Given a complex number $e^{ix}$, the nth root can be computed using Euler's formula:
$$e^{i(x + 2k\pi)/n} = \cos((x+2k\pi)/n)+i \sin((x+2k \pi)/n).$$
If $n$ is an irrational number, can $k$ vary from negative infinity to infinity without repeating any solutions?
If so are ALL the points on the complex circle part of the solution?
This might be a really easy question to answer, but I'm a bit stuck. Also I made it up.