Proof that the complex numbers form a field

Proof

We check the axioms of a field one by one, using the rules for the complex numbers.

1. Let $z = a + b i$, $w = c + d i$ and $u = e + f i$ be complex numbers, with $a, b, c, d, e, f$ real. The operations are $z + w = (a + c) + (b + d) i$ and $z w = (a c - b d) + (a d + b c) i$.
2. All computations with $a, \dots, f$ below take place in $\R$. They use only the field axioms of $\R$, by the reals form a field. Two complex numbers are compared through their real and imaginary parts, by equality of complex numbers.
3. Addition is commutative and associative. Indeed, both properties hold for the real parts and for the imaginary parts separately, since they hold in $\R$.
4. Let $0 = 0 + 0 i$. Then $z + 0 = (a + 0) + (b + 0) i = z$.
5. Let $-z = (-a) + (-b) i$. Then $z + (-z) = 0 + 0 i = 0$. So $(\C, +)$ is an abelian group with identity $0$.
6. Multiplication is commutative. Indeed, $w z = (c a - d b) + (c b + d a) i$, and this equals $z w$ because real multiplication is commutative.
7. Multiplication is associative. We have $z w = (a c - b d) + (a d + b c) i$. Multiplying by $u$ gives $$(z w) u = (a c e - b d e - a d f - b c f) + (a c f - b d f + a d e + b c e) i.$$
8. Also $w u = (c e - d f) + (c f + d e) i$. Multiplying $z$ by it gives $$z (w u) = (a c e - a d f - b c f - b d e) + (a c f + a d e + b c e - b d f) i.$$ The real parts in steps 7 and 8 agree, and so do the imaginary parts. So $(z w) u = z (w u)$.
9. Let $1 = 1 + 0 i$. Then $1 \cdot z = (1 a - 0 b) + (1 b + 0 a) i = z$. By step 6, also $z \cdot 1 = z$.
10. Multiplication distributes over addition. We have $w + u = (c + e) + (d + f) i$, so $$z (w + u) = (a c + a e - b d - b f) + (a d + a f + b c + b e) i.$$
11. Also $z w + z u = \bigl((a c - b d) + (a e - b f)\bigr) + \bigl((a d + b c) + (a f + b e)\bigr) i$. This has the same parts as step 10. So $z (w + u) = z w + z u$. By step 6, also $(w + u) z = w z + u z$.
12. Steps 3 to 11 show that $\C$ is a ring, and step 6 shows that it is a commutative ring.
13. We have $0 \neq 1$, because $\operatorname{Re} 0 = 0$ and $\operatorname{Re} 1 = 1$ differ.
14. Now let $z = a + b i \neq 0$. Then $a \neq 0$ or $b \neq 0$, by equality of complex numbers. Squares of real numbers are nonnegative, and the square of a nonzero real number is positive. So $s = a^2 + b^2 > 0$.
15. Let $v = \dfrac{a}{s} + \dfrac{-b}{s} i$, which makes sense because $s \neq 0$. Then $$z v = \left( a \cdot \frac{a}{s} - b \cdot \frac{-b}{s} \right) + \left( a \cdot \frac{-b}{s} + b \cdot \frac{a}{s} \right) i = \frac{a^2 + b^2}{s} + 0 i = 1.$$
16. So every nonzero $z$ has a multiplicative inverse $z^{-1} = v$. Together with steps 12 and 13, this proves that $\C$ is a field.