566 Chapter^8Example f(z) = tanz is meromorphic in lzl < 2. Its poles in this region
are at z = ±%7r (zeros of cotz = 1/tanz), and it is otherwise analytic
in lzl < 2.
Poles are discussed in detail in Sections 9.2 and 9.3.
A zero z 0 off is called a zero of infinite order iff no equation of the form
(8.8-1) holds for some positive integer m and g continuous [respectively, ofclass 'D or 1£) with g(zo) f O].
Examples
- The function that is identically zero in D has at each point of D a
zero of infinite order. - The function f(z) = z + z = 2 Re z, defined on C, has a zero at every
point of the imaginary axis, and it is different from zero elsewhere in
the complex plane. Since the zeros of this function are not isolated,
it is clear that an equation of the ,form (8.8-1) is not possible with g
continuous and g(iy 0 ) f 0. Hence all the zeros z = iy are of infinite
order.
Theorem 8.17 The zeros of finite order of a continuous function on an
open set D are isolated; i.e., in some deleted neighborhood of a zero f has
no other zeros.
Proof If z 0 E Dis a zero off of order m (m ;::: 1), we have
f(z) = (z - zorg(z)
where g is continuous on D and g(zo) f 0. By the definition of continuity
at a point, for every E > 0 there is a Ii > 0 such that lz - z 0 I < Ii implies
that lg(z) -g(zo)I < E. Let E =^1 / 2 lg(zo)I, and let /i 1 be the corresponding
Ii. Then we have
lg(z)-g(zo)I < %lg(zo)I
whenever jz - zo I < /i 1. This implies that g( z 1 ) f 0 for any z 1 such that
0 < lz1 --zol < li1. Otherwise, we would have at that point
lg(zo)I < %Jg(zo)I
which is impossible. Therefore, /(z1) = (z1 - zorg(z1) f 0 for any Z1 E
N8 1 (zo).Corollary 8.8 The set of zeros of finite order of a continuous function on
an open set is countable.
Theorem 8.18 The set of zeros of finite order of a continuous function