Question 8.20: Determine a function that maps the upper half of the z plane...
Determine a function that maps the upper half of the z plane onto each of the indicated regions in the w plane.
(a)
Let points P, Q, S, and T [Fig. 8-79] map, respectively, into P^{\prime}, Q^{\prime}, S^{\prime}, and T^{\prime} [Fig. 8-78]. We can consider P^{\prime} Q^{\prime} S^{\prime} T^{\prime} as a limiting case of a polygon (a triangle) with two vertices at Q^{\prime} and S^{\prime} and the third vertex P^{\prime} or T^{\prime} at infinity.
By the Schwarz-Christoffel transformation, since the angles at Q^{\prime} and S^{\prime} are equal to \pi / 2, we have
\frac{d w}{d z}=A(z+1)^{[(\pi / 2) / \pi]-1}(z-1)^{[(\pi / 2) / \pi]-1}=\frac{A}{\sqrt{z^{2}-1}}=\frac{K}{\sqrt{1-z^{2}}}
Integrating,
w=K \int \frac{d z}{\sqrt{1-z^{2}}}+B=K \sin ^{-1} z+B
When z=1, w=b. Hence
b=K \sin ^{-1}(1)+B=K \pi / 2+B (1)
When z=-1, w=-b. Hence,
-b=K \sin ^{-1}(-1)+B=-K \pi / 2+B (2)
Solving (1) and (2) simultaneously, we find B=0, K=2 b / \pi. Then
w=\frac{2 b}{\pi} \sin ^{-1} z
The result is equivalent to entry A-3(a) on page 248
A-3 Semi-infinite strip of width a w=\sin{\frac{\pi z}{a}}
if we interchange w and z, and let b=a / 2.
(b)
Let points P, O, Q[z=1] and S map into P^{\prime}, O^{\prime}, Q^{\prime}[w=b i] and S^{\prime}, respectively. Note that P, S, P^{\prime}, S^{\prime} are at infinity (as indicated by the arrows) while O and O^{\prime} are the origins [z=0] and [w=0] of the z and w planes. Since the interior angles at O^{\prime} and Q^{\prime} are \pi / 2 and 3 \pi / 2, respectively, we have by the Schwarz-Christoffel transformation,
\frac{d w}{d z}=A(z-0)^{[(\pi / 2) / \pi]-1}(z-1)^{[(3 \pi / 2) / \pi]-1}=A \sqrt{\frac{z-1}{z}}=K \sqrt{\frac{1-z}{z}}
Then
w=K \int \sqrt{\frac{1-z}{z}} d z
To integrate this, let z=\sin ^{2} \theta and obtain
\begin{aligned} w & =2 K \int \cos ^{2} \theta d \theta=K \int(1+\cos 2 \theta) d \theta=K\left(\theta+\frac{1}{2} \sin 2 \theta\right)+B \\ & =K(\theta+\sin \theta \cos \theta)+B=K\left(\sin ^{-1} \sqrt{z}+\sqrt{z(1-z)}\right)+B \end{aligned}
When z=0, w=0 so that B=0. When z=1, w=b i so that b i=K \pi / 2 or K=2 b i / \pi. Then the required transformation is
w=\frac{2 b i}{\pi}\left(\sin ^{-1} \sqrt{z}+\sqrt{z(1-z)}\right)
