Schwarz–Christoffel mapping

Remarks

Such an $f$ exists by the Riemann mapping theorem composed with a Möbius map from the disc to $\mathbb{H}$, and it extends to the boundary by Carathéodory's theorem. On 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. Three 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. Example: 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. The same formula, with vertices at infinity allowed, describes maps onto unbounded polygonal regions. For example, $\arcsin$ maps $\mathbb{H}$ onto the half-strip $\set{w : \abs{\operatorname{Re} w} < \pi/2,\ \operatorname{Im} w > 0}$. Non-example: a conformal map of $\mathbb{H}$ onto the unit disc is not a Schwarz–Christoffel mapping. The disc is not bounded by a polygon, and the image of the real axis has no straight pieces.