the orthogonality theorem
(1) The i-lines C(a,α,b), C(c,γ,d) are orthogonal if and only if ad + bc=α*γ + αγ*.
(2) If i-lines L and M are orthogonal and i is an inversive transformation,
then
i(L) and i(M) are orthogonal.
(3) If L and M are distinct i-lines, then iL(M) = M if and only if L and M
are orthogonal.
Proof
(1) Consider the i-lines azz* - αz* - α*z + b = 0, czz* - γz* - γ*z + d = 0
To get equations in x and y, put z = x+iy, α = f+ig, γ = f'+ig'.
We have three cases:
(a) two extended lines.
If a = c = 0, then we have lines with xy-equations -2fx - 2gy + b = 0,
and -2f'x - 2g'y +c = 0. These are orthogonal if and only if
fg'+f'g = 0,
i.e. α*γ+αγ* = 0. Since a = c = 0, this is the stated condition.
(b) one extended line.
If c = 0, then the i-lines are orthogonal if and only if the line C(c,γ,d)
is a diameter of, i.e. passes through the centre of the circle C(a,α,b).
The centre is α/a, so this condition is -γ(α/a)* - γ(α/a) + d = 0.
Since a = 0, this is the stated condition.
(c) two circles.
As in page 1, the circles have centres A = α/a and C = γ/c, and radii r, s,
where r2 = (|α|2 - ab)/a2 and s2 = (|γ|2 - cd)/c2. They are therefore
orthogonal if and only if r2 + s2 = |AC|2 = |α/a+γ/c|2. Substituting for
r and s, and noting that |α/a+γ/c|2 = (α/a+γ/c)(α/a+γ/c)* =
|α|/a2 + (α*γ+αγ*)/ac + |γ|/c2, we get the stated condition.
(2) Since the inversive group is generated by inversions in i-lines, it is
enough to consider inversion in a single i-line L.
Since orthogonality is independent of the choice of coordinates, it is
enough to consider the cases where L is the real axis or the unit circle.
(a) L is the real axis.
Then, as in Example 1 the images are C(a,α*,b) and C(c,γ*,d).
By (1) the images are orthogonal if and only if ad + bc = αγ*+α*γ.
Since the right hand side is equal to α*γ+αγ*, we have the result.
(b) L is |z| = 1.
Then the images are C(b,α,a) and C(d,γ,c), and these are orthogonal
if and only if bc + ad = α*γ+αγ*. Now the left hand side is equal to
ad + bc, so again we have the result.
(3) Once again, it is enough to consider cases where L is the real axis or
the unit circle.
Let M be the i-line C(a,α,b).
(a) L is the real axis.
Here, inversion in L is just reflection, so the image is equal to M if and only if
M is a line perpendicular to L or a circle with centre on L, i.e. is orthogonal to L.
(b) L is |z| = 1, i.e. is C(1,0,-1).
Then, by (1), L, M are orthogonal if and only if 1.a+(-1).b = 0, i.e. a = b.
The image of M is C(b,α,a), so is M if and only if C(b,α,a)=C(a,α,b),
so b = ka, α = kα and a = kb for some non-zero real k.
If α ≠ 0, this is equivalent to a = b.
If α = 0, then we require b = ka, a = kb, so b = k2b.
Thus either a = b = 0, so M is a line through 0 (orthogonal to L) or a≠ 0, so
M is a circle centre 0. In the latter case, if inversion in L maps this to itself,
then
L = M, so they are not distinct.
|