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  "body": "\n## Proof\n\nWe first prove $\\abs{\\int g} \\le \\int \\abs{g}$ for a continuous complex function $g$, then apply it piece by piece.\n\n1. Let $[p, q]$ be an interval and $g : [p, q] \\to \\C$ continuous. Its real and imaginary parts and $\\abs{g}$ are continuous, so they are [[Riemann integrable|06SESV7GQY]]. Put $I = \\int_p^q g(t) \\, dt$.\n2. For a complex constant $\\lambda$, $\\int_p^q \\lambda g = \\lambda I$. This follows from the [[linearity of the real integral|080RWVVSB6]] applied to $\\operatorname{Re}(\\lambda g)$ and $\\operatorname{Im}(\\lambda g)$.\n3. If $I = 0$, then $\\abs{I} \\le \\int_p^q \\abs{g}$, since $\\abs{g} \\ge 0$. Otherwise put $\\lambda = \\abs{I} / I$, so that $\\abs{\\lambda} = 1$ and $\\lambda I = \\abs{I}$.\n4. By step 2, $\\abs{I} = \\lambda I = \\int_p^q \\lambda g$. This number is real, so it equals $\\int_p^q \\operatorname{Re}(\\lambda g)$.\n5. We have $\\operatorname{Re}(\\lambda g) \\le \\abs{\\lambda g} = \\abs{g}$. By the [[monotonicity of the integral|0MBWQKM8QR]], $\\abs{I} \\le \\int_p^q \\abs{g(t)} \\, dt$.\n6. Now choose $a = t_0 \u003c t_1 \u003c \\dots \u003c t_m = b$ such that $\\gamma$ is continuously differentiable on each $[t_{j-1}, t_j]$. Put $g(t) = f(\\gamma(t)) \\gamma'(t)$, which is continuous on each such piece.\n7. By the [[definition of the contour integral|00FWNP9TXY]] and [[additivity over intervals|05RCDKHZM6]], $\\int_\\gamma f(z) \\, dz = \\sum_{j=1}^{m} \\int_{t_{j-1}}^{t_j} g(t) \\, dt$. In the same way, $L(\\gamma) = \\sum_{j=1}^m \\int_{t_{j-1}}^{t_j} \\abs{\\gamma'(t)} \\, dt$.\n8. By the [[triangle inequality|0FJZMBX0ZZ]] and step 5, $\\abs{\\int_\\gamma f(z) \\, dz} \\le \\sum_{j=1}^m \\int_{t_{j-1}}^{t_j} \\abs{f(\\gamma(t))} \\abs{\\gamma'(t)} \\, dt$.\n9. Since $\\gamma(t)$ lies on the image of $\\gamma$, $\\abs{f(\\gamma(t))} \\abs{\\gamma'(t)} \\le M \\abs{\\gamma'(t)}$. By monotonicity and linearity, the right side of step 8 is at most $M \\sum_{j} \\int_{t_{j-1}}^{t_j} \\abs{\\gamma'(t)} \\, dt = M L(\\gamma)$.\n\n## Idea\n\nRotating by the unit number $\\lambda$ turns the integral into a nonnegative real number, which is the integral of a real part. A real part is never larger than the modulus, so the integral of $\\abs{g}$ wins. Then bound $\\abs{f}$ by $M$.\n",
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