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.