Real and Complex Analysis (Rudin)

292 REAL AND COMPLEX ANALYSIS
also necessary; thus among the annuli there is a different conformal type associated
with each real number greater than 1.
14.22 Theorem A(r1' R 1) and A(r2' R 2) are conformally equivalent if and only
if Rt!r1 = R 2/r2·
PROOF Assume r 1 = r 2 = 1, without loss of generality. Put
and assume there exists J E H(A 1) such that J is one-to-one and J(A 1) = A2 .
Let K be the circle with center at 0 and radius r = .jR;. Since J -1: A2 -+ A 1
is also holomorphic,f - 1(K) is compact. Hence
for some E > O. Then V = J(A(1, 1 + E» is a connected subset of A2 which
does not intersect K, so that V c A(1, r) or V c A(r, R2). In the latter case,
replace J by R21f. So we can assume that V c A(1, r). If 1 < 1 Zn 1 < 1 + E and
1 Zn 1-+ 1, then J(zn) E V and {f(zJ} has no limit point in A2 (since J- 1 is
continuous); thus 1 J(z.) 1-+ 1. In the same manner we see that 1 J(z.) 1-+ R2 if
IZnl-+ R 1·
Now define
(1)
(2)
log R2
0(=--
log R1
(3)
and
u(z) = 2 log IJ(z)l- 20( log Izl (4)
Let a be one of the Cauchy-Riemann operators. Since aj = 0 and aJ = 1', the
chain rule gives
so that
a(2 log 1 J I) = a(log (l!» = I'If,
(5)
(au)(z) = I'(z) _ ~
J(z) z
(6)
Thus u is a harmonic function in A1 which, by the first paragraph of this
proof, extends to a continuous function on A1 which is 0 on the boundary of
A 1 • Since nonconstant harmonic functions have no local maxima or minima,
we conclude that u = O. Thus
(7)