Proof of the Algebraic Inversion Theorem

 The Algebraic Inversion Theorem Suppose that L and C are i-lines, and that iC denotes inversion with respect to C. Then iC(L) is an i-line, if A and B are inverse with respect to L, then iC(A) and iC(B) are inverse with respect to iC(L). iC° iL = iL'° iC ,where L' = iC(L). Proof We observe that the definition of inversion depends only on length, so does not depend on the choice of axes. Thus, if C is an extended line, then we may assume that it is the real axis, so iC(z)=z*. If C is a circle, we may assume it is |z|=r, so iC(z)=r2/z*. Suppose that A and B are inverse with respect to L, and have complex coordinates a and b respectively. Then, from Apollonius Theorem, L = Ak(A,B) for some k. Hence L has equation |z-a| = k|z-b|. If C is the real axis, then iC(L) has equation |z*-a| = k|z*-b| For any complex w, |w| = |w*|. If we apply this to both sides, we see that iC(L) has equation |z-a*| = k|z-b*|, i.e. |z-iC(a)| = k|z-iC(b)|. Thus iC(L) belongs to Ak(iC(A),iC(B)). It follows from Apollonius theorem that iC(L) is an i-line and that iC(A) and t(B) are inverse with respect to it. If C is the circle |z|=r, then iC(L) has equation |r2/z*-a| = k|r2/z*-b|. Now if we multiply each side by |z*|/|a|, and apply conjugation to each side, we get |r2/a*-z|=k|b*/a*||r2/b* -z|. Now, since |w| = |-w| and iC(z)=r2/z*, the equation becomes |z-iC(A)| = k|b*/a*||z-iC(B)|. Once again, iC(L) is an i-line, and iC(A) and iC(B) are inverse with respect to it. Suppose that B is the inverse of A with respect to L, i.e. that B = iL(A). The second part of the theorem says that iC(B) = iL'(iC(A)) Thus, as B = iL(A), iC(iL(A)) = iL'(iC(A)). Since this holds for any A, the third part follows.