What would be the solution to x^^x= i? (imaginary unit)

Question

(I don’t know proper math format sorry) I was looking at reverse operators. Let’s start with addition (reverse is subtraction) 1-2 is -1: so we extend our number field. Division next. We divide 3 by 2 and get 1.5 . Now we have rational numbers. Take exponents. Take the square root of a negative, and we invent imaginaries. Now, we know that i has a valid square root, 1/root 2 + i/root 2. However, what if we take the super-root of i? Now, I’m pretty sure that the ONLY way to extend the field is to set up the equation x tetrated to x is i. (quick idea, with my experiments (x^^a)^(x^^b)=x^^(a+b)) Would this result in an extension or would it be complex? Furthermore, if this exists as j, could we solve x pentated to x is “j”? Would this work forever? Is it bounded for how many units you can have? Would they be algebraic fields?

Answer 1

$$x^x=i\iff x\log x=\log x\,e^{\log x}=\log i$$

and

$$\log x=W(\log i)$$ where $W$ denotes the Lambert function. $\mathbb C$ is sufficient.