Complex Analysis - Maths KU

414 Chapter 7 Conformal Mappings
A
y
(a) Half-plane y ≥ 0
A′
–1 1
i
0
v
3π/2
π/2
B′
(b) Polygonal region for Example 2
Figure 7.24 Figure for Example 2
B
x
u
f ′ (z) =
the principal square root to rewrite (7) as
f ′ A
(z) =
(z2 .
1/2
− 1)
Furthermore, since the principal value of (−1) 1/2 = i, wehave
A
(z2 =
1/2
− 1)
A
[−1(1− z 2 )]
1/2 = A
i
1
(1 − z2 1
= −Ai
1/2
) (1 − z2 . (8)
1/2
)
From (7) of Section 4.4 we recognize that an antiderivative of (8) is given by
f(z) =−Ai sin −1 z + B, (9)
where sin −1 z is the single-valued function obtained byusing the principal
square root and principal value of the logarithm and where A and B are
complex constants. If we choose f(−1) = −i and f(1) = i, then the constants
A and B must satisfythe system of equations
−Ai sin −1 (−1) + B = Ai π
+ B = −i
2
−Ai sin −1 (1) + B = −Ai π
+ B = i.
2
Byadding these two equations we see that 2B = 0, or, B = 0. Now by
substituting B = 0 into either the first or second equation we obtain A =
−2/π. Therefore, the desired mapping is given by
f(z) =i 2
π sin−1 z.
This mapping is shown in Figure 7.23. The line segments labeled A and B
shown in color in Figure 7.23(a) are mapped by w = i 2
π sin−1 z onto the line
segments labeled A ′ and B ′ shown in black in Figure 7.23(b).
EXAMPLE 2 Using the Schwarz-Christoffel Formula
Use the Schwarz-Christoffel formula (6) to construct a conformal mapping
from the upper half-plane onto the polygonal region shown in gray in Figure
7.24(b).
Solution We proceed as in Example 1. The region shown in grayin Figure
7.24(b) is an unbounded polygonal region with interior angles α1 =3π/2 and
α1 = π/2 at the vertices w1 = i and w2 = 0, respectively. If we select x1 = −1
and x2 = 1 to map onto w1 and w2, respectively, then (6) gives
f ′ (z) =A (z +1) 1/2 (z − 1) −1/2 . (10)