If a rational exponent on a complex number $z^q$ is the representation of a finite number of roots, then if the exponent is irrational this mean that there are infinite countable roots?
If this is the case... the cardinality of the number of roots is the same for any irrational exponent? Thanks in advance.
NOTE: there are different questions about irrational exponents but no one answer what Im searching so please dont mark this as repeated.
Let's take a simple example: $\sqrt[\Large\pi]1=\Big(e^{2k\pi~i}\Big)^\frac1{\Large\pi}=e^{2ki}=\cos\big(2k\big)+i\sin\big(2k\big)$, for all $k\in\mathbb Z$.