## Motivation [motivation] ### The Gap Between Existence and Computation The Riemann mapping theorem guarantees that any simply connected proper subdomain of $\mathbb{C}$ is conformally equivalent to the open unit disc $\mathbb{D}$ — but the proof, which proceeds via normal families and an extremal argument, is entirely non-constructive. It produces no formula for the mapping function, no algorithm for evaluating it, and no insight into how [boundary](/page/Boundary) geometry influences the map. For the working analyst or engineer who needs to solve a boundary value problem on a complicated planar domain, the theorem is a starting point, not a solution: it says the conformal map exists but offers no way to write it down. This gap between existence and computation is not merely a practical inconvenience — it reflects a genuine theoretical obstruction. For a general Jordan domain, the conformal map from $\mathbb{D}$ (or the upper half-plane $\mathbb{H}$) is typically transcendental and cannot be expressed in closed form. The few domains for which explicit formulas are available — discs, half-planes, strips, sectors — are all bounded by straight lines or circular arcs, and even modest modifications to these geometries destroy the possibility of elementary expressions. ### Why Polygons Are Special Polygonal domains occupy a privileged position because their boundaries are piecewise linear: the boundary of a polygon $P$ consists of finitely many straight edges meeting at vertices where the direction changes abruptly. This simple geometric structure has a precise analytic consequence for the conformal map $f : \mathbb{H} \to P$. On the real axis between consecutive prevertices, the image $f(t)$ traces a straight edge, so $\arg f'(t)$ is constant on each such interval. The [derivative](/page/Derivative) $f'$ is therefore a holomorphic function whose argument is piecewise constant on the boundary, with prescribed jumps at finitely many points. A holomorphic function in $\mathbb{H}$ whose argument has a known piecewise-constant boundary profile is severely constrained. The jumps in $\arg f'$ pin down the logarithmic derivative $f''/f'$ up to a bounded holomorphic function, and the boundary condition (together with the growth at infinity) forces that remaining freedom to vanish. The result is an explicit product formula for $f'$ — and hence, upon integration, an integral formula for $f$ itself. This is the Schwarz–Christoffel formula. ### From Half-Plane to Disc to Exterior The original derivation works on the upper half-plane, where the prevertices lie on the real line and the formula takes its simplest algebraic form. A Möbius transformation $\mathbb{D} \to \mathbb{H}$ translates the result into a disc formula, where the prevertices lie on $\partial \mathbb{D}$. A less obvious but equally important variant addresses the *exterior* problem: mapping the complement of $\overline{\mathbb{D}}$ conformally onto the complement of a polygon. The exterior formula governs a different class of applications — external aerodynamic flows, capacitance problems, far-field scattering — and its derivation requires a separate argument because the domain is unbounded and the map has a pole at infinity. [/motivation] ## Definition The Schwarz–Christoffel theory begins with a precise specification of the target geometry. The definitions below fix conventions for polygonal domains, their angle data, and the prevertices that parametrise the mapping. A polygon is not merely a collection of vertices — its angles encode the turning behaviour of the boundary, and the distinction between interior angles and exterior turning angles is the source of much notational confusion in the literature. We adopt the convention that all angles are normalised by $\pi$, so that a "right angle" corresponds to $\alpha = 1/2$. [definition:Polygonal Domain] A **polygonal domain** is an open, bounded, simply connected subset $P \subset \mathbb{C}$ whose boundary $\partial P$ is a closed polygonal curve. That is, $\partial P$ consists of finitely many straight line segments (edges) $[w_1, w_2], [w_2, w_3], \ldots, [w_n, w_1]$, where $w_1, \ldots, w_n \in \mathbb{C}$ are distinct points (vertices) listed in counterclockwise order. The **interior angle** at vertex $w_k$ is $\alpha_k \pi$, where $\alpha_k \in (0, 2)$ is the fraction of a full turn measured inside $P$. A convex vertex has $\alpha_k \in (0, 1)$, a straight vertex has $\alpha_k = 1$, and a reflex vertex has $\alpha_k \in (1, 2)$. [/definition] The interior angles of any simple $n$-gon satisfy a classical constraint from Euclidean geometry: the angles of a polygon sum to $(n-2)\pi$. In our normalised notation, this becomes the angle-sum identity $\sum_{k=1}^n \alpha_k = n - 2$. This identity will appear throughout the theory — it governs the convergence of the Schwarz–Christoffel integral and the degree of its integrand. When traversing the boundary $\partial P$ counterclockwise, the tangent direction changes at each vertex. The magnitude and sign of this change are captured by the exterior turning angles. [definition:Exterior Turning Angle] Let $P$ be a polygonal domain with vertices $w_1, \ldots, w_n$ and interior angles $\alpha_1 \pi, \ldots, \alpha_n \pi$. The **exterior turning angle** at vertex $w_k$ is \begin{align*} \beta_k = 1 - \alpha_k. \end{align*} At a convex vertex, $\beta_k \in (0, 1)$ and the boundary turns left (counterclockwise). At a reflex vertex, $\beta_k \in (-1, 0)$ and the boundary turns right (clockwise). The angle-sum identity $\sum \alpha_k = n - 2$ translates to \begin{align*} \sum_{k=1}^{n} \beta_k = 2, \end{align*} reflecting the fact that the tangent direction completes one full counterclockwise revolution ($2\pi$) when traversing a simple closed polygon. [/definition] The Schwarz–Christoffel formula expresses the conformal map in terms of certain distinguished boundary points in the source domain. [definition:Prevertices] Let $P$ be a polygonal domain and let $f : \Omega \to P$ be a conformal bijection, where $\Omega$ is either the upper half-plane $\mathbb{H}$, the unit disc $\mathbb{D}$, or the exterior $\hat{\mathbb{C}} \setminus \overline{\mathbb{D}}$. The **prevertices** are the points $\zeta_1, \ldots, \zeta_n \in \partial \Omega$ satisfying $f(\zeta_k) = w_k$ — that is, the boundary points that map to the polygon's vertices. When $\Omega = \mathbb{H}$, the prevertices $x_1 < x_2 < \cdots < x_n$ lie on $\mathbb{R} \cup \{\infty\}$, with at most one prevertex at $\infty$. When $\Omega = \mathbb{D}$ or $\hat{\mathbb{C}} \setminus \overline{\mathbb{D}}$, the prevertices $\zeta_1, \ldots, \zeta_n$ lie on $\partial \mathbb{D}$ in counterclockwise order. [/definition] ## The Half-Plane Formula ### Deriving the Formula from Boundary Behaviour The key to the Schwarz–Christoffel formula is that the conformal map $f : \mathbb{H} \to P$ is uniquely determined (up to the Möbius symmetries of $\mathbb{H}$) by the Riemann mapping theorem, and the piecewise-linear boundary of $P$ forces $f'$ into a rigid product form. The derivation proceeds by analysing what $f'$ must look like on the real axis and extending this information into the upper half-plane. Consider $f : \mathbb{H} \to P$ conformal, extending [continuously](/page/Continuity) to $\overline{\mathbb{H}}$, with prevertices $x_1 < x_2 < \cdots < x_n$ on $\mathbb{R}$ (we allow $x_n = \infty$ for now). On the interval $(x_k, x_{k+1})$, the image $f(t)$ traces the edge $[w_k, w_{k+1}]$. Since this edge is a straight line segment, the derivative $f'(t)$ has constant argument on $(x_k, x_{k+1})$ — that is, $\arg f'(t)$ is a step function on $\mathbb{R}$, constant between prevertices. As $t$ increases through the prevertex $x_k$, the image direction turns by the interior angle. More precisely, $\arg f'$ has a jump of $(\alpha_k - 1)\pi = -\beta_k \pi$ at $x_k$. This is because the new edge direction differs from the old edge direction by the supplement of the interior angle: the tangent rotates by $\pi - \alpha_k \pi = -\beta_k \pi$ when measured as a signed exterior turning angle. Now consider the function \begin{align*} \frac{f''(z)}{f'(z)} = \frac{d}{dz} \log f'(z). \end{align*} Since $\arg f'$ jumps by $(\alpha_k - 1)\pi$ at $x_k$ and $f'$ has a zero or singularity of order $\alpha_k - 1$ at $x_k$ (the map bends by angle $\alpha_k \pi$), the logarithmic derivative has a simple pole at each finite prevertex: \begin{align*} \frac{f''(z)}{f'(z)} = \sum_{k=1}^{n} \frac{\alpha_k - 1}{z - x_k} + \text{(holomorphic remainder)}. \end{align*} The remainder is holomorphic in $\mathbb{H}$, extends continuously to $\mathbb{R}$, and takes real values on $\mathbb{R}$ (since $\arg f'$ is constant between prevertices). By the Schwarz reflection principle, the remainder extends to an entire function. The growth condition at infinity — which follows from the angle-sum identity $\sum(\alpha_k - 1) = -2$, ensuring that $f'(z) = O(z^{-2})$ as $z \to \infty$ — forces the remainder to vanish by Liouville's theorem. Therefore $f''/f' = \sum_{k} (\alpha_k - 1)/(z - x_k)$, and exponentiating yields \begin{align*} f'(z) = c_1 \prod_{k=1}^{n} (z - x_k)^{\alpha_k - 1} \end{align*} for some constant $c_1 \in \mathbb{C} \setminus \{0\}$. Integrating from a base point $z_0 \in \mathbb{H}$ produces the Schwarz–Christoffel integral. The following theorem records the precise statement, including the full well-posedness: not only does the formula give the right derivative, but the resulting map is globally injective and surjects onto the interior of $P$. [quotetheorem:684] Several features of this result deserve emphasis. First, the formula involves $n - 3$ free real parameters once we fix three prevertices (using the three-parameter Möbius group of $\mathbb{H}$), plus the two complex constants $c_1$ and $c_2$. The prevertices are not given in advance — they are determined implicitly by the requirement that the image polygon have the correct vertices $w_1, \ldots, w_n$. This is the *parameter problem*, which we address below. Second, the exponents $\alpha_k - 1$ lie in $(-1, 1)$, so the integrand has [integrable](/page/Integral) singularities at the prevertices — the integral converges despite the apparent blow-up. At a convex vertex ($\alpha_k < 1$), the exponent is negative and $f'$ vanishes at $x_k$; at a reflex vertex ($\alpha_k > 1$), the exponent is positive and $f'$ has an integrable divergence. Third, when one prevertex is placed at $\infty$ — say $x_n = \infty$ — the corresponding factor $(z - x_n)^{\alpha_n - 1}$ is absorbed into the constant $c_1$, and the product runs over only $n - 1$ factors. This is a common normalisation choice that simplifies the integral. ### Why the Exponents Must Be $\alpha_k - 1$ The exponents $\alpha_k - 1$ in the Schwarz–Christoffel product are sometimes presented as a definition to be memorised, but they have a transparent geometric meaning. Consider the conformal map near a single vertex. [example:Power Map At A Vertex] The simplest conformal map that bends a straight line into a corner of angle $\alpha \pi$ is the power map \begin{align*} \phi : \mathbb{H} &\to W_\alpha \\ z &\mapsto z^\alpha, \end{align*} where $W_\alpha = \{r e^{i\theta} : r > 0, \, 0 < \theta < \alpha \pi\}$ is the sector of opening $\alpha \pi$. The positive real axis maps to the positive real axis, the negative real axis maps to the ray $\{r e^{i\alpha\pi} : r > 0\}$, and the origin is the vertex. The derivative is \begin{align*} \phi'(z) = \alpha z^{\alpha - 1}. \end{align*} The exponent $\alpha - 1$ appears because the derivative of $z^\alpha$ is $\alpha z^{\alpha - 1}$. In the Schwarz–Christoffel formula, each factor $(z - x_k)^{\alpha_k - 1}$ reproduces exactly this local vertex behaviour near $x_k$: the map $f$ looks like $(z - x_k)^{\alpha_k}$ near $x_k$, and its derivative inherits the exponent $\alpha_k - 1$. [/example] This local picture is the entire content of the formula — the Schwarz–Christoffel integral is nothing more than the statement that $f'$ is a product of the local vertex contributions, with no additional holomorphic factor. ## The Disc Formula ### Reduction to the Half-Plane via Möbius Transformation Many applications prefer the unit disc $\mathbb{D}$ as the canonical domain, especially in numerical work where the prevertices lie on the compact [set](/page/Set) $\partial \mathbb{D}$ rather than the unbounded real line. The disc formula follows immediately from the half-plane formula by composing with a Möbius transformation. The map \begin{align*} M : \mathbb{D} &\to \mathbb{H} \\ z &\mapsto i \, \frac{1 - z}{1 + z} \end{align*} is a conformal bijection from $\mathbb{D}$ to $\mathbb{H}$, sending $\partial \mathbb{D}$ to $\mathbb{R} \cup \{\infty\}$. If $g : \mathbb{D} \to P$ is the conformal map from the disc to the polygon, then $f = g \circ M^{-1} : \mathbb{H} \to P$ is the half-plane map. Applying the chain rule to $g = f \circ M$ and substituting the half-plane formula for $f'$ yields the disc formula after absorbing the Möbius derivative into the constant. The algebraic manipulation replaces each factor $(M(z) - x_k)^{\alpha_k - 1}$ with a factor involving $z$ and the disc prevertex $\zeta_k = M^{-1}(x_k) \in \partial \mathbb{D}$. After simplification, the factors take the form $(1 - z/\zeta_k)^{\alpha_k - 1}$, and the Möbius derivative contributes an overall power that is absorbed into $c_1$. The result is the following theorem. [quotetheorem:685] The disc formula has a direct geometric interpretation identical to the half-plane version: on each arc of $\partial \mathbb{D}$ between consecutive prevertices, the image under $g$ traces a straight edge, and $\arg g'$ is piecewise constant with jumps of $(\alpha_k - 1)\pi$ at the prevertices. The derivation via the Möbius transformation shows that no new mathematical content is needed — the disc and half-plane formulas are related by a change of coordinates. A practical advantage of the disc formula is that all prevertices lie on the unit circle, so their positions are parametrised by $n$ angles. The Möbius group of $\mathbb{D}$ has three real parameters (one rotation and a two-parameter family of hyperbolic translations), so fixing three prevertices eliminates this symmetry. The remaining $n - 3$ prevertex positions, together with $c_1$ and $c_2$, are determined by matching the polygon's vertices. [example:Regular N Gon From The Disc] The conformal map from $\mathbb{D}$ to a regular $n$-gon is the most symmetric case. By the rotational symmetry of both $\mathbb{D}$ and the regular $n$-gon, the prevertices must be equally spaced: $\zeta_k = e^{2\pi i k/n}$ for $k = 1, \ldots, n$. All interior angles are equal: $\alpha_k = 1 - 2/n$ (since the interior angle of a regular $n$-gon is $(1 - 2/n)\pi$). The exponents are $\alpha_k - 1 = -2/n$. The Schwarz–Christoffel integrand becomes \begin{align*} \prod_{k=1}^{n} \left(1 - \frac{z}{\zeta_k}\right)^{-2/n} = \prod_{k=1}^{n} \left(1 - z \, e^{-2\pi i k/n}\right)^{-2/n}. \end{align*} The product $\prod_{k=1}^n (1 - z \, e^{-2\pi i k/n}) = 1 - z^n$ (this is the factorisation of the cyclotomic polynomial), so the integrand simplifies to \begin{align*} (1 - z^n)^{-2/n}. \end{align*} The conformal map is therefore \begin{align*} g(z) = c_1 \int_0^z (1 - w^n)^{-2/n} \, dw. \end{align*} For $n = 3$ (equilateral triangle), this becomes $\int_0^z (1 - w^3)^{-2/3} \, dw$, which is an incomplete Beta function. For $n = 4$ (square), the integrand is $(1 - w^4)^{-1/2}$, and the map is an elliptic integral. For general $n$, the integral is a Schwarz–Christoffel integral with a single-term integrand — a dramatic simplification that illustrates the power of symmetry in reducing the parameter problem. [/example] ## The Exterior Formula ### Reversing Orientation: From Interior to Exterior The interior Schwarz–Christoffel formula maps a simply connected domain ($\mathbb{H}$ or $\mathbb{D}$) onto the interior of a polygon. Many applications require the complementary construction: a conformal map from the exterior of a reference domain onto the exterior of a polygon. The exterior of a bounded polygon $P$ is the unbounded simply connected domain $\hat{\mathbb{C}} \setminus \overline{P}$ in the Riemann sphere, and the natural source domain is $\hat{\mathbb{C}} \setminus \overline{\mathbb{D}}$, the exterior of the closed unit disc. The exterior map $g : \hat{\mathbb{C}} \setminus \overline{\mathbb{D}} \to \hat{\mathbb{C}} \setminus \overline{P}$ differs from the interior map in two essential ways. First, $g$ has a simple pole at $\infty$ (since it maps $\infty \mapsto \infty$ with $g(z) \sim c_1 z$ for large $|z|$), whereas the interior map is bounded. Second, the orientation of $\partial \mathbb{D}$ as seen from the exterior is clockwise — the exterior domain lies to the right of $\partial \mathbb{D}$ when traversed counterclockwise — so the turning angles at the prevertices are the *negated* exterior turning angles $-\beta_k \pi$. These two differences conspire to produce a formula where the exponents are $-\beta_k$ rather than $\alpha_k - 1 = -\beta_k$ — in fact, the exponents are the same quantity with opposite sign from the interior formula. More precisely, the interior exponents are $\alpha_k - 1 = -\beta_k$, while the exterior exponents are $-\beta_k$ applied to a different base: $(1 - \zeta_k/z)^{-\beta_k}$ rather than $(1 - z/\zeta_k)^{-\beta_k}$. The factor $1/z$ in the argument, rather than $z$, reflects the fact that the map is holomorphic outside the disc. [quotetheorem:688] The structural parallel between the interior and exterior formulas is illuminating. In both cases, the derivative $g'$ is a product of factors indexed by the prevertices, each contributing a power-law singularity that produces the correct turning angle at the corresponding vertex. The key difference is the sign of the exponents and the form of the factors: $(z - x_k)^{\alpha_k - 1}$ in the half-plane interior formula, $(1 - z/\zeta_k)^{\alpha_k - 1}$ in the disc interior formula, and $(1 - \zeta_k/z)^{-\beta_k}$ in the disc exterior formula. The constraint $\sum \beta_k = 2$ ensures that $g'(z) \to c_1$ as $z \to \infty$, giving the simple-pole normalisation. A notable feature of the exterior formula is that it applies only to bounded polygons — the exterior of an unbounded polygonal region (such as a half-strip or a channel) is not simply connected in $\hat{\mathbb{C}}$ and requires different techniques. ### Comparison of Interior and Exterior Exponents The relationship between the three formulas — half-plane interior, disc interior, and disc exterior — is clarified by examining the exponents. [example:Exponent Comparison For A Square] Consider a square with interior angles $\alpha_k = 1/2$ for each $k = 1, 2, 3, 4$. The exterior turning angles are $\beta_k = 1 - 1/2 = 1/2$. In the **half-plane interior** formula, the integrand is $\prod_{k=1}^4 (z - x_k)^{-1/2}$, which has a square-root singularity at each prevertex — the map bends by a right angle ($\pi/2$) at each vertex. In the **disc interior** formula, the integrand is $\prod_{k=1}^4 (1 - z/\zeta_k)^{-1/2}$, identical in structure. In the **disc exterior** formula, the derivative is $g'(z) = c_1 \prod_{k=1}^4 (1 - \zeta_k/z)^{-1/2}$. Near each prevertex, $g$ still bends by a right angle, but now the map sends arcs of $\partial \mathbb{D}$ to the polygon edges traversed in the opposite sense (as seen from the exterior). The check: $\sum \beta_k = 4 \cdot (1/2) = 2$, consistent with the exterior angle-sum identity. [/example] ## The Parameter Problem ### Degrees of Freedom The Schwarz–Christoffel formula expresses the conformal map in terms of the prevertices $x_1, \ldots, x_n$ (or $\zeta_1, \ldots, \zeta_n$) and the constants $c_1, c_2$. The polygon's geometry — its vertices $w_1, \ldots, w_n$ — is encoded implicitly: the vertices are obtained by evaluating $f$ at the prevertices. The question of *which* prevertices produce a *given* polygon is the **parameter problem**, and it is the central computational difficulty of the Schwarz–Christoffel method. The parameter count is as follows. In the half-plane formulation, the $n$ prevertices $x_1, \ldots, x_n \in \mathbb{R}$ give $n$ real parameters. The Möbius group $\operatorname{Aut}(\mathbb{H})$ has three real parameters, which we can use to fix three prevertices (e.g., $x_{n-2} = 0$, $x_{n-1} = 1$, $x_n = \infty$, which also eliminates the corresponding factor from the integrand). This leaves $n - 3$ real parameters for the remaining prevertices. The constants $c_1 \in \mathbb{C} \setminus \{0\}$ and $c_2 \in \mathbb{C}$ contribute four real parameters (two each), but $c_2$ is simply a translation (determined by the location of one vertex) and $|c_1|$, $\arg c_1$ control the overall scale and rotation of the polygon. The polygon $P$ has $n$ vertices, contributing $2n$ real parameters. However, the shape of $P$ up to similarity (translation, rotation, and scaling) is determined by $2n - 4$ real parameters. The interior angles $\alpha_1, \ldots, \alpha_n$ (constrained by $\sum \alpha_k = n - 2$) account for $n - 1$ of these, and the $n - 1$ independent edge-length ratios account for the remaining $n - 3$ (since one edge length is fixed by the overall scale). This matches: the $n - 3$ free prevertices determine the $n - 3$ edge-length ratios, while the angles are prescribed by the exponents. ### Symmetry and Special Polygons For highly symmetric polygons, the parameter problem simplifies dramatically or disappears entirely. If the polygon has a rotational symmetry of order $m$, then the prevertices inherit the same symmetry: they are equally spaced in [groups](/page/Group) related by rotation by $2\pi/m$. This reduces the number of free parameters. [example:Equilateral Triangle From The Half-Plane] For the equilateral triangle with vertices at $w_1 = 0$, $w_2 = 1$, and $w_3 = e^{i\pi/3}$, we have $\alpha_k = 1/3$ for each $k$ and $n = 3$. There are $n - 3 = 0$ free prevertices — the three Möbius parameters exhaust the freedom, so we may choose $x_1 = 0$, $x_2 = 1$, $x_3 = \infty$. The formula gives \begin{align*} f(z) = c_1 \int_0^z \zeta^{-2/3} (\zeta - 1)^{-2/3} \, d\zeta. \end{align*} The constant $c_1$ is determined by requiring $f(1) = w_2 = 1$: \begin{align*} c_1 = \frac{1}{\displaystyle\int_0^1 t^{-2/3}(1-t)^{-2/3} \, dt} = \frac{1}{B(1/3, 1/3)}, \end{align*} where $B(1/3, 1/3) = \Gamma(1/3)^2 / \Gamma(2/3)$ is the Beta function. This is a complete solution — no parameter problem arises because the triangle has exactly three vertices. [/example] For polygons with four or more vertices, the parameter problem is a system of nonlinear equations that must generally be solved numerically. The standard approach is to formulate the side-length conditions \begin{align*} |f(x_{k+1}) - f(x_k)| = |w_{k+1} - w_k|, \qquad k = 1, \ldots, n-1, \end{align*} as a system in the unknown prevertices, and solve by Newton's method or a variant. ### Numerical Methods The numerical solution of the parameter problem has been the subject of extensive algorithmic development. The Schwarz–Christoffel Toolbox (SC Toolbox) of Driscoll and Trefethen, implemented in MATLAB, solves the parameter problem via a combination of Newton iteration for the prevertex positions and adaptive Gauss–Jacobi quadrature for evaluating the Schwarz–Christoffel integral. The key numerical challenge is that the integrand has algebraic singularities at the prevertices, which require specialised quadrature rules (Gauss–Jacobi rather than Gauss–Legendre) for accurate evaluation. For the exterior formula, the parameter problem has the same structure: the $n - 3$ free prevertex positions (after fixing three by the Möbius symmetry of $\hat{\mathbb{C}} \setminus \overline{\mathbb{D}}$, which has a three-real-parameter automorphism group when $\infty$ is fixed) are determined by the side-length matching conditions. The SC Toolbox handles both interior and exterior problems. ## Degenerate Vertices and Unbounded Polygons ### Infinite Vertices and Half-Strips The Schwarz–Christoffel formula extends naturally to unbounded polygonal domains — regions bounded by finitely many line segments and rays extending to infinity. Such domains arise when one or more "vertices" are at $\infty$: a vertex at infinity corresponds to two adjacent edges that are parallel rays, so the "interior angle" at the infinite vertex is $\alpha = 0$ (the boundary goes straight through without turning) or $\alpha = 2$ (a cusp), or more commonly $\alpha = 0$ for a strip-like end. In the half-plane formula, a vertex at infinity with $\alpha_k = 0$ contributes the factor $(z - x_k)^{-1}$ to the integrand. The edge on each side of this "vertex" is a ray extending to infinity, and the vertex is the point at infinity where these two parallel edges "meet." [example:Half-Strip Via Schwarz–Christoffel] Consider the half-strip $S = \{w \in \mathbb{C} : 0 < \operatorname{Re}(w) < 1, \, \operatorname{Im}(w) > 0\}$. This is an unbounded polygon with three finite vertices: $w_1 = 0$, $w_2 = 1$, and a vertex at infinity ($w_3 = i\infty$, formally). The interior angles are $\alpha_1 = 1/2$ (right angle at $0$), $\alpha_2 = 1/2$ (right angle at $1$), and $\alpha_3 = 0$ (the "angle" at infinity, where the two vertical edges are parallel). The check: $\sum \alpha_k = 1/2 + 1/2 + 0 = 1 = 3 - 2$. Placing prevertices at $x_1 = -1$, $x_2 = 1$, $x_3 = \infty$ (sending the vertex at infinity to $\infty$), the factor for $k = 3$ is absorbed into $c_1$. The formula gives \begin{align*} f(z) = c_1 \int_{z_0}^z (\zeta + 1)^{-1/2}(\zeta - 1)^{-1/2} \, d\zeta + c_2 = c_1 \int_{z_0}^z \frac{d\zeta}{\sqrt{\zeta^2 - 1}} + c_2. \end{align*} The antiderivative is $\cosh^{-1}(z)$ (or equivalently $\log(z + \sqrt{z^2 - 1})$). Applying the boundary conditions $f(-1) = 0$ and $f(1) = 1$ determines $c_1$ and $c_2$: \begin{align*} f(z) = \frac{1}{\pi} \cosh^{-1}(z). \end{align*} This recovers the standard conformal map from $\mathbb{H}$ to the half-strip $\{w : 0 < \operatorname{Re}(w) < 1, \, \operatorname{Im}(w) > 0\}$. Equivalently, the inverse map $z = \cosh(\pi w)$ sends the half-strip to the upper half-plane. [/example] ### Slit Domains and Degenerate Angles At the other extreme, a vertex with interior angle $\alpha_k = 2$ corresponds to a slit: the boundary reverses direction at the vertex, creating a cut. The exponent $\alpha_k - 1 = 1$ in the Schwarz–Christoffel product means the factor $(z - x_k)^1$ is a regular (non-singular) linear factor. Slit domains arise naturally in free-boundary problems and in the study of flow around obstacles with sharp trailing edges. ## Common Techniques ### Composition with Möbius Transformations The Schwarz–Christoffel formula maps a canonical domain ($\mathbb{H}$, $\mathbb{D}$, or $\hat{\mathbb{C}} \setminus \overline{\mathbb{D}}$) to a polygon. To map between two non-canonical domains — for instance, from one polygon to another, or from a polygon to a circular-arc domain — one composes the Schwarz–Christoffel map with additional [conformal maps](/page/Conformal%20Maps). A particularly common technique is to apply a Möbius transformation *before* the Schwarz–Christoffel map, changing the normalisation of the prevertices. For instance, in the half-plane formulation, the three-parameter Möbius group of $\mathbb{H}$ allows us to send any three prescribed prevertices to any three prescribed real points (or $\infty$). The standard choices are: send one prevertex to $\infty$ (simplifying the product by removing one factor), or place two prevertices at $0$ and $1$ for computational convenience. ### Exploiting Reflection Symmetry If the target polygon has a line of symmetry — for instance, a rectangle or an isosceles triangle — then the conformal map satisfies a Schwarz reflection relation. The prevertices inherit the symmetry: if $P$ is symmetric about the real axis, then the prevertices on $\partial \mathbb{D}$ appear in conjugate pairs. This halves the number of unknowns in the parameter problem. [example:Rectangle Via Schwarz–Christoffel] A rectangle with vertices $w_1, w_2, w_3, w_4$ has all interior angles $\alpha_k = 1/2$ and two lines of symmetry. Placing prevertices at $x_1 = -1/\kappa$, $x_2 = -1$, $x_3 = 1$, $x_4 = 1/\kappa$ for some $\kappa \in (0, 1)$ (exploiting the symmetry $x_k \mapsto -x_k$ and $x_k \mapsto 1/x_k$), the Schwarz–Christoffel integral becomes \begin{align*} f(z) = c_1 \int_0^z \frac{d\zeta}{\sqrt{(1 - \zeta^2)(1 - \kappa^2 \zeta^2)}} + c_2. \end{align*} This is a Legendre elliptic integral of the first kind $F(z \,|\, \kappa)$. The parameter $\kappa$ is determined by the aspect ratio of the rectangle: the ratio of the half-periods $K(\kappa)$ and $K'(\kappa) = K(\sqrt{1 - \kappa^2})$ of the complete elliptic integrals equals the aspect ratio. This is the classical connection between rectangles and elliptic [functions](/page/Function) — the Schwarz–Christoffel formula makes it explicit. The single free parameter $\kappa$ (after exhausting the Möbius symmetry) corresponds to the one degree of freedom in the shape of a rectangle up to similarity: its aspect ratio. [/example] ## References Driscoll, T. A. and Trefethen, L. N., *Schwarz–Christoffel Mapping* (2002). Nehari, Z., *Conformal Mapping* (1952). Ablowitz, M. J. and Fokas, A. S., *Complex Variables: Introduction and Applications* (2003). Henrici, P., *Applied and Computational Complex Analysis, Vol. 3* (1986). Crowdy, D. G., *The Schwarz–Christoffel mapping to bounded multiply connected polygonal domains*, *Proceedings of the Royal Society A* (2005).