1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

60 CHAPTER, 2 • COMPLEX FUNCTIONS

Solution The inverse transformation is z = w3~4;2', so if we designate the

range of f as B, then

w = I ( z) E B <=> 1- • ( w) = z E D1 ( - 1 - i)

w -6 - 2i

<=>

3

_

4

i E Dt( -1 - i)

<=> I w ; ~ ~i 2i + 1 + ii < 1.

l

w -

6

• 2
  <=> 3 - 4i i + 1 + i ·^113 - 4i ·1 < 1.^13 - 4i ·1

<=> lw - 6 - 2i + (1 + i) (3 - 4i)I < s

<=> lw + 1 - 3il < 5.

Hence the disk with center - 1 - i and radius 1 is mapped one-to--One and onto

the disk with center -1+3i and radius 5 as shown in Figure 2.10.

w =S(z)

~

v

Figure 2.10 The mapping w = S (z) = (3 - 4i) z + 6 + 2i.

• EXAMPLE 2.11 Show that the image of the right half-plane Re (z) 2 1
  under the linear transformation w = (- 1 + i) z -2+3i is the half-plane v 2 u + 7.
  Solution The inverse transformation is given by

_ w+2- 3i _ u +2+i(v-3)

z- - '

- l+i - l + i

which we write as

. - u + v - 5. - u -v + 1
x+iy=
2
+i
2

Substituting x = (- u-i;v- s ) into R.e(z) = x 2 1 gives (- u-i;v- S) 2 1, which

simplifies to v 2 u + 7. Figure 2.11 illustrates the mapping.