Complex Analysis - Maths KU

102 Chapter 2 Complex Functions and Mappings
w = 1/z
C′
C
1 2
Figure 2.42 The reciprocal mapping
Reciprocal Mapping The reciprocal function f(z) =1/z can be
written as the composition of inversion in the unit circle and complex conjugation.
Using the exponential forms c(z) =re −iθ and g(z) =e iθ /r of these
functions we find that the composition c ◦ g is given by:
�
1
c(g(z)) = c
r eiθ
�
= 1
r e−iθ .
Bycomparing this expression with (1), we see that c(g(z)) = f(z) =1/z.
This implies that, as a mapping, the reciprocal function first inverts in the
unit circle, then reflects across the real axis.
Image of a Point under the Reciprocal Mapping
Let z0 be a nonzero point in the complex plane. If the point w0 = f(z0) =
1/z0 is plotted in the same copy of the complex plane as z0, then w0 is
the point obtained by:
(i) inverting z 0 in the unit circle, then
(ii) reflecting the result across the real axis.
EXAMPLE 1 Image of a Semicircle under w =1/z
Find the image of the semicircle |z| =2,0≤ arg(z) ≤ π, under the reciprocal
mapping w =1/z.
Solution Let C denote the semicircle and let C ′ denote its image under
w =1/z. In order to find C ′ , we first invert C in the unit circle, then we
reflect the result across the real axis. Under inversion in the unit circle, points
with modulus 2 have images with modulus 1
2 . Moreover, inversion in the unit
circle does not change arguments. So, the image of the C under inversion in
the unit circle is the semicircle |w| = 1
2 ,0≤ arg(w) ≤ π. Reflecting this set
across the real axis negates the argument of a point but does not change its
modulus. Hence, the image after reflection across the real axis is the semicircle
given by |w| = 1
2 , −π ≤ arg(w) ≤ 0. We represent this mapping in Figure
2.42 using a single copyof the complex plane. The semicircle C shown in color
is mapped onto the semicircle C ′ shown in black in Figure 2.42 by w =1/z.
Using reasoning similar to that in Example 1 we can show that the reciprocal
function maps the circle |z| = k, k �= 0, onto the circle |w| =1/k. As
the next example illustrates, the reciprocal function also maps certain lines
onto circles.