Mathematical Methods for Physics and Engineering : A Comprehensive Guide

COMPLEX VARIABLES

φ 1 φ 2

φ 3

x 1 x 2

− 11

y

x

w 1 w 2

ib w^3

−aa

s

r

w=g(z)

Figure 24.6 Transforming the upper half of thez-plane into the interior of

a triangle in thew-plane.

We can also map the upper half of thez-plane into an infiniteopenpolygon

by considering it as the limiting case of some closed polygon.

Find a transformation that maps the upper half of thez-plane into the triangular region

shown in figure 24.6 in such a way that the pointsx 1 =− 1 andx 2 =1are mapped

into the pointsw=−aandw=a, respectively, and the pointx 3 =±∞is mapped into

w=ib. Hence find a transformation that maps the upper half of thez-plane into the region

−a<r<a,s> 0 of thew-plane, as shown in figure 24.7.

Let us denote the angles atw 1 andw 2 in thew-plane byφ 1 =φ 2 =φ,whereφ=tan−^1 (b/a).
Sincex 3 is taken at infinity, we may omit the corresponding factor in (24.30) to obtain

w=

{

A

∫z

0

(ξ+1)(φ/π)−^1 (ξ−1)(φ/π)−^1 dξ

}

+B

=

{

A

∫z

0

(ξ^2 −1)(φ/π)−^1 dξ

}

+B. (24.31)

The required transformation may then be found by fixing the constantsAandBas
follows. Since the pointz= 0 lies on the line segmentx 1 x 2 , it will be mapped onto the line
segmentw 1 w 2 in thew-plane, and by symmetry must be mapped onto the pointw=0.
Thus settingz=0andw= 0 in (24.31) we obtainB= 0. An expression forAcan be
found in the form of an integral by setting (for example)z=1andw=ain (24.31).
We may consider the region in thew-plane in figure 24.7 to be the limiting case of the
triangular region in figure 24.6 with the vertexw 3 at infinity. Thus we may use the above,
but with the angles atw 1 andw 2 set toφ=π/2. From (24.31), we obtain

w=A

∫z

0

dξ

√

ξ^2 − 1

=iAsin−^1 z.

By settingz=1andw=a, we findiA=2a/π, so the required transformation is

w=

2 a

π

sin−^1 z.