Many of you may recall "An obvious pattern to $i\uparrow\uparrow n$ that is eluding us all?", an old question of mine, and just recently, I saw this new question that poses a simple extension to tetration at non-integer values:

$$a\uparrow\uparrow b=\begin{cases}a^b,&b\in[0,1]\\a^{a\uparrow\uparrow(b-1)},&b\in(1,+\infty)\\\log_a(a\uparrow\uparrow(b+1)),&b\in(-\infty,0)\end{cases}$$

Combining this with my old question, I made the following graph of $i\uparrow\uparrow x$ for $x\in(-2,9)$, using $z=re^{i\theta},\theta\in[0,2\pi)$:

enter image description here

(x-axis is the real axis and y-axis is the imaginary axis)

Is there anything special to be said about this?

About how each look always connects to the previous 'branch' and then steps off the branch until it hits the next one. Can we prove this is indeed the case?

Prove or disprove that the tangent line at each interception is equivalent for the original branch and the branch coming out that heads towards the center.

It also appears the branches connect perpendicularly.

Are these shapes similar to one another?

The pattern is rather intriguing, don't you think?

It appears $(-1)\uparrow\uparrow x$ is likewise interesting to look at:

Link to graph.

It starts off how one might expect it to start off:

enter image description here

It makes a loop:

enter image description here

Then another loop:

enter image description here

And then it blows up into a circular shape that reaches about 35 units away from the origin:

enter image description here

Closer image:

enter image description here

And then it goes on past $10^{30}$:

enter image description here

Closer image:

enter image description here

Medium zoom:

enter image description here

Another loop:

enter image description here

Any explanation for why this is so much more 'chaotic' than $i\uparrow\uparrow x$? Perhaps we can define chaotic or not as follows:

\begin{align}\lim_{x\to\infty}a\uparrow\uparrow x\ \text{converges}\implies\text{stable/non-chaotic}\end{align}

\begin{align}\lim_{x\to\infty}a\uparrow\uparrow x\ \text{diverges}\implies\text{unstable/chaotic}\end{align}

Once again we see apparently perpendicular connections, though trivially all at $(-1,0)$.

There are new patterns though. We get almost cardioids, but not quite. We saw two interesting loop looking shapes as well. Any idea what these are? It appears these loops get really long and form the quasi-cardioids.

Is it the case that these loops always connect back to $(-1,0)$ from the same direction from which they came?

And is there a 4-turned patter? The first line connecting to $(-1,0)$ came from above, the second line connecting to $(-1,0)$ came from the right, the third came from below, the fourth from the left, and if we keep graphing more of these, the fifth comes from above once again.

Can we do an analysis to these different shapes? My graphing calculator isn't the best, and I have to do these one by one... Particularly, what can we say about $z\uparrow\uparrow x$ for $|z|=1$ and $x\in\mathbb R$?

After quite a few graphs, I've come to the following conclusion that:

When $|z|=1,\operatorname{arg}(z)=\theta\in(-,\pi]$, then $\lim\limits_{x\to\infty}z\uparrow\uparrow x$ tends to exist for $\theta<\theta_0$ and diverges for $\theta>\theta_0$. What is this $\theta_0$? And is my conclusion correct?

It also appears that it may be far from trivial proving the lines in $i\uparrow\uparrow x$ actually connect. Though they are close, I have found that $\sqrt i\uparrow\uparrow x$ clearly does not connect:enter image description here

Here is the general graph. It appears as though $\theta_0=\pi/2$.

Simply Beautiful Art
  • 71,916
  • 11
  • 112
  • 250
  • Wonder what caused the downvote. – Simply Beautiful Art May 20 '17 at 00:14
  • 1
    Touching comes from the fact that $i^i = e^{-\pi/2} \in (0, 1)$. It is also a bit of interesting that paths of the same color intersects tangentially and paths of different colors intersects perpenticularly (if they intersect). – Sangchul Lee May 20 '17 at 00:23
  • Huh, and as I'm trying to construct $(-i)↑↑x$, I find it... quite strangely shaped. – Simply Beautiful Art May 20 '17 at 00:31
  • If you use the principal branch cut for the complex logarithm, then you should have $\overline{(-i)\uparrow\uparrow x} = i \uparrow\uparrow x$, which means that the graph of one should be the $x$-axis reflection of the graph of the other. – Sangchul Lee May 20 '17 at 00:35
  • Ah, okay, I'm actually using $\ln(-i)=\frac{3\pi i}2$ and drawing my shapes much differently. – Simply Beautiful Art May 20 '17 at 00:39
  • I haven't voted either way, but "is there anything special to be said about this" is vague and very open. Is there a specific conjecture based on the picture that you would like to see proved/disproven? I also find the "$a^b$ when $0\le b<1$" part extremely unnatural: It makes $f(x)=a\uparrow\uparrow x$ on the reals a line segment on $[-1,0]$, and not differentiable at any integer. – Mark S. May 20 '17 at 00:42
  • @MarkS. That is true, but I did not choose this definition. I merely found it, applied it, and found something interesting. There are, notably, multiple questions at the end of the post which should be more objective. – Simply Beautiful Art May 20 '17 at 00:45
  • @SimplyBeautifulArt Ah, I was working on my comment before you edited in the more specific question, and then was distracted by things other than MSE. – Mark S. May 20 '17 at 00:46
  • @MarkS. It took you 12 minutes to write that comment? O.o? – Simply Beautiful Art May 20 '17 at 00:46
  • @SangchulLee So I chose my branch such that $(-i)^x=e^{\frac{3\pi x}2i}$, which does not give me $\overline{(-i)\uparrow\uparrow x} = i \uparrow\uparrow x$, but rather almost the 'opposite' of that, per se, as you can see if you graph it. – Simply Beautiful Art May 20 '17 at 01:02
  • @SangchulLee Yes, indeed, if you look at the last graph, it follows that $\overline{z\uparrow\uparrow x} = \overline z \uparrow\uparrow x$ if we choose the branch correspondingly. – Simply Beautiful Art May 20 '17 at 13:44
  • This is a very interesting topic. I have made a programm to draw this (thanks to VB's System.Numerics, which enables complex numbers) and played a bit around with the value of $z$. However, there are cases when the process creates what seems like a cycle (I think $1 + 2 i$ was one of them, but I'm not sure). Still, this is pretty cool. – G. Ünther May 20 '17 at 20:16
  • @G.Ünther do share :D – Simply Beautiful Art May 20 '17 at 20:35
  • @Simply Beautiful Art I'll do, but I am currently unable to do so (I am only 15) and need some sleep ;D. Please be patient, I'll share my programm tomorrow (I expect it to be online in about 15 hours). Thank you for your interest :D – G. Ünther May 20 '17 at 20:51
  • at the last remark: I found one likely value $\theta_0= 1.2486079913755423879 \cdot \pi/2$ where the iteration towards fixpoints changes from convergence (=1-step-periodic) fixpoint towards divergence (multi-step-periodic) points . This is due the fact that the value for $u$ with $\exp(u \cdot \exp(-u))=z$ increases over $u>1$ and for $u \lt 1$ it is known that the iteration converges by D.Shell & W.Thron. (this is just one critial $\theta_0$, maybe there are more because $z=f(u)$ is not linear; didn't check this so far) – Gottfried Helms May 12 '19 at 19:21
  • I did my computations with Pari/GP and default precision of 200 (dec) digits. A quick plot `plot(th=0,2, abs(u= - LambertW(-(c=I*th*Pi)) ))` shows a somehow logarithm-curve and thus that there is only ***one*** $\theta_0$ where with $u=f(\theta_0)$ we can have $|u|=1$ So we should have only *one* $\theta_0$ where this change from convergence to divergence/periodic fixpoints occurs. In the plot-command the $c$-variablename is meant as $\log(z)$ and $u$ as used in the previous comment. – Gottfried Helms May 12 '19 at 19:34
  • You really deserve your user name. This is pure beauty ! Thanks – Claude Leibovici May 13 '19 at 10:11
  • It is confusing to claim you are working with tetration here. Defining a domain where exponential and tetration are the same should have given everyone pause who have looked at this page. – Daniel Geisler May 14 '19 at 12:04

3 Answers3


$i\uparrow\uparrow t$


All of the relevant patterns here seem to involve only nonnegative inputs, so define $f:[0,\infty)\to \mathbb C$ by $f\left(t\right)=\begin{cases}i^{t} & \text{ if }t\in[0,1)\\i^{f\left(t-1\right)} & \text{ if }t>1\end{cases}$. To get a sense for the graph of $f$, we can use different colors for $[0,1)$, $[1,2)$,...


Here is a graph of $f(t)$ for $t\in[0,9)$: spiral image

It starts at $1$, then follows an indigo quarter-circle to $i$, then an orange sort of vertical sigmoid to $e^-\pi/2$, and then a green sigmoid up towards what appears to be a point on the initial quarter-circle, and spiraling inwards with colors like red, purple, brown, blue, yellow, and then pink.

For notational convenience, set $p=\dfrac{\pi}{2}$ and $g\left(t\right)=i^{t}=\left(e^{ip}\right)^{t}=e^{ipt}=\exp\left(ipt\right)$. This makes $f\left(t\right)=\begin{cases}g\left(t\right) & \text{ if }t\in[0,1)\\g\left(f\left(t-1\right)\right) & \text{ if }t>1\end{cases}$.

Do these actually connect? Yes

Note that $f$ is continuous (i.e. the graph is connected) since $1=g\left(0\right)$. For example, $f\left(2\right)=g\left(g\left(g\left(2-2\right)\right)\right)=g\left(g\left(2-1\right)\right)={\displaystyle \lim_{t\to2^{-}}}f\left(t\right)$. Therefore, there no breaks in the graph at $i$, $e^{-p}$, etc.

There is another sequence of apparent connections "in the middle", with the first being where the third arc (green) meets the original quarter-circle (indigo) at $\exp\left(ipe^{-p}\right)$. For this particular one, note that $f\left(e^{-p}\right)=g\left(e^{-p}\right)=g\left(g\left(i\right)\right)=g\left(g\left(f\left(1\right)\right)\right)=f\left(3\right)$. The others are simply the result of applying $g$ to both sides: For example, since $f\left(3-\varepsilon\right)\approx f\left(e^{-p}\right)$ and there's an intersection around $f\left(4-\varepsilon\right)=g\left(f\left(3-\varepsilon\right)\right)\approx g\left(f\left(e^{-p}\right)\right)=f\left(1+e^{-p}\right)$ as well, etc.

"the branches connect perpendicularly" True.

Now we will show that the intersections in the diagram that look perpendicular all really are. To start, let's look at two particular ones.

Intersection at $i$

On $\left[0,1\right]$, we have $f\left(t\right)=g\left(t\right)$, which draws a quarter-circle in the complex plane, and it's horizontal at $i$ ($t=1$). On $\left[1,2\right]$, we have $f\left(t\right)=g\left(g\left(t-1\right)\right)$. Taking the derivative of this and evaluating at $t=1$ (to take the limit of the derivative of $f$ as $t$ approaches $1$ from above), we get $\boxed{-ip^{2}}$, so $f$ is moving vertically for $t$ just above $1$. As vertical is perpendicular to horizontal, we do have a right angle at $i$.

Intersection at $\exp\left(ipe^{-p}\right)$

Note that $f\left(e^{-p}\right)=g\left(e^{-p}\right)=g\left(g\left(i\right)\right)=g\left(g\left(f\left(1\right)\right)\right)=f\left(3\right)$. The derivative of $g\left(g\left(g\left(t-2\right)\right)\right)$ as $t$ approaches $3$ is $p^{3}e^{-p}\sin\left(p\left(1+e^{-p}\right)\right)-i\left(p^{3}e^{-p}\cos\left(p\left(1+e^{-p}\right)\right)\right)$. Since $\cos\left(p+x\right)=-\sin x$ and $\sin\left(p+x\right)=\cos x$, this simplifies to $\boxed{p^{3}e^{-p}\left(\cos\left(pe^{-p}\right)+i\sin\left(pe^{-p}\right)\right)}$.

And the derivative of $f\left(t\right)=g\left(t\right)$ at $t=e^{-p}$ is $-p\sin\left(pe^{-p}\right)+i\left(p\cos\left(pe^{-p}\right)\right)=\boxed{p\left(-\sin\left(pe^{-p}\right)+i\cos\left(pe^{-p}\right)\right)}$.

These two complex numbers are perpendicular as vectors, so this is a right angle as well.

The other intersections

To get all of the other intersections, note that $g\left(t\right)$ has a complex derivative of $ipe^{ipt}\ne0$, so it is conformal (see, for example, this MSE question), meaning that $g$ preserves the local angles in the diagram. Because of the recursive definition of $f$, this means that the two intersections considered above propagate along the diagram. For example, since there's a perpendicular intersection around $f\left(3-\varepsilon\right)\approx f\left(e^{-p}\right)$, there's a perpendicular intersection around $f\left(4-\varepsilon\right)\approx f\left(1+e^{-p}\right)$.

"Are these shapes similar to one another?" No.

I'm not certain how to interpret this question, but everything that comes to mind has the answer "no". For example, the first part of the graph of f is a quarter circle, but the fourth part (red) is certainly not.

$z\uparrow\uparrow t$

I think these questions about $z\uparrow\uparrow t$, which were added significantly after the original posting, deserve their own separate Math(s) StackExchange question. That said, here are some observations that are a bit too long for a comment.

When $|z|=1$, $\theta_0=\pi/2$? Probably.

Suppose we started with $g\left(t\right)=\exp\left(i\theta t\right)$ instead of $g\left(t\right)=\exp\left(i\frac{\pi}{2}t\right)$? Then $f\left(3\right)=g\left(g\left(g\left(1\right)\right)\right)$ is at $\exp\left(i\theta e^{i\theta e^{i\theta}}\right)$ $=\exp\left(i\theta e^{i\theta\left(\cos\theta+i\sin\theta\right)}\right)$ $=\exp\left(i\theta e^{i\theta\cos\theta-\theta\sin\theta}\right)$ $=\exp\left(i\theta e^{-\theta\sin\theta}e^{i\theta\cos\theta}\right)$ $=\exp\left(i\theta e^{-\theta\sin\theta}\left(\cos\left(\theta\cos\theta\right)+i\sin\left(\theta\cos\theta\right)\right)\right)$ $=\exp\left(\theta e^{-\theta\sin\theta}\left(i\cos\left(\theta\cos\theta\right)-\sin\left(\theta\cos\theta\right)\right)\right)$. So in absolute value, this is $\exp\left(-\theta e^{-\theta\sin\theta}\sin\left(\theta\cos\theta\right)\right)$. To determine if $f\left(3\right)$ is in the unit disc, we need to check that $h\left(\theta\right)=-\theta e^{-\theta\sin\theta}\sin\left(\theta\cos\theta\right)\le0$. Certainly $h\left(0\right)=h\left(\frac{\pi}{2}\right)=0$, and $h$ is negative in between, so when $\theta$ is slightly more than $\frac{\pi}{2}$, $f\left(3\right)$ leaves the unit disc. The next greatest zero of $h$ is at $\frac{3\pi}{2}$ (and there is another at approximately $5.34$).

However, a similar analysis shows that $f\left(2\right)$ leaves the unit disc for $\theta\in\left(\pi,2\pi\right)$ (the corresponding $h\left(\theta\right)$ is just $-\theta\sin\theta$). Therefore, $f\left(2\right)$ and $f\left(3\right)$ are both in the unit disc (considering $\theta\in[0,2\pi)$) only for $\theta\in\left[0,\frac{\pi}{2}\right]$.

I would consider this evidence suggestive that that would be the "stable range". If you were interested in $\theta\in\left[-\pi,\pi\right]$ instead, note that both $h$'s are even functions, and so we get a conjectured "stable range" of $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, which seems to agree with numerical experiment.

Mark S.
  • 21,408
  • 2
  • 42
  • 98
  • 1
    Hello Mark, concerning your last consideration you might like to see my new answer below, having a more precise value for $\theta_0$ and conjectures the divergence/convergence behave again with reference to the Shell/Thron-analysis of that problem. – Gottfried Helms May 14 '19 at 05:54

This is not an answer to the question, but a programm to play around with.

As I stated in a comment above, I made a programm where you can test out different values for the complex tetration.

The important part of the programm is:

Private Function tetration(a As Complex, b As Double) As Complex
    If b > 1 Then
        Return Complex.Pow(a, tetration(a, b - 1))
    ElseIf b < 0 Then
        Return Complex.Log(tetration(a, b + 1)) / Complex.Log(a)
        Return Complex.Pow(a, b)
    End If
End Function

The other bits are important for the GUI and other stuff, but I will not explain those in greater detail.

You can download my programm here: https://www.mediafire.com/?w39h6ahd0ppjl53

EDIT: I fixed a bug. Now it should not chrash anymore: http://www.mediafire.com/file/vwoirrp6lmqtcpc/ComplexTetration2.exe

It chrashes if the numbers get to quick, but I haven't found a way around it yet, so there is the risk of chrashing the programm, keep that in mind.

If you want to take a look in the source code, I'll try to share it as well (but I don't recommend it, as it is more than messy).

Here are some interesting plots for your function: (red is the start, blue is the end)

$$z = 1 + 2i$$ $$t \in [0, 5]$$


$$z = 1 + 2i$$ $$t \in [0, 30]$$


This seems to orbit around, but doesn't approach a fixpoint, as far as my experiments went.

$$z = 1.5 i$$ $$t \in [0, 50]$$


This seems to approach a fixpoint.

$$z = -0.5 + 0.1 i$$ $$t \in [0, 10]$$


This looks chaotic and chrashed my programm via the huge numbers it spat out.

An experiment about the stability of $z ↑ ↑ t$

I got curious and wanted to know when your function is stable and when it violently goes everywhere.

But I didn't knew how to test this behaviour correctly.

So I just called the plot messy, if $|z ↑ ↑ t|$ exceeds some limiting value called $\text{Max}$.

Now I tested a giant array of points for $z$ and colored a pixel white if the plot was not messy, and made it colorful, depending on how long it took to become messy.

The calculation took 55 minutes, but the result is a pretty cool fractal!

Of course, this gives only an idea of how the actual picture should look - This is just an approximation and I went with a very simple condition to determine if a plot is chaotic. This might exclude some points that are actually stable and might include some points that are actually messy.

The view-window is $-5$ to $5$ on both axis.

$$t \in [0, 100]$$

$$\Delta t = 0.01$$

$$\text{Max} = 100$$

When is the funtion not messy?

I think this is quite a beautiful picture. The fractal structure makes sense, because the tetration was defined recursivly - in a way, this is iteration. Some of the structures look similar to the fractal generated by Newtons' method for $e^x + 1$.

Code: For every point the following calculations are done:

            For y1 = 0 To 1 / xStep
                x1 = y1 * xStep
                z2 = Complex.Pow(z1, x1)
                i = 1
                While i < xRangeUpper
                    z2 = Complex.Pow(z1, z2)
                    i += 1
                    If Complex.Abs(z2) > MAX Or Double.IsNaN(Complex.Abs(z2)) = True Then
                        Exit While
                    End If
                End While
                If Complex.Abs(z2) > MAX Or Double.IsNaN(Complex.Abs(z2)) = True Then
                    Exit For
                End If

            If Complex.Abs(z2) > MAX Or Double.IsNaN(Complex.Abs(z2)) = True Then
                BMP.SetPixel(x, y, myOwnColor(i)) 'messy
                BMP.SetPixel(x, y, Color.White)   'not messy
            End If

If you want to play around with all the values, then you can download it here: http://www.mediafire.com/file/bjn9pwnvfdcu9qf/TetrationFractal.exe

You can zoom with scrolling, and clicking on a point centers the view-window around that point.

G. Ünther
  • 647
  • 7
  • 16
  • What interesting graphs. Thanks! – Simply Beautiful Art May 21 '17 at 11:47
  • Thanks! I forgot to mention that you can zoom in if you scoll with the mouse-wheel, that way zooming is easier. – G. Ünther May 21 '17 at 12:00
  • Yeah, I realized that :-) Also, it tends to crash when I ask for $z=-1$ and $t\in[0,500]$ XD – Simply Beautiful Art May 21 '17 at 12:03
  • I wonder how I get around this... hmmm. If I somehow find a solution, I'll update my programm, but for now I leave it. – G. Ünther May 21 '17 at 12:16
  • Well, it gets pretty chaotic at $z=-1$, so....yeah – Simply Beautiful Art May 21 '17 at 12:18
  • I think I found it. The numbers where so big, it just returned NaN. Luckily, there is a function called Double.IsNaN() :D I'll add the new link to my answer. – G. Ünther May 21 '17 at 12:26
  • $z=1.5I$ seems to converge. When you plot it on $t\in[0,200]$ you get [this](https://i.stack.imgur.com/Bui5O.png). And for $z=-1/2+i/10$ $t\in[0,10]$ at four levels of zoom we have [this](https://i.stack.imgur.com/kPqxE.png), [this](https://i.stack.imgur.com/XWVvo.png), [this](https://i.stack.imgur.com/GnZw1.png), and [this](https://i.stack.imgur.com/fT0LT.png). Unfortunately, my images were made with Mathematica, so I can't share a standalone executable. However, I could probably make a CDF that can be run with [this huge program](https://www.wolfram.com/cdf-player/) if someone is interested. – Mark S. May 22 '17 at 12:54
  • 1
    I made a diagram that showes if the plot for $z$ is messy. However, this is an approximation. How can we define a better condition? When is the plot chaotic? – G. Ünther May 22 '17 at 18:10
  • I like the fractal, but suspect a much lower "Max" will suffice. Maybe 2 (as in the Mandelbrot set) or something a vit higher like 4 (just to be safe). @Simply as they will be interested too. – Mark S. May 22 '17 at 18:15
  • @MarkS. Very interested indeed – Simply Beautiful Art May 22 '17 at 21:41
  • 4
    The white area looks like the "Shell-Thron-region" (see wikipedia/tetration) which mark ***the bases*** *b* which have a convergent trajectory when iterating infinitely. – Gottfried Helms May 22 '17 at 22:40
  • 2
    In your third picture, surely you mean $z=1.5 i $ (the imaginary base, not the real one)? – Gottfried Helms May 22 '17 at 22:45
  • Indeed I do. Made a typo there. – G. Ünther May 23 '17 at 04:41
  • Concerning the fractal: Intutively I understand *"messy"* as shorthand for *"escaping-to-infinity"* or "*...to-unreasonably-high*". But in the left side there are plenty of points colored, which means, they are "messy" - giving the impression to me they escape to infinity (by measuring the time for arriving at some meaningful upper bound = MAX). Now using Pari/GP with arbitrary precision ability I find, that most of that points $z$ give *periodic* orbits, most oscillating between only $3 $ "accumulation points" - so I think, they should not become colored as "*messy*" . – Gottfried Helms May 23 '17 at 11:52
  • They do oscillate! But only if you take $\Delta t = 1$. The function however is defined for real numbers as well, and for the values between two integers the numbers do get unreasonably high! For example look at the basis $-1$. Iterating this with $\Delta t = 1$ just gives $-1$ over and over again. If you choose a smaller $\Delta t = 0.01$ you see that between the integer values for $t$ the function blows up to way beyond $10^{30}$. – G. Ünther May 23 '17 at 12:07
  • Ah well, this might be a source of incompatibility. However I didn't observe in your code-snippet iterations by fractal steps... you do `i+=1` and apply with this always one time the exponential with base $z_1$ ... hmm... – Gottfried Helms May 23 '17 at 12:34
  • There is. See the variable y1? There is the snippet: For y1 = 0 To 1 / xStep" and after that "x1 = y1 * xStep". This gernerates values for x1 from 0 to 1 with the step size xStep. Then I iterate. This gets me to the values more effectively. – G. Ünther May 23 '17 at 12:59

You ask:

When $|z|=1,\operatorname{arg}(z)=\theta\in(-,\pi]$, then $\lim\limits_{x\to\infty}z\uparrow\uparrow x$ tends to exist for $\theta<\theta_0$ and diverges for $\theta>\theta_0$. What is this $\theta_0$? And is my conclusion correct?

For the question of fixpoints of the iteration of $x \to z^x$ the observation is relevant that we can introduce the variables $t$ and $u=\log(t)$ writing $z=t^{1/t}$ and $z=\exp(u \cdot \exp(-u))$
For this we have a proof of D. Shell & W. Thron that

  • for $|u|<1$ the iteration converges to the fixpoint $t$,
  • for $|u|>1$ the iteration diverges, and
  • for $|u|=1$ it converges or diverges depending on the algebraic character of $\arg(u)/\pi$

$\qquad $Note 1: For the second case "divergence" can also mean: convergence to a set of periodic points (heuristically).
$\qquad $Note 2: there is at least one older discussion of that property of $|u|$ here on MSE; I can add the link when I've found it

Now, if on the other hand we have by your demand $|z|=1$ and thus $z=\exp(î \theta)$, then we have $\log(z) = u \exp(-u) = î \theta$.
There is a nice formula using the LambertW-function to find $u$ from $\log(z)$ such that we have $$ u = - \text{LambertW}(- \log(z)) \\ = - \text{LambertW}(- î \theta) $$ If we define a function $$f(\theta) =|u| = | - \text{LambertW}(- î \theta) | \qquad \qquad \text {for } 0 \le \theta \le 2 \pi $$ then we find a monoton increasing curve: image
It shows that there is only one point where $f(\theta_0)=|u|=1$ and smaller $\theta$ lead to smaller $|u|$ and thus convergence in the iteration, and larger $\theta$ lead to larger $|u|$ and thus to divergence (set-of-periodic-points) in the iteration according to the mentioned categorization of Shell/Thron.

About $\theta_0$ such that $|z|=1$ and $|u|=1$ (third case of above list)

I found the approximation for $\theta_0$ using the solver in Pari/GP
$\qquad \qquad $ th0 = solve(th=1.9,2.0, abs(-LambertW(-I*th))-1)
getting $$\theta_0 = 1.961308846... = 0.6243039957... \cdot \pi $$

Using this $\theta_0$ I show the scatterplot of some first iterates of $z_0=1$:
The blue dots are the first 1000 iterates and by the grey connection-lines we see some very chaotic behave. To have a better grip I've also marked the first four iterates with red color.
But we can do better; the "border" of the figure seems (well:seems) to be interpolatable. And in fact we get smoother multi-section trajectories if we document only each 3rd or 11 or so dot and add an interpolating line. The following image shows the iterates in a 68-fold multi-sectioned trajectory .
In the picture I have colored the first 5 sections of the full trajectory up to some 200 steps for each section and with their dots connected. They give nicer lines, seemingly closing the figure, and are partially (nearly) overlaid.
This can be done even more dense - in fact even arbitrarily dense!

This can be done if we use values from the convergents of the continued fraction of $\arg(u)/\pi$. Then at multi-sections accordingly with $s=68$, $s=797$, $s=5715$ ,$s=2 \cdot 15517$, $s=...$ stepwidths we get increasingly dense sectional trajectories (while of course they are always sets of discrete/disconnected complex points).

That trajectories seem to not-to-contract, so we don't get convergence.
This is of course conjectured by visual inspection, but can be backed by the heuristic, that increasing values from the convergents of the continued fraction provides smaller distances between the steps in a sectioned trajectory. And this must then mean that we arrive arbitrarily near at $z_{-1}=0$ and then arbitrarily "near" at $z_{-2} = \log_z(0)$; so we shall always have extremely far-away dots in the picture.

The latter consideration is only based on observation and conclusion - it would be good if one could make a proof for this....

Gottfried Helms
  • 32,738
  • 3
  • 60
  • 134
  • A related newer question/answer: https://math.stackexchange.com/questions/3323851/proof-or-hints-towards-proof-for-asymptotic-shape-of-orbit-0-to-1-to-b-to – Gottfried Helms Aug 21 '19 at 01:29