds only for real coefficients.

$\blacksquare$