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  "body": "\n## Proof\n\nWe use the polar form of $z$ and de Moivre's formula.\n\n1. Let $n \\ge 1$. For an integer $k$, let $\\omega_k = \\cos \\frac{2 \\pi k}{n} + i \\sin \\frac{2 \\pi k}{n}$.\nFor real arguments the power series of $\\cos$ and $\\sin$ are the real [[sine and cosine|0WB57D9488]].\nSo [[Euler's formula|04R3CY41VJ]] gives $\\omega_k = e^{2 \\pi i k / n}$.\n2. First let $0 \\le k \u003c n$. By [[de Moivre's formula|03MTBVG3VE]], $\\omega_k^n = \\cos(2 \\pi k) + i \\sin(2 \\pi k)$.\n3. Since $\\sin$ and $\\cos$ are [[$2\\pi$-periodic|0P4NQD746K]], applied $k$ times, $\\cos(2 \\pi k) = \\cos 0 = 1$ and $\\sin(2 \\pi k) = \\sin 0 = 0$.\nSo $\\omega_k^n = 1$, and $\\omega_k$ is an $n$th [[root of unity|0S61N75H45]].\n4. Conversely, let $z$ be a complex number with $z^n = 1$. Then $z \\neq 0$, since $0^n = 0 \\neq 1$.\n5. Let $r = |z| \u003e 0$ be the [[modulus|0KW2X0TFBX]]. Since the [[modulus is multiplicative|0HYNZE940T]], induction on $n$ gives $|z^n| = r^n$.\nSo $r^n = |1| = 1$.\n6. Claim: $r = 1$. If $0 \u003c r \u003c 1$, then $0 \u003c r^m \u003c 1$ for all $m \\ge 1$, by induction: $r^{m+1} = r \\cdot r^m \u003c r \u003c 1$.\nIf $r \u003e 1$, then in the same way $r^m \u003e 1$ for all $m \\ge 1$. Both contradict $r^n = 1$.\n7. Let $\\theta = \\operatorname{Arg} z$. By the [[polar form|0BRXFNQ5JV]] and step 6, $z = \\cos \\theta + i \\sin \\theta$.\n8. By de Moivre's formula, $1 = z^n = \\cos(n \\theta) + i \\sin(n \\theta)$.\nBy [[equality of complex numbers|05V7J4WD1C]], $\\cos(n \\theta) = 1$ and $\\sin(n \\theta) = 0$.\n9. Let $m = \\floor{(\\pi - n \\theta) / (2 \\pi)}$, using the [[floor function|0YVRNEDSVG]], and let $t = n \\theta + 2 \\pi m$.\nThen $2 \\pi m \\le \\pi - n \\theta \u003c 2 \\pi m + 2 \\pi$, so $-\\pi \u003c t \\le \\pi$.\n10. By periodicity, applied $\\abs{m}$ times, $\\cos t = 1$ and $\\sin t = 0$. So $t$ is an [[argument|0ZZRSX91QR]] of $1$ in $(-\\pi, \\pi]$.\n11. Also $0$ is an argument of $1$ in $(-\\pi, \\pi]$. By uniqueness of the [[principal value of the argument|0SQSQP748R]], $t = 0$.\n12. Let $j = -m$. Then $n \\theta = 2 \\pi j$, so $\\theta = 2 \\pi j / n$ with $j$ an integer.\n13. By the [[division algorithm|0CHQKZFK5S]], $j = n q + k$ with integers $q$ and $k$ and $0 \\le k \u003c n$.\nThen $\\theta = \\frac{2 \\pi k}{n} + 2 \\pi q$.\n14. By periodicity, applied $\\abs{q}$ times, $\\cos \\theta = \\cos \\frac{2 \\pi k}{n}$ and $\\sin \\theta = \\sin \\frac{2 \\pi k}{n}$.\nSo $z = \\omega_k$ with $0 \\le k \u003c n$.\n15. By steps 1 to 3 and steps 4 to 14, $z^n = 1$ holds if and only if $z = e^{2 \\pi i k / n}$ for some integer $k$ with $0 \\le k \u003c n$.\n\n## Idea\n\nAn $n$th root of unity has modulus $1$, so it is $\\cos \\theta + i \\sin \\theta$ for some angle $\\theta$.\nBy de Moivre's formula, its $n$th power has angle $n \\theta$, and this power is $1$ exactly when $n \\theta$ is a multiple of $2 \\pi$.\nReducing the multiple modulo $n$ leaves the $n$ angles $2 \\pi k / n$ with $0 \\le k \u003c n$.\n",
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