Orbits of $z_{n+1} = z_n ^2 - 1$ Consider the sequence $z_{n+1} = z_n ^2 - 1$ defined for an arbitrary complex number $z_0$. I am trying to determine all $z_0$ such that the sequence eventually becomes periodic.
Here is my progress so far:
If $|z_0|> \frac{1+\sqrt{5}}{2}$ the sequence is absolutely increasing since
$$|z_{n+1}|=|z_n^2 - 1|\geq|z_n|^2-1 > |z_n| ,$$
which is true since $ |z_n| > \frac{1+\sqrt{5}}{2}$ holds inductively.
If $z_0=\frac{1+\sqrt{5}}{2}$ the sequence would be the constant sequence of $\frac{1+\sqrt{5}}{2}$ and hence periodic.
Similar to the first case if $|z_0|<\frac{1}{2}$ or more precisely if $|z_0|$ less than the root of $\alpha^3 +2\alpha -1 = 0,$ the sequence $\{z_{2i} \}$ becomes strictly decreasing since
$$|z_{2i}| = |z_{2i-1}^2 - 1 |= |z_{2i-2}^4 - 2z_{2i-2}^2|<  |z_{2i-2}| \Leftrightarrow |z_{2i-2}^3 - 2z_{2i-2}| < 1.$$
Which holds true if
$$ |z_{2i-2}^3 - 2z_{2i-2}| \leq  |z_{2i-2}^3|+| 2z_{2i-2}| <1.$$
And as a result $\{z_i\}$ cannot be periodic.

Let us find the alternative solutions.
Firstly, let us consider the case of the exact periodicity.
Assuming the periods $1,2,3,\dots,$ easily to get

*

*$z=z^2-1,\quad z_{11}=\phi=\dfrac{1+\sqrt5}2,\quad z_{12}=-\dfrac1\phi=\dfrac{1-\sqrt5}2;$

*$z=(z^2-1)^2-1,\quad z(z+1)(z^2-z-1)=0,$
$\qquad z_{21}=z_{11},\quad z_{22}=z_{12},\quad z_3=0,\quad z_4=1;$

*

*$z=((z^2-1)^2-1)^2-1,\quad (z^2-z-1)(z^6+z^5-2z^4-z^3+z^2+1)=0$
$\qquad z_{31}=z_{11},\quad z_{32}=z_{12},$
$\quad z_{33,34}\approx -1.42203\pm 0.114188i,\quad z_{35,36}\approx-0.0871062\pm0.655455i,\quad z_{37,38}=1.00914\pm0.324759i;$

*

*$\dots$
Also, the roots of the equations
$$z^2-1=z_{mn},$$
$$((z^2-1)^2-1)=z_{mn},\dots$$
leads to the required sequences.
