For $k$ , $n\in \Bbb Z$, with some abuse of notation $ 1^{1/k}={e^{-j{(2\pi n /k)}}}.$
What is the set of points which are roots of unity of degree $\pi$? (With the same abuse of notation $1^{1/\pi}$).
(1) If one would continue just like before, they may end up with ${e^{-j{(2n)}}}$, $n\in \Bbb Z$. Is this the complete set of solutions?
(2) $\{z\,|\,z={e^{-j{(2n)}}}, n\in \Bbb Z \}$ and $\{z\,|\, \operatorname{abs}(z)=1 \}\}$ are two different sets, right? (As one is countably infinite other is not.) Then it would be incorrect to say that $\{z\,|\,\operatorname{abs}(z)=1\}$ is the complete set of points which are roots of unity of degree $\pi$.
Thanks, I hope it was not a pointless question and (as far as I checked) not a duplicate.
Yes, this is the right conceptual generalization. The set you are looking for is the first set that you indicated, and you're correct that it is not the same thing as $\{x:|x|=1\}$. That set is the entire unit circle, and includes all the roots of unity of any degree. However, your set is actually dense in the unit circle.
A notional comment though: using $j^2=-1$ is extremely unusual in mathematics. It's sometimes used in physics or engineering, but virtually never in mathematics.