{
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  "body": "\n## Counterexample\n\nLet $f : \\mathbb{C} \\to \\mathbb{C}$ be the [[complex conjugate|0BJV24C4SY]] map, $f(z) = \\bar z$.\nThe function $f$ is continuous, but it is not [[complex differentiable|0D1C9Y1EXA]] at any point $z_0 \\in \\mathbb{C}$.\nThus $f$ is not [[holomorphic|0DG0Z72CQD]] on any nonempty open set.\nProof: the difference quotient at $z_0$ is\n$$\\frac{f(z_0 + h) - f(z_0)}{h} = \\frac{\\bar h}{h}.$$\nIt is $1$ for each real $h \\neq 0$ and $-1$ for each purely imaginary $h \\neq 0$. So it has no limit as $h \\to 0$.\n\n## Remarks\n\nThe real and imaginary parts $u = x$ and $v = -y$ are differentiable, but $u_x = 1$ and $v_y = -1$. So the [[Cauchy–Riemann equations|0T71SSYZ39]] fail at each point.\nAs a map of $\\mathbb{R}^2$, $f$ is the reflection in the real axis. It is real linear but not complex linear.\nFor the same reason, $\\operatorname{Re} z$, $\\operatorname{Im} z$ and $|z|$ are not holomorphic on any nonempty open set.\n\n## Explanations\n\n### L1 human\n\nAs a map of the plane, conjugation is a mirror, and it is very smooth.\nBut a complex derivative must be the same in each direction, and conjugation fixes the real direction and reverses the imaginary direction.\nSo the quotient gives $1$ in one direction and $-1$ in the other, and there is no complex derivative.\n",
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    "name": "The complex conjugate is not holomorphic",
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        "locator": "Independence of the complex conjugate",
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  "msc": [
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  "name": "The complex conjugate is not holomorphic",
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