We have the the PoincareMinkowski Map
 The map m is given by
m(x,y) = (2x/(1x^{2}y^{2}),2y/(1x^{2}y^{2}),
(1+x^{2}+y^{2})/(1x^{2}y^{2})).
 The map m maps an hline to the
intersection of H with a plane through O.
 The inverse is given by
m^{1}(x,y,(1+x^{2}+y^{2})^{½}) =
(x/(1+(1+x^{2}+y^{2})^{½}),
y/(1+(1+x^{2}+y^{2})^{½}).
The elements of H(2) are best explained in terms of the complex variable
z = x+iy. The formulae above implicitly involve z, so it is natural to work
using complex numbers. We try to avoid using "z" for a coordinate.
A point P(α,β,γ) lies on H provided that
γ^{2} = 1+α^{2}+β^{2}, and γ > 0.
We write z = α+iβ, so γ = (1+z^{2})^{½}.
Then m^{1}(P) = z/(1+γ).
We also observe that for complex Z with Z < 1, m(Z) is the point Q.
Since Q is on H, it is of the form (a,b,c) with c = (1+a^{2}+b^{2})^{½}. This is
determined by a and b, and hence by the complex W = a+ib. The above
result shows that W = 2Z/(1Z^{2}), and c = (1+Z^{2})/(1Z^{2}).


The case hεH^{+}(2).
Then h(z) = κ(za)/(a*z1), where κ = 1 and a < 1.
Thus, hom^{1}(P) = Z = κ(z/(1+γ)a)/(a*(z/(1+γ)1)
Clearing fractions, Z = κ(za(1+γ))/(a*z(1+γ)).
In preparation for the calculation of m(Z), we look at 1±Z^{2} and Z.
 1Z^{2}
= 1za(1+γ)^{2}/a*z(1+γ)^{2}
= (a*z(1+γ)^{2}za(1+γ^{2})/a*z(1+γ)^{2}
= (a^{2}z^{2} + (1+γ)^{2} z^{2} a^{2}(1+γ)^{2})/a*z(1+γ)^{2}.
= (((1+γ)^{2}z^{2})(1a^{2})/a*z(1+γ)^{2}.
Now, (1+γ)^{2} = 1+2γ+γ^{2} = 1 +2γ + (1+z^{2}) = 2+2γ+z^{2}.
So 1Z^{2}
= 2(1+γ)(1a^{2})/a*z(1+γ)^{2}.
 1+Z^{2}
= (a*z(1+γ)^{2}+za(1+γ)^{2})/a*z(1+γ)^{2}
= (a^{2}z^{2}+(1+γ)^{2}+z^{2}+a^{2}(1+γ)^{2}
2(1+γ)(a*z+az*)/a*z(1+γ)^{2}.
= ((1+a^{2})(z^{2}+1+γ)^{2})2(1+γ)(a*z+az*))/a*z(1+γ)^{2}.
Now, z^{2}+(1+γ)^{2} = 2(1+z^{2}+γ) = 2γ(1+γ).
So 1+Z^{2} = 2(1+γ)((1+a^{2})γ  (a*z+az*))/a*z(1+γ)^{2}.
 Z
= κ(za(1+γ))/(a*z(1+γ))
= κ(za(1+γ))(az*(1+γ))/a*z(1+γ)^{2}
= κ(az^{2}z(1+γ)a^{2}z*+a(1+γ)^{2})/a*z(1+γ)^{2}
= κ(a(z^{2}+(1+γ)^{2})(1+γ)(z+a^{2}z*))/a*z(1+γ)^{2}.
As before, z^{2}+(1+γ)^{2} = 2γ(1+γ).
So Z = κ(1+γ)(2γ(z+a^{2}z*))/a*z(1+γ)^{2}.
Thus W
= 2Z/(1Z^{2})
= 2κ(1+γ)(2γ(z+a^{2}z*))/2(1+γ)(1a^{2})
= κ(2γ(z+a^{2}z*))/(1a^{2}).
Since 1a^{2} is real, the real and imaginary parts of W
(i.e. the x and y coordinates of the final image) are
real linear combinations of α, β and γ.
Also, (1+Z^{2})/(1Z^{})
= 2(1+γ)((1+a^{2})γ  (a*z+az*))/2(1+γ)(1a^{2})
= ((1+a^{2})γ (a*z+az*))/(1a^{2}).
Here, 1a^{2} and 1+a^{2} are real and (a*z+az*) is a
real number (it is of the form w+w*), so this number
is also a real linear combination of α, β and γ.
Thus, overall mohom^{1} maps x to Ax for a real matrix A.
The case h indirect.
But then h(z) = h'(z*), with h'εH^{+}(2).
Replacing z by z* in the above analysis, we obtain the result
for this case as well.

z* denotes the complex
conjugate of z
γ^{2} = 1+z^{2} 