Proof Page

Proof of Theorem E2

Theorem E2
If t is a direct isometry, then t(z) = az + b, where a, b are in C, and |a| = 1.
If t is an indirect isometry, then t(z) = az* + b, where a, b are in C, and |a| = 1.

Proof
Suppose that t is an isometry with t(0) =b and t(1) =g.

Suppose that s maps z to az + b. Then s(0) = b, s(1) = a + b.
Let a = g - b, so s(1) = a + b = g.
Note that, as t preserves distance, |g - b| =|1-0| = 1, so |a| = 1.

The idea is to construct a map
with the same effect as t
on 0 and 1.

We must check that s is an isometry.
This is easy since |s(z) - s(w)| = |(az+b)-(aw+b)| = |a(z-w)| = |z-w|.

The case where s(z) =az*+b
is very similar

A simple calculation shows that r = s^-1°t fixes 0 and 1.
By Lemma 1, r is the identity or r_x (reflection in the x-axis).
Hence t = s or s°r_x, so has the required form.

By Lemma 1. s^-1°t is
more or less determined.

euclidean group

Theorem E2 If t is a direct isometry, then t(z) = az + b, where a, b are in C, and \|a\| = 1. If t is an indirect isometry, then t(z) = az* + b, where a, b are in C, and \|a\| = 1.
Proof Suppose that t is an isometry with t(0) =b and t(1) =g. Suppose that s maps z to az + b. Then s(0) = b, s(1) = a + b. Let a = g - b, so s(1) = a + b = g. Note that, as t preserves distance, \|g - b\| =\|1-0\| = 1, so \|a\| = 1.	The idea is to construct a map with the same effect as t on 0 and 1.
We must check that s is an isometry. This is easy since \|s(z) - s(w)\| = \|(az+b)-(aw+b)\| = \|a(z-w)\| = \|z-w\|.	The case where s(z) =az*+b is very similar
A simple calculation shows that r = s^-1°t fixes 0 and 1. By Lemma 1, r is the identity or r_x (reflection in the x-axis). Hence t = s or s°r_x, so has the required form.	By Lemma 1. s^-1°t is more or less determined.