The trace of a matrix is the sum of the entries on the leading diagonal. So for

a 2*2 matrix a11 a12

a21 a22

the trace is a11+a22.

Let a and b be matrices in SL(2,C) i.e. 2*2 matrices with entries that are

complex numbers and with determinant 1. Let us call the inverse matrices A,B

Preliminaries: show that trace is invariant under conjugation and inversion i.e. that trace(B*a*b) = trace(a)=trace(A)

Show that trace(a*b) + trace(a*B) = trace(a)*trace(b)

Show that if a*a = -I if and only if a has trace 0 (where I is the identity matrix).

Show that a^3 = I if and only if a has trace -1 and a^6=I if and only if

a has trace 1.

Let x = trace(a),y = trace(b),z = trace(a*b),w = trace(a*B)

Show that trace(a*b*A*B) = x*x+y*y+z*z-x*y*z-2

= x*x+y*y+w*w-x*y*w-2

= x*x+y*y-z*w-2

(The above relations, especially the first, are very important for geometry and in the theory of Kleinian Groups)

Now the hard part:

Suppose x^2 <> 4. Show that for n>=1 (trace(a^n*b*A^N*B)-2)/(trace^2(a^n)-4) is constant.

I got this result from the book "Outer Circles" which says that you can prove it for n=2 and then use induction. But I could not do it by induction.

Instead I found a general expression for calculating trace(a^n*w) given knowledge of trace(a), trace(w) and trace(a*w) by using a difference equation, and then proved it by brute force.