Are there algebras over real numbers (with exponentiation), such that there is such $z$ that does not include components in $\mathbb{C}$ (or in a subset isomorphic to $\mathbb{C}$), for which $(-1)^z\in \mathbb{R}$ and irrational?
What about such $z$ that $z^z\in\mathbb{R}$ and irrational?


I mean, we can rise $(-1)$ to an imaginary power and have an irrational real number. As I know, neither in split-complex numbers, nor tessarines, nor in dual numbers, nor in quaternions you have elements with this property except those that have components in subrings, isomorphic to $\mathbb{C}$.


Are there such examples in other extensions of $\mathbb{R}$, matrices, vectors, tensors, power series, whatever?


Maybe my question is not perfectly formulated and one would be able to come up with some trivial examples, but I hope you understood the idea.



Are there algebras over reals besides complex numbers, where identities, analoguous to $(-1)^i=e^{-\pi}$ and $i^i=e^{-\pi/2}$ hold?
I will address the question in the case $f(z) = (-1)^z = \exp(z \log(-1))$ and $A$ is a finite-dimensional commutative (associative unital) $\mathbb{R}$-algebra, where $\log(-1)$ is a suitable choice of the logarithm in $A$.


Wlog. $A$ is a local $\mathbb{R}$-algebra $(A,\mathfrak{m})$. There are two possibilities. First possibility is $A / \mathfrak{m} \cong \mathbb{C}$, in which case $A$ has a $\mathbb{C}$-structure extending the $\mathbb{R}$-structure, and the second possibility is $A / \mathfrak{m} \cong \mathbb{R}$. Either way, $A \cong \mathbb{K} \oplus \mathfrak{m}$ as vector spaces (**not** as algebras though!), where $\mathbb{K} \in \{\mathbb{R},\mathbb{C}\}$.


*1. Claim:* $\log(-1)$ exists iff $A$ has a complex structure.


*Proof:* Write $w = r + X \in A$ for $r\in \mathbb{K}$ and $X \in \mathfrak{m}$. We need $\exp(w) = -1$, i.e. $\exp(X) = -\exp(-r) \in \mathbb{K}^\times$. However
$$
\exp(X) = 1 + \sum\_{k=1}^n \frac{X^k}{k!} \in 1 + \mathfrak{m} \subset A^\times,
$$
where $n \in \mathbb{N}$ is the smallest integer such that $X^n = 0$, as all elements of $\mathfrak{m}$ are nilpotent. Hence, in particular
$$
-\exp(-r) = 1,
$$
which is possible iff $r \in \mathbb{C}$. Moreover, we have $X = 0$ for the following reason:
$$
0 = \sum\_{k=1}^n \frac{X^k}{k!} = X \left( 1 + \sum\_{k=2}^n \frac{X^{k-1}}{k!} \right),
$$
but the expression in the brackets is a unit because $\sum\_{k=2}^n \frac{X^{k-1}}{k!} \in \mathfrak{m}$. This argument is important for our second claim.


Thus the function $(-1)^z$ makes sense only if $A$ has a complex structure.


*2. Claim:* there is no $z \in A$ without complex components such that $(-1)^z \in \mathbb{R}$


*Proof:* Fix a value/branch $\lambda:= \log(-1) \in \mathbb{C}^\times$. Since $z$ has no complex components, we have $z = X \in \mathfrak{m}$, i.e. $z$ is nilpotent, and $(-1)^z = \exp(\lambda X) =: \exp(Y)$ for $Y:= \lambda X \in \mathfrak{m}$. Same argument as above shows that $\exp(Y) \in \mathbb{R}$ if and only if $Y = 0$.


Thus there aren't any non-trivial $z \in A$ such that $(-1)^z \in \mathbb{R}$.


I suspect one could show something similar in the case of infinite-dimensional commutative (associative unital) Banach algebras.


I don't have an answer for non-commutative algebras, finite-dimensional or not.


