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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 $\pi$?

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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.

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