Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs

Theorem
Let $f \left({z}\right) = a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0$ 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.

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

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

Suppose $\alpha$ is wholly real.

Then by Complex Number equals Conjugate iff 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:
 * $f \left({\alpha}\right) = \overline{f \left({\alpha}\right)}$

From Conjugate of Polynomial is Polynomial of Conjugate:
 * $\overline{f \left({\alpha}\right)} = f \left({\overline \alpha}\right)$

from which it follows that:
 * $f \left({\overline \alpha}\right) = 0$

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