Modification of Schwarz-Christoffel integral I found two different formulations of the Schwarz-Christoffel formula (e.g. Link1, p.20 and Link2, p. 9). The first is
\begin{align*}
z=w(\zeta)=&A+C\int\limits^{z}\prod\limits_{k=1}^n\left(\zeta-z_k\right)^{\alpha_k-1}d\zeta
\end{align*}
The second is
\begin{align*}
z=w(\zeta)=&A+C\int\limits^{z}\prod\limits_{k=1}^n\left(1-\frac{\zeta}{z_k}\right)^{\alpha_k-1}d\zeta
\end{align*}
In both equations


*

*$A$ and $C$ are complex constants,

*$z_k$ are the coordinates of point $k$ on the unit circle corresponding to the vertex $k$ of the rectangle,

*$\alpha_k$ are the interior angles of the vertices by means of multiples of $\pi$

*and $\zeta$ is a point outside the unit circle such that $|\zeta| > 1$.


In the second link on page 20 it is said that:
"Composing the first equation with standard conformal maps leads to variations of the Schwarz-Christoffel formula for mapping from other fundamental domains [...]. The simplest such modification has the unit disk as domain. The vertices then lie  in counterclockwise order on the unit circle and equation one can be transformed to equation two."
Why is this transformation possible for the upper complex half-plane and how is it carried out?
 A: First of all, there is no real difference between the two formulas in your post: in the second, $C$ includes $\prod (-z^k)^{\alpha_k-1}$, which is a constant. 
It is somewhat remarkable that the transformation between the disk and the plane preserves the general appearance of the S-C formula. (Not so for other domains, such as an infinite strip). The steps of this transformation can be found on pages 192-193 of Conformal mapping by Zeev Nehari (published by Dover, highly recommended). I paraphrase the text with occasional relabeling of indices, as I hate seeing $\nu$ used as an index... For the halfplane, Nehari has
$$f(z)=\alpha\int_0^z \prod_{k=1}^n (z-a_k)^{-\mu_k}\,dz+\beta \tag{50}$$
By the linear transformation
$$z=i\frac{1+\zeta}{1-\zeta},\quad \zeta=\frac{z-i}{z+i}$$
which maps $|\zeta|<1$ onto $\operatorname{Im}z>0$, we can also obtain from (50) a formula for the conformal map of the unit circle onto the polygon $D$. We have 
$$(z-a_k)^{\mu_k} = \left[ i\frac{1+\zeta}{1-\zeta} -a_k\right]^{\mu_k}
=\left(\frac{a_k+i}{1-\zeta}\right)^{\mu_k} \left[ \zeta - \frac{a_k-i}{a_k+i} \right]^{\mu_k} \tag{a}$$
and 
$$dz=\frac{2i\,d\zeta}{(1-\zeta)^2}\tag{b}$$
Let $$b_k=\frac{a_k-i}{a_k+i} $$ which is the point on the unit circle corresponding to $a_k$. Since $\sum \mu_k=2$, the denominators $(1-\zeta)$ in (a) cancel the denominator in  (b). We end up with
$$f(\zeta)=\alpha_2\int_0^\zeta \prod_{k=1}^n (\zeta-b_k)^{-\mu_k}\,d\zeta+\beta_2  $$
where   $\alpha_2,\beta_2$ are constants.
