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  "body": "\n## Proof\n\n1. Let $z = a + b i$ be a [[complex number|09FTAET5Z4]], with $a$ and $b$ real.\n2. Its [[conjugate|0BJV24C4SY]] is $\\bar{z} = a + (-b) i$.\n3. By the rule for multiplying complex numbers,\n$$z \\bar{z} = \\bigl(a \\cdot a - b \\cdot (-b)\\bigr) + \\bigl(a \\cdot (-b) + b \\cdot a\\bigr) i = (a^2 + b^2) + 0 i.$$\n4. By the definition of the [[modulus|0KW2X0TFBX]], $|z| = \\sqrt{a^2 + b^2}$.\nHere $\\sqrt{a^2 + b^2}$ is the nonnegative real number whose square is $a^2 + b^2$, by [[existence of square roots|068BZBXE0C]].\n5. So $|z|^2 = a^2 + b^2$. With step 3 and [[equality of complex numbers|05V7J4WD1C]], this gives $z \\bar{z} = |z|^2$.\n6. Since $a^2 \\ge 0$ and $b^2 \\ge 0$, the number $z \\bar{z} = a^2 + b^2$ is a nonnegative real number.\n\n## Idea\n\nThe product $(a + b i)(a - b i)$ is a difference of squares, $a^2 - (b i)^2 = a^2 + b^2$.\nThe imaginary parts cancel, and what remains is the square of the distance from $z$ to $0$.\n",
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