Complex Number is Algebraic over Real Numbers

Theorem
Let $z \in \C$ be a complex number.

Then $z$ is algebraic over $\R$.

Proof
Let $z = a + i b$.

Let $\overline z = a - i b$ be the complex conjugate of $z$.

From Product of Complex Number with Conjugate:
 * $z \overline z = a^2 + b^2$

From Sum of Complex Number with Conjugate:
 * $z + \overline z = 2 a$

Thus from Viète's Formulas, both $z$ and $\overline z$ are roots of the polynomial:
 * $X^2 - 2 a X + \paren {a^2 + b^2}$

Hence the result by definition of algebraic over $\R$.