Complex Analysis - Maths KU

366 Chapter 6 Series and Residues
v
1
|w – 1| = 1
C′
Figure 6.16 Image of C lies within the
disk |w − 1| < 1.
u
no zeros on the contour C.From |f(z) − g(z)| = |g(z) − f(z)|, we see that by
dividing the inequality by |f(z)| we have, for all z on C,
|F (z) − 1| < 1, (32)
where F (z) =g(z)/f(z).The inequality in (32) shows that the image C ′ in
the w-plane of the curve C under the mapping w = F (z) is a closed path and
must lie within the unit open disk |w − 1| < 1 centered at w = 1.See Figure
6.16. As a consequence, the curve C ′ does not enclose w = 0, and therefore
1/w is analytic in and on C ′ .By the Cauchy-Goursat theorem,
�
1
dw =0
w
or
�
F ′ (z)
dz =0,
F (z)
(33)
C ′
since w = F (z) and dw = F ′ (z) dz.From the quotient rule,
we get
C
F ′ (z) = f(z)g′ (z) − g(z)f ′ (z)
[f(z)] 2 ,
C
F ′ (z)
F (z) = g′ (z)
g(z) − f ′ (z)
f(z) .
Using the last expression in the second integral in (33) then gives
� � ′ g (z)
g(z) − f ′ �
(z)
dz =0
f(z)
or
�
g ′ �
(z)
dz =
g(z)
f ′ (z)
f(z) dz.
It follows from (28) of Theorem 6.20, with Np = 0, that the number of zeros
of g inside C is the same as the number of zeros of f inside C. ✎
EXAMPLE 7 Location of Zeros
Locate the zeros of the polynomial function g(z) =z 9 − 8z 2 +5.
Solution We begin by choosing f(z) =z 9 because it has the same number
of zeros as g.Since f has a zero of order 9 at the origin z = 0, we begin
our search for the zeros of g by examining circles centered at z = 0.In other
words, if we can establish that |f(z) − g(z)| < |f(z)| for all z on some circle
|z| = R, then Theorem 6.21 states that f and g have the same number of
zeros within the disk |z|