Real and Complex Analysis (Rudin)

CONFORMAL MAPPING 283
Under these conditions, the inverse of cP is hoi om orphic in Q 2 (Theorem 10.33)
and hence is a conformal mapping of Q 2 onto Ql.
lt follows that conformally equivalent regions are homeomorphic. But there
is a much more important relation between conform ally equivalent regions: If cP
is as above,f--4 f 0 cP is a one-to-one mapping of H(Q 2) onto H(Q 1 ) which preserves
sums and products, i.e., which is a ring isomorphism of H(Q2) onto H(Q 1 ).
If Q 1 has a simple structure, problems about H(Q 2 ) can be transferred to problems
in H(Q 1 ) and the solutions can be carried back to H(Q 2) with the aid of the
mapping function cpo The most important case of this is based on the Riemann
mapping theorem (where Q 2 is the unit disc U), which reduces the study of H(Q)
to the study of H(U), for any simply connected proper subregion of the plane. Of
course, for explicit solutions of problems, it may be necessary to have rather
precise information about the mapping function.
14.8 Theorem Every simply connected region Q in the plane (other than the
plane itself) is conformally equivalent to the open unit disc U.
Note: The case of the plane clearly has to be excluded, by Liouville's
theorem. Thus the plane is not conformally equivalent to U, although the two
regions are homeomorphic.
The only property of simply connected regions which will be used in the
proof is that every holomorphic function which has no zero in such a region has
a holomorphic square root there. This will furnish the conclusion" U) implies (a)"
in Theorem 13.11 and will thus complete the proof of that theorem.
PROOF Suppose Q is a simply connected region in the plane and let Wo be a
complex number, Wo ¢ Q. Let 1: be the class of all rjJ E H(Q) which are oneto-one
in Q and which map Q into U. We have to prove that some rjJ E 1:
maps Q onto U.
We first prove that 1: is not empty. Since Q is simply connected, there
exists a cp E H(Q) so that cp2(Z) = Z - Wo in Q. If cp(Z 1) = CP(Z2), then also
cp2(ZI) = cp2(Z2)' hence ZI = Z2; thus cp is one-to-one. The same argument
shows that there are no two distinct points ZI and Z2 in Q such that CP(Zl) =
-CP(Z2). Since cp is an open mapping, cp(Q) contains a disc D(a; r), with
o < r < 1 a I. The disc D( - a; r) therefore fails to intersect cp(Q), and if we
define rjJ = r/(cp + a), we see that rjJ E 1:.
The next step consists in showing that if rjJ E 1:, if rjJ(Q) does not cover all
of U, and if Zo E Q, then there exists a rjJ 1 E 1: with
It will be convenient to use the functions CPa defined by
Z-Q(
CPa(Z) = -1 --•
- Q(Z