Roots of Unity problem Let m and n be positive integers have that have no common factor. Prove that the set of numbers $(z^\frac{1}{n})^m$ is the same as the set of numbers $(z^m)^\frac{1}{n}$.We denote this common set of numbers by $z^\frac{m}{n}$. Show that $$z^\frac{m}{n} = \sqrt[n]{|z|^m}\left(\cos\left(\frac{m}{n}(\theta +2k\pi)\right)+i\sin\left(\frac{m}{n}(\theta +2k\pi)\right)\right)$$, $k = 0,1,....,n-1$
So I began to solve this by $(z^\frac{1}{n})^m$ is the same as $(z^m)^\frac{1}{n}$, and they equal each other. By turning into polar form and using the rule of exponents to get the other one.
where I get stuck is solving for this $z^\frac{m}{n} = \sqrt[n]{|z|^m}$ ($\cos(\frac{m}{n}(\theta +2k\pi))+i\sin(\frac{m}{n}(\theta +2k\pi))$)
so I started to solve this problem by
Let  $w = (z^m)$ and $w^n = (z^n)$ by using $z = re^{i\theta}$.
I have $r_0^{n} e^{in\beta} = r^me^{im\theta}$ , then I get $r_0 = r^{\frac{m}{n}}$ and the angle is $n\beta = m\theta + 2k\pi$ and k $\in \mathbb{Z}$ therefore $\beta = \frac{m\theta +2k\pi}{n}$. Now this is where I go in deep confusing, when the question asks you,"Let m and n be positive integers have that have no common factor" and the answer key has transformed this $\beta = \frac{m\theta +2k\pi}{n}$ into $\beta = \frac{m\theta +2mk\pi}{n}$ , is it because $\frac{2m\pi}{n}$ doesn't change the root of unity because $m$ and $n$ are relatively prime. If they had common factor the roots will be repeating over and over again.
my question is how did the answer key "transformed" $\beta = \frac{m\theta +2k\pi}{n}$ into $\beta = \frac{m\theta +2mk\pi}{n}$? After this I believe I can finish it off.
 A: This result is valid more generally over arbitrary commutative fields. Consider such a field $K$, two relatively prime nonzero natural numbers $m, n \in \mathbb{N}^{*}$ and an element $a \in K$ which has $n$ radicals of $n$-th order (in other words such that the polynomial $X^n-a \in K[X]$ decomposes completely over $K$ with simple roots). For arbitrary $r \in \mathbb{N}^{*}$ and $x \in K$ let us write $R_r(x)\colon=\{t \in K|\ t^r=x\}$ for the set of $r$-th order radicals of $x$. It is then the case that:
$$R_n(a^m)=\{x^m\}_{x \in R_n(a)}.$$
We begin by remarking that if $a=0_K$ things are trivially clear, so we continue under the hypothesis $a \neq 0_K$. Let us denote the set on the right-hand side of the above relation by $T$. It is clear that $T \subseteq R_n(a^m)$. As to the converse inclusion, remark that in general $R_r(x)$ is precisely the set of roots of the $r$-degree polynomial $X^r-x$ and since fields are in particular integral domains it follows that $|R_r(x)| \leqslant r$. Also notice that the map:
$$\begin{align*}
R_n(a) &\to T\\
x &\mapsto x^m 
\end{align*}$$
is by definition a surjection. It will suffice to prove that it is also injective, for then we can infer that $|T|=|R_n(a)|=n \leqslant |R_n(a^m)| \leqslant n$, which signifies that $R_n(a^m)$ also has precisely $n$ elements and must thus be equal to $T$ (since any proper subset of a finite set has cardinality smaller than the ambient set).
Consider thus two $n$-th roots of $a$ such that $x^m=y^m$. Since $a \neq 0_K$ and $n \neq 0$ it is obvious that $0_K \notin R_n(a)$ and from $x^n=y^n=a$ we may therefore infer that $\left(\frac{x}{y}\right)^n=1_K$ together with the analogous relation $\left(\frac{x}{y}\right)^m=1_K$. In the abelian multiplicative group $K^{\times}$ the element $u\colon=\frac{x}{y}$ thus admits both numbers $m, n$ in its annihilator. Since the annihilator is an ideal of the ring $\mathbb{Z}$, it will also contain the sum $m\mathbb{Z}+n\mathbb{Z}=\mathbb{Z}$ (since $m$ and $n$ are relatively prime, $1$ is a linear combination of the two). This means that in particular $1$ is in the annihilator of $u$, which explicitly means that $u^1=u=1_K$ and entails $x=y$. The map in question is thus proved to be injective.

In the specific case of your problem, $\mathbb{C}$ is algebraically closed and of characteristic $0$, which means that any nonzero element $z \in \mathbb{C}^{\times}$ has precisely $n$ radicals of order $n$ for any $n \in \mathbb{N}^*$ (because the binomial $X^n-z$ is separable, being coprime with its derivative $nX^{n-1}$).
