The Complex Numbers: Pairs of Real Numbers with Their Addition and Multiplication, and the Imaginary Unit

In the setting of The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness.

¶ The set of complex numbers is $\mathbb{C}=\mathbb{R}\times\mathbb{R}$.

¶ For $(a,b),(c,d)\in\mathbb{C}$, their sum and product are

$$(a,b)+(c,d)=(a+c,\,b+d),\qquad (a,b)\cdot(c,d)=(ac-bd,\,ad+bc);$$

here $\mathbb{R}$ is a set by The Real Numbers §reals, so $\mathbb{C}$ is a set by Membership in a Cartesian Product, and the Cartesian Product of Two Sets Is a Set §set, every element of $\mathbb{C}$ is a pair $(a,b)$ with $a,b\in\mathbb{R}$ by The Cartesian Product of Two Classes §product, and the pairs $(a,b)$ and $(c,d)$ determine $a$, $b$, $c$ and $d$ by The Characteristic Property of Ordered Pairs and Nested Tuples of Sets §characteristic. For $a,b,c,d\in\mathbb{R}$, the entries $a+c$, $b+d$, $ac-bd$ and $ad+bc$ lie in $\mathbb{R}$, as $+$, $\cdot$ and differences are operations on the field $\mathbb{R}$ by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals and Negatives, Differences, Reciprocals and Quotients §negative, so the pairs on the right lie in $\mathbb{C}$ by Membership in a Cartesian Product, and the Cartesian Product of Two Sets Is a Set §membership. Hence $+$ and $\cdot$ are maps $\mathbb{C}\times\mathbb{C}\to\mathbb{C}$, that is, binary operations on $\mathbb{C}$, by Maps and Relations Given by Formulas §binary.

¶ The imaginary unit is $i=(0,1)$, which lies in $\mathbb{C}$ by Membership in a Cartesian Product, and the Cartesian Product of Two Sets Is a Set §membership, as $0,1\in\mathbb{R}$.