Lemma
For 0 < r < 1, tεH(2) maps the circle
Cr = {z : |z| =r}
to the locus
{z : |z - t(0)|/|t(0)*z - 1| = r}.
Proof
Let t(0) = γ. Then t-1 maps γ to 0.
Suppose first that t (and hence t-1) is direct.
As t-1 maps γ to 0, the hyperbolic group page
shows that t-1(z) = κ(z-γ)/(γ*z-1).
| z ε t(Cr) |
if and only if |
t-1(z) ε Cr,
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i.e. |
|t-1(z)| = r.
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i.e. |
|κ(z-γ)/(γ*z-1)| = r, |
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i.e. |
|z-γ|/|γ*z-1| = r, |
as |κ| = 1. |
The result follows as γ = t(0).
Now suppose that t is indirect.
As t-1is now indirect, the argument above shows that
z ε t(Cr) if and only if |(κ(z-γ)/(γ*z-1))*| = r.
But |(κ(z-γ)/(γ*z-1))* = |κ(z-γ)/(γ*z-1)|.
Thus, we have the result in this case also.
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