Question 8.24: Find a function that maps the upper half of the z plane onto...

Consider the upper half of the z plane shaded in Fig 8-83. Let P[z=0] and Q[z=1] of the x axis map into P^{\prime}[w=0] and Q^{\prime}[w=1] of the triangle, while the third point R[z=\infty] maps into R^{\prime}.

By the Schwarz-Christoffel transformation,

\frac{d w}{d z}=A z^{\alpha / \pi-1}(z-1)^{\beta / \pi-1}=K z^{\alpha / \pi-1}(1-z)^{\beta / \pi-1}

Then, by integration,

w=K \int\limits_{0}^{z} \zeta^{\alpha / \pi-1}(1-\zeta)^{\beta / \pi-1} d \zeta+B

Since w=0 when z=0, we have B=0. Also, since w=1 when z=1, we have

1=K \int\limits_{0}^{1} \zeta^{\alpha / \pi-1}(1-\zeta)^{\beta / \pi-1} d \zeta=\frac{\Gamma(\alpha / \pi) \Gamma(\beta / \pi)}{\Gamma\left(\frac{\alpha+\beta}{\pi}\right)}

using properties of the beta and gamma functions [see Chapter 10]. Hence

K=\frac{\Gamma\left(\frac{\alpha+\beta}{\pi}\right)}{\Gamma(\alpha / \pi) \Gamma(\beta / \pi)}

and the required transformation is

w=\frac{\Gamma\left(\frac{\alpha+\beta}{\pi}\right)}{\Gamma(\alpha / \pi) \Gamma(\beta / \pi)} \int_{0}^{z} \zeta^{\alpha / \pi-1}(1-\zeta)^{\beta / \pi-1} d \zeta

Note that this agrees with entry A-13 on page 252,

A-13 Interior of triangle                       w=\int\limits_{0}^{z}t^{\alpha /\pi-1}(1-t)^{\beta/\pi-1}d t

since the length of side A^{\prime} B^{\prime} in Fig. 8-39 is

\int\limits_{0}^{1} \zeta^{\alpha / \pi-1}(1-\zeta)^{\beta / \pi-1} d \zeta=\frac{\Gamma(\alpha / \pi) \Gamma(\beta / \pi)}{\Gamma\left(\frac{\alpha+\beta}{\pi}\right)}