Linear Fractional Transformations 1.0  Introduction A linear fractional transformation (LFT) is defined as a function of the form where a,b,c,d are complex constants. Every LFT defines a one-to-one mapping of the extended complex plane (C È ¥) onto itself. Two LFTs f and g are similar iff there exists a purely linear transformation L(z) = a z + b where a and b are complex constants such that for every complex z. (The condition is reciprocal, because if L is a linear function, then so is the inverse of L.) This condition implies a mapping between the iterates of f and the iterates of g as shown schematically in Figure 1. In essence, letting zk denote the iterates of f for any given initial value z0, and letting wk denote the iterates of g with the corresponding initial value w0 = L(z0), it follows that if f and g are similar, then wk = L(zk) for all k. In other words, the iterates of f can be mapped to the iterates of g by the linear function L, and since a linear transformation of the complex plane consists of (at most) a simple translation, a rotation about the origin, and a uniform scale change, it follows that the configuration of points L(zk) is geometrically similar to the configuration of points zk. To derive necessary and sufficient conditions for similarity, suppose the two LFTs are similar. Then by definition there exists a linear function L(z) = a z + b such that L(f(z)) = g(L(z)), so we have Substituting for f(z) and L(z) into this equation gives If a was zero then L would simply be a constant, and if c or C was zero the respective LFT would simply be a linear function, so we set these degenerate cases aside and consider the general case with a,C,C all non-zero. We can therefore normalize each side of the preceding equation by dividing through by c or Ca respectively (so the coefficients of z in the denominators are zero), and then equate the three remaining corresponding coefficients, which gives the three conditions Solving the first and third conditions for a and b we get Substituting this expression for b into the second condition and solving for a gives (assuming ad - bc ¹ 0). Equating this with the previous expression for a, we arrive at a necessary and sufficient condition for the LFTs f and g to be similar: This shows that for any given LFT (az+b)/(cz+d), the similarity class is determined solely by the parameter (a+d)2/(ad-bc). Hereafter we will refer to this as the similarity parameter, and denote it by s. Thus, for any LFT (az+b)/cz+d) we define Two LFTs f and g are said to be conjugate if there exists an LFT h such that h(f(z)) = g(h(z)) for all z. This is identical to the definition of similarity, except that similarity requires h to be purely linear. Thus, similarity would seem to be a more restrictive condition than conjugacy. However, the necessary and sufficient condition for f and g to be conjugate is exactly the same as the condition for similarity (cf. Complex Functions, p 32, Jones and Singerman, 1987). The only strength gained by allowing h to be an LFT instead of a linear function is in the "singular" cases when the coefficients of the linear coupling function L are not well defined, such as when c, C, (a+d), or (A+D) are equal to 0. However, when (a+d) = 0 the only conjugate LFTs are those with A+D = 0. Thus, returning to the previous conditions and inserting (a+d) = (A+D) = 0, we find that in this case it is necessary and sufficient for a to be given by (1). Thus, the only case when conjugacy can exist without a well-defined linear similarity correspondence relationship between the two LFTs is if c or C equal zero. The first of the three conditions above implies that is either c or C is zero, then the other is also zero, unless a is zero, which would imply a degenerate linear function.) In these cases, by the inclusion of the "point at infinity" (which is the same expedient used to ensure closure of the LFT iterates) we can include among the linear similarity relations those with infinite coefficients (and, conversely, zero coefficients). 2.0  Closed Form Expressions For the nth Iterate To find a closed-form expression for the nth iterate zn of the linear fractional transformation f(z) = (az+b)/(cz+d), it is most convenient to consider a similar function, g, having the sequence of iterates w0, w1, w2, ... where wk is related to zk by a linear function for some complex constants a and b. Our objective is to find a linear transformation such that a closed-form expression for wn (the nth iterate of g) is easy to determine. In general, the original LFT can be expressed in terms of w as follows Solving for wk+1 gives where Notice that if c is not equal to zero, we cannot force C to equal zero (except by putting a = 0, which is singular). This rules out any chance of finding coefficients a,b for which wk+1 = Kwk. However, there are several other "nice" LFT forms for which a closed-form solution can easily be found. One such, which we will call the "bi-polar" form, is Every LFT (az+b)/(cz+d) with two distinct fixed points is similar to an LFT of this special form. The coefficients a and b of the linear similarity function are determined by setting B = 0, which implies that We then impose the condition (C+D)/A = 1, which implies Substituting the previous expression for b and solving for a gives This is determined by mapping the two fixed points to the points 0 and 1 on the complex plane. We will refer to this as the bi-polar form of the given LFT similarity class. The value of K here is given by Notice that this implies The "K form" has fixed points at 0 and 1, and it can be shown by induction that zn has the convenient closed-form expression In this case the periodicity requirement is simply . It's interesting that to convert a purely real LFT to this "bi-polar K" form, the real axis is mapped to the line 0.5 + mi, which figures in the Riemann Hypothesis. The real and imaginary components of zn = xn + iyn can be expressed in closed form in terms of the components of K and z0. If we set then the components of zn are given by where f = nq. This shows that the action of the LFT on the complex plane is entirely dependent on the magnitude and phase angle of the constant K, which, as we saw previously, is given by If a,b,c,d are all real, then s is real, in which case either K is real (s>4 or s<0) or K is complex (0