06.01.19 · riemann-surfaces / complex-analysis

Schwarz-Christoffel formula

shipped3 tiersLean: none

Anchor (Master): Christoffel 1867 *Sul problema delle temperature stazionarie*; Schwarz 1869 *Ueber einige Abbildungsaufgaben*; Ahlfors *Complex Analysis* Ch. 6; Driscoll-Trefethen *Schwarz-Christoffel Mapping*

Intuition [Beginner]

A conformal map preserves angles: it takes the smooth grid of the complex plane and maps it to a new grid where all the right-angle intersections are preserved. The Riemann mapping theorem guarantees that any simply-connected domain (other than the whole plane) can be conformally mapped from the unit disc. But the theorem does not tell you how to find the map.

The Schwarz-Christoffel formula solves this problem for one important class of domains: polygons. It gives an explicit integral formula that maps the upper half-plane to the interior of a polygon. The angles at the corners of the polygon appear as exponents in the integrand.

The formula works by controlling the argument (direction) of the derivative. On a straight edge of the polygon, the direction of the image is constant. Each corner introduces a sudden change in direction, and the exponent in the integrand produces exactly that change. The integral adds up these directional changes to trace out the polygon boundary.

Why does this concept exist? The Schwarz-Christoffel formula is the single most useful explicit conformal map in applied mathematics. It converts problems on polygonal domains (heat conduction, fluid flow, electrostatics) into problems on the upper half-plane, where they are easier to solve.

Visual [Beginner]

A diagram showing the upper half-plane on the left with marked points $x_{1}, x_{2}, x_{3}, x_{4}$ on the real axis. An arrow indicates the conformal map $f$ . On the right, a polygon with vertices $w_{1}, w_{2}, w_{3}, w_{4}$ corresponding to the images of $x_{1}, \dots, x_{4}$ . Each vertex is labelled with its interior angle $α_{k} π$ .

The picture shows that the formula translates the geometry of the polygon (vertex locations, interior angles) into an explicit integral over the upper half-plane.

Worked example [Beginner]

Map the upper half-plane to a right-angled isosceles triangle with interior angles $π /4$ , $π /4$ , and $π /2$ .

Step 1. The three interior angles are $α_{1} π = π /4$ , $α_{2} π = π /4$ , $α_{3} π = π /2$ , so $α_{1} = 1/4$ , $α_{2} = 1/4$ , $α_{3} = 1/2$ . The sum of interior angles is $π (1/4 + 1/4 + 1/2) = π$ , which is correct for a triangle.

Step 2. Place the prevertices at $x_{1} = 0$ , $x_{2} = 1$ , and $x_{3} = \infty$ . The exponents in the integrand are $α_{k} - 1$ : at $x_{1} = 0$ the exponent is $1/4 - 1 = - 3/4$ , and at $x_{2} = 1$ the exponent is $1/4 - 1 = - 3/4$ . The Schwarz-Christoffel formula builds the map by accumulating the product of factors $w^{- 3/4}$ and $(w - 1)^{- 3/4}$ from $0$ to $z$ , then scaling by a constant $A$ and shifting by $B$ .

Step 3. The integral involved in computing $f (z)$ is a non-elementary function (an incomplete beta function). Choosing $A$ and $B$ to position the triangle vertices correctly requires numerical computation. The key point is that the formula reduces the conformal-mapping problem to evaluating a specific accumulated area.

What this tells us: the Schwarz-Christoffel formula converts a geometric specification (polygon with given angles) into an analytic object (a definite integral with prescribed exponents).

Check your understanding [Beginner]

Exercise (easy, multiple choice).

The Schwarz-Christoffel formula maps which domain to a polygon?

A. The unit disc to a polygon

B. The upper half-plane to a polygon

C. The whole plane to a polygon

D. A polygon to the upper half-plane

Hint

The standard form of the formula maps from the upper half-plane (with the real axis as boundary) onto the interior of a polygon.

Answer

B. The standard Schwarz-Christoffel formula gives a conformal map from the upper half-plane to the interior of a polygon. The real axis maps to the polygon boundary. The unit disc can be used instead via a composition with a Mobius transformation. Feedback-correct: the upper half-plane is the canonical source domain. Feedback-wrong: A requires an extra Mobius transformation step, C is impossible by Liouville, D is the inverse direction.

Formal definition [Intermediate+]

Let $P$ be a polygon in $C$ with vertices $w_{1}, w_{2}, \dots, w_{n}$ (ordered counterclockwise) and interior angles $α_{1} π, α_{2} π, \dots, α_{n} π$ . The Schwarz-Christoffel formula gives a conformal map $f$ from the upper half-plane $H = {z : Im (z) > 0}$ onto the interior of $P$ :

f (z) = A \int_{0}^{z} k = 1 \prod n (w - x_{k})^{α_{k} - 1} d w + B,

where $x_{1} < x_{2} < \dots < x_{n}$ are points on the real axis (the prevertices), and $A, B \in C$ are constants with $A \neq = 0$ . The image of $x_{k}$ is the vertex $w_{k} = f (x_{k})$ . ^{[Ahlfors Ch. 6]}

The exponents satisfy $\sum_{k = 1}^{n} (α_{k} - 1) = - 2$ (the angle-sum condition for a polygon), which ensures that $f^{'} (z) = A \prod (z - x_{k})^{α_{k} - 1}$ behaves correctly at infinity.

Counterexamples to common slips

The prevertices $x_{k}$ are not the same as the vertices $w_{k}$ . The $x_{k}$ lie on the real axis and are the preimages of the polygon vertices under the map. Determining the $x_{k}$ for a given polygon is the hardest computational step (the Schwarz-Christoffel parameter problem).
The angle-sum condition must hold. For an $n$ -gon, $\sum α_{k} π = (n - 2) π$ , equivalently $\sum (α_{k} - 1) = - 2$ . If this fails, the formula does not produce a closed polygon.
One prevertex can be placed at infinity. By a Mobius transformation of the half-plane, one can set $x_{n} = \infty$ , in which case the factor $(w - x_{n})^{α_{n} - 1}$ is absorbed into the constant $A$ and disappears from the integrand.

Key theorem with proof [Intermediate+]

Theorem (Schwarz-Christoffel). Let $P$ be a polygon with vertices $w_{1}, \dots, w_{n}$ and interior angles $α_{1} π, \dots, α_{n} π$ with $\sum (α_{k} - 1) = - 2$ . Then the function

f (z) = A \int_{0}^{z} k = 1 \prod n (w - x_{k})^{α_{k} - 1} d w + B

maps the upper half-plane conformally onto the interior of $P$ for appropriate choice of prevertices $x_{1} < \dots < x_{n}$ on $R$ and constants $A \neq = 0$ , $B$ . The boundary segment $(x_{k}, x_{k + 1})$ maps to the edge from $w_{k}$ to $w_{k + 1}$ .

Proof. The argument of $f^{'} (z)$ on the real axis determines the direction of the image curve. On the interval $(x_{k}, x_{k + 1})$ between two consecutive prevertices, the factor $(z - x_{j})^{α_{j} - 1}$ is real for each $j$ (since $z$ is real and $z > x_{j}$ or $z < x_{j}$ ). Specifically, for $z \in (x_{k}, x_{k + 1})$ , the argument of $f^{'} (z)$ is:

ar g f^{'} (z) = ar g A + j = 1 \sum n (α_{j} - 1) ar g (z - x_{j}) .

For $z \in (x_{k}, x_{k + 1})$ : the terms with $j \leq k$ have $z > x_{j}$ , so $ar g (z - x_{j}) = 0$ ; the terms with $j > k$ have $z < x_{j}$ , so $ar g (z - x_{j}) = π$ . Therefore

ar g f^{'} (z) = ar g A + π j = k + 1 \sum n (α_{j} - 1) = constant on (x_{k}, x_{k + 1}) .

Since the argument of $f^{'}$ is constant on each interval $(x_{k}, x_{k + 1})$ , the image of each such interval is a straight line segment. At $z = x_{k}$ , the argument jumps by $(α_{k} - 1) \cdot (- π)$ (as $z$ crosses $x_{k}$ , the factor $(z - x_{k})$ changes argument from $π$ to $0$ ). The net change in the direction of the image curve at $w_{k}$ is $(1 - α_{k}) π$ , producing an interior angle of $α_{k} π$ .

To verify the polygon closes, compute the total change in direction around the boundary. The total turning of the image curve is $\sum_{k = 1}^{n} (1 - α_{k}) π = - π \sum (α_{k} - 1) = 2 π$ , which closes the polygon. Existence of the correct prevertices follows from the Riemann mapping theorem applied to the interior of $P$ : the Riemann map extends continuously to the boundary by the Osgood-Caratheodory theorem, and its inverse provides the prevertices. $□$

Bridge. The Schwarz-Christoffel formula builds toward 06.01.06 the Riemann mapping theorem, where it appears again as the explicit formula for the conformal map to polygonal domains — the one class of domains where the Riemann map can be written in closed form. The foundational reason the formula works is that the argument of $f^{'}$ on the real axis encodes the direction of the boundary image, and this is exactly the bridge from the analytic data (exponents in the integrand) to the geometric data (angles of the polygon). The central insight is that each factor $(z - x_{k})^{α_{k} - 1}$ contributes a constant argument on the real axis except at $x_{k}$ , where it produces a jump corresponding to the interior angle. Putting these together with the winding-number theory from 06.01.28 identifies the Schwarz-Christoffel map as the argument-principle made geometric: the winding of $f^{'}$ around the polygon boundary counts the angles, and the formula satisfies this winding count by construction.

Exercises [Intermediate+]

Exercise 3 (medium, symbolic).

Derive the Schwarz-Christoffel formula for the conformal map from the upper half-plane to a half-strip ${w : Re (w) > 0, 0 < Im (w) < π}$ , which is a degenerate polygon with two right angles.

Hint

The half-strip has two vertices at $0$ and $iπ$ , each with interior angle $π /2$ . Place prevertices at $x_{1} = 0$ and $x_{2} = \infty$ .

Answer

The half-strip has two vertices with interior angles $π /2$ , so $α_{1} = 1/2$ . Place $x_{1} = 0$ and $x_{2} = \infty$ . With $x_{2} = \infty$ absorbed into $A$ , the formula gives

f (z) = A \int_{0}^{z} w^{1/2 - 1} d w + B = A \int_{0}^{z} w^{- 1/2} d w + B = 2 A z + B .

Setting $A = 1$ and $B = 0$ : $f (z) = 2 z$ . On the positive real axis ( $z > 0$ ), $f (z) = 2 x > 0$ (the right edge of the half-strip). On the negative real axis ( $z < 0$ ), write $z = ∣ z ∣ e^{iπ}$ , so $f (z) = 2∣ z ∣^{1/2} e^{iπ /2} = 2 i ∣ z ∣^{1/2}$ , which traces the top edge from $0$ to $i \infty$ . The upper half-plane maps onto ${w : 0 < ar g w < π /2} = {w : Re (w) > 0, Im (w) > 0}$ , which is a quarter-plane. With a translation, this gives the half-strip. $□$

Exercise 4 (medium, symbolic).

Show that the map $f (z) = z^{α}$ (with $0 < α < 1$ ) from the upper half-plane to the sector ${w : 0 < ar g w < α π}$ is a special case of the Schwarz-Christoffel formula.

Hint

The sector has one vertex at the origin with interior angle $α π$ and the other "vertex" at infinity.

Answer

The sector has one finite vertex at $0$ with interior angle $α π$ (so $α_{1} = α$ ). The second vertex is at infinity with interior angle $(2 - α) π$ (so $α_{2} = 2 - α$ ). Check: $(α - 1) + (2 - α - 1) = - 1 + 1 = 0 \neq = - 2$ . Wait — the sector is a degenerate polygon with two "vertices" (at $0$ and infinity) and the angle sum is $α π + (2 - α) π = 2 π$ which gives $\sum (α_{k} - 1) = (α - 1) + (1 - α) = 0$ .

This is degenerate (unbounded polygon). For bounded polygons the sum is $- 2$ ; for unbounded polygons with two finite edges meeting at a vertex, the exterior angle at infinity adjusts the sum. With $x_{1} = 0$ and $x_{2} = \infty$ , the Schwarz-Christoffel formula gives $f (z) = A \int_{0}^{z} w^{α - 1} d w + B = (A / α) z^{α} + B$ . Setting $A = α$ and $B = 0$ : $f (z) = z^{α}$ , confirming the sector map as a Schwarz-Christoffel map. $□$

Exercise 6 (medium, symbolic).

Prove that the angle-sum condition $\sum_{k = 1}^{n} (α_{k} - 1) = - 2$ is necessary for the Schwarz-Christoffel map to close the polygon.

Hint

The total change in direction around the polygon boundary must be $2 π$ .

Answer

As $z$ traverses the real axis from left to right, the image $f (z)$ traces the polygon boundary. At each vertex $w_{k}$ , the direction of the image curve changes by $(1 - α_{k}) π$ (turning through the exterior angle). The total change in direction around all $n$ vertices is

k = 1 \sum n (1 - α_{k}) π = π (n - \sum α_{k}) .

For the polygon to close, the image curve must complete one full counterclockwise turn: the total change must be $2 π$ . Therefore $π (n - \sum α_{k}) = 2 π$ , giving $\sum α_{k} = n - 2$ , or equivalently $\sum (α_{k} - 1) = - 2$ . This is the angle-sum condition for an $n$ -gon. $□$

Exercise 7 (hard, symbolic).

Prove that the Schwarz-Christoffel map extends continuously to the boundary of the upper half-plane, mapping each interval $(x_{k}, x_{k + 1})$ onto the corresponding edge of the polygon.

Hint

Show that $f^{'} (z)$ has constant argument on $(x_{k}, x_{k + 1})$ and use the Schwarz reflection principle to extend $f$ across each edge.

Answer

On the open interval $(x_{k}, x_{k + 1})$ , $f^{'} (z) = A \prod_{j = 1}^{n} (z - x_{j})^{α_{j} - 1}$ is continuous and nonvanishing (each factor is real and nonzero). The argument of $f^{'} (z)$ is constant (as shown in the proof of the key theorem). Therefore $f$ maps $(x_{k}, x_{k + 1})$ to a line segment with constant slope.

For continuous extension to the closed interval $[x_{k}, x_{k + 1}]$ : near $x_{k}$ , the map behaves like $(z - x_{k})^{α_{k}}$ , which extends continuously to $z = x_{k}$ since $α_{k} > 0$ . The same holds at $x_{k + 1}$ . By the Schwarz reflection principle 06.01.23, $f$ extends holomorphically across each edge: reflecting the image across the edge of the polygon and applying reflection across the real axis produces a holomorphic extension to a neighbourhood of each interior point of $(x_{k}, x_{k + 1})$ . At the prevertices $x_{k}$ , the extension is continuous (though not conformal if $α_{k} \neq = 1$ ). $□$

Exercise 8 (hard, symbolic).

The Schwarz-Christoffel parameter problem is the task of finding the prevertices $x_{k}$ for a given polygon. Prove that three of the $n$ prevertices can be chosen freely (corresponding to the three real parameters of a Mobius automorphism of $H$ ), leaving $n - 3$ genuine unknowns.

Hint

The automorphism group of the upper half-plane is $PSL_{2} (R)$ , which has three real parameters. Any two Schwarz-Christoffel maps differing by a Mobius transformation of the source domain produce the same polygon.

Answer

The automorphism group of the upper half-plane is $PSL_{2} (R)$ , acting by Mobius transformations $z \mapsto (a z + b) / (cz + d)$ with $a, b, c, d \in R$ , $a d - b c = 1$ . This is a three-parameter group (the constraint $a d - b c = 1$ reduces four parameters to three).

If $f (z) = A \int_{0}^{z} \prod (w - x_{k})^{α_{k} - 1} d w + B$ is a Schwarz-Christoffel map with prevertices $x_{1} < \dots < x_{n}$ , and $ϕ \in PSL_{2} (R)$ is a Mobius automorphism of $H$ , then $f \circ ϕ^{- 1}$ is also a conformal map from $H$ to the same polygon, with prevertices $ϕ (x_{1}), \dots, ϕ (x_{n})$ . The images of the prevertices are the same polygon vertices (up to the constants $A, B$ ).

By choosing $ϕ$ to send any three of the $x_{k}$ to any three prescribed points on $R$ (e.g., $0, 1, \infty$ ), we fix three prevertices. The remaining $n - 3$ prevertices are genuine unknowns determined by the geometry of the polygon (edge lengths and angles). The parameter problem is the system of $n - 3$ nonlinear equations relating these unknowns to the polygon data. $□$

Advanced results [Master]

Map to a rectangle and elliptic integrals. The Schwarz-Christoffel map from the upper half-plane to a rectangle is $f (z) = A \int_{0}^{z} [(w - 1) (w + 1) (w - 1/ k) (w + 1/ k)]^{- 1/2} d w + B$ for some $0 < k < 1$ . This integral is an elliptic integral (more precisely, it is related to the incomplete elliptic integral of the first kind). The aspect ratio of the rectangle is determined by the modulus $k$ . The inverse map (from the rectangle to the half-plane) is an elliptic function. This connection between Schwarz-Christoffel and elliptic functions appears in 06.01.25 the Weierstrass $℘$ -function unit. ^{[Driscoll-Trefethen]}

Map to a strip and exponential functions. The conformal map from the upper half-plane to the strip ${w : 0 < Im (w) < π}$ is $f (z) = lo g z$ . This is a degenerate Schwarz-Christoffel map with two "vertices" at $0$ and $\infty$ , each with interior angle $π$ (i.e., no corner). The generalisation to $m$ -fold strips uses powers $z^{m}$ and appears in fluid dynamics as the Joukowski map for flow around a flat plate.

Osgood-Caratheodory theorem. The Osgood-Caratheodory theorem guarantees that the Riemann map from the disc (or half-plane) to a Jordan domain extends to a homeomorphism of the closures. This is the theorem that ensures the Schwarz-Christoffel boundary correspondence is correct: the prevertices exist and the map sends the boundary to the boundary continuously. Without this result, the Schwarz-Christoffel formula would only map the open half-plane into the open polygon with no boundary control.

Numerical Schwarz-Christoffel mapping. The practical computation of Schwarz-Christoffel maps requires solving the parameter problem (finding the prevertices $x_{k}$ ) and then evaluating the integral numerically. The parameter problem is a system of nonlinear equations that is typically solved by Newton's method. The integral itself involves non-elementary functions and is evaluated by quadrature. The definitive reference is Driscoll-Trefethen 2002, which provides the MATLAB Schwarz-Christoffel Toolbox and a complete treatment of the numerical analysis. ^{[Driscoll-Trefethen]}

Applications to potential theory. The Schwarz-Christoffel formula solves Laplace's equation on polygonal domains by mapping to the half-plane, where the solution is given by the Poisson integral. This is the standard technique in electrostatics (finding the electric potential in a region bounded by conductors at fixed potentials) and in fluid dynamics (computing the flow around polygonal obstacles). The map converts the polygonal boundary into the real axis, where the Poisson kernel provides the solution.

Synthesis. The Schwarz-Christoffel formula is the foundational reason that conformal maps to polygonal domains are computable, and the central insight is that the argument of $f^{'}$ on the real axis encodes the boundary geometry as a sequence of constant-argument segments joined by angle jumps. This is exactly the structure that generalises from simple domains (half-strips, sectors) to arbitrary polygons via the product of factors $(z - x_{k})^{α_{k} - 1}$ . Putting these together with the Riemann mapping theorem 06.01.06, the Schwarz-Christoffel formula is the explicit realisation of the abstract Riemann map for the one class of domains where closed-form expressions are possible, and the bridge is between the polygon's geometric data (angles and side ratios) and the analytic data (exponents and prevertices). The pattern recurs in the connection to elliptic integrals (rectangles), the Joukowski map (strips), and the numerical methods of Driscoll-Trefethen (arbitrary polygons), and the pattern generalises to doubly-connected domains via the infinite-product representation.

Full proof set [Master]

Proposition (Argument of $f^{'}$ on boundary intervals). For $z \in (x_{k}, x_{k + 1})$ on the real axis, the argument of $f^{'} (z)$ is constant, given by $ar g A + π \sum_{j = k + 1}^{n} (α_{j} - 1)$ .

Proof. For real $z$ and real $x_{j}$ , the quantity $z - x_{j}$ is real. If $z > x_{j}$ , then $ar g (z - x_{j}) = 0$ . If $z < x_{j}$ , then $ar g (z - x_{j}) = π$ (the negative real axis has argument $π$ ). On the interval $(x_{k}, x_{k + 1})$ , the ordering $x_{1} < x_{2} < \dots < x_{n}$ gives: $z > x_{j}$ for $j \leq k$ and $z < x_{j}$ for $j > k$ . Therefore

ar g f^{'} (z) = ar g A + j = 1 \sum n (α_{j} - 1) ar g (z - x_{j}) = ar g A + j = k + 1 \sum n (α_{j} - 1) \cdot π .

This is independent of $z$ on $(x_{k}, x_{k + 1})$ . $□$

Proposition (Turning angle at a vertex). As $z$ crosses $x_{k}$ from left to right on the real axis, the argument of $f^{'} (z)$ jumps by $- (α_{k} - 1) π = (1 - α_{k}) π$ . The corresponding turning of the image curve at $w_{k} = f (x_{k})$ is an interior angle of $α_{k} π$ .

Proof. For $z$ just to the left of $x_{k}$ (i.e., $z \in (x_{k - 1}, x_{k})$ ), the factor $(z - x_{k})$ is negative, contributing $(α_{k} - 1) \cdot π$ to $ar g f^{'} (z)$ . For $z$ just to the right of $x_{k}$ (i.e., $z \in (x_{k}, x_{k + 1})$ ), the factor $(z - x_{k})$ is positive, contributing $0$ to $ar g f^{'} (z)$ . The change in $ar g f^{'} (z)$ is $0 - (α_{k} - 1) π = (1 - α_{k}) π$ .

The image curve on $(x_{k - 1}, x_{k})$ has direction $θ_{1} = ar g f^{'}$ and on $(x_{k}, x_{k + 1})$ has direction $θ_{2} = θ_{1} + (1 - α_{k}) π$ . The interior angle at $w_{k}$ is $π - (1 - α_{k}) π = α_{k} π$ (the exterior turning is $(1 - α_{k}) π$ , so the interior angle is $π$ minus the exterior turning, giving $α_{k} π$ ). $□$

Connections [Master]

Riemann mapping theorem 06.01.06. The Schwarz-Christoffel formula is the explicit form of the Riemann map for polygonal domains. The Riemann mapping theorem guarantees existence of the conformal map; the Schwarz-Christoffel formula provides the explicit integral representation. The Osgood-Caratheodory theorem ensures the boundary correspondence is a homeomorphism.
Index / winding number 06.01.28. The proof of the Schwarz-Christoffel formula relies on tracking the argument of $f^{'} (z)$ along the real axis, which is a winding-number computation. The total change in argument around the polygon boundary is $2 π$ , confirming closure via the angle-sum condition. The winding number of the image curve around an interior point equals $1$ , consistent with the index theory.
Weierstrass $℘$ -function 06.01.25. The Schwarz-Christoffel map from the upper half-plane to a rectangle is an elliptic integral, and its inverse is an elliptic function. The Weierstrass $℘$ -function provides the canonical doubly-periodic realisation of this inverse map. The connection between Schwarz-Christoffel and elliptic functions is the bridge between conformal mapping and the theory of doubly-periodic functions.

Historical & philosophical context [Master]

Christoffel 1867 ^{[Christoffel 1867]}, in Sul problema delle temperature stazionarie e la rappresentazione conforme in the Annali di Matematica Pura ed Applicata, derived the formula in the context of solving Laplace's equation for steady-state temperatures on polygonal domains. Christoffel was motivated by the physical problem of heat conduction and recognised that conformal mapping to the upper half-plane reduces the Dirichlet problem to the Poisson integral.

Schwarz 1869 ^{[Schwarz 1869]}, in Ueber einige Abbildungsaufgaben in the Journal fur die reine und angewandte Mathematik, independently discovered the formula from the perspective of conformal mapping theory. Schwarz was studying the problem of mapping the upper half-plane to a polygon and developed the reflection principle 06.01.23 as a tool for extending the map across the boundary. The modern computational treatment is due to Trefethen and collaborators, systematised in Driscoll-Trefethen 2002 ^{[Driscoll-Trefethen]}, which provides the standard numerical algorithms and software for Schwarz-Christoffel mapping.

Bibliography [Master]

@article{Christoffel1867,
  author = {Christoffel, Elwin Bruno},
  title = {Sul problema delle temperature stazionarie e la rappresentazione conforme},
  journal = {Annali di Matematica Pura ed Applicata},
  volume = {1},
  year = {1867},
  pages = {89--103},
  note = {Schwarz-Christoffel formula for conformal maps to polygons}
}

@article{Schwarz1869,
  author = {Schwarz, Hermann Amandus},
  title = {Ueber einige Abbildungsaufgaben},
  journal = {Journal f\"ur die reine und angewandte Mathematik},
  volume = {70},
  year = {1869},
  pages = {105--120},
  note = {Independent discovery of the Schwarz-Christoffel formula}
}

@book{Ahlfors1979,
  author = {Ahlfors, Lars V.},
  title = {Complex Analysis},
  publisher = {McGraw-Hill},
  year = {1979},
  edition = {3rd},
  note = {Chapter 6: conformal mapping, Schwarz-Christoffel}
}

@book{DriscollTrefethen2002,
  author = {Driscoll, Tobin A. and Trefethen, Lloyd N.},
  title = {Schwarz-Christoffel Mapping},
  publisher = {Cambridge University Press},
  year = {2002},
  note = {Definitive numerical treatment of Schwarz-Christoffel mapping}
}

@book{SteinShakarchi2003,
  author = {Stein, Elias M. and Shakarchi, Rami},
  title = {Complex Analysis},
  publisher = {Princeton University Press},
  year = {2003},
  volume = {II},
  note = {Princeton Lectures in Analysis, Chapter 8}
}

Prerequisites

06.01.28

Tier anchors

beginner: Needham *Visual Complex Analysis* Ch. 5 informal (conformal maps with corners); Stewart *Calculus* integration review
intermediate: Ahlfors *Complex Analysis* Ch. 6; Stein-Shakarchi *Complex Analysis* Vol. II Ch. 8; Conway *Functions of One Complex Variable* Ch. VI
master: Christoffel 1867 *Sul problema delle temperature stazionarie*; Schwarz 1869 *Ueber einige Abbildungsaufgaben*; Ahlfors *Complex Analysis* Ch. 6; Driscoll-Trefethen *Schwarz-Christoffel Mapping*

References

TODO_REF
Christoffel 1867 — Sul problema delle temperature stazionarie e la rappresentazione conforme · Ann. Mat. Pura Appl. 1, pp. 89--103
TODO_REF
Schwarz 1869 — Ueber einige Abbildungsaufgaben · J. reine angew. Math. 70, pp. 105--120
TODO_REF
Ahlfors — Complex Analysis · McGraw-Hill 1979, Ch. 6 (conformal mapping, Schwarz-Christoffel)
TODO_REF
Stein-Shakarchi — Complex Analysis (Vol. II) · Princeton Lectures in Analysis II, Ch. 8 (conformal mapping)
TODO_REF
Driscoll-Trefethen — Schwarz-Christoffel Mapping · Cambridge University Press 2002

Reviewer

TBD

Estimated time

beginner: 15m
intermediate: 35m
master: 70m