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  "body": "\n## Definition\n\nLet $P$ be the bounded open region enclosed by a simple polygon with $n \\ge 3$ vertices $w_1, \\ldots, w_n$, listed in the positive direction.\nLet $\\pi\\alpha_k$ be the interior angle of $P$ at $w_k$, with $0 \u003c \\alpha_k \u003c 2$; then $\\sum_{k=1}^{n} (\\alpha_k - 1) = -2$.\nLet $\\mathbb{H}$ be the [[upper half-plane|X6XGGH5FPW]].\nA *Schwarz–Christoffel mapping* for $P$ is a [[conformal map|0VHNKP1N3K]] $f$ of $\\mathbb{H}$ onto $P$.\n\nThe *Schwarz–Christoffel formula* gives the form of every such $f$:\n$$f(z) = A + C \\int_{z_0}^{z} \\prod_{k=1}^{n} (\\zeta - x_k)^{\\alpha_k - 1}\\, d\\zeta, \\qquad z \\in \\mathbb{H},$$\nwith real *prevertices* $x_1 \u003c x_2 \u003c \\cdots \u003c x_n$, a point $z_0 \\in \\mathbb{H}$ and constants $A, C \\in \\C$.\nThe powers are the branches that are holomorphic on $\\mathbb{H}$.\nHere $f$ extends continuously to the real axis with $f(x_k) = w_k$.\nIf one prevertex is placed at $\\infty$, its factor is omitted from the product.\n\n## Remarks\n\nSuch an $f$ exists by the [[Riemann mapping theorem|02WHMEG5ZR]] composed with a Möbius map from the disc to $\\mathbb{H}$, and it extends to the boundary by [[Carathéodory's theorem|AZ6H4ZKK55]].\nOn each interval $(x_k, x_{k+1})$ the argument of $f'$ is constant, so the interval is mapped onto a straight segment; at $x_k$ the argument jumps by $\\pi(1 - \\alpha_k)$, the exterior angle.\nThree of the prevertices can be chosen freely, because $\\mathbb{H}$ has a three-parameter group of conformal automorphisms; the others are determined by $P$ and usually found numerically.\nExample: for the right isosceles triangle, one may take prevertices $0, 1, \\infty$ with angles $\\pi/2, \\pi/4, \\pi/4$, giving $f(z) = \\int_0^z \\zeta^{-1/2} (\\zeta - 1)^{-3/4}\\, d\\zeta$ up to the constants.\nThe same formula, with vertices at infinity allowed, describes maps onto unbounded polygonal regions.\nFor example, $\\arcsin$ maps $\\mathbb{H}$ onto the half-strip $\\set{w : \\abs{\\operatorname{Re} w} \u003c \\pi/2,\\ \\operatorname{Im} w \u003e 0}$.\nNon-example: a conformal map of $\\mathbb{H}$ onto the unit disc is not a Schwarz–Christoffel mapping.\nThe disc is not bounded by a polygon, and the image of the real axis has no straight pieces.\n",
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