We have a matrix C such that the map c(x) = Cx fixes Z(0,0,1),
and has the property that CTKC = K, where K = diag(1,1,-1).
Since c maps Z to Z, the third column of C must be (0,0,1)T.
where 0 is the zero vector in R2,
u is some vector in R2, and
A direct calculation shows that
But this is K, so u = 0, and hence UTU = I, so that U is orthogonal,
the hyperbolic-projective theorem