How can I find parameter of the rational map? I have a family of rational maps $f$ ( the Blaschke fraction ) with one complex parameter  $\rho$:
$$f(z) = \rho z^2 \frac{z-3}{1-3z}$$
I want to find $\rho$ such that map $f$ has a parabolic period 3 cycle on the unit circle.
So $\rho$ should be a solution of system of equations:
$$ \begin{cases}f^3(z_0) = z_0 \\ 
|(f^3)'(z_0)| = 1
\end{cases}$$
where

*

*$f^3(z) = f(f(f(z)0)))$ is 3-rd iteration of function f

*$(f^3)'(z_0)$ is the first derivative of $f^3$ with respect to $z$ at $z_0$

*$z_0$ is the point from period 3 parabolic cycle

but computing derivative of iterated function is hard
The image of dynamic plane with Julia set:

It is fig 3 from paper : Near Parabolic Renormalization for Unicritical Holomorphic Maps by Arnaud Chéritat
I know that $\rho$ has a modulus equal to one (complex number):
$$ |\rho| = 1 $$
My first try was easy solution ( point on the unit circle with angle in turns = 1/3)
$$ \rho =  e^{2 \pi i / 3} =  -0.5 +0.8660254037844386 i$$
Then the period 3 cycle is on the unit cirlce but the cycle is not parabolic and the image is different.
Q. How can I find such $\rho$ using algebra or numerical root finding ? How can I solve above system of equations ?
 A: It is an approximation of the answer using brute force and visual analysis.
First analyze the map f

*

*find critical points

*compute Mandelbrot set

*find parameter $\rho$ : point on the unit circle which is parabolic point : cusp or a root

So there are 3 critical points :

*

*2 finite critical  points : z=1.0 i z= 0.0

*infinity

Here is parameter plane ( rho plane)

Here zoom around cusp point ( black)  on the unit circle ( white)

So $\rho \approx -0.6170144002709304 +0.7869518599370003*I$
Here is dynamic plane ( z plane ) with Julia and Fatou sets

There are 3 Fatou components:

*

*(white = 255) basin of attraction to infinity

*( light gray = 195) basin of attraction to fixed point z=0 ( superattracting) with critical point z=0

*(dark gray = 135) basin of parabolic period 3 cycle ( critical point z= 1)

Increasing precision of $\rho$ value : there are two 3 point cycle. Choose rho such that 2 cycles join into one cycle.
$$\rho = e^{2 \pi i t}$$
t is changing from 0.3558289901601067 to 0.3558800000000000
Final rho (t)  = e^(2piti) is
rho =  -0.6172665900123702 +0.7867540637673888I  with  turn(rho) =0.3558800000000000  and   radius = 1.0000000000000000




On the last image critical orbit goss to limit cycle from both sides . I think that it is numerical error .
Rho value seem to be good
