TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-13

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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$C$ is the real coordinate plane, with coordinate arithmetic

Statement

The map $Φ : C ⟶ R^{2}, Φ (a + b i) = (a, b),$ is a bijection. Under it, $Φ ((a + b i) + (u + v i)) = (a + u, b + v)$ and $Φ ((a + b i) (u + v i)) = (a u - b v, a v + b u) .$

Facts & Assumptions

Given: The complex field in its quotient construction.

[F1]

Every complex number has a unique form $a + b i$ , and addition and multiplication have the displayed coordinate formulas ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

Proof

technique · direct

1.1F1

Uniqueness in [F1] makes $Φ$ well-defined and injective.

1.2F1

For every $(a, b) \in R^{2}$ , the complex number $a + b i$ maps to $(a, b)$ , so $Φ$ is surjective.

2.1F1step 1.1

Applying $Φ$ to the addition formula in [F1] gives the first coordinate identity.

3.1F1step 1.1∎

Applying $Φ$ to the multiplication formula in [F1] gives the second coordinate identity.

Depends on

$C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$

Dependency tree · 2 levels

Sources

R. K. Srivastava, Complex Analysis lecture notes

CC is the real coordinate plane, with coordinate arithmetic