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  "body": "\n## Proof\n\n1. Let $z = a + b i$ and $w = c + d i$ with $a, b, c, d$ real.\n2. We first show that the [[conjugate|0BJV24C4SY]] of $z w$ is $\\bar{z} \\bar{w}$.\nWe have $z w = (a c - b d) + (a d + b c) i$, so $\\overline{z w} = (a c - b d) - (a d + b c) i$.\n3. Also $\\bar{z} \\bar{w} = (a - b i)(c - d i) = \\bigl(a c - (-b)(-d)\\bigr) + \\bigl(a (-d) + (-b) c\\bigr) i = (a c - b d) - (a d + b c) i$.\nBy [[equality of complex numbers|05V7J4WD1C]], $\\overline{z w} = \\bar{z} \\bar{w}$.\n4. By [[a complex number times its conjugate|0ZB7J071WQ]] applied to $z w$, then step 3, we get $|z w|^2 = (z w)(\\bar{z} \\bar{w})$.\n5. Multiplication in $\\C$ is commutative and associative, since $\\C$ is a [[field|0FRSR10DWF]]. So $(z w)(\\bar{z} \\bar{w}) = (z \\bar{z})(w \\bar{w})$.\n6. Applying [[a complex number times its conjugate|0ZB7J071WQ]] to $z$ and to $w$ gives $(z \\bar{z})(w \\bar{w}) = |z|^2 |w|^2 = (|z| \\, |w|)^2$.\n7. By steps 4 to 6, the real numbers $|z w|$ and $|z| \\, |w|$ have the same square.\n8. Both numbers are nonnegative, since the [[modulus|0KW2X0TFBX]] is nonnegative.\nA nonnegative real number has only one nonnegative square root, by [[existence of square roots|068BZBXE0C]].\nSo $|z w| = |z| \\, |w|$.\n\n## Idea\n\nSquaring removes the square root in the modulus and turns $|z|^2$ into $z \\bar{z}$.\nConjugation respects products, so $|z w|^2$ splits as $|z|^2 |w|^2$.\n",
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