1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

406 CHAPTER 10 • CONFORMAL MAPPING

Using the fa.ct that 1 1:..ii = t• we rewrite this equation as

z+2 l+w

i;-=1- w·

We now expand the equation and obtain z + 2 -zw -2w = iz + izw, which can

be solved for w in terms of z, giving the desired solution

w = S (z) = (1 - i) z + 2.

(1+i)z+ 2

We let D be a region in the z plane that is bounded by either a circle or a

straight line C. We further let z 1 , Z2, and z3 be three d istinct points that lie on

C and have the property that an observer moving along C from z 1 to z3 through

z 2 finds the region D to be on the left. If C is a circle and D is the interior

of C, then we say that C is positively oriented. Conversely, the ordered triple

(z1> z2, za) uniquely determines a region that lies to the left of C.

We let G be a region in the w plane that is bounded by either a circle or

a straight line K. We further let w 1 , w 2 , and wa be three distinct points that

lie on K such that an observer moving along K from w1 to W3 through w2 finds

the region G to be on the left. Because a bilinear transformation is a conformal

mapping that maps the class of circles and straight lines onto itself, we can use

the implicit formula to construct a bilinear transformation w = S (z) that is a

one-t~one mapping of D onto G.

• EXAMPLE 10.6 Show that the mapping

w = S (z) = (1 - i) z + 2

(1+i)z+ 2

maps the d isk D : lz + l I < 1 onto the upper half- plane Im ( w) > 0.

Solution For convenience, we choose the ordered triple z 1 = - 2, z 2 = - 1 - i,

and Z3 = 0, which gives the circle C : lz + ll = 1 a positive orientation and the

disk D a left orientation. From Example 10.5, the corresponding image points

are

w 1 = S(z1) = - 1,

Because the ordered triple of points WlJ w2, and w3 lie on the u-axis, it follows

that the image of circle C is the u-axis. T he points w 1 , w2, and w3 give the

upper half-plane G : Im (w) > 0 a left orientation. Therefore, w = S (z) maps

the disk D onto the upper half-plane G. To check our work, we choose a point

zo that lies in D and find the half-plane in which its image, wo, lies. The choice

z 0 = -1 yields w 0 = S(-1) = i. Hence the upper half-plane is the correct

image. This situation is illustrated in Figure 10.6.