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