Does an iterated exponential $z^{z^{z^{...}}}$ always have a finite period

Solution 1:

This is an attempt at proving statement 1. Showing that the conclusion holds for any $y \in \mathbb{R}$ with $|y| > y^\star$ basically boils down to proving $\lim\limits_{|y| \to \infty}(x+yi)^{x+yi} = 0$ for fixed $x$. Without loss of generality we may assume $y >0$ since $\bar z^{\bar z} = \overline{z^z}$. We may also ignore the argument of $(x+yi)^{x+yi}$ and show that $|(x+yi)^{x+yi}| \to 0$

$|(x+yi)^{x+yi}| = (x^2+y^2)^{x/2}e^{y\arctan(x/y)-y\pi/2}$. Since $x$ is fixed $(x^2+y^2)^{x/2} \sim y^{x}$ and since $y\arctan(x/y) \to x$, as $y \to \infty$ we have $e^{y\arctan(x/y)-y\pi/2} \sim e^{-y} \implies (x+yi)^{x+yi} \sim y^{x}e^{-y} \to 0$ since $e^{-y} \to 0$ way faster than $y^{x} \to \infty$.

I am still not sure how to show there is a smallest value $y^\star$ for which the conclusion is true. Nor do I know of a formula, or even an algorithm, to find it, other than a lot of numerical experimentation.