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Let $f:\mathbb C\to \mathbb C$ be defined by $f(z)=e^z-1$. Let $f^n$ denote the $n$-fold composition of $f$.

In my new paper Erdős space in Julia sets I show that $$Z:=\{z\in \mathbb C:\lvert\operatorname{Im}(f^n(z))\rvert\to\infty\}$$ contains a homeomorphic copy of the set of points in Hilbert space $\ell^2$ which have all rational coordinates. But I do not know of a specific complex number $z$ which belongs to this set. It is easy to find $z$'s for which the real part goes to infinity; $\operatorname{Re}(f^n(1+0i))\to\infty$ as $n\to\infty$, but the imaginary part of $f^n(1+0i)$ is always $0$. So the question is, can you give the precise coordinates of a point in the complex plane which belongs to $Z$?

How about an answer to this question for $f(z)=e^z$?

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    $\begingroup$ Would you be amenable to linking to the arXiv rather than to ResearchGate? (You've done so; thanks!) I always prefer interacting with the former. Anyway, what does it mean to say that a subset of $\mathbb C$ contains $\ell^2(\mathbb Q)$? $\endgroup$
    – LSpice
    Commented May 8, 2020 at 17:14
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    $\begingroup$ @LSpice Ok I changed the link to arXiv. I mean that $\ell^2(\mathbb Q)$ is homeomorphic to a subset of $Z$. $\endgroup$
    – D.S. Lipham
    Commented May 8, 2020 at 17:17

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It depends on what you mean by precise coordinates. I am not sure that I would expect to find a number that has a specific closed form. But then, I do not know how to find a "precise" point where the real part tends to infinity under iteration either, had you not chosen a real parameter.

To find such a point up to arbitrary precision, on the other hand, is quite easy. For instance, let $N\geq 0$, define $$ z_{N,N} := \log N + 2\pi i N. $$ and inductively let $z_{N,j}$ (for $j<N$) be the preimage of $z_{N,j+1}$ in the strip at imaginary parts between $(2j-1)\pi$ and $(2j+1)\pi$.

Then $z_{N,0}$ converges to a point $z_0$ such that $f^j(z_0)$ has imaginary parts between $(2j-1)\pi$ and $(2j+1)\pi$, and in particular the imaginary parts converge to infinity.

Here is some simple python code:

>>> n=100
>>> orbit = [0]*(n+1)
>>> orbit[n] = math.log(n) + 2*math.pi*1j*n
>>> for j in range(n):
          orbit[n-j-1] = cmath.log(orbit[n-j]+1) + (n-j-1)*2*math.pi*1j

I get approximately $z_0 = 2.1302059107690132+1.1190548923421213j$. Of course, given the very strong expansion and instability, if you iterate forward this will only follow the desired orbit for a small number of iterations (for me, it stays close for 10 iterations), but Gottfried Helms has given a higher-precision estimate below.

EDIT. My notation was somewhat messed up above in the original post; it should be better now I hope. Note that the $\log N$ can be omitted in the definition of $z_{N,N}$, and you will converge to the same value.

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    $\begingroup$ I've tried my procedures adapted to the function $f:z \to \exp(z)-1$ and approximated your given value of $z_0$ further. Using z0= 2.130205910769013348261983648 975380557561596052319 0123657053290821840652119357270 + 1.11905489234212128966 4680770516034645067307 226057015401033623172 3557445701337388*I one can iterate 24 times, and then it gets periodic (of course with increasing error, which can be reduced when internal precision for computations is for instance 800 dec digits). Your value agrees convincingly to the result given a $K$-vector in my notation as $K=[11,10,...1,0]$ $\endgroup$
    – Gottfried Helms
    Commented Aug 13, 2020 at 8:14
  • $\begingroup$ Hi Lasse, I'm still interested in this approximation. Just wanted to see if there are any typos above because I don't quite understand the z's. $\endgroup$
    – D.S. Lipham
    Commented Feb 21, 2021 at 19:21
  • $\begingroup$ Also does this technique allow you to specify the external address beforehand? $\endgroup$
    – D.S. Lipham
    Commented Feb 21, 2021 at 19:30
  • $\begingroup$ @D.S.Lipham Yes, something was definitely wrong! Hopefully it makes more sense now. Yes, you can specify the external address, but you have to be a little bit careful about where you place the real part, and whether you want to get a point on the ray or the endpoint. Essentially you can use my model from my paper "Topological dynamics of exponential maps on their escaping sets" to tell where the n-th image of your desired point should roughly be, then pull back under the actual exponential map under consideration. If that makes any sense. $\endgroup$
    – Lasse Rempe
    Commented Feb 25, 2021 at 0:12
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I have one idea, but am not sure whether this is really leading somewhere respective your question.

This is derived from a recent experimental work of mine, where I discuss a method to create $n$-periodic points for the iterated function $f: z \to \exp(z)$.

Giving a vector $K$ of $n$ branchindexes for the (iterated) logarithm as parameter, the implemented function gives back an initial value $z_0$ which can be iterated $n$ times with function $f$ and whose imaginary part increases (roughly by $2\pi î$) over that iterations.

For instance calling z0 = find([6,5,4,3,2,1,0]) find a value $z_0$ which is $7$-periodic. This does of course not solve your problem of an imaginary part growing to infinity, but might give an idea... :

The vector of branchindexes can be made arbitrarily long, for instance I've tried with length n=128, K=[127,126,...,0] and found the following result z0=find(K):

   2.090728841+  1.235766664*I   <-- z0 (only 10 signif. digits displayed)
   2.660235518+  7.640965408*I      ... list of iterates ....
   3.023074791+ 13.97645147*I              ...
   3.289206714+ 20.28951149*I              ...
   3.499351193+ 26.59232200*I
   3.672981867+ 32.88951626*I
   3.820915018+ 39.18326906*I
   3.949779448+ 45.47474630*I
   4.063931345+ 51.76463340*I
   4.166386797+ 58.05336160*I
   4.259320451+ 64.34121681*I
     ... 
   6.676175218+786.9605354*I
   6.682733533+793.2510338*I
   0.8873292332+798.4983613*I
   2.090728841+  1.235766664*I    (getting periodic after 128 iterations)

Comparing the results $z_{0,n}$ where $n$ is the length of the vector $K$ of branchindexes with $n \in \{16,32,64,128 \}$

display reduced to 60 dec digits, internal precision 800 dec digits
-----------------------------------------------------------------------
length(K)       real part of z0
 16           2.09072884145065670358930701024074056461462449074482469887391
 32           2.09072884145065670358930571871821763780774559404297853095181
 64           2.09072884145065670358930571871821763780774559404297853095179 
128           2.09072884145065670358930571871821763780774559404297853095179

length(K)      imag part of z0
 16         + 1.23576666409482263688534788841502804976771359539736086858255*I
 32         + 1.23576666409482263688534750071480578600396895244789044041851*I
 64         + 1.23576666409482263688534750071480578600396895244789044041849*I
128         + 1.23576666409482263688534750071480578600396895244789044041849*I

... we get the impression, that there is some asymptotic for $z_0$ towards $n \to \infty$ which we can approximate arbitrarily near. (Note however, that for $n=128$ I used $800$ dec digits internal precision). (A visual example is in the paragraph about "aperiodic points" in the manuscript linked to below)

(If this seem to be useful for you at all, then a small manuscript at my webspace is more informative, where I explain this procedure a bit more. For readability I've introduced in the answer here the name find() for the main-function.
This is all written in Pari/GP and if you need this I can provide the scripts)

update By configuring the vector $K$ more exessive, for instance K=[2^31,2^30,2^29,...,2^0] we get initial values $z_0$ whose iterations gives exponentially growing imaginary values (until periodicity occurs).
Example:

     2.665280329+7.622847729*I    <-- z0 
     3.292353128+13.98978765*I     ... iterates over f ...
     3.951062797+26.61442216*I           ...
     4.627032158+51.78428650*I
     5.311751681+102.0721534*I
     6.000773471+202.6161507*I
     6.691894308+403.6854928*I
     7.384041486+805.8134993*I
     8.076693109+1610.063510*I
     8.769593941+3218.560203*I
     9.462618394+6435.551762*I
     10.15570435+12869.53388*I
     10.84882095+25737.49759*I
     11.54195286+51473.42471*I
     12.23509241+102945.2788*I
     12.92823577+205888.9869*I
     13.62138104+411776.4031*I
     14.31452727+823551.2354*I
     15.00767397+1647100.900*I
     15.70082092+3294200.229*I
     16.39396798+6588398.887*I
     17.08711510+13176796.20*I
     17.78026225+26353590.84*I
     18.47340941+52707180.10*I
     19.16655659+105414358.6*I
     19.85970376+210828715.7*I
     20.55285094+421657429.8*I
     21.24599812+843314858.1*I
     21.93914530+1686629715.*I
     22.63229248+3373259428.*I
     23.32543966+6746518854.*I
  2.088818122+1.349303771E10*I     ... after this periodicity occurs ...

The initial value $z_0$ to $60$ dec digits is

z0 = 2.66528032862300130094954352169380883320313130819912077261863
   + 7.62284772864970968721488615058188954049634904915456329976660*I

It is perhaps of interest, that the values in the list above correspond to the $1$-periodic fixpoints, which can of course be adressed by the Lambert-W-function with its complex branches. See the following list

 2.665280329+7.622847729*I   -W(-1,-(2^1+1))=     2.653191974+13.94920833*I
 3.292353128+13.98978765*I   -W(-1,-(2^2+1))=     3.287768612+26.58047150*I
 3.951062797+26.61442216*I   -W(-1,-(2^3+1))=     3.949522742+51.76012200*I
    ....                         ...                   ...
 18.47340941+52707180.10*I  -W(-1,-(2^24+1))=     18.47340941+105414358.6*I
 19.16655659+105414358.6*I  -W(-1,-(2^25+1))=     19.16655659+210828715.7*I
 19.85970376+210828715.7*I  -W(-1,-(2^26+1))=     19.85970376+421657429.8*I
 20.55285094+421657429.8*I  -W(-1,-(2^27+1))=     20.55285094+843314858.1*I
 21.24599812+843314858.1*I  -W(-1,-(2^28+1))=     21.24599812+1686629715.*I
 21.93914530+1686629715.*I  -W(-1,-(2^29+1))=     21.93914530+3373259428.*I
 22.63229248+3373259428.*I  -W(-1,-(2^30+1))=     22.63229248+6746518854.*I
 23.32543966+6746518854.*I  -W(-1,-(2^31+1))=  23.32543966+1.349303771E10*I
2.088818122+1.349303771E10*I  -W(-1,-(2^32+1))=  24.01858684+2.698607541E10*I

It looks like when two consecutive W-values are taken as the edge-points of the rectangle in the complex plane, and they mark the anti diagonal, then the values of the list mark the top left of the main-diagonal, and approach the true main-diagonal edges more and more when the index of the entry in the list increases. Perhaps this gives a description for an asymptotic behaviour of the resulting coordinate if the index (and the complex component) approaches infinity.

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    $\begingroup$ The structure of periodic points of the exponential map (and cases such as $e^z-1$ is well-understood, this goes back to Devaney & Krych. There you can also find examples of points whose imaginary parts tend to infinity. I have posted an answer at math.stackexchange.com/a/3789416/16629 . $\endgroup$
    – Lasse Rempe
    Commented Aug 13, 2020 at 9:47

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