Schwarz-Christoffel transformation: point at infinity In this paper how can one simply write $\displaystyle A = A' \frac{1}{x_n^{\phi_n/\pi - 1}}$ on $(21.6)$. Can anyone hint me what is meant by "with it then
being dropped from the list of points in the transformation"? Are we making transformation on $A$ so that we don't have to transform the point $x_n$ at infinity?
 A: One can simply write $ A = A' \frac{1}{x_n^{\phi_n/\pi - 1}}$ because this formula introduces $A'$. This is the same as to say: let $A'=A\, x_n^{\phi_n/\pi - 1}$. 
They make this notational change for the  ease of subsequent computation in $(21.7)-(21.8)$, which show the effect of letting $x_n\to \infty$. As a result of passing to the limit, the formula $(21.8)$ is obtained, in which the product has one fewer term than $(21.2)$. This is informally expressed by saying that moving $x_n$ to infinity simplifies the formula, because $x_n$ is not present in $(21.8)$. 
Conceptually, the Schwarz-Christoffel formula for $f:H\to P$ ($H$ is upper halfplane, $P$ a polygon) is based on keeping track of $\arg f'$ on the boundary of $H$. The geometry of $P$ dictates that $\arg f'$  must be piecewise constant, with jumps at the points that get mapped to the vertices of $P$. The factors $(x-x_k)^{\phi_k/\pi-1}$ are introduced to create these jumps. But a jump between $-\infty$ and $+\infty$ can happen naturally, without such a factor. 
Let's consider an example from real analysis (unrelated to conformal maps). Say, we want a piecewise constant function on $\mathbb R$ that takes values $0,3,5$ on some intervals. If all jumps are finite points, we end up with something like
$$f(x)=\begin{cases} 0, \quad & x\le 0 \\ 3, &0<x\le 1 \\ 5, & 1<x\le 2 \\ 0, & x>2 \end{cases} \tag1$$
(Here the jumps are at $x=0$, $x=1$, and $x=2$). But (1) looks unnecessarily complicated: the function
$$g(x)=\begin{cases} 0, \quad & x\le 0 \\ 3, &0<x\le 1 \\ 5, & x>1 \end{cases} \tag2$$
also does what we want, and is easier to write down, because one of three jumps happens at infinity.
