Real and Complex Analysis (Rudin)

286 REAL AND COMPLEX ANALYSIS
PROOF (a) is clear. To prove (b), write f(z) = zcp(z). Then cp E H(U), cp(O) = 1,
and cp has no zero in U, sincefhas no zero in U - {O}. Hence there exists an
h E H(U) with h(O) = 1, h2(Z) = cp(z). Put
(z E U). (2)
Then g.2(Z) = z2h2(Z2) = Z2cp(Z2) = f(Z2), so that (1) holds. It is clear that
g(O) = 0 and g'(O) = 1. We have to show that g is one-to-one.
Suppose z and W E U and g(z) = g(w). Since f is one-to-one, (1) implies
that Z2 = w 2 • So either z = w (which is what we want to prove) or z = -w.
In the latter case, (2) shows that g(z) = -g(w); it follows that g(z) = g(w) = 0,
and since g has no zero in U - {O}, we have z = w = O.
IIII
14.13 Theorem IfF E H(U - {O}), F is one-to-one in U, and
1 co
F(z) = - + L IXn Zn
Z n=O
(z E U), (1)
then
(2)
This is usually called the area theorem, for reasons which will become apparent in
the proof.
PROOF The choice of 1X0 is clearly irrelevant. So assume 1X0 = O. Neither the
hypothesis nor the conclusion is affected if we replace F(z) by AF(AZ)
(I AI = 1). So we may assume that IXI is real.
Put U r = {z: Izl < r}, C r = {z: Izl = r}, and V. = {z: r < Izl < I}, for
0< r < 1. Then F(U r) is a neighborhood of 00 (by the open mapping
theorem, applied to 1IF); the sets F(U r ), F(C r ), and F(V.) are disjoint, since F
is one-to-one. Write
1
F(z) = - + IXIZ + cp(z)
z
(z E U), (3)
F = u + iv, and
(4)
For z = re i8 , we then obtain
u = A cos e + Re cp and v = - B sin e + 1m cpo (5)