224 Chapter 5
If z 2 = oo, then T is given by
Z -Z3
Tz = ---= (z,oo,z3,z4) = (z3,z4,z)Z-Z4
Similarly, if z 3 = oo, thenand if Z4
Tz = ---Z2 -Z4 = (z,z2,oo,z4) = 1: ( z,z2,z4 )
Z -Z4
oo, then
Z -Z3
Tz = ---= (z,z2,z3,oo) = (z,z2,z3)
Z2 -Z3(5.5-5)(5.5-6)(5.5-7)If S were another bilinear transformation having the same property,
the composition Ts-^1 would leave the points 1, 0, oo invariant. But as
will be seen in Section 5.6, a bilinear function leaves no more than two
points invariant unless it is the identical transformation. This can easily
be verified directly in the present situation, because if w = ( az + b) / ( cz + d)
maps 1, 0, oo into 1, 0, oo, we get the equations
a+ b = c+d, b= o, c=O
so that a= d and w = z. Hence, denoting the identical transformation by
I, we have Ts-^1 =I, or T = S, which shows that Tis uniquely determined.
More generally, a bilinear transformation is uniquely determined when the
images w2, w 3 , W4 of any three distinct points z 2 , z 3 , z 4 are prescribed.
This is a consequence of the invariance of the cross ratio, for if w is then
the image of the point z, we must have
(5.5-8)
and we need only to solve (5.5-8) for w.Ifin either of (5.5-4)-(5.5-7) we replace z by z 1 , we get
(5.5-9)
which shows that the cross ratio (z 1 , z 2 , z 3 , z 4 ) is numerically equal to
the image of z 1 under the bilinear transformation that maps z 2 , z 3 , z 4 into
1, 0, 00.
This property can be used to produce an elegant and more general proof
of the invariance of the cross ratio of four distinct points under bilinear
transformations. In fact, let L be any bilinear transformation, and let T
be, as above, the bilinear transformation that carries z 2 , z 3 , z 4 into 1, 0,
oo, respectively. Then TL-^1 carries. Lz2, L~3, Lz4 into 1, 0, oo. Hence,
in view of (5.5-9), we obtain
(Lz1,Lz2,Lz3,Lz4) = TL-^1 (Lz1) =.Tz1 = (zi,z2,z3,z4)