750 Chapter9
9.16 MAPPING PROPERTIES OF ANALYTIC
FUNCTIONS. INVERSE FUNCTION THEOREMS
Theorem 9.23 If the nonconstant function f is analytic at z = a, then
in some neighborhood N6(a), f(z) assumes every value bin a sufficiently
small neighborhood of J(a).
Proof The function F(z) = f(z)-f(a) is analytic at a and has a zero at a
of order m ~ 1. Since F( z) does not vanish identically in a neighborhood
of a, the zero z = a is isolated. Thus we can choose a disk lz - al ::; 8 in
which a is the only zero of F(z), and on which F(z) is analytic (Fig. 9.33)
Let -y: lz - al = 8 and e = minze-y IF(z)I. Since F(z) -.:/: 0 for z E -y, we
have e > 0. Consider the neighborhood Ne(f(a)), and let b be any point
in this neighborhood, i.e., such that
IJ(a)-bl <-e
If we set g(z) = J(a) - b, then
lg(z)I < IF(z)Ifor z E -y, smce
IJ(a) - bl< e =min zE-y IF(z)I::; IF(z)I
By Rouche's theorem, F(z) and F(z)+g(z) = f(z)-J(a)+ J(a)-b = f(z)-
b have the same number of zeros inside -y, namely m. Hence f(z) - b = 0,
or f(z) = b for m values or z in N 0 (a).
Remark If m = 1, there is just one value of z such that f(z) = b. If
m > 1, with a suitable choice of 8 it can be asserted that the value b is
attained at m distinct points z in N 0 (a). In fact, in this case f'(z) has at a
a zero of order m - l and it is not identically zero in a deleted neighborhood
of a. Hence there is a closed disk lz - al ::; 81 such that neither f nor f'
y v0 x (^0) u
Fig. 9.33