Complex Tetration Does anyone know how to calculate complex tetrations?
There are formulas for tetrations but I always end up in a complex tetration:
$^{n} a = a^{(^{n-1} a)}$
Now, if we use this formula for complex numbers, we get:
$^{i} 2 = 2^{(^{i-1} 2)}$
Then we must apply the formula again:
$2^{(^{i-1} 2)} = 2^{(2^{(^{i-2} 2)})}$
And it will go on infinitely since we cannot subtract or add a complex number with a real number...
So, does anyone know a method for this?
Thanks for answers.
 A: (Disclaimer: This answer is a (slightly edited) full copy from a question in mathoverflow and might serve as a general hint how to understand imaginary iteration heights in tetration)
I'm discussing this from the view of iterated exponentiation (although the technical process should be the same with other functions as well).
If you can use the Schroeder-function for the continuous iteration, then the iteration-height-parameter (say "h") goes into the exponent of some basis (the log of the fixpoint, often denoted as $ \small \lambda$ ). Imaginary heights h then switch the value of the Schroeder-function to the negative; this allows then to extend the iteration, in some sense, "beyond infinite height".

For instance, use base $ \small b = \sqrt 2 $ for iterated exponentiation, $ \small z_0=x, z_1=b^x , z_2=b^{b^x}, \ldots $. Then

*

*if you begin at, say, $ \small z_0=x=1$ you can iterate to infinite height to approach the limit at $ \small z_\infty = 2^{\small ^-}$ .

*if you start at $ \small z_0=x=3$ you can approach $ \small z_\infty = 2^{\small ^+}$ or even $ \small z_{-\infty}=4^{\small ^-}$ .

*if you start at $ \small z_0=x=5$ you can approach $ \small z_{-\infty} = 4^{\small ^+}$ or even $ \small z_{\infty}=  \infty$ .

$ \implies $ You cannot iterate from a value $ \small  z_m<2 $ to a value $ \small  2 < z_w < 4 $ using real heights, even when infinite.
But if you use the imaginary unit height you iterate directly from $ \small z_m=1$ to something like $  \small z_{m+i}=2.4 $:

*

*Assume again $ \small z_0=1$. Then the value of the Schröder-function (which is assumed to be normed to have the powerseries
$ \small \sigma(x)= 1x+\sum_{k>1} a_k x^k $ ) is about $ \small s=-0.316049330525 $.

*Then with height say $h=1$ gives $ \small \sigma°^{-1}( \lambda^1 s)\cdot 2 +2=b=\sqrt 2$ because that is the iteration of height 1 (in the exponent of $ \small \lambda$ ). (Remark: the "circle"-super-postfix $\sigma°^{-1}$ means       the functional inverse, not the reciprocal)

*If we replace that exponent by
$  \small h_w = i \cdot {\pi \over  \ln \lambda } $ then we get $  \small \sigma°^{-1}( \lambda^{h_w} s) \cdot 2 +2=2.46791405022...$ which is thus, in some sense, "beyond infinity" with respect to the iteration height.


late update I add a picture to illustrate the previous statements.
This is picture, where I studied the application of imaginary heights, using the base for exponentiation $b=\sqrt2$. It has the attracting real fixpoint $t=2$ and $\lambda \approx 0.69...$.
As an example, look at the left side, with $z_0=1 + 0\cdot î$. Using iteration with real heights (here in steps of $1/10$ ) we move rightwards to $z_1=b^{z_0}=b = 1.414...$ and by more iterations more towards the fixpoint $t =2+ 0 î$. This is indicated by the orange arrows.
Note that because $t=2$ is a fixpoint, we cannot arrive at points on the real axis more to the right hand!
But using imaginary heights, iterations move from $z_0 $ to $z_h$ on the indicated circular curve (computed data for the big dots and the other markers are in steps of $0.1 { \pi \over \ln \lambda} î$ see legend), which is indicated by the blue arrow.
This iteration does not go towards the fixpoint, but repeats to cycle around it. On that cycling the trajectory crosses the real axis beyond the fixpoint.
(Legend-1: the circular curves which connect the computed iteration-values of imaginary heights are Excel-cubic-splines based on $200$ exactly computed iteration values per "circle" and thus are now good approximations of the true continuous iterations; update 2
Legend-2: I've used $\beta$ in the picture instead of $\lambda$ because in Excel the $\beta$ is on the keyboard but $\lambda$ must be made with some difficulties)

