We define the (incomplete) elliptic integral of the first kind with elliptic modulus $0 < k < 1$ as
$$ F(z; k) = \int_{0}^{z} \frac{dw}{\sqrt{(1-w^2)(1-k^2w^2)}}, \text{ Im}(z) > 0 $$
It is well-known that $F$ maps the complex upper half-plane conformally onto some rectangle $R(k)$. One method for proving this starts out with showing that $F$ maps the extended real line one-to-one onto the boundary $\delta R(K)$. But I do not quite understand how $F(z; k)$ accomplishes this with respect to the turning angles involved. I can see how 90 degree counterclockwise turns are "generated" at $w = -1$ and $w = 1$, but not at $w = -1/k$ and $w = 1/k$, where the root gets real again.
Should the denominator in the integrand not be $\sqrt{1-w^2}\sqrt{1-k^2w^2}$ for that to happen? And if so, why is the elliptic integral always shown without this additional separated square root? As it is now it seems not even to be analytic in $\mathbb{H}$. And furthermore, if we wish to have continuity in the closure of the upper half-plane $\overline{\mathbb{H}}$ --- something which is explicitly used in a couple of arguments I have seen ---, do we not really need the even further separated $k\sqrt{1+w}\sqrt{1-w}\sqrt{\frac{1}{k}+w}\sqrt{\frac{1}{k}-w}$ as denominator, like in the Schwarz-Christoffel mapping formula? For me this is all a bit confusing.