Julia set of finite Blaschke product

Question

I want to compute the Julia set of finite Blaschke product $B_3$ from the second version of the paper "PARABOLIC AND NEAR-PARABOLIC RENORMALIZATION FOR LOCAL DEGREE THREE" by FEI YANG,, see Figure 14 from page 52

$$B_3 = (\frac{z+1/2}{z/2+1})^3$$

First I should analyze the map so I compute first derivative :

$$ (3*(z+1/2)^2)/(z/2+1)^3-(3*(z+1/2)^3)/(2*(z/2+1)^4) $$

and critical point ( z for which derivative is equal to zero)

$$ z = -1/2 $$

Results of the analysis:

• $B_3$ is a Blaschke product whose Julia set is the unit circle
• The point z = 1 is a 1-parabolic fixed point with two attracting petals
• there is only one critical point z=-1/2
• the unit disk $D$ and $C \setminus D$ are two immediate basins of parabolic fixed point
• poles: z = -2 (order 3 pole)
• zeros: z = -1/2 (order 3 zero)

Now I can create attracting petal ( trap). It is a a disc inside component with parabolic fixed point in its boundary

Then I try to compute the dynamical chessboard ( I use binary decomposition of attracting petal)

My image

is different then that from the paper. It looks like 1/3 of original image.

Where is my error ? Are there other critical points?

Here is my program

Answer 1

I have found the answer in another paper on similar topic : Near Parabolic Renormalization for Unicritical Holomorphic Maps by Arnaud Chéritat

It is dynamical chessboard of not $B_3$ but of normalized $B_3$ denoted as $\tilde{B_3}$

In general:

$$\tilde{B_d}= \frac{z^d + a}{1 + az^d }$$

where $$a = \frac{d − 1}{d + 1}$$

so for d=3

$$\tilde{B_3}= \frac{z^3 + \frac{1}{2}}{1 + \frac{z^3}{2} }$$

Now images :are good for "every" $d >= 2$

code