Theorem 4
- If W lies in the h-segment UV, then D(U,W) < D(U,V). and D(W,V) < D(U,V),
- there is a unique M in the h-segment UV with D(U,M) = D(M,V).
Proof
Suppose that W lies between U and V.
As usual, we simplify the problem by applying the Origin Lemma.
Let h be the h-inversion mapping U to O.
Suppose that h(V) = V' and h(W) = W'.
Then, by Theorem 2, W' lies between O and V'.
Since it includes O, the h-segment OV' is a euclidean segment.
| D(O,W')
|
= D(0,tz) = |tz| |
as D(0,w) = |w| |
|
= t|z| = tD(0,z) |
|
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= tD(O,V') < D(O,V') |
as t < 1 |
But O = h(U), W' = h(W) and V' = h(V), so that
D(U,W) < D(U,V).
|
D(z,w) = |z-w|/|w*z-1|,
so that
D(z,0) = |z|/|-1| = |z|,
and
D(0,w| = |-w|/|-1| = |w|.
|
If, instead, we map V to O, and observe that D(z,0) = |z|,
we get D(W,V) < D(U,V).
Observe that, as W moves along UV,
D(U,W) increases continuously from 0 to D(U,V) and
D(W,V) decreases continuously from D(U,V) to 0.
An application of the Intermediate Value Theorem to
f(W) = D(W,V) - D(U,W) shows the existence of M.
Since f is strictly decreasing, M is unique.
We can avoid the appeal to continuity in the proof by
a calculation which acually gives the point M.
Suppose that we apply the h-inversion h as in the proof.
Then, the point V' and W' on OV' have the coordinates z and tz.
Now D(O,W') = t|z|,
and D(W',V') = ||tz -z|/|z*tz-1| = |z|(1-t)/(1-t|z|2)
Thus,
| D(O,W') = D(W',V') |
if and only if |
t|z| = |z|(1-t)/(1-t|z|2), |
|
i.e. |
t2|z|2 -2t + 1 = 0 |
After a little simplification, the roots are
t = 1/(1 + √(1-|z|2)) and 1/(1 - √(1-|z|2)).
A further calculation shows that only the first has tz in D.
Thus M = h(z/(1+ √(1-|z|2))).
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