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# Schwarz-Christoffel transformation (circular version)

The complex-variables function

$f(z)=\int_{0}^{z}\prod_{{k=1}}^{n}(\zeta-z_{k})^{{\alpha_{k}-1}}d\zeta\,,$ |

maps the closed unit disc $\overline{D}=\{\lvert z\rvert\leq 1\}$ in the complex plane conformally onto a polygon with $n$ sides, interior angles $0<\alpha_{k}\pi<2\pi$, and vertices $f(z_{k})$. (The polygon is assumed to be not self-intersecting.) 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 Schwarz-Christoffel 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 $g_{k}(z_{k})\neq 0$. 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 half-disc. For the same reason, the fractional power used in the formula is to be a single-valued branch continuous on the half-disc. 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 $0<r\leq 1$.

We describe the method used to compute the figure. Points in the domain $\overline{D}$ are first parameterized as $z=re^{{i\theta}}$, with $0\leq r\leq 1$ and $0\leq\theta<2\pi$ ranging over a discrete grid, shown schematically in Figure 4. The integral defining the function $f$ is path-independent, 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:

$f\bigl((r+\Delta r\bigr)e^{{i\theta}})=f(re^{{i\theta}})+\int_{{re^{{i\theta}}% }}^{{(r+\Delta r)e^{{i\theta}}}}\prod_{{k=1}}^{n}(\zeta-z_{k})^{{\alpha_{k}-1}% }\,d\zeta\,,$ |

so that $f(z)$ is found by summing a previously-computed value and a new integral to be computed. And the new integral is computed using 32-point Gauss quadrature after reparameterizing the path with $d\zeta=e^{{i\theta}}\,dr$.

The computation of the integrand

$\prod_{{k=1}}^{n}(\zeta-z_{k})^{{\alpha_{k}-1}}=\exp\Bigl(\sum_{{k=1}}^{n}(% \alpha_{k}-1)\log(\zeta-z_{k})\Bigr)$ |

is straightforward, though we must be careful to respect the branch cuts prescribed above. The $\log$ function in most computer languages takes a branch cut on the negative axis. To get the single-valued branches we need in this situation, we must instead compute $\zeta\mapsto\log(\zeta-z_{k})$ via the expression

$\zeta\mapsto\log iz_{k}+\log\frac{\zeta-z_{k}}{iz_{k}}\,,$ |

where $iz_{k}$ 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 $z=re^{{i\theta}}$, 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 $z_{k}=e^{{2\pi i(k-1)/n}}$ 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:

$\alpha_{1}=\tfrac{1}{4},\quad\alpha_{2}=\tfrac{1}{2},\quad\alpha_{3}=\tfrac{1}% {4}\,.$ |

# 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):

$\displaystyle\alpha_{1}=0.2422\,,\quad\alpha_{2}=\alpha_{1}=1.3263\,,\quad% \alpha_{3}=\alpha_{9}=0.3026\,,$ | ||

$\displaystyle\alpha_{4}=\alpha_{8}=1.3026\,,\quad\alpha_{5}=\alpha_{7}=0.2754% \,,\quad\alpha_{6}=1.3440\,.$ |

# 0.3 Demonstration computer programs

# References

- 1 Lars V. Ahlfors. Complex Analysis, third edition. McGraw-Hill, 1979.

## Mathematics Subject Classification

31A99*no label found*30C20

*no label found*

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## Comments

## These pictures look like water waves...

After sitting back and looking at the pictures that I made, they seem remarkably similar to what we expect to observe, in a pool with the given boundary has a small pebble dropped vertically on the centre. This behavior can be "explained" if one assumes that the wavefront is always orthogonal to the direction of propagation (and thus this orthogonality is preserved under the conformal mapping from a disc).

I'm pretty clueless about physics, so I would like to ask: is there some sort of physical principle that justifies this?

Thanks for any enlightenment,

// Steve

## Re: These pictures look like water waves...

I am amazed that you got these pictures to show up at all. It's as if you had some kind of magical touch that makes the computer want to comply with your commands. I always get error messages, different error messages, but error messages nonetheless.