14.12.2012 Views

Complex Analysis - Maths KU

Complex Analysis - Maths KU

Complex Analysis - Maths KU

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

64 Chapter 2 <strong>Complex</strong> Functions and Mappings<br />

C′<br />

w = iz<br />

i<br />

–1 1<br />

–i<br />

Figure 2.5 The mapping w = iz<br />

C′<br />

C<br />

w = z 2<br />

4i<br />

2i<br />

C<br />

2 4<br />

Figure 2.6 The mapping w = z 2<br />

byrotating the half-plane S through an angle π/2 radians counter-clockwise<br />

about the origin. This observation about the mapping w = iz will be verified<br />

in Section 2.3. In the following examples, we represent a complex mapping<br />

using a single copyof the complex plane.<br />

EXAMPLE 3 Image of a Parametric Curve<br />

Use (11) to find the image of the line segment from 1 to i under the complex<br />

mapping w = iz.<br />

Solution Let C denote the line segment from 1 to i and let C ′ denote its<br />

image under f(z) =iz. Byidentifying z0 = 1 and z1 = i in (7), we obtain a<br />

parametrization z(t) =1− t + it, 0≤ t ≤ 1, of C. The image C ′ is then given<br />

by(11):<br />

w(t) =f(z(t)) = i(1 − t + it) =−i(1 − t) − t, 0 ≤ t ≤ 1.<br />

With the identifications z0 = −i and z1 = −1 in (7), we see that w(t) isa<br />

parametrization of the line segment from −i to −1. Therefore, C ′ is the line<br />

segment from −i to −1. This mapping is depicted in Figure 2.5 using a single<br />

copyof the complex plane. In Figure 2.5, the line segment shown in color is<br />

mapped onto the line segment shown in black by w = iz.<br />

EXAMPLE 4 Image of a Parametric Curve<br />

Find the image of the semicircle shown in color in Figure 2.6 under the complex<br />

mapping w = z 2 .<br />

Solution Let C denote the semicircle shown in Figure 2.6 and let C ′ denote<br />

its image under f(z) =z 2 . We proceed as in Example 3. Bysetting z0 =0<br />

and r = 2 in (10) we obtain the following parametrization of C:<br />

Thus, from (11) we have that:<br />

z(t) =2e it , 0 ≤ t ≤ π.<br />

w(t) =f(z(t)) = � 2e it� 2 =4e 2it , 0 ≤ t ≤ π, (12)<br />

is a parametrization of C ′ . If we set t = 1<br />

2<br />

parametrization of C ′ :<br />

s in (12), then we obtain a new<br />

W (s) =4e is , 0 ≤ s ≤ 2π. (13)<br />

From (10) with z0 = 0 and r = 4, we find that (13) defines a circle centered at<br />

0 with radius 4. Therefore, the image C ′ is the circle |w| =4. We represent<br />

this mapping in Figure 2.6 using a single copyof the plane. In this figure, the<br />

semicircle shown in color is mapped onto the circle shown in black by w = z 2 .

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!