Can $e^{k \cdot \pi}$ be algebraic for any real algebraic number $k \neq 0$?

Question

Specifically, I am trying to solve for t in the following system of constraints :

1. $e^{(a + \iota b) t} = r$, and
2. $e^{(a - \iota b) t} = s$, and
3. $t > 0$

where $a$ and $b$ are known to be real algebraic numbers, and $r$ and $s$ are known to be rational numbers.

One can conclude (by dividing the LHS and RHS of 1. and 2., and because $r\cdot s > 0$). that $r = s$, and that $2bt = 2n\pi$ for some integer $n$. This means that $e^{\frac{a}{b}n\pi} = r$.

I wish to characterize the cases when this can happen.

EDIT : I see that $a = 0$ (and hence $r = 1$) is one possibility (in which case there are many possible values of $t$, one for each $n \in \mathbb{Z}$). Are there other possibilities ?

Answer 1

$e^{\pi}$ is a trascendental number by the Gelfond-Schneider theorem, and $e^{k\pi}=(e^{i\pi})^{-ik}$ is a trascendental number for any algebraic $k\neq 0$.