7 48 Chapter 9for all n :> N and all z E 'Y· Let Yn(z) = fn(z) - f(z). Then combining
(9.15-8) and (9.15-9), we have
IYn(z)I < lf(z)I
for all z E 7, provided that n > N. Hence, by Rouche's theorem, f(z)
and f(z) + Yn(z) = fn(z) have the same number of zeros inside 'Y· Thus
if z 0 is not a zero of f(z), the functions fn(z) for n > N have no. zeros
in a neighborhood of z 0 , and if z 0 is a zero of order m (m ;::: 1) of f(z),then each fn(z) has, for n > N, exactly m zeros inside 1 (which may or
may not coincide with z 0 ). ·Remark That the condition f(z) f 0 is necessary can be readily seen
by an example. For instance, if fn(z) = (cosz)/n and A= {z: lzl < 1},
we have fn(z) ~ 0 as n ~ oo, the convergence being uniform on compact
subsets of A. However, none of the functions fn(z) vanishes in A.
Example Let fn(z) = [(n + 1)/n]z - 1/n. Then fn(z) ~ z as n ~ oo,the convergence being uniform on compact subsets of lzl < R. The limit
function f(z) = z has a zero at z = 0 which is an accumulation point of
the zeros of fn(z), namely, the points Zn = 1/(n + 1).
Exercises 9.9
- Use formula (9.14-3) to evaluate each of the following integrals.
)- z^6 dz
(a) z1+1' where c+: z = 2eit, 0::; t::; 271"
c+
(b) -.--dz, where c+: z = 2.5ezt, 0::; t::; 271"
JCOS11"Z ·
Sln7l"Z
c+J
ez dz ·
(c) ez +
1, where c-: z = 4e-it, O::; t::; 271"
c-
- Use formula (9.14-7) to evaluate fa+ z[f'(z)/ f(z)] dz, where f(z) =
( z^2 + 4 )^2 / ( z^2 + z + 1 )^3 and c+: z = 3eit, o ::; t ::; 271". - Use the argument principle to show that the number of zeros of the
function w = z^2 m + az^2 m-l + b^2 , with a and b real, a^2 + b^2 f O, m a
positive integer, the zeros having a positive real part, ism whenever m
is even, and m - 1 or m + 1 if m is odd, depending on whether a > 0
or a< 0. (Hint: Consider ~c+ argf(z) where c+ = r+ + [iR, -iR],
r+: z = Reit, -%11" ::; t ::; %71", with R ---+ oo.)- Use Rouche's theorem to find the number of roots of each of the