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The set of real numbers is closed under multiplication, but not under exponentiation (Eg. square root of negative numbers). That is, $\exists a, b \in R \mid {a^b} \notin R$. Then we introduced complex numbers and it is closed under exponentiation.

Now that I came across tetration in Wikipedia, is the set of complex numbers closed under tetration? or can tetration between any two complex numbers does not exist in C? That is, $\exists a, b \in C \mid {^ba} \notin C$ is true for any $a,b$?

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    $\begingroup$ Can you elaborate on why you are not satisfied with the information provided by the Wikipedia page you linked? In my view, Wikipedia seems to answer your question, so explaining why it doesn't would help us clear things up for you. $\endgroup$
    – Mark S.
    Mar 28, 2022 at 11:36
  • $\begingroup$ @MarkS. see the changes. Also, I wasn't able to find closure property of tetration in Wikipedia. $\endgroup$
    – Sourav Kannantha B
    Mar 28, 2022 at 13:12
  • $\begingroup$ You are mistaken. The complex numbers are not closed under exponentiation. Exponentiation is not even a binary operation on the complex numbers. There are no axiom that even define what such an operation is supposed to be. Sure, $z^n$ is well defined for all $z\in\mathbb{C}$ and $n\in\mathbb{N},$ but that only gives you a function $\mathbb{N}\times\mathbb{C}\to\mathbb{C},$ not a function $\mathbb{C}^2\to\mathbb{C}.$ There is no meaningful way to the extend the former to the latter. Tetration is a function $\mathbb{N}^2\to\mathbb{N}.$ $\endgroup$
    – Angel
    Mar 28, 2022 at 14:21
  • $\begingroup$ Perhaps a better question is to ask whether it is possible to define a binary function $u:\mathbb{C}^2\to\mathbb{C}$ such that $u(\alpha,\beta+1)=\exp[\alpha\cdot{u}(\alpha,\beta)]$ for $\alpha,\beta\in\mathbb{C},$ and this is, in a very loose sense, close to the notion of tetration, but it is by no means a natural extension of the concept. I would call it a related concept, rather than an extension. $\endgroup$
    – Angel
    Mar 28, 2022 at 14:25
  • $\begingroup$ @Angel There is extension of exponentiation in C2→C. It is multivalued, but a principal value is defined. And so is logarithm. See this Wikipedia section. $\endgroup$
    – Sourav Kannantha B
    Mar 28, 2022 at 19:31

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