# Can the Schwarz-Christoffel Transformation only transform polygons with rational angles?

In the text I have read that introduces the Schwarz-Christoffel Conformal Transformation, the transformation is described to act over polygons with interior angles given by $$\alpha \pi$$ for $$0<\alpha\leq 2$$. The issue is that none of what I have read indicates whether $$\alpha \in \mathbb{Q}$$ or $$\alpha \in \mathbb{R}$$. Intuitively, raising the exponents in the Schwarz-Christoffel Equation to irrational powers doesn't make much sense, so it makes sense if $$\alpha \in \mathbb{Q}$$, but I am wondering if someone can confirm this for me?

Further, if it is indeed required that $$\alpha \in \mathbb{Q}$$, then does there exist some conformal mapping that maps the interior of a closed polygon to the upper half plane where the interior angles of the closed polygon may be irrational multiples of $$\pi$$?

## 1 Answer

The only requirement on values of $$\alpha$$ is that it be in the real interval $$(0, 2)$$. From the book Schwarz-Christoffel Mapping by Driscoll and Trefethen:

Indeed, arbitrary real exponents can meaningfully appear in the Schwarz-Christoffel equation, although the resulting region may overlap itself and not be bounded by a polygon in the usual sense of the term.

As such, yes $$\alpha$$ can take an irrational value.