Define the function $$F(w)=\int\limits_0^w \frac{d\xi}{\sqrt{(1-\xi^2)(1-k^2\xi^2)}}$$
I would like to prove this map takes the upper half plane $\mathscr{H}$ to the rectangle $\mathscr{R}$.
The function has points of singularities are $\xi=\pm 1/k, \pm 1$ for $k\in (0,1)$ and we know that the roots of the function are -1/2, so we are rotating intervals by $\pi/2 $ (or $-\pi/2$). How do I show explicitly that this happens? ie, is there a way to traverse the real line and show that my boundaries of $\mathscr{H}$ map to $\mathscr{R}$ in the way we have described?
Furthermore, if I reflect $\mathscr{H}$ over the real line, with branch cuts from $(-\infty, -1)$ and $(1,\infty)$, how do I show what happens if I take $\xi=\xi_0+re^{i\theta}$ for $\xi_0=1$ and rotate $\xi$ for $\theta\in(0,2\pi)$?
It is the Jacobi form of the elliptic integral of the first kind (https://en.wikipedia.org/wiki/Elliptic_integral)
This mapping from $\mathscr{H}$ to $\mathscr{R}$ is a particular case of Schwarz-Christoffel mapping (http://mathfaculty.fullerton.edu/mathews/c2003/SchwarzChristoffelMod.html), due to the decomposition of the denominator into
$$k(z+1)^{1/2}(z-1)^{1/2}\left(z+\frac1k\right)^{1/2}\left(z-\frac1k\right)^{1/2}$$
The Schwarz-Christoffel "lecture grid" proves that, looking at:
the exponents: the "turning angles" are all $+\frac{\pi}{2}$, because of exponents $\frac{1}{2}$.
the roots: the points where these turns occur are $\pm1,\pm \frac1k.$