# Non-conformal Schwarz-Christoffel integral

Using "conformal" to mean a holomorphic bijection, the Riemann Mapping theorem guarantees the existence of a conformal map from the upper half-plane $\mathbb{H}=\{z=x+iy\in\mathbb{C}:y>0\}$ to the interior of any polygon $P$, and I know that such maps are given by a basic transformation of a Schwarz-Christoffel integral $$S(z)=\int_0^z\frac{d\zeta}{(\zeta-A_1)^{\beta_1}\dots(\zeta-A_n)^{\beta_n}}$$ where the $A_k$'s are distinct and increasing, $\beta_k<1$ for each $k$, and $1<\sum_{k=1}^n\beta_k\leq 2$. More precisely, if $F:\mathbb{H}\to\Omega$ is conformal where $\Omega$ is the interior of a polygon, then $F(z)=c_1S(z)+c_2$ for some Schwarz-Christoffel integral $S(z)$ and $c_1,c_2\in\mathbb{C}$.

I'm interested in the reverse of this idea that conformal mappings onto polygons are essentially given by Schwarz-Christoffel integrals. That is, are Schwarz-Christoffel integrals conformal in general? From Stein and Shakarchi's Complex Analysis, Proposition 4.1 on p. 236, I know that a Schwarz-Christoffel integral maps to a polygon with vertices $S(A_k)$ (and possibly $S(\infty)$, if $\sum_{k=1}^n\beta_k<2$). As the text states, this polygon does not have to be a simple polygon, and so it's easy to see that $S(z)$ is not necessarily conformal for the case when it maps the real line to a self-intersecting curve.

However, the book goes further to state (without proof) that even when the mapped polygon is simple, $S$ still might not be conformal. Are there any illustrative examples of Schwarz-Christoffel integrals that map the real line to simple polygons but are not conformal?

The answer depends on the missing lower bound on $\beta_k$. The Schwarz-Chrisoffel formula requires $-1\le\beta_k<1$.

The integrand in the formula has some constant argument along the boundary segment $[A_{k-1},A_k]$. Along the next segment $[A_k,A_{k+1}]$ it has a new direction. The change in the argument is $\beta_k\pi$. This will be the external angle of the resulting polygon. So we need $-1<\beta_k<1$ (or $-1\le\beta_k<1$ if we allow the polygon to turn back by $180^\circ$).

If $\beta_k<-1$ then the map will not be conformal around $A_k$. Take a small a small semicircle around $A_k$; it will be mapped to the a disk slice with central angle $(1-\beta_k)\pi>2\pi$, so the function cannot be injective anymore.

Now let us see what happens if we have $-1<\beta_k<1$ for $k=1,\ldots,n$.

To obtain a bounded polygon, we need $1<\sum\beta_k$. Then the integral in the formula has the magnitude order $|\zeta|^{-\sum\beta_k}$, so the function provided has some finite limit $P_0$ at $\infty$. The images of the boundary segments $[A_k,A_{k+1}]$ are always line segments, because the argument of the integrand changes only at the points $A_k$. Therefore, the image of the boundary line is a closed polygon $P_0P_1\ldots P_n$ where $P_0$ is the image of $\infty$.

The angle change (oriented external angle) at vertex $P_k$ of the polygon is $\beta_k\pi$ for $k=1,\ldots,n$. The angle change at $P_0$ is in $(-\pi,+\pi)$; if the polygon is simple then the sum of all changes must be $\pm2\pi$. Due to $\sum\beta_k>1$, the sum cannot be negative, so the sum if the angle changes is $2\pi$ and therefore the polygon has positive orientation. (This also shows that positively oriented simple polygons are possible only for $1<\sum\beta_k<3$.)

If $-1<\beta_k<1$, $1<\sum\beta_k$ and the polygon is simple then it follows from the argument principle that the map is conformal: every interior point of the polygon is the image of a single point in the upper half-plane and vice versa.

• Is there really a simple way to apply the argument principle? Stein and Shakarchi goes to some length to prove the elliptic integral $\int_0^z\frac{d\zeta}{(1-\zeta^2)^{\frac{1}{2}}(1-k^2\zeta^2)^{\frac{1}{2}}}$ is conformal onto a rectangle even though it satisfies the conditions you state. – Michael M May 16 '15 at 1:28
• Since the function has a finite limit at infinity, we can consider the real line the closed boundary of the upper half plane; then the argument principle works. (If you find this approach is too wild, take a composition with the map $i\frac{1-z}{1+z}$ that maps the unit disk to the upper hall plane. The composition maps the unit circle to the rectangle; then the argument principle shows that is is conformal.) – Géza Kós May 17 '15 at 11:49