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23.2 Construction of Conformal Mappings 769
y
v
Re(z) > 0
i
i
1
x u
0
−i
FIGURE 23.24 Mapping the right half-plane onto the
unit disk in Example 23.13.
Put these z j ’s and w j ’s into equation (23.1):
(1 − w)(−1 − i)(i)(−i − z) = (i − z)(−i)(1 − i)(−1 − w).
Solve for w to get
z − 1
w = T (z) =−i .
z + 1
To ensure that T maps S onto the unit disk (not its complement), pick a point in S,say z =1, and
compute
T (1) = 0
as the center of the unit disk. Thus, we have a bilinear transformation (hence a conformal
mapping) of the right half-plane onto the unit disk.
Later (Example 23.16) we will want a mapping from the unit disk to the right half-plane.
This is just the inverse of the mapping of Example 23.13, where we went the other way (inverse
mapping) from the right half-plane to the unit disk. To find this inverse, solve for z in terms of w
in Example 23.13 to obtain
i − w
z = .
i + w
This is the mapping in the reverse direction from that of the last example. If we started from
scratch, and wanted a mapping from the unit disk in the z-plane to the right half-plane in the
w-plane, then we would interchange z and w in this mapping to write
i − z
w = .
i + z
EXAMPLE 23.14
We will find a conformal mapping of the upper half-plane onto the exterior of the unit disk
|w|= 1inthe w-plane. Again, since the boundaries are a line (the real axis) and a circle, we can
attempt a bilinear transformation.
Choose three points on the boundary of the upper half-plane, which is the real axis oriented
from left to right (so the upper half-plane is over the left shoulder as we walk along the line). Say
we choose z 1 =−1, z 2 = 0, and z 3 = 1, as in Figure 23.25. Choose three points on the boundary
of the disk of radius 1, but select them in order clockwise so that, as we walk along the circle in
this direction, the exterior of the circle is over our left shoulder. Say we pick w 1 =−1,w 2 = i,
and w 3 = 1. Now solve for w in
(−1 − w)(1 − i)(−1)(1 − z) = (−1 − z)(1)(−1 − i)(1 − w)
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