62 Chapter 1
If we compute the distances NP and NQ from N(O,O, 1) to the points
P(x,y,0) and Q(a,/3,"(), respectively, we find thatN p2 = x2 + y2 + 1 = JzJ2 + 1
NQ2 = a2 + /32 + ('Y -1)2 = 1 - 'Y = JzJ2\ 1
(1.17-12)(1.17-13)by using (1.17-7) and (1.17-8). Hence it follows that NP· NQ = 1, which
shows that the points P and Q are inverses of each other with respect to
a sphere of unit radius with center at N.
If we consider in the C plane two points P and P' representing the
complex numbers z and z^1 , respectively, as well as the corresponding images
Q and Q^1 on the Riemann sphere (Fig. 1.15), we also have NP'· NQ' = 1,
so that NP· NQ =NP'· NQ', or
NP NP'
=
NQ'Hence the triangles NP P' a,nd N Q' Q, which have in common the angle
with the vertex at N formed by proportional sides, are similar, so that
=
NQ'Solving for Q'Q, using (1.17-12), the analog of (1.17-13), and notingthat P'P = Jz - z'J, we obtain
Fig. 1.15
-- P'P·NQ'
Q'Q=-N-P-