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  "body": "\n## Proof\n\nWe use the local power series of $f$ and differentiate it term by term.\n\n1. Fix $c \\in U$. Since $U$ is open, choose $r \u003e 0$ with $D = \\set{z : \\abs{z - c} \u003c r} \\subseteq U$. Since [[holomorphic functions are analytic|0826SGV3R0]], there are $a_n$ with $f(z) = \\sum_{n \\ge 0} a_n (z - c)^n$ for all $z \\in D$.\n2. Let $0 \u003c s \u003c r$. The series converges at $z = c + s$, so its terms tend to $0$. So $(\\abs{a_n} s^n)$ is bounded. By the definition of the [[radius of convergence|045EWAXY9P]] $R$, $s \\le R$. Hence $R \\ge r$.\n3. Let $S(z)$ be the sum of the series on its disc of convergence, which contains $D$. Since [[power series are holomorphic in the disc of convergence|0YMP521ZR5]], $S'(z) = \\sum_{n \\ge 1} n a_n (z - c)^{n-1}$, a power series with the same radius $R$.\n4. For $z \\in D$, $f$ and $S$ agree on the open set $D$ around $z$. The difference quotients at $z$ agree for small $h$, so $f'(z) = S'(z)$.\n5. By step 3 and the same theorem applied to the series of $S'$, the function $S'$ is complex differentiable at each point of $D$. By step 4, $f' = S'$ on $D$, so $f'$ is [[complex differentiable|0D1C9Y1EXA]] at $c$.\n6. Since $c \\in U$ was arbitrary, $f'$ is [[holomorphic|0DG0Z72CQD]] on $U$. Applying this to $f', f'', \\dots$ in turn, by induction each $f^{(n)}$ exists and is holomorphic on $U$.\n7. Let $h = U_h + i V_h$ be holomorphic on $U$, with real and imaginary parts $U_h, V_h$. By the [[Cauchy–Riemann equations|0T71SSYZ39]], at each point the [[partial derivatives|05AXG3QRKR]] exist and $h' = (U_h)_x + i (V_h)_x$.\n8. Also $(U_h)_y + i (V_h)_y = -(V_h)_x + i (U_h)_x = i h'$. So, taking partial derivatives on real and imaginary parts, $\\partial h / \\partial x = h'$ and $\\partial h / \\partial y = i h'$.\n9. We show by induction on $k \\ge 0$ that every partial derivative of $f$ of order $k$ exists on $U$. It equals $i^m f^{(k)}$, where $m$ counts the derivatives in $y$. For $k = 0$ the only one is $f = i^0 f^{(0)}$.\n10. For the inductive step, apply step 8 to the holomorphic function $h = i^m f^{(k)}$, whose derivative is $i^m f^{(k+1)}$. We get $\\partial_x h = i^m f^{(k+1)}$ and $\\partial_y h = i^{m+1} f^{(k+1)}$.\n11. Each $f^{(k)}$ is complex differentiable, hence continuous: $f^{(k)}(z + h) - f^{(k)}(z) = h \\cdot (\\text{difference quotient}) \\to 0$. So all partial derivatives of $f$ of all orders exist and are continuous on $U$.\n\n## Idea\n\nNear each point, $f$ is a power series, and the derivative of a power series is again a power series with the same radius. So $f'$ is again holomorphic, and we can repeat forever. The Cauchy–Riemann equations turn each partial derivative into a complex derivative times a power of $i$.\n",
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