Can we define $z^{\frac{m}{n}}$, where $z\in\mathbb{C}$ and $m,n\in\mathbb{Z}$? I was thinking that it might has to be $m$ and $n$ coprimes, but I don't have a consolidated idea of how I can prove it. Incidentally, how could I prove that it doesn't work for any integers? (is there any counterexample? I was thinking about $z^{\frac{1}{2}}=\pm z$).
So, my first question is, Can we define $z^{\frac{m}{n}}$, where $z\in\mathbb{C}$ and $m,n\in\mathbb{Z}$?
After that, if the answer is "no", can we say something using that fact that i said previously?
PS: I need to prove that statement without using exponential definition of complex numbers. So, what I need to use is:
Find a $z$ that satisfies $z^n=z_0$ with:
$$z=\sqrt[n]{|z_0|}\left(\cos\left(\frac{\theta_0+2k\pi}{n}\right)+i\sin\left(\frac{\theta_0+2k\pi}{n}\right)\right),\text{ for all }k\in\mathbb{Z}.$$
 A: You first need to write $\frac mn$ such that it is reduced and $n$ is a positive integer. Then write $z^{m/n}=(z^m)^{1/n}$ and apply the given formula with $z_0=z^m$. You get $n$ solutions as long as $z\ne0$ or $m\ge0$.
A: *

*Regarding the necessity of the coprime condition, consider that
$$\{-1, e^{\pm\frac\pi3}\}=(-1)^{\frac13}=\left(i^6\right)^{\frac13}\neq
i^{\frac63}=i^2=-1.$$


*Let $z\in\mathbb C\setminus\{0\}, \,m\in\mathbb Z, \,n\in\mathbb
Z^+$ such that $\gcd(m,n)=1.$ Then $$z^{\frac mn}:=\exp\left(\frac
mn\log(z)\right)\\
=\exp\left(\frac mn\left[\ln|z|+i\arg(z)\right]\right)\\
=\exp\left({\frac mn\ln|z|}\right)\exp\left(i\frac{m\arg(z)}n\right)\\
=|z|^{\frac mn}\exp\left(i\frac{m\mathrm{Arg}(z)+2k\pi}n\right)\\
=\sqrt[n]{|z|^m}\exp\left(i\frac{\mathrm{Arg}(z^m)+2k\pi}n\right)\\
=\sqrt[n]{|z^m|}\exp\left(i\frac{\mathrm{arg}(z^m)}n\right)\\
=\left(z^m\right)^{\frac1n}.$$ In other words, under the given conditions, it is valid to handle $z^{\frac mn}$ by rewriting it as $\left(z^m\right)^{\frac1n}$, then applying De Moivre’s theorem within the parenthesis, then finally taking the $n$th root as described in your final paragraph.
P.S. The coprime condition ensures that $$\frac{m\arg(z)}n\mod 2\pi$$ does not return extraneous values.
P.P.S. For $z\in\mathbb C\setminus\{0\}, \,m\in\mathbb Z, \,n\in\mathbb Z^+,$ $$\quad z^{\frac mn}=\left(z^{\frac1n}\right)^m.$$ In this case, the coprime condition is unnecessary.
