Real and Complex Analysis (Rudin)

ANALYTIC CONTINUATION 327
16.15 Tbeorem Suppose 0 is a simply connected region, (f, D) is a function
element, D c: 0, and (f, D) can be analytically continued along every curve in 0
that starts at the center of D. Then there exists g E H(O) such that g(z) = f(z)
for all ZED.
PROOF Let r 0 and r 1 be two curves in 0 from the center ex of D to some
point {J E O. It follows from Theorems 16.13 and 16.14 that the analytic continuations
of(f, D) along ro and r 1 lead to the same element (gp, Dp), where
Dp is a disc with center at {J. If Dpl intersects Dp, then (gpl' Dpl) can be
obtained by first continuing (f, D) to {J, then along the straight line from {J to
{Jl' This shows that gPI = gp in Dpl (1 Dp.
The definition
g(z) = gp(z)
is therefore consistent and gives the desired holomorphic extension off.
IIII
16.16 Remark Let 0 be a plane region, fix w ¢ 0, let D be a disc in O. Since
D is simply connected, there exists f E H(D) such that exp [f(z)] = z - w.
Note thatf'(z) = (z - W)-l in D, and that the latter function is holomorphic
in all of O. This implies that (f, D) can be analytically continued along every
path y in 0 that starts at the center ex of D: If y goes from ex to {J, if Dp =
D({J; r) c: 0, if
and if
r" = y -+ [{J, z] (1)
gp(z) = r «( - W)-l d( + f(ex)
Jr.
(z E Dp), (2)
then (gp, Dp) is the continuation of(f, D) along.Y.
Note that gp(z) = (z - W)-l in Dp.
Assume now that there exists g E H(O) such that g(z) = f(z) in D. Then
g'(z) = (z - w) -1 for all z E O. If r is a closed path in 0, it follows that
Indr (w) = -2 1 . r g'(z) dz = O.
1tzjr
(3)
We conclude (with the aid of Theorem 13.11) that the monodromy theorem
fails in every plane region that is not simply connected.