What is the result of ${^{i}i}$?

Question

$i+i=2i$, $i\cdot i=-1$, $i^i=e^{-\pi/2}$, but what is the result of ${^{i}i}$? That is, what is the $i$-th tetration of $i$?

Answer 1

In order to answer your question we first need to extend tetration to allow complex bases and heights. To the best of my knowledge there is no currently accepted way to do this, so your question is not really answerable.

Keep in mind that priori there is no reason for a function $\mathbb{N}\rightarrow\mathbb{N}$ to have a "natural" extension to a larger domain like $\mathbb{Q},\mathbb{R}$, or $\mathbb{C}$. Of course many functions do have this property (addition, multiplication, exponentiation with some subtleties, the factorial, ...), but this sort of generalizability should not be assumed at the outset.

There has been some work on extending tetration to complex bases and heights, although like I said above I don't believe that any of it reaches expressions like ${}^ii$. See e.g. here or here.

Answer 2

At this time, we don't even have a definition of tetration with rational heights, let alone complex ones. (If we did, we might try to treat real heights by limits, then complex heights by some as yet unimaginable generalization.) Why not?

I can define $x^{3/2}$ as the solution of $y^2=x^3$, because the associativity of multiplication ensures this $y$ also satisfies $y^{2n}=x^{3n}$, as we'd expect of $x^{(3n)/(2n)}$. But the same trick doesn't work with ${}^{3/2}x$, since exponentiation doesn't associate (or commute, for that matter).

See also here.