Schwarz triangle function

Remarks

Changing the pair of solutions replaces $s$ by $(\alpha s + \beta)/(\gamma s + \delta)$ with $\alpha\delta - \beta\gamma \neq 0$, so $s$ is well defined up to a Möbius transformation.

Let $a, b, c$ be real and set $\lambda = 1 - c$, $\mu = c - a - b$, $\nu = a - b$. If $0 \le \lambda, \mu, \nu < 1$, then $s$ is a conformal map from $\mathbb{H}$ onto a triangle bounded by circular arcs or segments. Its angles are $\pi\lambda$, $\pi\mu$, $\pi\nu$ at the images of $0$, $1$, $\infty$.

Example: for $a = b = \frac12$, $c = 1$ all three angles are $0$. Here $c \in \Z$, so the ratio is formed with the logarithmic second solution. Up to a Möbius transformation it equals $\tau = i \, K(1 - z)/K(z)$. Here $K(m) = \frac{\pi}{2} {}_2F_1(\frac12, \frac12; 1; m)$ is the complete elliptic integral of the first kind. Its inverse is the modular lambda function.

Non-example: a ratio $w_2 / w_1$ of two proportional solutions is constant, so it is not a triangle function; independence of $w_1$ and $w_2$ is needed.