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  "body": "\n## Proof\n\nWe show that $g(z) = (f(z) - f(a))/(z - a)$ has a primitive on a disc around the circle. Then its circle integral is $0$.\n\n1. There is $R \u003e r$ with $D = \\set{z : \\abs{z - c} \u003c R} \\subseteq U$. Otherwise, for each $n$ there is $z_n \\notin U$ with $\\abs{z_n - c} \u003c r + 1/n$.\n2. The sequence $(z_n)$ in step 1 is bounded. Applying the [[Bolzano–Weierstrass theorem|0KDJD249DT]] to real and imaginary parts, a subsequence tends to some $z^*$ with $\\abs{z^* - c} \\le r$. So $z^* \\in U$, and $U$ is open, so a disc around $z^*$ lies in $U$. This disc contains some $z_n$, a contradiction.\n3. Since $f$ is [[holomorphic|0DG0Z72CQD]], it is [[complex differentiable|0D1C9Y1EXA]] and hence continuous on $U$. Define $g(z) = \\frac{f(z) - f(a)}{z - a}$ for $z \\in D$, $z \\ne a$, and $g(a) = f'(a)$. Then $g$ is continuous on $D$.\n4. Let $z_0 \\in D$, $z_0 \\ne a$. For $z \\ne z_0$ near $z_0$, a direct computation gives\n   $$\\frac{g(z) - g(z_0)}{z - z_0} = \\frac{1}{(z - a)(z_0 - a)} \\left( \\frac{f(z) - f(z_0)}{z - z_0} (z_0 - a) - (f(z_0) - f(a)) \\right).$$\n   As $z \\to z_0$ this has a limit. So $g$ is holomorphic on the open set $D \\setminus \\set{a}$.\n5. For a closed rectangle $P = \\set{x + iy : p_1 \\le x \\le q_1, \\ p_2 \\le y \\le q_2} \\subseteq D$ with $p_1 \u003c q_1$, $p_2 \u003c q_2$, put $B(P) = \\oint_{\\partial P} g(z) \\, dz$, counterclockwise. As in [[Goursat's theorem|0BRG85WXYC]], [[additivity over intervals|05RCDKHZM6]] and a [[substitution|0D37QFPEZE]] on the backward sides give\n   $$B(P) = \\int_{p_1}^{q_1} g(s + i p_2) \\, ds + i \\int_{p_2}^{q_2} g(q_1 + it) \\, dt - \\int_{p_1}^{q_1} g(s + i q_2) \\, ds - i \\int_{p_2}^{q_2} g(p_1 + it) \\, dt.$$\n6. If $P$ is cut by vertical and horizontal lines into smaller rectangles, $B(P)$ is the sum of their $B$ values. Indeed, in the sum of the formulas of step 5, each inner edge appears twice with opposite signs.\n7. Claim: $B(P) = 0$ for each such $P \\subseteq D$. If $a \\notin P$, then $P$ lies in the open set $D \\setminus \\set{a}$, where $g$ is holomorphic by step 4. Goursat's theorem gives $B(P) = 0$.\n8. Let $a = \\alpha + i \\beta \\in P$ and $\\eta \u003e 0$. Cut $P$ by those of the lines $x = \\alpha \\pm \\eta$ and $y = \\beta \\pm \\eta$ that meet the inside of $P$.\n9. One piece is $P_\\eta = P \\cap \\set{x + iy : \\abs{x - \\alpha} \\le \\eta, \\ \\abs{y - \\beta} \\le \\eta}$, which contains $a$. Every other piece lies in one of the half-planes $x \\le \\alpha - \\eta$, $x \\ge \\alpha + \\eta$, $y \\le \\beta - \\eta$, $y \\ge \\beta + \\eta$. So it does not contain $a$.\n10. By steps 6, 7 and 9, $B(P) = B(P_\\eta)$. The rectangle $P$ is closed and bounded, hence [[compact|07PFYV65MB]], so the continuous $\\abs{g}$ is [[bounded on $P$|0V4DET59NW]] by some $M$.\n11. The perimeter of $P_\\eta$ is at most $8 \\eta$. By the [[estimation lemma|0DJMGPN8K9]], $\\abs{B(P)} = \\abs{B(P_\\eta)} \\le 8 \\eta M$. Since $\\eta \u003e 0$ is arbitrary, $B(P) = 0$. This proves the claim.\n12. Write $c = x_c + i y_c$. For $z = x + iy \\in D$, the rectangle $Q_z$ of points $s + it$ with $s$ between $x_c$ and $x$ and $t$ between $y_c$ and $y$ lies in $D$. Indeed $\\abs{(s + it) - c} \\le \\abs{z - c}$.\n13. Define, with oriented integrals,\n   $$F(z) = \\int_{x_c}^{x} g(s + i y_c) \\, ds + i \\int_{y_c}^{y} g(x + it) \\, dt, \\qquad G(z) = i \\int_{y_c}^{y} g(x_c + it) \\, dt + \\int_{x_c}^{x} g(s + iy) \\, ds.$$\n14. The difference $F(z) - G(z)$ is the right side of step 5 with $p_1 = x_c$, $q_1 = x$, $p_2 = y_c$, $q_2 = y$. It is $0$ if $x = x_c$ or $y = y_c$. Swapping $p_1$ and $q_1$, or $p_2$ and $q_2$, changes its sign, since $\\int_\\alpha^\\beta = -\\int_\\beta^\\alpha$. With both pairs in increasing order, it is $B(Q_z)$.\n15. So $F(z) - G(z) = \\pm B(Q_z)$ or $0$. By step 11, $F = G$ on $D$.\n16. By [[the integral of a continuous function is an antiderivative|0NC2A3HWGT]], applied to real and imaginary parts, $\\partial F / \\partial y = i g$ from the formula for $F$. In the same way $\\partial F / \\partial x = g$ from the formula for $G$. Each integral runs over an open interval of points in $D$.\n17. Write $F = U + iV$. Then $U_x = \\operatorname{Re} g$, $V_x = \\operatorname{Im} g$, $U_y = -\\operatorname{Im} g$, $V_y = \\operatorname{Re} g$. These are continuous, so [[$U$ and $V$ are differentiable|0KXSK8TEEB]], and they satisfy the Cauchy–Riemann equations.\n18. By the [[Cauchy–Riemann equations|0T71SSYZ39]], $F$ is holomorphic on $D$ with $F' = U_x + i V_x = g$.\n19. The circle $\\gamma(\\theta) = c + r e^{i \\theta}$, $\\theta \\in [0, 2\\pi]$, is a closed continuously differentiable path in $D$. By the [[fundamental theorem of contour integration|0YQ458D7PF]], $\\oint_{\\abs{z - c} = r} g(z) \\, dz = 0$.\n20. For $z$ on the circle, $\\abs{z - c} = r \u003e \\abs{a - c}$, so $z \\ne a$ and $g(z) = \\frac{f(z)}{z - a} - \\frac{f(a)}{z - a}$.\n21. By step 19 and linearity of the [[contour integral|00FWNP9TXY]], $\\oint_{\\abs{z - c} = r} \\frac{f(z)}{z - a} \\, dz = f(a) \\oint_{\\abs{z - c} = r} \\frac{dz}{z - a}$.\n22. Since $\\abs{a - c} \u003c r$, the [[integral of $1/(z - a)$ over the circle|0SXZ68HV6F]] is $2 \\pi i$. So $\\oint_{\\abs{z - c} = r} \\frac{f(z)}{z - a} \\, dz = 2 \\pi i f(a)$, which is the formula for the [[circle integral|0T7ABX4CWE]].\n\n## Idea\n\nThe difference quotient $g(z) = (f(z) - f(a))/(z - a)$ is holomorphic except at $a$, where it is still continuous. A continuous bad point cannot spoil Goursat's theorem, because a tiny rectangle around it has a tiny integral. So $g$ has a primitive on a disc, its circle integral is $0$, and what is left is $f(a)$ times the integral of $1/(z - a)$.\n",
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