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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.

Matin Yousefi
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    This is a question about the Julia set of $z \mapsto z^2-1$: if you haven't come across that term, you should look it up first. Pretty fractals! – HTFB Aug 30 '21 at 20:28
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    See a picture at https://upload.wikimedia.org/wikipedia/commons/4/4f/Julia-1.png – lhf Aug 30 '21 at 20:35
  • Even more fun to look at video zooms of fractals. https://youtu.be/YOneAeBz8BQ – Thomas Andrews Aug 30 '21 at 20:51
  • How is the absolute values converging to some value not indicate a period result? The absolute values are still non-negative, so decreasing doesn’t mean not converging. – Thomas Andrews Aug 30 '21 at 20:55
  • In theory, increasing doesn’t mean it isn’t periodic, either, but the sequence is increasing exponentially, which is why the increasing case does not converge to a periodic sequence. – Thomas Andrews Aug 30 '21 at 21:00
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    When $|z_0|<1/2, $ the sequence actually converges to the periodic $0,-1,0,-1,\dots.$ – Thomas Andrews Aug 30 '21 at 21:02
  • Specifically, each attracting periodic point is in the interior of one of those red blobs, and each red blob has one attracting periodic point in its interior. The repulsive periodic points are dense in the boundary between red and green. – David E Speyer Sep 12 '21 at 11:35

1 Answers1

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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.

Yuri Negometyanov
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