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  "body": "\n## Proof\n\n1. Let $z = a + b i$ with $a, b$ real and $z \\neq 0$. Let $\\bar{z} = a - b i$ be its [[conjugate|0BJV24C4SY]] and $|z|$ its [[modulus|0KW2X0TFBX]].\n2. Since $z \\neq 0$, we have $a \\neq 0$ or $b \\neq 0$, by [[equality of complex numbers|05V7J4WD1C]].\nThe square of a nonzero real is positive, so $a^2 + b^2 \u003e 0$. By the definition of the modulus, $|z|^2 = a^2 + b^2$.\nSo $|z|^2 \u003e 0$, and the real number $1 / |z|^2$ exists.\n3. Let $v = \\dfrac{\\bar{z}}{|z|^2}$, that is, the product of the real number $1/|z|^2$ and $\\bar{z}$.\n4. By [[a complex number times its conjugate|0ZB7J071WQ]], $z \\bar{z} = |z|^2$.\n5. Using commutativity and associativity of multiplication in the [[field|0FRSR10DWF]] $\\C$, we get $z v = \\dfrac{1}{|z|^2} (z \\bar{z}) = \\dfrac{|z|^2}{|z|^2} = 1$.\n6. Inverses in a field are unique. Indeed, $z z^{-1} = 1$ and $z v = 1$ give $v = v (z z^{-1}) = (v z) z^{-1} = z^{-1}$.\nSo $z^{-1} = \\bar{z} / |z|^2$.\n7. In coordinates, a real number $t$ times $a - b i$ is $t a - t b i$. With $t = 1/(a^2 + b^2)$, step 6 gives\n$$z^{-1} = \\frac{a}{a^2 + b^2} - \\frac{b}{a^2 + b^2} \\, i.$$\n\n## Idea\n\nThe product $z \\bar{z}$ is the positive real number $|z|^2$.\nDividing this identity by $|z|^2$ shows that $\\bar{z} / |z|^2$ undoes multiplication by $z$.\n",
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