Wilson Stothers' Inversive Geometry and CabriJava Pages

The Algebraic Inversion Theorem

We use complex numbers as "coordinates" in E.
The point A(a,b) is described by the complex number a =a + ib.
Then the locus A_k(A,B) = {P : PA = k.PB} has equation |z -a| = k.|z - b|,
where a, b are the coordinates A, B.
For typographical reasons, we will use z* to denote the complex conjugate of z.

Immediately we see that inversion in the i-line given by the real axis is
reflection in the real axis, i.e. maps z to z* (and ∞ to itself).
We shall simply say that
A is the point a

Inversion in the i-line |z| = r is also easy to describe,
it maps z to r²/z* (non-zero z), and interchanges 0 and ∞.

The simple proof

We are now in a position to give a neat proof that inversion maps i-lines
to i-lines. As a bonus, we get information about the images of inverse points.
The Algebraic Inversion Theorem
Suppose that L and C are i-lines, and that i_C denotes inversion with respect to C.
Then

i_C(L) is an i-line,
if A and B are inverse with respect to L, then
i_C(A) and i_C(B) are inverse with respect to i_C(L).
i_C° i_L = i_L'° i_C ,where L' = i_C(L).

Proof of The Algebraic Inversion Theorem

We can now prove a result fundamental to the creation of hyperbolic geometry.
Observe that, if L and M are lines, then i_L is just reflection in L.
It follows that i_L(M) = M if and only if L and M are orthogonal.

In fact, this is true for any i-lines. As you may guess, the proof amounts
to showing that the general case can be reduced to this special case by
applying inversion.
The Mirror Property
Suppose that L and M are distinct i-lines, then
i_L(M) = M if and only if L and M are orthogonal.
Proof

In fact, when , L and M are orthogonal, more is true.
If L and M are orthogonal i-lines, then i_L(Mº) = Mº,
where Mº is the interior of M.

In the CabriJava pane on the right, S is a point on the arc of L lying inside M.
P is the centre of L, and PS cuts M at R and T. Then i(R) = T since i_L
maps M to M, and the i-line PS to itself. Also i_L(S) = S (on L).
A lies on the segment RS, so its inverse A' lies on ST.
Thus the points of Mº inside L map to those outside and vice versa.
You can experiment by dragging S or A.

We can think of this as saying that the two regions of the disc Mº are
"reflections" of one another in a circular "mirror" L. This is not optically sound,
but it is a good analogy.

Main inversive page

We use complex numbers as "coordinates" in E. The point A(a,b) is described by the complex number a =a + ib. Then the locus A_k(A,B) = {P : PA = k.PB} has equation \|z -a\| = k.\|z - b\|, where a, b are the coordinates A, B. For typographical reasons, we will use z* to denote the complex conjugate of z. Immediately we see that inversion in the i-line given by the real axis is reflection in the real axis, i.e. maps z to z* (and ∞ to itself).	We shall simply say that A is the point a
Inversion in the i-line \|z\| = r is also easy to describe, it maps z to r²/z* (non-zero z), and interchanges 0 and ∞.	The simple proof
We are now in a position to give a neat proof that inversion maps i-lines to i-lines. As a bonus, we get information about the images of inverse points. The Algebraic Inversion Theorem Suppose that L and C are i-lines, and that i_C denotes inversion with respect to C. Then i_C(L) is an i-line, if A and B are inverse with respect to L, then i_C(A) and i_C(B) are inverse with respect to i_C(L). i_C° i_L = i_L'° i_C ,where L' = i_C(L). Proof of The Algebraic Inversion Theorem

We can now prove a result fundamental to the creation of hyperbolic geometry. Observe that, if L and M are lines, then i_L is just reflection in L. It follows that i_L(M) = M if and only if L and M are orthogonal. In fact, this is true for any i-lines. As you may guess, the proof amounts to showing that the general case can be reduced to this special case by applying inversion. The Mirror Property Suppose that L and M are distinct i-lines, then i_L(M) = M if and only if L and M are orthogonal. Proof
In fact, when , L and M are orthogonal, more is true. If L and M are orthogonal i-lines, then i_L(Mº) = Mº, where Mº is the interior of M. In the CabriJava pane on the right, S is a point on the arc of L lying inside M. P is the centre of L, and PS cuts M at R and T. Then i(R) = T since i_L maps M to M, and the i-line PS to itself. Also i_L(S) = S (on L). A lies on the segment RS, so its inverse A' lies on ST. Thus the points of Mº inside L map to those outside and vice versa. You can experiment by dragging S or A.
We can think of this as saying that the two regions of the disc Mº are "reflections" of one another in a circular "mirror" L.	This is not optically sound, but it is a good analogy.