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  "body": "\n## Proof\n\nWe show that $\\operatorname{Log}$ is continuous on $S$ and then invert the derivative of $\\exp$.\n\n1. The set $(-\\infty, 0]$ is closed in $\\C$, so the [[slit plane|06JYC06JHY]] $S$ is open. For $z \\in S$, the [[principal branch|0YKY4M3KRQ]] is $\\operatorname{Log} z = \\ln \\abs{z} + i \\operatorname{Arg} z$, with $\\operatorname{Arg} z \\in (-\\pi, \\pi]$ the [[principal argument|0SQSQP748R]].\n2. Let $z \\in S$ and $\\theta = \\operatorname{Arg} z$. If $\\theta = \\pi$, then by the [[definition of an argument|0ZZRSX91QR]] and [[$\\cos \\pi = -1$, $\\sin \\pi = 0$|01A8JGR4JM]], $z = -\\abs{z} \\in (-\\infty, 0]$. This is false, so $\\theta \\in (-\\pi, \\pi)$.\n3. Let $z \\in S$. By the [[product of exponentials|0WM7DJCM6G]], $e^{\\operatorname{Log} z} = e^{\\ln \\abs{z}} e^{i \\operatorname{Arg} z}$. Here $e^{\\ln \\abs{z}} = \\abs{z}$, because the [[exponential series is the real exponential|06JZGMYTYD]] at real points and [[$e^{\\ln t} = t$|0RKW2X63YR]].\n4. By [[Euler's formula|04R3CY41VJ]], with the real [[cosine|08W5RN375X]] and [[sine|02KTRHQMHN]] at real points, $e^{i\\theta} = \\cos \\theta + i \\sin \\theta$. So $e^{\\operatorname{Log} z} = \\abs{z} (\\cos \\theta + i \\sin \\theta) = z$.\n5. Write $z = x + iy \\in S$ with $x, y$ real, and let $\\varphi = \\theta / 2 \\in (-\\pi/2, \\pi/2)$. By the [[angle addition formulas|028XEQCE88]] and the [[Pythagorean identity|03H7WZ4T3K]], $\\sin \\theta = 2 \\sin \\varphi \\cos \\varphi$ and $\\cos \\theta = 2 \\cos^2 \\varphi - 1$.\n6. Since $x = \\abs{z} \\cos \\theta$ and $y = \\abs{z} \\sin \\theta$, step 5 gives $\\abs{z} + x = 2 \\abs{z} \\cos^2 \\varphi$ and $y = 2 \\abs{z} \\sin \\varphi \\cos \\varphi$.\n7. We have $\\abs{z} + x \u003e 0$ on $S$. Indeed, if $y \\ne 0$ then $\\abs{z} \u003e \\abs{x} \\ge -x$. If $y = 0$ then $x \u003e 0$, because $z \\in S$.\n8. By steps 6 and 7, $\\cos \\varphi \\ne 0$ and $\\frac{y}{\\abs{z} + x} = \\frac{\\sin \\varphi}{\\cos \\varphi} = \\tan \\varphi$, by the [[definition of the tangent|06T0HS14PP]].\n9. Since $\\varphi \\in (-\\pi/2, \\pi/2)$, the [[definition of the arctangent|065XW2V6C3]] gives $\\varphi = \\arctan \\frac{y}{\\abs{z} + x}$. So $\\operatorname{Arg} z = 2 \\arctan \\frac{y}{\\abs{z} + x}$ for each $z \\in S$.\n10. The functions $x$, $y$ and $\\abs{z}$ are continuous in $z$, and the denominator in step 9 is positive on $S$. The arctangent is [[differentiable|05P4B4Q22V]], hence [[continuous|07ZMX4ANJS]]. The function $\\ln$ is [[continuous on $(0, \\infty)$|0RKW2X63YR]].\n11. By step 10 and the [[continuity of compositions|0NRQPQX3VA]], $\\operatorname{Arg}$ and $\\ln \\abs{\\cdot}$ are continuous on $S$. So $\\operatorname{Log}$ is continuous on $S$.\n12. Fix $z_0 \\in S$ and put $w_0 = \\operatorname{Log} z_0$. For $z \\in S$ with $z \\ne z_0$, put $w = \\operatorname{Log} z$. By step 4, $e^w = z \\ne z_0 = e^{w_0}$, so $w \\ne w_0$.\n13. Then $\\frac{\\operatorname{Log} z - \\operatorname{Log} z_0}{z - z_0} = \\frac{w - w_0}{e^w - e^{w_0}} = \\frac{1}{Q(w)}$, where $Q(w) = \\frac{e^w - e^{w_0}}{w - w_0}$ for $w \\ne w_0$.\n14. By the [[derivative of the complex exponential|019H1Z6419]], $Q(w) \\to e^{w_0} = z_0$ as $w \\to w_0$ with $w \\ne w_0$.\n15. Let $\\varepsilon \u003e 0$. Choose $\\eta \u003e 0$ with $\\abs{Q(w) - z_0} \u003c \\varepsilon$ for $0 \u003c \\abs{w - w_0} \u003c \\eta$. By step 11, choose $\\delta \u003e 0$ with $\\abs{\\operatorname{Log} z - w_0} \u003c \\eta$ for $z \\in S$, $\\abs{z - z_0} \u003c \\delta$.\n16. By steps 12 and 15, $\\abs{Q(\\operatorname{Log} z) - z_0} \u003c \\varepsilon$ for $z \\in S$ with $0 \u003c \\abs{z - z_0} \u003c \\delta$. So $Q(\\operatorname{Log} z) \\to z_0$ as $z \\to z_0$.\n17. We have $z_0 \\ne 0$. When $\\abs{Q - z_0} \u003c \\abs{z_0}/2$, we get $\\abs{Q} \u003e \\abs{z_0}/2$ and $\\abs{1/Q - 1/z_0} = \\abs{Q - z_0} / (\\abs{Q} \\abs{z_0}) \u003c 2 \\abs{Q - z_0} / \\abs{z_0}^2$.\n18. By steps 13, 16 and 17, the difference quotient of $\\operatorname{Log}$ at $z_0$ tends to $1/z_0$. So $\\operatorname{Log}'(z_0) = 1/z_0$.\n19. This holds at each point of the open set $S$, so $\\operatorname{Log}$ is [[holomorphic|0DG0Z72CQD]] on $S$ with $\\operatorname{Log}'(z) = 1/z$.\n\n## Idea\n\nOn the slit plane, the principal argument has the explicit formula $2 \\arctan(y/(\\abs{z} + x))$, which shows that $\\operatorname{Log}$ is continuous. Since $\\operatorname{Log}$ inverts the [[exponential|054VXHMCA3]], its difference quotient is the reciprocal of a difference quotient of $\\exp$, which tends to $e^{w_0} = z_0$.\n",
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