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  "body": "\n## Proof\n\nWe compare the complex linear approximation of $f$ with the real linear approximations of $u$ and $v$.\n\n1. For $h = s + it \\ne 0$ with $s, t$ real and $z_0 + h \\in U$, write $\\Delta u(h) = u(x_0 + s, y_0 + t) - u(x_0, y_0)$ and $\\Delta v(h) = v(x_0 + s, y_0 + t) - v(x_0, y_0)$. Then $f(z_0 + h) - f(z_0) = \\Delta u(h) + i \\Delta v(h)$, by the definition of [[real and imaginary parts|02ZE6JHJ7G]].\n2. For real numbers $A, B$, we have $(A + iB) h = (As - Bt) + i (Bs + At)$.\n3. Hence, for $A, B$ real,\n$$f(z_0 + h) - f(z_0) - (A + iB) h = \\big(\\Delta u(h) - As + Bt\\big) + i \\big(\\Delta v(h) - Bs - At\\big).$$\nCall the two real brackets $\\varepsilon_1(h)$ and $\\varepsilon_2(h)$.\n4. We have $|\\varepsilon_1|, |\\varepsilon_2| \\le |\\varepsilon_1 + i \\varepsilon_2| \\le |\\varepsilon_1| + |\\varepsilon_2|$. So $(\\varepsilon_1 + i\\varepsilon_2)(h)/|h| \\to 0$ as $h \\to 0$ if and only if $\\varepsilon_1(h)/|h| \\to 0$ and $\\varepsilon_2(h)/|h| \\to 0$.\n5. Also $|h| = \\norm{(s, t)}$. Dividing by $h$ instead of $|h|$ does not change whether a quantity tends to $0$, because $\\abs{h / |h|} = 1$.\n6. Suppose first that $f$ is [[complex differentiable|0D1C9Y1EXA]] at $z_0$. Write $f'(z_0) = A + iB$ with $A, B$ real.\n7. By the definition of $f'(z_0)$, the quotient $\\big(f(z_0 + h) - f(z_0) - f'(z_0) h\\big)/h$ tends to $0$. By steps 3 to 5, $\\varepsilon_1(h)/\\norm{(s,t)} \\to 0$ and $\\varepsilon_2(h)/\\norm{(s,t)} \\to 0$.\n8. So $u$ is [[differentiable|0135XZZQ8M]] at $(x_0, y_0)$, and its derivative is the linear map $(s, t) \\mapsto As - Bt$. Thus $u_x = A$ and $u_y = -B$ there.\n9. In the same way, $v$ is differentiable at $(x_0, y_0)$ with derivative $(s, t) \\mapsto Bs + At$. Thus $v_x = B$ and $v_y = A$ there.\n10. Hence $u_x = A = v_y$ and $u_y = -B = -v_x$ at $(x_0, y_0)$. Also $f'(z_0) = A + iB = u_x + i v_x$.\n11. Conversely, suppose that $u$ and $v$ are differentiable at $(x_0, y_0)$ and satisfy the two equations there. Put $A = u_x(x_0, y_0)$ and $B = v_x(x_0, y_0)$.\n12. The equations give $u_y = -B$ and $v_y = A$ at $(x_0, y_0)$. So the derivatives of $u$ and $v$ are the maps $(s, t) \\mapsto As - Bt$ and $(s, t) \\mapsto Bs + At$.\n13. By the definition of differentiability, $\\varepsilon_1(h)/|h| \\to 0$ and $\\varepsilon_2(h)/|h| \\to 0$ as $h \\to 0$.\n14. By steps 3 to 5, $\\big(f(z_0 + h) - f(z_0)\\big)/h - (A + iB) \\to 0$ as $h \\to 0$.\n15. So $f$ is complex differentiable at $z_0$, and $f'(z_0) = A + iB = u_x(x_0, y_0) + i v_x(x_0, y_0)$.\n\n## Idea\n\nMultiplication by $A + iB$ is the real linear map of the plane with matrix $\\begin{pmatrix} A \u0026 -B \\\\ B \u0026 A \\end{pmatrix}$. So a complex derivative is a real derivative whose matrix has this special form. The Cauchy–Riemann equations say exactly that the Jacobian matrix of $(u, v)$ has this form.\n",
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