Question 8.25: (a) Find a function which maps the upper half of the z plane...
(a) Find a function which maps the upper half of the z plane of Fig. 8-85 onto the shaded region in the w plane of Fig. 8-84.
(b) Discuss the case where b \rightarrow 0.
(a) The interior angles at Q and T are each \pi-\alpha, while the angle at S in 2 \pi-(\pi-2 \alpha)=\pi+2 \alpha. Then, by the Schwarz-Christoffel transformation, we have
\frac{d w}{d z}=A(z+1)^{(\pi-\alpha) / \pi-1} z^{(\pi+2 \alpha) / \pi-1}(z-1)^{(\pi-\alpha) / \pi-1}=\frac{A z^{2 \alpha / \pi}}{\left(z^{2}-1\right)^{\alpha / \pi}}=\frac{K z^{2 \alpha / \pi}}{\left(1-z^{2}\right)^{\alpha / \pi}}
Hence, by integration
w=K \int\limits_{0}^{z} \frac{\zeta^{\alpha / \pi}}{\left(1-\zeta^{2}\right)^{\alpha / \pi}} d \zeta+B
When z=0, w=a i; then B=a i and
w=K \int\limits_{0}^{z} \frac{\zeta^{2 \alpha / \pi}}{\left(1-\zeta^{2}\right)^{\alpha / \pi}} d \zeta+a i (1)
The value of K can be expressed in terms of the gamma function using the fact that w=b when z=1 [Problem 8.102]. We find
K=\frac{(b-a i) \sqrt{\pi}}{\Gamma\left(\frac{\alpha}{\pi}+\frac{1}{2}\right) \Gamma\left(1-\frac{\alpha}{\pi}\right)} (2)
(b) As b \rightarrow 0, \alpha \rightarrow \pi / 2 and the result in (a) reduces to
w=a i-a i \int\limits_{0}^{z} \frac{\zeta d \zeta}{\sqrt{1-\zeta^{2}}}=a i \sqrt{1-z^{2}}=a \sqrt{z^{2}-1}
In this case, Fig. 8-84 reduces to Fig. 8-86. The result for this case can be found directly from the Schwarz-Christoffel transformation by considering P^{\prime} Q^{\prime} S^{\prime} T^{\prime} U^{\prime} as a polygon with interior angles at Q^{\prime}, S^{\prime}, and T^{\prime} equal to \pi / 2,2 \pi, and \pi / 2, respectively.
