Complex Number is Algebraic over Real Numbers

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Theorem

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

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


Proof

Let $z = a + i b$.

Let $z = a - i b$ 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:

$z^2 - 2 a + \paren {a^2 + b^2}$

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

$\blacksquare$


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