1550251515-Classical_Complex_Analysis__Gonzalez_

756 Chapter9

z in N 6 (0) satisfying w = f(z). Each of these points lies in an analytic

single-valued branch of z = 1-^1 (w).

Proof The assumption implies that w = f(z) can be represented, in some

neighborhood of the origin, in the form

w = amzm + am+tZm+l + · · · = amzm[l + 1/i(z)]

where am'# 0and1/1(0) = 0. Hence for lzl small enough we have 11/i(z)I < 1,

and using the binomial expansion we find that

W = Az[l + 1fi(z)]^1 fm = Az ~ (17;) [1/i(z)]k

= Az[l + g(z)] = F(z) (9.16-11)

where A denotes o~e of the mth roots of am and wm = w. Equation (9.16-

11) can now be solved for z by (9.16-1) with Win some neighborhood of

w = 0 and c+ contained in that neighborhood, so that

z = p-l(W) = _1 J (F'(() d(

27ri F(()-W

c+

or alternatively, in the form

z = B1W +B 2 W^2 + ···, IWl<f

where

Bn= 1 / d(

27l"in [F(()]n

c+

Either way we see that z is an analytic function of W for IWI sufficiently

small. From

we get

or

J[F-l(wlfm)] = w

In (9.16-11) the coefficient A is determined up to an mth root of unity. If in

this equation A is replaced by wkA, where Wk = e^2 kn:i/m, then W becomes
Wk W and w^1 /m becomes Wkwlf m. With a corresponding enumeration or
the inverse images we have
k = 1,2, ... ,m