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 $\overline z = a - i b$ 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 $\R$.
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Example $75$