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 Ak(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 iC denotes inversion with respect to C.

  • 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 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 iL is just reflection in L.
It follows that iL(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
iL(M) = M if and only if L and M are orthogonal.


In fact, when , L and M are orthogonal, more is true.

If L and M are orthogonal i-lines, then iL() = ,
where 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 iL
maps M to M, and the i-line PS to itself. Also iL(S) = S (on L).
A lies on the segment RS, so its inverse A' lies on ST.
Thus the points of 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 are
"reflections" of one another in a circular "mirror" L.
This is not optically sound,
but it is a good analogy.

Main inversive page