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  "body": "\n## Proof\n\nWe multiply the two trigonometric forms and use the angle addition formulas.\n\n1. Let $z \\neq 0$ and $w \\neq 0$. Let $r = |z|$ and $s = |w|$ be their [[moduli|0KW2X0TFBX]], so $r \u003e 0$ and $s \u003e 0$.\n2. Let $\\theta$ be an [[argument|0ZZRSX91QR]] of $z$ and $\\varphi$ an argument of $w$.\nBy definition, $z = r (\\cos \\theta + i \\sin \\theta)$ and $w = s (\\cos \\varphi + i \\sin \\varphi)$.\n3. Multiplication in $\\C$ is commutative and associative, since [[$\\C$ is a field|0FRSR10DWF]]. So $z w = r s \\, (\\cos \\theta + i \\sin \\theta)(\\cos \\varphi + i \\sin \\varphi)$.\n4. Expanding with $i^2 = -1$, the last product is $(\\cos \\theta \\cos \\varphi - \\sin \\theta \\sin \\varphi) + (\\sin \\theta \\cos \\varphi + \\cos \\theta \\sin \\varphi) i$.\n5. By the [[angle addition formulas|028XEQCE88]], this equals $\\cos(\\theta + \\varphi) + i \\sin(\\theta + \\varphi)$.\n6. Since the [[modulus is multiplicative|0HYNZE940T]], $|z w| = r s \u003e 0$. In particular $z w \\neq 0$.\n7. By steps 3 to 6, $z w = |z w| \\bigl(\\cos(\\theta + \\varphi) + i \\sin(\\theta + \\varphi)\\bigr)$. So $\\theta + \\varphi$ is an argument of $z w$.\n8. It remains to show that each argument $\\psi$ of $z w$ differs from $\\chi = \\theta + \\varphi$ by a multiple of $2 \\pi$.\n9. Both $\\psi$ and $\\chi$ are arguments of $z w$. Divide both forms by the real number $|z w| \u003e 0$ and compare parts.\nBy [[equality of complex numbers|05V7J4WD1C]], $\\cos \\psi = \\cos \\chi$ and $\\sin \\psi = \\sin \\chi$.\n10. The power series of [[sine and cosine|0WB57D9488]] give $\\cos(-x) = \\cos x$ and $\\sin(-x) = -\\sin x$, since $(-x)^{2n} = x^{2n}$ and $(-x)^{2n+1} = -x^{2n+1}$.\n11. Let $\\delta = \\psi - \\chi$. By the angle addition formulas and step 10, $\\cos \\delta = \\cos \\psi \\cos \\chi + \\sin \\psi \\sin \\chi$.\nBy step 9 and the [[Pythagorean identity|03H7WZ4T3K]], this is $\\cos^2 \\chi + \\sin^2 \\chi = 1$.\n12. In the same way, $\\sin \\delta = \\sin \\psi \\cos \\chi - \\cos \\psi \\sin \\chi = \\sin \\chi \\cos \\chi - \\cos \\chi \\sin \\chi = 0$.\n13. Let $m = \\floor{(\\pi - \\delta) / (2 \\pi)}$, using the [[floor function|0YVRNEDSVG]]. Then $2 \\pi m \\le \\pi - \\delta \u003c 2 \\pi m + 2 \\pi$.\nSo $\\delta' = \\delta + 2 \\pi m$ satisfies $-\\pi \u003c \\delta' \\le \\pi$.\n14. Since $\\sin$ and $\\cos$ are [[$2\\pi$-periodic|0P4NQD746K]], applied $\\abs{m}$ times, $\\cos \\delta' = 1$ and $\\sin \\delta' = 0$.\n15. So $1 = 1 \\cdot (\\cos \\delta' + i \\sin \\delta')$, and $|1| = 1$. Thus $\\delta'$ is an argument of $1$ in $(-\\pi, \\pi]$.\n16. Also $0$ is an argument of $1$ in $(-\\pi, \\pi]$, since $\\cos 0 = 1$ and $\\sin 0 = 0$ by the power series.\n17. The [[principal value of the argument|0SQSQP748R]] is unique, so $\\delta' = 0$. Hence $\\psi = \\theta + \\varphi - 2 \\pi m$.\n18. So $\\theta + \\varphi$ is an argument of $z w$, and every argument of $z w$ is congruent to $\\theta + \\varphi$ modulo $2 \\pi$.\nThis is $\\arg(z w) \\equiv \\arg z + \\arg w \\pmod{2 \\pi}$.\n\n## Idea\n\nWriting $z$ and $w$ in trigonometric form, their product has modulus $r s$.\nThe angle addition formulas turn the product of the two unit factors into the unit factor of the angle $\\theta + \\varphi$.\nTwo arguments of the same number always differ by a multiple of $2\\pi$, which gives the congruence.\n",
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