Question 8.17: Establish the validity of the Schwarz-Christoffel transforma...
Establish the validity of the Schwarz-Christoffel transformation.
We must show that the mapping function obtained from
\frac{d w}{d z}=A\left(z-x_{1}\right)^{\alpha_{1} / \pi-1}\left(z-x_{2}\right)^{\alpha_{2} / \pi-1} \cdots\left(z-x_{n}\right)^{\alpha_{n} / \pi-1} (1)
maps the real axis of the z plane onto a given polygon of the w plane [Figs. 8-76 and 8-77].
To show this, observe that from (1) we have
As z moves along the real axis from the left toward x_{1}, let us assume that w moves along a side of the polygon toward w_{1}. When z crosses from the left of x_{1} to the right of x_{1}, \theta_{1}=\arg \left(z-x_{1}\right) changes from \pi to 0 while all other terms in (2) stay constant. Hence \arg d w decreases by \left(\alpha_{1} / \pi-1\right) \arg \left(z-x_{1}\right)=\left(\alpha_{1} / \pi-1\right) \pi=\alpha_{1}-\pi or, what is the same thing, increases by \pi-\alpha_{1} [an increase being in the counterclockwise direction].
It follows from this that the direction through w_{1} turns through the angle \pi-\alpha_{1}, and thus w now moves along the side w_{1} w_{2} of the polygon.
When z moves through x_{2}, \theta_{1}=\arg \left(z-x_{1}\right) and \theta_{2}=\arg \left(z-x_{2}\right) change from \pi to 0 while all other terms stay constant. Hence, another turn through angle \pi-a_{2} in the w plane is made. By continuing the process, we see that as z traverses the x axis, w traverses the polygon, and conversely.
We can actually prove that the upper half plane is mapped onto the interior of the polygon (if it is closed) by (1) [see Problem 8.26].
