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  "body": "\n## Proof\n\n1. Let $U \\subseteq \\C$ be open, let $f : U \\to \\C$ be [[holomorphic|0DG0Z72CQD]], and let $z_0 \\in U$ with $f'(z_0) \\ne 0$. Identify $\\C$ with $\\R^2$ by $x + iy \\leftrightarrow (x, y)$.\n2. Since $f$ is holomorphic, it is [[complex differentiable|0D1C9Y1EXA]] at $z_0$. Write $a = f'(z_0)$ and $a = \\alpha + i\\beta$ with $\\alpha, \\beta \\in \\R$.\n3. Let $L : \\R^2 \\to \\R^2$ be the map $L(h) = a h$, a product of complex numbers. It is $\\R$-linear, because complex multiplication distributes over sums and commutes with real scalars.\n4. For $h \\ne 0$ with $z_0 + h \\in U$, we have $\\frac{\\abs{f(z_0 + h) - f(z_0) - a h}}{\\abs{h}} = \\abs{\\frac{f(z_0 + h) - f(z_0)}{h} - a}$. Here $\\abs{\\cdot}$ is the [[modulus|0KW2X0TFBX]], which is the Euclidean norm on $\\R^2$.\n5. By step 2, the right side of step 4 tends to $0$ as $h \\to 0$. So $f$ is [[differentiable at $z_0$|0135XZZQ8M]] as a map from an open subset of $\\R^2$ to $\\R^2$, with derivative $L$.\n6. Write $f = u + iv$ with $u, v$ real valued. Restricting $h$ in step 2 to real numbers $t \\ne 0$ gives $\\lim_{t \\to 0} \\frac{f(z_0 + t) - f(z_0)}{t} = a$.\n7. Taking real and imaginary parts in step 6 gives the partial derivatives $u_x(z_0) = \\alpha$ and $v_x(z_0) = \\beta$.\n8. Restricting $h$ in step 2 to $h = it$ with real $t \\ne 0$ gives $\\lim_{t \\to 0} \\frac{f(z_0 + it) - f(z_0)}{t} = i a = -\\beta + i\\alpha$.\n9. Taking real and imaginary parts in step 8 gives $u_y(z_0) = -\\beta$ and $v_y(z_0) = \\alpha$.\n10. By steps 7 and 9, the [[Jacobian matrix|0YF0VXKXGB]] of $f$ at $z_0$ is\n$$J = \\begin{pmatrix} \\alpha \u0026 -\\beta \\\\ \\beta \u0026 \\alpha \\end{pmatrix}.$$\n11. Let $\\lambda = \\abs{a} = \\sqrt{\\alpha^2 + \\beta^2}$. Since $a \\ne 0$, we have $\\lambda \u003e 0$.\n12. Let $Q = \\lambda^{-1} J$, so $J = \\lambda Q$. Then $Q^T Q = \\lambda^{-2} \\begin{pmatrix} \\alpha^2 + \\beta^2 \u0026 0 \\\\ 0 \u0026 \\alpha^2 + \\beta^2 \\end{pmatrix} = I_2$, so $Q$ is [[orthogonal|0QRBVFJ50X]].\n13. Also $\\det Q = \\lambda^{-2}(\\alpha^2 + \\beta^2) = 1$. So $Q$ is a rotation matrix.\n14. By steps 5, 10, 12 and 13, $f$ is [[conformal|0VHNKP1N3K]] at $z_0$.\n15. If $f'(z) \\ne 0$ for each $z \\in U$, then steps 1 to 14 apply at each $z \\in U$. So $f$ is conformal on $U$.\n\n## Idea\n\nThe complex derivative says that near $z_0$ the map $f$ is close to multiplication by the number $a = f'(z_0)$. As a real linear map, multiplication by $a$ has the matrix with rows $(\\alpha, -\\beta)$ and $(\\beta, \\alpha)$. This matrix is $\\abs{a}$ times a rotation, which is exactly the definition of conformal.\n",
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