1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

88 CHAPTER 2 • COMPLEX FUNCTIONS

but not conversely. That is, Equation (2-34) makes sense when (u, v) = (O, 0),
whereas Equation (2-33) does not. Yet Figure 2.23 seems to indicate that f maps
Re (z) ;:::: ~ onto the entire disk Di (0), including the point (0, 0). Actually, it
does not, because (0, 0) has no preimage in the complex plane. The way out
of this dilemma is to use the complex point at infinity. It is that quantity that
gets mapped to the point (u, v) = (O, O), for as we have already indicated in
Equation (2-32), the preimage of 0 under the mapping ~ is indeed oo. •

•EXAMPLE 2.23 For the transformation ~.find the image of the portion of

the half-plane Re(z) <:: ~that is inside the closed disk D1 (~) = {z: lz - ~I

~ 1}.

Solution Using the result of Example 2.22, we need only find the image of the

disk D1 U) and intersect it with the closed disk D 1 (1). To begin, we note that

D 1 (~) = { (x, y): x^2 + y2 - x ~ ~}.

Because z = 1-^1 (w) =~.we have, as before,

u -v - (1)

<==> u (^2) +v 2 + i u (^2) +v 2 E D1 -2
(
u )
2
( - v )
2
u < 3
<==> u2 + v2 + u2 + v2 u2 + v2 - 4
1 u 3
<==> u2 + v 2 - u2 + v2 < -- 4
<==> (u+ ~r +v
2
::::: (~)
2
,
which is an inequality that determines the set of points in thew plane that lie on
and outside the circle c i ( -i) = { w : lw + i I = n. Note that we do not have
to deal with the point at infinity this time, as the last inequality is not satisfied
when (u, v) = (0, 0).
When we intersect this set with D 1 (1), we get the crescent-shaped region
shown in Figure 2.24.