# 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.

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 $$nX^{n-1}$$).

• thanks! but I still don't understand how they go from $\beta = \frac{m\theta+2k\pi}{n}$ into $\beta = \frac{m\theta+2mk\pi}{n}$. – EM4 Sep 22 '20 at 14:46
• @EM4 It seems there is a mistake in the first relation you set forth in your posting. If $z=r\mathrm{e}^{\mathrm{i}\theta}$ in polar form, then the general form of an $n$-th root of $z^m$ is $\sqrt[n]{r^m}\mathrm{e}^{\mathrm{i}\frac{m\theta+2k\pi}{n}}$, with $0 \leqslant k \leqslant n-1$. – ΑΘΩ Sep 22 '20 at 16:39
• this is what I was thinking then my book says $e^{\frac{m\theta +2mk\pi}{n}}$ , so I was like hmmm why is m in involved for for $2mk\pi$. – EM4 Sep 22 '20 at 16:41
• @EM4 that would be the form which the argument of an $n$-th root of $z$ takes after elevation to the $m$-th, and it is by virtue of $m, n$ being coprime that $\left\{\mathrm{e}^{\mathrm{i}\frac{m\theta+2k\pi}{n}}\right\}_{0 \leqslant k \leqslant n-1}=\left\{\mathrm{e}^{m\mathrm{i}\frac{\theta+2k\pi}{n}}\right\}_{0 \leqslant k \leqslant n-1}$. There is also the realisation to be made that some times (perhaps more often than one would like it..) textbooks can commit either mistakes or other forms of argumentative sloppiness. – ΑΘΩ Sep 23 '20 at 2:02
• oh okay, because I was wondering $\frac{2mk\pi}{n}$ doesn't change it if they don't share the exact factor. – EM4 Sep 23 '20 at 2:06