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  "body": "\n## Proof\n\nWe cut the rectangle into four, keep the worst quarter, and repeat. Near the limit point, $f$ is almost linear.\n\n1. Since $f$ is [[holomorphic|0DG0Z72CQD]], it is [[complex differentiable|0D1C9Y1EXA]] at each $z_1 \\in U$. So $f(z_1 + h) - f(z_1) = h \\cdot \\frac{f(z_1 + h) - f(z_1)}{h} \\to 0 \\cdot f'(z_1) = 0$. So $f$ is continuous on $U$.\n2. For a closed rectangle $P = \\set{x + iy : p_1 \\le x \\le q_1, \\ p_2 \\le y \\le q_2} \\subseteq U$ with $p_1 \u003c q_1$ and $p_2 \u003c q_2$, write $B(P) = \\oint_{\\partial P} f(z) \\, dz$, the [[contour integral|00FWNP9TXY]] along $\\partial P$ counterclockwise.\n3. The bottom side is $s + i p_2$ with $s \\in [p_1, q_1]$, and the right side is $q_1 + it$ with $t \\in [p_2, q_2]$. The top and left sides are traversed backwards. Using [[additivity over intervals|05RCDKHZM6]] and the [[substitution|0D37QFPEZE]] $s \\mapsto p_1 + q_1 - s$ for backward sides, we get the formula below.\n   $$B(P) = \\int_{p_1}^{q_1} f(s + i p_2) \\, ds + i \\int_{p_2}^{q_2} f(q_1 + it) \\, dt - \\int_{p_1}^{q_1} f(s + i q_2) \\, ds - i \\int_{p_2}^{q_2} f(p_1 + it) \\, dt.$$\n4. Cut $P$ by the lines $x = m_1$ and $y = m_2$, where $m_1 = (p_1 + q_1)/2$ and $m_2 = (p_2 + q_2)/2$. This gives four rectangles $P^{(1)}, \\dots, P^{(4)}$.\n5. Add the four formulas of step 3. Each inner half-segment appears twice with opposite signs, and the outer half-segments combine by additivity over intervals. So $B(P) = \\sum_{k=1}^4 B(P^{(k)})$.\n6. Let $\\mathrm{p}(P)$ be the perimeter of $P$ and $d(P)$ its diagonal length. Any two points of $P$ are at distance at most $d(P)$. Each $P^{(k)}$ has perimeter $\\mathrm{p}(P)/2$ and diagonal $d(P)/2$.\n7. By step 5 and the [[triangle inequality|0FJZMBX0ZZ]], some $k$ has $\\abs{B(P^{(k)})} \\ge \\abs{B(P)}/4$.\n8. Put $Q_0 = Q$. Given $Q_n$, let $Q_{n+1}$ be a quarter of $Q_n$ chosen as in step 7. By induction, $\\abs{B(Q)} \\le 4^n \\abs{B(Q_n)}$, $\\mathrm{p}(Q_n) = 2^{-n} \\mathrm{p}(Q)$ and $d(Q_n) = 2^{-n} d(Q)$.\n9. Each $Q_n$ is closed and bounded, so it is [[compact|07PFYV65MB]]. The $Q_n$ are nonempty and nested. By [[Cantor's intersection theorem|0EZBG8SKJM]], there is a point $z_0 \\in \\bigcap_{n} Q_n$. Then $z_0 \\in Q \\subseteq U$.\n10. Define $\\psi(z) = \\frac{f(z) - f(z_0)}{z - z_0} - f'(z_0)$ for $z \\in U$, $z \\ne z_0$, and $\\psi(z_0) = 0$. Then $\\psi(z) \\to 0$ as $z \\to z_0$.\n11. Put $\\ell(z) = f(z_0) + f'(z_0)(z - z_0)$. Then $f(z) = \\ell(z) + \\psi(z)(z - z_0)$ for all $z \\in U$.\n12. The function $\\Phi(z) = f(z_0) z + f'(z_0) (z - z_0)^2/2$ satisfies $\\Phi' = \\ell$ on $\\C$. Indeed $\\frac{(z + h - z_0)^2 - (z - z_0)^2}{h} = 2(z - z_0) + h$.\n13. Since $\\partial Q_n$ is a closed path, the [[fundamental theorem of contour integration|0YQ458D7PF]] gives $\\oint_{\\partial Q_n} \\ell(z) \\, dz = 0$. By step 11, $B(Q_n) = \\oint_{\\partial Q_n} \\psi(z)(z - z_0) \\, dz$. The integrand $f - \\ell$ is continuous.\n14. Let $\\varepsilon \u003e 0$. By step 10 choose $\\delta \u003e 0$ with $\\abs{\\psi(z)} \\le \\varepsilon$ for $z \\in U$, $\\abs{z - z_0} \u003c \\delta$. Choose $n$ with $d(Q_n) \u003c \\delta$.\n15. For $z \\in \\partial Q_n$, both $z$ and $z_0$ lie in $Q_n$. So $\\abs{z - z_0} \\le d(Q_n) \u003c \\delta$ and $\\abs{\\psi(z)(z - z_0)} \\le \\varepsilon \\, d(Q_n)$.\n16. The path $\\partial Q_n$ has length $\\mathrm{p}(Q_n)$. By the [[estimation lemma|0DJMGPN8K9]], $\\abs{B(Q_n)} \\le \\varepsilon \\, d(Q_n) \\, \\mathrm{p}(Q_n) = \\varepsilon \\, 4^{-n} d(Q) \\, \\mathrm{p}(Q)$.\n17. By step 8, $\\abs{B(Q)} \\le 4^n \\abs{B(Q_n)} \\le \\varepsilon \\, d(Q) \\, \\mathrm{p}(Q)$.\n18. Since $\\varepsilon \u003e 0$ is arbitrary, $B(Q) = 0$. That is, $\\oint_{\\partial Q} f(z) \\, dz = 0$.\n\n## Idea\n\nEach subdivision loses at most a factor $4$ in the integral, while the product of diameter and perimeter also shrinks by a factor $4$. Near the limit point $f$ equals a linear function, whose integral vanishes, plus an error that is small compared to $\\abs{z - z_0}$. So the error wins the race and the integral is $0$.\n",
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