SchwarzChristoffel transformation (circular version)
The complexvariables function
$$f(z)={\int}_{0}^{z}\prod _{k=1}^{n}{(\zeta {z}_{k})}^{{\alpha}_{k}1}d\zeta ,$$ 
maps the closed unit disc^{} $\overline{D}=\{z\le 1\}$ in the complex plane conformally onto a polygon^{} with $n$ sides, interior angles^{} $$, and vertices $f({z}_{k})$. (The polygon is assumed to be not selfintersecting.) The parameters ${z}_{k}$ lie on the unit circle, and depend, generally in a complicated way, on the length of the sides of the polygon.
The fractional powers ${(\zeta {z}_{k})}^{{\alpha}_{k}1}$ serve to clamp up an arc of the circle into a pointy angle of measure ${\alpha}_{k}\pi $. Indeed, the proof of the SchwarzChristoffel formula shows that the function $f$ can be decomposed near ${z}_{k}$ as
$$f(z)=f({z}_{k})+{(z{z}_{k})}^{{\alpha}_{k}}{g}_{k}(z),$$ 
where ${g}_{k}$ is an analytic function^{} with ${g}_{k}({z}_{k})\ne 0$. See Figure 1.
Note that the exponent^{} is ${\alpha}_{k}$ — not ${\alpha}_{k}/2$ — because the neigbourhood of a point ${z}_{k}$ in the domain space looks like a halfdisc. For the same reason, the fractional power used in the formula is to be a singlevalued branch continuous^{} on the halfdisc. Finally, the extra $1$ exponents that appear in the integral representation for $f$ come from the power rule^{} for differentiation^{}.
0.1 Example: $n=3$
Figure 2 illustrates a mapping from the disc to a triangle ($n=3$). The contours are the approximate images, under $f$, of circles of radius $$.
We describe the method used to compute the figure. Points in the domain $\overline{D}$ are first parameterized as $z=r{e}^{i\theta}$, with $0\le r\le 1$ and $$ ranging over a discrete grid, shown schematically in Figure 4. The integral defining the function $f$ is pathindependent, and a natural choice for the paths are rays emanating from the origin. When computing the integrals along each ray, we exploit the additivity of the complex path integral:
$$f((r+\mathrm{\Delta}r){e}^{i\theta})=f(r{e}^{i\theta})+{\int}_{r{e}^{i\theta}}^{(r+\mathrm{\Delta}r){e}^{i\theta}}\prod _{k=1}^{n}{(\zeta {z}_{k})}^{{\alpha}_{k}1}d\zeta ,$$ 
so that $f(z)$ is found by summing a previouslycomputed value and a new integral to be computed. And the new integral is computed using 32point Gauss quadrature^{} after reparameterizing the path with $d\zeta ={e}^{i\theta}dr$.
The computation of the integrand
$$\prod _{k=1}^{n}{(\zeta {z}_{k})}^{{\alpha}_{k}1}=\mathrm{exp}\left(\sum _{k=1}^{n}({\alpha}_{k}1)\mathrm{log}(\zeta {z}_{k})\right)$$ 
is straightforward, though we must be careful to respect the branch cuts prescribed above. The $\mathrm{log}$ function in most computer languages takes a branch cut on the negative axis. To get the singlevalued branches we need in this situation, we must instead compute $\zeta \mapsto \mathrm{log}(\zeta {z}_{k})$ via the expression
$$\zeta \mapsto \mathrm{log}i{z}_{k}+\mathrm{log}\frac{\zeta {z}_{k}}{i{z}_{k}},$$ 
where $i{z}_{k}$ is the direction of the tangent^{} to the circle at the point ${z}_{k}$.
Finally, after having obtained a discrete set of image points $f(z)$ traced along each circle $z=r{e}^{i\theta}$, the contours in the figure are obtained by interpolating a curved Bézier spline through the image points.
If a triangle is prescribed with the vertex locations, it is not immediately obvious what the parameters ${z}_{k}$ should be to obtain that triangle. In the examples here, we simply avoid this difficulty by arbitrarily choosing the parameters ${z}_{k}={e}^{2\pi i(k1)/n}$ to be equally spaced on the unit circle, and hope that nice figures result.
The ${\alpha}_{k}$ parameters are easily determined from the angles of the desired figure; they are, in this example:
$${\alpha}_{1}=\frac{1}{4},{\alpha}_{2}=\frac{1}{2},{\alpha}_{3}=\frac{1}{4}.$$ 
0.2 Example: $n=10$
Figure 5 shows an example with $n=10$ points. The strategy for computing this figure is similar^{} to that of the triangle.
The parameters for this star are (rounded to four decimal places):
${\alpha}_{1}=0.2422,{\alpha}_{2}={\alpha}_{1}=1.3263,{\alpha}_{3}={\alpha}_{9}=0.3026,$  
${\alpha}_{4}={\alpha}_{8}=1.3026,{\alpha}_{5}={\alpha}_{7}=0.2754,{\alpha}_{6}=1.3440.$ 
0.3 Demonstration computer programs

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http://svn.goldsaucer.org/repos/PlanetMath/SchwarzChristoffelTransformationCircularVersion/schwarzchristoffel.pyPython source code for producing images of the SchwarzChristoffel transformation

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http://svn.goldsaucer.org/repos/PlanetMath/SchwarzChristoffelTransformationCircularVersion/explanation.pyPython source code for the explanatory diagrams
References
 1 Lars V. Ahlfors. Complex Analysis, third edition. McGrawHill, 1979.
Title  SchwarzChristoffel transformation (circular version) 

Canonical name  SchwarzChristoffelTransformationcircularVersion 
Date of creation  20130322 16:52:57 
Last modified on  20130322 16:52:57 
Owner  stevecheng (10074) 
Last modified by  stevecheng (10074) 
Numerical id  8 
Author  stevecheng (10074) 
Entry type  Result 
Classification  msc 31A99 
Classification  msc 30C20 