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770 CHAPTER 23 Conformal Mappings and Applications

y

= −1 z 2 = 0 z = 1 x
z 1 3
v
w = i
2

w = 1
3
= −1
w 1 u

FIGURE 23.25 Mapping the upper half-plane onto the
exterior of the unit disk in Example 23.14.

to get
(1 + i)z − 1 + i
w = T (z) = .
(−1 + i)z + 1 + i
To ensure that the upper half-plane maps to the exterior (not the interior) of the unit circle, pick
a point in the upper half-plane and verify that it maps to the exterior (not the interior) of the
unit circle. For example, choosing z = 2i, we find that T (2i) =−3i with a magnitude greater
than 1.

Thus far, we have gotten by with bilinear transformations. These are limited by the fact that
they map circles and lines to circles and lines.

EXAMPLE 23.15
We will map the horizontal infinite strip S :−π/2 < Im(z)<π/2 onto the unit disk |w| < 1.
The boundary of S is not a line but consists of two lines, so a bilinear transformation is out
of the question. To get our hands on a beginning, recall from Example 23.2 that the exponential
z
function w = e maps horizontal lines to half-rays from the origin. The boundary of S consists of
two horizontal lines: Im(z) =−π/2 and Im(z) = π/2. On the lower boundary line, z = x −iπ/2,
so
x −iπ/2
x
z
w = e = e e =−ie ,
which varies over the negative imaginary axis as x takes on all real values. On the upper boundary
line of S, z = x + iπ/2 and
x
z
x iπ/2
w = e = e e = ie ,
which varies over the positive part of the imaginary axis as x varies over the real line. Now the
imaginary axis forms the boundary of the right half-plane Re(w) > 0inthe w-plane and also the

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October 14, 2010 15:39 THM/NEIL Page-770 27410_23_ch23_p751-788