Proof that the modulus is multiplicative

Proof

1. Let $z = a + b i$ and $w = c + d i$ with $a, b, c, d$ real.
2. We first show that the conjugate of $z w$ is $\bar{z} \bar{w}$. We have $z w = (a c - b d) + (a d + b c) i$, so $\overline{z w} = (a c - b d) - (a d + b c) i$.
3. Also $\bar{z} \bar{w} = (a - b i)(c - d i) = \bigl(a c - (-b)(-d)\bigr) + \bigl(a (-d) + (-b) c\bigr) i = (a c - b d) - (a d + b c) i$. By equality of complex numbers, $\overline{z w} = \bar{z} \bar{w}$.
4. By a complex number times its conjugate applied to $z w$, then step 3, we get $|z w|^2 = (z w)(\bar{z} \bar{w})$.
5. Multiplication in $\C$ is commutative and associative, since $\C$ is a field. So $(z w)(\bar{z} \bar{w}) = (z \bar{z})(w \bar{w})$.
6. Applying a complex number times its conjugate to $z$ and to $w$ gives $(z \bar{z})(w \bar{w}) = |z|^2 |w|^2 = (|z| \, |w|)^2$.
7. By steps 4 to 6, the real numbers $|z w|$ and $|z| \, |w|$ have the same square.
8. Both numbers are nonnegative, since the modulus is nonnegative. A nonnegative real number has only one nonnegative square root, by existence of square roots. So $|z w| = |z| \, |w|$.