For $z \in \mathbb{C}$ consider the sequence $z, z^z, z^{z^z} ... ,$ that is, the iterated exponential with base $z$. If $z$ belongs to the Shell-Thron region the iterated exponential will converge. The boundary of the Shell-Thron region is given by the parametric equations
$ x = \Re(\exp(\exp(it-e^{it})))$
$y =\Im(\exp(\exp(it-e^{it})))$
for $t \in \mathbb{R}.$ I am interested in its image under the natural log, that is, the parametric curve
$ x = \Re(\exp(it-e^{it}))$
$y =\Im(\exp(it-e^{it}))$
I graphed this curve on my calculator and it appears to be a cardioid. Now $\exp(it-e^{it})) = \cos(\sin t - t)e^{-\cos t} + i\sin(\sin t - t)e^{-\cos t}$. My idea was to use the double angle formula (or another trig identity) to find an $a \in \mathbb{R}$ and a function $s = f(t)$ such that
$a(2\cos s - \cos 2s) = \cos(\sin t - t)e^{-\cos t}$
$a(2\sin s - \sin 2s) = \sin(\sin t - t)e^{-\cos t}$
but I can't seem to get anywhere.
Edit: I would like to see a proof that the equations
$ x = \Re(\exp(it-e^{it}))$
$y =\Im(\exp(it-e^{it}))$
cannot and do not represent a cardioid. What I mean is a result along the lines of:
It is impossible to find an $a \in \mathbb{R}$ and a function $s = f(t)$ such that
$a(2\cos s - \cos 2s) = \cos(\sin t - t)e^{-\cos t}$
$a(2\sin s - \sin 2s) = \sin(\sin t - t)e^{-\cos t}$
