1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $nX^{n-1}$).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ thanks! but I still don't understand how they go from $\beta = \frac{m\theta+2k\pi}{n}$ into $\beta = \frac{m\theta+2mk\pi}{n}$. $\endgroup$ – EM4 Sep 22 at 14:46
  • $\begingroup$ @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 $0 \leqslant k \leqslant n-1$. $\endgroup$ – ΑΘΩ Sep 22 at 16:39
  • $\begingroup$ 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 $2mk\pi$. $\endgroup$ – EM4 Sep 22 at 16:41
  • $\begingroup$ @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. $\endgroup$ – ΑΘΩ Sep 23 at 2:02
  • $\begingroup$ oh okay, because I was wondering $\frac{2mk\pi}{n}$ doesn't change it if they don't share the exact factor. $\endgroup$ – EM4 Sep 23 at 2:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.