Proof that a complex number times its conjugate is the square of its modulus

Proof

1. Let $z = a + b i$ be a complex number, with $a$ and $b$ real.
2. Its conjugate is $\bar{z} = a + (-b) i$.
3. By the rule for multiplying complex numbers, $$z \bar{z} = \bigl(a \cdot a - b \cdot (-b)\bigr) + \bigl(a \cdot (-b) + b \cdot a\bigr) i = (a^2 + b^2) + 0 i.$$
4. By the definition of the modulus, $|z| = \sqrt{a^2 + b^2}$. Here $\sqrt{a^2 + b^2}$ is the nonnegative real number whose square is $a^2 + b^2$, by existence of square roots.
5. So $|z|^2 = a^2 + b^2$. With step 3 and equality of complex numbers, this gives $z \bar{z} = |z|^2$.
6. Since $a^2 \ge 0$ and $b^2 \ge 0$, the number $z \bar{z} = a^2 + b^2$ is a nonnegative real number.