Consider Mandelbrot generated by the following iteration

z_n+1 = z_n^w + c

where c is a complex number and w is as well.

The value of z may approach a limit cycle of period 2 in which it value oscillates between Z₁ and Z₂. This means that

Z₂ = Z₁^w + c
Z₁ = Z₂^w + c

Subtracting the second equation from the first gives

Z₂ − Z₁ = Z₁^w − Z₂^w
which may be rearranged to
Z₂ + Z₂^w = Z₁ + Z₁^w

Let the common value of each side of the equation be denoted as λ. Thus

Z₂ + Z₂^w = λ
Z₁ + Z₁^w = λ

Thus Z₁ and Z₂ are solutions to the equation

Z + Z^w = λ

For w≠2 there may be more than two solutions, in which case Z₁ and Z₂ would be any pair of the set of solutions.

Conditions for a Stable Limit Cycle

The deviations from the limit values satisfy the following three equations

z_n+2 − Z₂ = z_n+1^w − Z₁^w
z_n+1 − Z₁ = z_n^w − Z₂^w
z_n − Z₂ = z_n-1^w − Z₁^w

The ratio |z_n+2 − Z₂|/|z_n − Z₂| can be expressed in the form

[|z_n+2 − Z₂|/|(z_n+1 − Z₁)|][|(z_n+1 − Z₁)|//|z_n − Z₂|]]

Then z_n+2 − Z₂ can be replaced by its equal value of z_n+1^w − Z₁^w. In the second term of the product z_n+1 − Z₁ can be replaced by its equal value of z_n^w − Z₂^w.

Thus in order for |z_n+2 − Z₂| to be less than |z_n − Z₂| it is necessary that

|[(z_n+1^w − Z₁^w)/(z_n+1 − Z₁)][(z_n^w − Z₂^w)/(z_n − Z₂)]| < 1

At the boundary between the stable and unstable values of c the inequality becomes an equality so

|[(z_n+1^w − Z₁^w)/(z_n+1 − Z₁)][(z_n^w − Z₂^w)/(z_n − Z₂)]| = 1

Now consider the limits as n→∞. Both terms go to 0/0 so l'Hospital's rule applies and hence

lim |[(z_n+1^w − Z₁^w)/(z_n+1 − Z₁)| = |wZ₁^w-1|
and
lim [(z_n^w − Z₂^w)/(z_n − Z₂)]|= |wZ₂^w-1|

Thus the condition to be satisfied for a limit 2-cycle is

|w²(Z₁Z₂)^w-1| = 1

The complex numbers whose absolute value is equal to 1 are those given by exp(iφ) for 0≤φ≤2π. Thus

Z₁Z₂ = (1/w²)exp(iφ/(w-1))

The Case of a Positive Integer Exponent

As stated previously, the values of Z₁ and Z₂ are solutions to the equation

Z + Z^w − λ = 0

When w is a positive integer the above equation is a monic polynomial. The product of the roots of a monic polynomial is equal to the constant term in the equation. For w=2 then

Z₁Z₂ = −λ

Also for a monic polynomial the sum of the roots is equal to the coefficient of the term with the next to largest exponent. For w=2 this is 1 and for w>2 it is equal to 0. For w=2 then

Z₁ + Z₂ = 1
and hence
Z₂ = 1 − Z₁

For w=2

(1/w²)exp(iφ/(w-1)) = (1/4)exp(iφ)

Thus

Z₁(1-Z₁) = −(1/4)exp(iφ)
and therefore
Z₁² − Z₁ − (1/4)exp(iφ) = 0

This equation can be solved for Z₁ as a function of φ, say Z₁(φ).

Since for this case

Z₁² + c = Z₂ = 1 − Z₁
and hence
c = 1 − (Z₁(φ) + Z₁(φ)²)

(To be continued.)