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(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?

StarrryEyedIdiot
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    As for the question in your title, see the answer given by Yves Daoust. $$x^x=i$$

    As for the stuff in the rest of the question, you don't need to extend the number system to solve $x^x=i$. However, there are some interesting extensions ... you may want to read about quaternions and octonions. Have fun!

     – Zubin Mukerjee Apr 24 '20 at 19:57
  • You should note also that the complex logarithm is multi-valued, because the complex exponential is not straightforwardly invertible. You will need an arbitrary branch cut. – Zubin Mukerjee Apr 24 '20 at 19:59
  • You should not change question after it was answered. – user Apr 24 '20 at 20:21
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    If you are using ^ and ^^ for exponentiation and tetration in the usual way, then (x^^2)^(x^^2)=$(x^x)^{(x^x)}=x^{x^{x+1}}\not=$x^^4. – Barry Cipra Apr 24 '20 at 20:22
  • I’m voting to close this question because it needs more details. A proper definition of $x~\widehat~\widehatx$ is not given, the question has been altered, and the post is somewhat unclear. – Simply Beautiful Art Apr 26 '20 at 13:08

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

and

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