Real and Complex Analysis (Rudin)

ANALYTIC CONTINUATION 329
We claim that Q is afundamental domain ofr. This means that statements (a)
and (b) of the following theorem are true.
16.19 Theorem Let rand Q be as above.
(a) If qJl and qJ2 E rand qJl #: qJ2' then qJl(Q) n qJ2(Q) = 0·
(b) UtperqJ(Q)=IJ+.
(c) r contains all transformations qJ E G of the form
qJ(Z) = az + b
cz + d
(1)
for which a and d are odd integers, band c are even.
PROOF Let r 1 be the set of all qJ E G described in (c). It is easily verified that
r 1 is a subgroup of G. Since (1 E r 1 and T E r 1, it follows that r c r l' To
show that r = r 1, i.e., to prove (c), it is enough to prove that (a/) and (b)
hold, where (a/) is the statement obtained from (a) by replacing r by r l' For
if (a/) and (b) hold, it is clear that r cannot be a proper subset of r l'
We shall need the relation
1m Z
1m qJ(z) = I cz + d 12 (2)
which is valid for every qJ E G given by (1). The proof of (2) is a matter of
straightforward computation, and depends on the relation ad - be = 1.
We now prove (a/). Suppose qJl and qJ2 E r 1, qJl #: qJ2' and define qJ =
qJl 1 0 qJ2' If Z E qJl(Q) n qJ2(Q), then qJl 1 (z) E Q n qJ(Q). It is therefore
enough to show that
Q n qJ(Q) = 0 (3)
if qJ E r 1 and qJ is not the identity transformation.
The proof of (3) splits into three cases.
If c = 0 in (1), then ad = 1, and since a and d are integers, we have
a = d = ± 1. Hence qJ(z) = Z + 2n for some integer n #: 0, and the description
of Q makes it evident that (3) holds.
If c = 2d, then c = ± 2 and d = ± 1 (since ad - be = 1). Therefore
qJ(z) = u(z) + 2m, where m is an integer. Since (1(Q) c D(t; !), (3) holds.
If c#:O and c #: 2d, we claim that I cz + d I > 1 for all Z E Q. Otherwise,
the disc D( -die; 1/1 c I) would intersect Q. The description of Q shows that if
IX #: -t is a real number and if D(IX; r) intersects Q, then at least one of the
points -1, 0, 1 lies in D(IX; r). Hence lew + d I < 1, for w = -1 or 0 or 1. But
for these w, cw + d is an odd integer whose absolute value cannot be less
than 1. So I cz + d I > 1, and it now follows from (2) that 1m qJ(z) < 1m Z for