1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

2.5 • THE RECIPROCAL TRANSFORMATION w = ~ 8 7

continuous mapping from the extended z plane onto the extended w plane. We
leave the details to you.

• EXAMPLE 2.22 Show that the image of the half-plane A = { z : Re (z) 2: !}

under the mapping w =~is the closed disk D1 (1) = {w: lw - ll ~ 1}.

Solution Proceeding as we did in Example 2.6, we get the inverse mapping of
u+iv=w=f(z)=~asz=f-^1 (w)=;!;. Then

u+iv = w EB{'=} r^1 (w) =z=x+iy EA

= --.-^1 = x + iy. E A

u+w

{'=} - 2--u 2 + t - 2. --- v 2 = x + iy. E A

u +v u + v

<==::> _ u_ =x=Re(x+iy) >!

u2 +v2 - 2

u 1

= ---u2 +v 2 >-- 2

==> u^2 - 2u + 1 + v^2 :$ 1

<==::> (u - 1)^2 + (v -0)^2 :$ 1,

(2-33)

(2-34)

which describes the disk D 1 (1). As the reciprocal transformation is one-to-
one, preimages of the points in the disk D 1 (1) wi.11 lie in the right half-plane
Re(z) <:: ~· Figure 2.23 illustrates this result.

)'

x

w=l z

v

Figure 2.23 T he image of Re( z) ;::: ~ under the mapping w = ~.

u

Remark 2.5 Alas, there is a fly in the ointment here. As our notation indicates,

Equations (2-33) and (2-34) are not equivalent. The former implies the latter,