Take a branch of the function $f(t) = \sqrt{1-t^2}$ on the closed upper-half plane ($\bar{H}$) so that $f(i) = \sqrt{2} > 0$. Then, we can define
$\phi(z) = \int_0^{z} \frac{dt}{f(t)}$, where the path is taken in $\bar{H}$.
This function maps $\bar{H}$ to a half-infinite rectangular strip $[-\pi/2,\pi/2] \times [0,\infty)$.
I don't completely understand why this is the case.
Wouldn't
$\psi(z) = \int_0^{z} f(t) dt$ map $\bar{H}$ onto $[-\pi/4,\pi/4] \times (-\infty,0]$?
The argument for this is the same for that of $\phi$: The function $\psi$ maps $[-1,1]$ to $[-\pi/4,\pi/4]$. Every time $z$ passes a branch point along the real axis, the argument changes by $\pm\pi$, so this results in a $\pm\frac{\pi}{2}$ turn.
So it appears $\psi$ essentially does what $\phi$ does. Then is there any reason to look at $\phi$ rather than $\psi$? (More generally, why does the Scwarz Christoffel mapping need all those things appear in the denominator rather than in the numerator?)
p.s. I don't really understand why the the turns happen in the directions that they do rather than making like a stair-case like shape in the case of $\phi$ or $\psi$.
EDIT: Here are some references that helped me clarify this: