1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

10.2 • BILINEAR TRANSFORMATIONS 403

(10-13) and c = 0, then S reduces to a linear transformation, which carries lines
onto lines and circles onto circles. If c =F 0, then we can wr ite S in the form

S(z) = a(cz+d) +be-ad=~+ bc-ad_l __

c(cz + d) c c cz + d

(10-15)

The condition ad =F be precludes the possibility that S reduces to a constant.
Equat ion (10- 15 ) indicates that S can be considered as a composition of func-
tions. It is a linear mapping~= cz+d, followed by the reciprocal transformation
Z = ~'followed by w = ~ + be-;;ad Z. In Chapter 2 we showed that eacl1 function
in this composition maps t he class of circles and lines onto itself; it follows that
the bilinear transformation S has this property. A half-plane can be considered
to be a family of parallel lines and a dis k as a family of circles. Therefore, we
conclude that a bilinear transformation maps the class of half-p lanes and disks
onto itself. Example 10.3 illustrates this idea.

• EXAMPLE 10 .3 Show that w = S (z} = i({; ;) maps the unit disk lzl < 1
  one-to-one and onto the upper half-plane Im (w) > 0.

Solution We first consider the unit circle C: lzl = 1, which forms the boundary

of the disk and find its image in thew plane. If we write S (z} = -;~ii, then we

see that a = - i, b = i, c = 1, and d =~l. Using Equation (10-14), we find that

the inverse is given by

Z-- s- 1 ( ) -W - -dw + b -- -w + .. i

cw- a w + i

(10-16}

If lzl = 1, then Equation ( 10 -16) implies t hat the images of points on the unit

circle satisfy the equation

lw+i l = 1-w+il. (10-17}

Squaring both sides of Equation (10-17), we obtain u^2 + (1 + v)^2 = u^2 + (1 - v)^2 ,

which can be simplified to yield v = 0, which is the equation of the u -axis in t he

w plane.

The circle C divides the z plane into two portions, and its image is the u -

axis, which divides thew plane into two port ions. The image of the point z = 0

is w = S (O) = i, so we expect t hat t he interior of the circle C is mapped onto

the portion of the w plane that lies ab ove the u -axis. To show that t his outcome

is true, we let lz l < 1. Then Equation (10-16) implies that the image values

must satisfy the inequality 1- w +ii < lw +ii, which we write as

d i = lw - ii < lw -(-i)I = dz.

If we interpret d 1 as the distance from w to i and d2 as t he distance from w

to -i, then a geometric argument shows that the image point w must lie in t he

upper half-plane Im (w) > 0, as shown in F igure 10.5. As S is one-to-one and