I was trying to clarify some questions I had about elliptic integrals using
http://websites.math.leidenuniv.nl/algebra/ellcurves.pdf
There they define the map $\phi\colon w\mapsto \int_0^w\frac{\mathrm{d}z}{\sqrt{1-z^2}}$ on $\mathbb{C}\setminus[-1,1]$ to get $\phi$ well-defined up to periods of the integral. The choice of the interval $[-1,1]$ is made so that $\sqrt{1-z^2}$ admits a single-valued branch.
Now, I know that the principal branch of the square root $\sqrt{z}$ is discontinuous on the half-line $(-\infty,0)$, so to get a holomorphic map we restrict to $\mathbb{C}\setminus (-\infty,0]$. Substituting $1-z^2$ for $z$ we get that the appropriate branch cuts for the above mapping $\sqrt{1-z^2}$ would be $(-\infty,-1]$ and $[1,\infty)$, which is somewhat the opposite of the suggested interval $[-1,1]$.
From that I conclude that they didn't choose the principal branch, otherwise for e.g. $z=2$ the map would be discontinuous.
My question is: Are both choices possible? Then there must be some way to choose another branch of $\sqrt{1-z^2}$. Is there a good way to see how to choose "elegant" branch cuts and the corresponding holomorphic branches?
A thought of my own: It should be possible to instead integrate on the Riemann sphere, using $\infty$ and not $0$ as a starting point. Then the two intervals would "swap roles". But I don't see how to formalize this.