Although we can use the Poincare model with euclidean tools, it is instructive
to see how we can work in hyperbolic geometry directly.
From the hyperbolic group pages, we know that an element h of H(2) has one
of the following forms, with c < 1, and κ = 1:
if h is direct, then h(z) = κ(zc)/(c*z1), and
if h is indirect, then h(z) = κ(z*c*)/(cz*1).
It is not difficult to see that the map h is determined by the point which maps
to 0  this is c  and the value of h(0)  which determines κ.
A useful transformation
For c < 1, the map h_{c}(z) = (zc)/(c*z1) satisfies
(1) h_{c} ε H(2),
(2) h_{c}(0) = c, h_{c}(c) = 0,
(3) h_{c} has order two.
proof
(1) and (2) are obvious.
(3) can be verified by calculation :
h_{c}(h_{c}(z)) = ((zc)/(c*z1)c)/(c*(zc)/(c*z1)1) =
z(1cc*)/((1cc*) = z.
Of course, the Origin Lemma guarantees the existence of an inversion h
which maps c to 0. As an inversion has order two, h also maps 0 to c.
As h is indirect, and maps c to 0, it has the form h(z) = κ(z*c*)(cz*1).
Since h(0) = c, c = κ(c*)/(1) = κc*, so κ = c/c*. We therefore have
the origin lemma  algebraic version
For 0 < c < 1, the inversion mapping c to 0 is h(z) = c(z*c*)/c*(cz*1).
Provided 0 < c < 1, the above map h is inversion in a hyperbolic line H,
so it fixes precisely the points of H. Thus, zεH if and only if
h(z) = z
i.e. c(z*c*)/c*(cz*1) = z. Thus, H has equation cc*zz*c*zcz*+cc* = 0.
If we put γ = 1/c*, then γ > 1, and the equation is zz*+γ*z+γz*+1 = 0.
Since this is a hyperbolic line, it is orthogonal to the unit circle C : z = 1.
Also, if γ > 1, then c = 1/γ* has c < 1, so every circle of this form will
give rise to a hyperbolic line, so must represent a circle orthogonal to C.
Of course, the points of H are those within the disk D, i.e. with z < 1.
The diameters of the unit circle C are also hyperbolic lines. They do not
occur in the Origin Lemma since inversion in any diameter maps 0 to 0.
A diameter H is determined by a point α ≠ 0, with α < 1. A point z lies
on H if and only if z = tα, with t real. This amounts to (z/α) = (z/α)*, or
α*zαz* = 0. if we write γ = iα, the equation is γ*z+γz* = 0.
We can unify these results as follows
the equation of a hyperbolic line
A hyperbolic line has equation azz*γ*zγz*+a = 0, where a is real and
γ > a.
The condition ensures that, if a = 0, then γ ≠ 0, so we get a diameter,
while, if a ≠ 0, then the equation becomes zz*(γ/a)*z(γ/a)z*+1 = 0.
This gives a circle orthogonal to C since γ/a > 1.
Maps of order two are easy to compute with. For example, if h has order
two, and F is a curve, then zεh(F) if and only if
h(z)εh(h(F)) = F. Thus,
if F has equation f(z) = 0, then h(F) has equation f(h(z)) = 0.
parametrization of a hyperbolic line
For distinct points α and β in D, the hyperbolic line joining α and β
has parametric equation {(γtα)/(α*γt1) : 1/γ < t < 1/γ},
where γ = (βα)/(α*β1).
The points of the hyperbolic segment αβ are given by tε[0,1].
justification
Let H denote the hyperbolic line through α and β.
Since α and β are distinct, we may as well assume that α ≠ 0.
Let h be the map h_{α}(z) = (z*c*)/(cz*1) considered above.
Now h maps α to 0, so maps H to a hyperbolic line through 0.
The image is therefore a diameter H'. Now β lies on H, so that
γ = h(β) = (βα)/(α*β1) is on H'. We observe that the points
of H' are of the form γt with t real and γt < 1, i.e. t < 1/γ.
Since h has order 2, the points of H are the points h(z), zεH'.
Now z = γt, so h(z) = (γtα)/(α*γt1), as required.
The points of αβ correspond to those of the euclidean segment
0γ. These are given by tε[0,1].
It is possible to adapt the proof to give :
The hyperbolic line αβ has equation azz*λ*zλz*+a = 0, where
γ = (βα)/(α*β1),
a = i(γ*αγα*), and
λ = i(γααγ*). We can get
equations in other, more interesing, ways.
further calculations

