752 Chapter9
Proof Since l'(z 0 ) -:/:- 0 the function F(z) = l(z) - wo has a simple zero
at zo. As in Theorem 9.23, we can find a o > 0 and an e > 0 such that
for each win N.(w 0 ) there is just one z E N 0 (zo) such that l(z) = w. Let
V = N,(w 0 ) and U = 1-^1 (V) with I restricted to N 0 (z 0 ). Then the map
of U into V by I is bijective, and since I is continuous, U is a neighborhood
of z 0 (Fig. 9.34). By the open mapping theorem, I is an open mapping
and this implies that 1-^1 : V -> U is continuous. To prove that 1-^1 is
analytic in V, it suffices to apply (9.14-8) to the function l(z) -w, which
has a simple root inside a= {(: I( -zol = 6}. This gives
z = 1-^1 cw) = -^1 J (!'(() d(
27ri 1(()-w
(9.16-1)
c+and it follows that 1-^1 ( w) is an analytic function of w in V by
Theorem 7.28. In fact, we have
d -1 1 I (!'(()
dw I (w) = 27ri [I(() - w]2 d(
c+
1 I d(= 27ri I ( () - w.
c+the last form being obtained from the first by integration by parts with
u = (, dv = f'(()d(/[f(() - w]2. More generally, we have
.!!_ 1 -1 w = (n -1)! I d( ,
d wn ( ) 27ri [I(() - W n ] n = 1,2,3, ...
c+Corollary 9.8 Let
w = l(z) = ao + al(z -zo) + az(z - z 0 )^2 + · · ·
or
w - Wo = al(z - zo) + az(z -z 0 )^2 + · · · (9.16-2)
where al = l'(zo)-:/:-0, be a Taylor representation of l(z) in Jz - zol ::=; o.
Then the inverse function z = 1-^1 (w) has the Taylor representation
z = zo + b1 ( w - wo) + b2 ( w -w 0 )^2 + · · · (9.16-3)
where
bn =~I (f'(()d(
27ri [I(() - wo]n+l 'n = 1,2, ... (9.16-4)
c+