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  "body": "\n## Proof\n\nWe take the $n$th root of the modulus and divide the argument by $n$.\n\n1. Let $z$ be a complex number and $n \\ge 1$ an integer. If $z = 0$, then $w = 0$ works, since $0^n = 0$.\n2. Now let $z \\neq 0$. Let $r = |z| \u003e 0$ be its [[modulus|0KW2X0TFBX]] and $\\theta = \\operatorname{Arg} z$.\nBy the [[polar form|0BRXFNQ5JV]], $z = r (\\cos \\theta + i \\sin \\theta)$.\n3. Let $s = \\exp\\bigl(\\ln(r) / n\\bigr)$, a real number, with $\\exp$ the real [[exponential function|0RMVS7RXZX]].\nHere $\\ln r$ exists because $r \u003e 0$.\n4. Claim: $\\exp(y)^m = \\exp(m y)$ for real $y$ and integers $m \\ge 0$. For $m = 0$ both sides are $1$, since all terms of the series for $\\exp(0)$ after the first vanish.\nIf it holds for $m$, the [[exponential of a sum|0E8HNVBCSP]] gives $\\exp(y)^{m+1} = \\exp(m y) \\exp(y) = \\exp((m + 1) y)$.\n5. By step 4 and [[the logarithm is the inverse of the exponential|0RKW2X63YR]], $s^n = \\exp(\\ln r) = r$.\n6. Let $w = s \\bigl(\\cos(\\theta / n) + i \\sin(\\theta / n)\\bigr)$.\n7. Multiplication in $\\C$ is commutative and associative, since [[$\\C$ is a field|0FRSR10DWF]].\nSo $w^n = s^n \\bigl(\\cos(\\theta / n) + i \\sin(\\theta / n)\\bigr)^n$.\n8. By [[de Moivre's formula|03MTBVG3VE]] with $x = \\theta / n$, the last power is $\\cos \\theta + i \\sin \\theta$.\n9. By steps 5, 7 and 8, $w^n = r (\\cos \\theta + i \\sin \\theta) = z$ by step 2.\nSo $z$ has an $n$th root $w$.\n\n## Idea\n\nIn polar form, raising to the $n$th power raises the modulus to the $n$th power and multiplies the angle by $n$.\nSo to undo it, take the positive real $n$th root of the modulus and divide the angle by $n$.\n",
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