Power Tower Sequence

Clearly, "No, it converges for none" is the wrong answer. Most people guessed that "it will converge for all".

However, a simple counter example is the complex cube root of unity $\omega$ . Since $\omega^2 = \omega^2$ and $(\omega^2)^2 = \omega$ , we see that the sequence will simply toggle back and forth between $\omega$ and $\omega^2$ .

Let's investigate this further. How can we describe all $z_1, |z_1| = 1$ which will lead to a convergent sequence with $z_{n+1} = z_n ^2$ ? First, let's write $z = e^{2\pi i \theta }$ with $0 \leq \theta < 1$ .
Writing $\theta$ in base 2, suppose we have $\theta_1 = 0.b_1 b_2 b_3 b_4 \ldots_2$ .
Then, if we define $\theta_n = 0.b_n b_{n+1} b_{n+2} \ldots_2$ , we have $z_{n+1} = z_n^2 = e^{2 \pi i \theta_n \times 2 } = 2^{ 2 \pi i \theta_{n+1} }$ .
As such, $z_n$ will converge if and only if $b_n$ is an eventually constant sequence.

Going back to $\omega = e^{ 2 \pi i \frac{1}{3} }$ , since the base 2 represntation of $\frac{1}{3} = 0.010101\ldots_2$ , we conclude that the sequence starting $\omega$ will not convege. In fact, we can also tell that the sequence will oscillate, based on $0.010101\ldots$ .

Furthermore, we can conclude that the sequence will converge if and only if $z = e^{ 2 \pi i r }$ where $r = \frac{ k}{2^l}$ for some integers $k$ and $l$ . As such, almost all $|z| =1$ will not lead to a convergent sequence.

Let, $x=a\pm bi$ . If we square it, we get $a^{2}-b^{2}\pm2abi$ which also have absolute value 1. Assume, $a^{2}-b^{2}=0$ and $2ab=1$ . Then, the sequence will converge. As an example, $\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$ converges. For many other arrangements we'll keep getting new complex numbers. As an example, $\frac{7}{\sqrt{58}}+i\frac{3}{\sqrt{58}}$ doesn't converge.