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  "body": "\n## Definition\n\nLet $\\mathbb{D}$ be the [[unit disc|8G5ABXRK85]], and let $\\mathcal{F}$ be the set of [[holomorphic|0DG0Z72CQD]] functions $f \\colon \\mathbb{D} \\to \\C$ with $f'(0) = 1$.\nFor $f \\in \\mathcal{F}$:\n\n1. $L_f$ is the supremum of the radii $r$ such that $f(\\mathbb{D})$ contains an open disc of radius $r$.\n2. $B_f$ is the supremum of the radii $r$ of the open discs $f(\\Omega)$, where $\\Omega \\subseteq \\mathbb{D}$ is open and $f$ is [[injective|0PVG1CRAJN]] on $\\Omega$.\n\n*Landau's constants* are the three numbers\n$$B = \\inf_{f \\in \\mathcal{F}} B_f, \\qquad L = \\inf_{f \\in \\mathcal{F}} L_f, \\qquad A = \\inf \\set{L_f : f \\in \\mathcal{F},\\ f \\text{ injective}}.$$\nThe number $B$ is the *Bloch constant*, $L$ is the *Landau constant*, and $A$ is the *univalent Landau constant*.\n\n## Remarks\n\nSince a disc covered injectively is a disc contained in the image, $B_f \\le L_f$ for all $f$, so $B \\le L$.\nThe infimum $A$ is taken over a smaller set of functions than $L$, so $L \\le A$.\nExample: for $f(z) = z$, the image is $\\mathbb{D}$, so $L_f = B_f = 1$.\nFor an injective $f \\in \\mathcal{F}$, the [[Koebe quarter theorem|WX0HMW4GT7]] gives a disc of radius $1/4$ in the image, so $A \\ge 1/4$.\nNon-example: $f(z) = 2z$ is not in $\\mathcal{F}$, because $f'(0) = 2$; without the normalisation the infimum would be $0$, as $f(z) = \\varepsilon z$ shows.\n[[Bloch's theorem|KW31KDX7TB]] is the statement $B \u003e 0$.\n",
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