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23.2 Construction of Conformal Mappings 775

v
y

x −c c u

FIGURE 23.30 Mapping the upper half-plane onto
a vertical strip in Example 23.18.

This integral reminds us of the real integral representation of the inverse sine function. Indeed,
write

w = A arcsin(z)
to mean that
w

z = sin .
A
To choose A so that −1 maps to −c and1to c, we need
c

sin = 1.
A
Thus, choose c/A = π/2, so A = 2c/π. Then
2c
w = arcsin(z).
π
If we choose c = π/2, this mapping is exactly w = arcsin(z), mapping U onto the strip Im(z)>
0,−π/2 < Re(w) < π/2. This is consistent with Example 23.3.

SECTION 23.2 PROBLEMS

In each of Problems 1 through 6, find a bilinear transfor- 9. Show that the Schwarz-Christoffel transformation
mation of the first domain onto the second. z
f (z) = 2i (ξ + 1) −1/2 (ξ − 1) −1/2 −1/2 dξ
ξ
1. |z| < 3 onto |w − 1 + i| < 6 0
maps the upper half-plane onto the rectangle
2. |z| < 3 onto |w − 1 + i| > 6
with vertices 0,c,c + ic,and ic where c =
3. |z + 2i| < 1 onto |w − 3| > 2 (1/2) (1/4)/ (3/4). For this problem, it is nec-
4. Re(z)> 1 onto Im(w) > −1 essary to know the integral formula
1
5. Re(z)< 0 onto |w| < 4 x−1 y−1
B(x, y) = u (1 − u) du
6. Im(z)> −4 onto |w − i| > 2 0
for the beta function B(x, y), and also to know that, in
7. Find a conformal mapping of the upper half-plane onto
terms of the gamma function,
the wedge 9 < arg(w) < π/3.
(x) (y)
8. Let w = f (z) = log(z) be defined by restricting the B(x, y) = .
(x + y)
argument of z to lie in [0,2π]. Show that f takes the
upper half-plane onto the strip 0 < Re(w) < π. Hint: See Section 15.3 (Problems 39 and 40).

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October 14, 2010 15:39 THM/NEIL Page-775 27410_23_ch23_p751-788