Complex numbers and Roots of unity

Question

I have no clue how to begin these problems. How do I start? I don't think I should pound em out...Thanks.

Let P be the set of $42^{\text{nd}}$ roots of unity, and let Q be the set of $70^{\text{th}} $ roots of unity. How many elements do P and Q have in common?

Let P be the set of $42^{\text{nd}} $roots of unity, and let Q be the set of $70^{\text{th}} $roots of unity. What is the smallest positive integer n for which all the elements in P and all the elements in Q are $n^{\text{th}}$ roots of unity?

For the second question, are you asking for the smallest integer $n$ such that $P\cup Q$ is the set of $n$-th roots of unity? — user31415926535, Jan 22 '14 at 02:58
See http://math.stackexchange.com/questions/614012/proof-that-mathbb-g-n-bigcap-mathbb-g-m-mathbb-g-mn. — lhf, Jan 22 '14 at 12:51

score 2 · Accepted Answer · answered Jan 22 '14 at 12:21

2

The first question is the same as asking for which positive integers $k<42$ and $n<70$ do we have $\dfrac{2\pi}{42}k=\dfrac{2\pi}{70}n\iff5k=3n$. The second is equivalent to determining the least common multiple of $42$ and $70$, which is $210$.

answered Jan 22 '14 at 12:21

Lucian

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I tried the first question and had 13 positive integers, but that is incorrect. Any further hints? – Math is Life Apr 14 '15 at 17:30
@MathisLife: I assume you got a bit confused by the word positive. In my country, which has historically been influenced by the French system, it includes $0$. – Lucian Apr 14 '15 at 23:52
Oh, I see. If we include $0$...would there be 14 positive integers then? – Math is Life Apr 15 '15 at 00:58
@Lucian: I meant to say *12 positive integers. – Mathy Person Apr 22 '15 at 00:23
@MathyPerson: I suggest that you revise your computations. – Lucian Apr 22 '15 at 06:43
@Lucian: I recalculated. 9 positive integers, including $0$? – Math is Life Apr 27 '15 at 01:16
@MathisLife: It is beyond me how on earth you obtain such odd results. Can you at least write down the general solution for the Diophantine equation $5k=3n$ ? This way maybe we'll be able to tell what went wrong with your computations. – Lucian Apr 27 '15 at 08:46

score 1 · Answer 2 · answered Jan 22 '14 at 03:02

The $n$-th roots of unity $\omega^{(k)}_n$ are \begin{equation} \omega^{(k)}_n = \exp\left(2\pi i k\over n\right) \mathrm{~for~} k=0,1,\ldots, n-1~. \end{equation} As some examples, for $n=2$, we have the roots $\omega^{(0)}_2=1$ and $\omega^{(1)}_2=-1$. Now, note for $n=4$, we have $\omega^{(2)}_4=-1$. There is a root in common here because $1/2 = 2/4$, i.e. the ration of $k/n$ is the same for these two roots (actually there is another common root, since $\omega^{(0)}_n = 1$ for all $n$). In fact, its clear from the formula for the $\omega$ that \begin{equation} \omega^{(k)}_{n} = \omega^{(l)}_m \end{equation} whenever ${k\over n} = {l \over m}$. So, given $n=72$ and $m=40$, you just have to figure out how many ways you can satisfy this relationship with integers $0 \le k < n$ and $0\le l < m$. Hopefully that's enough to get you started.

score 1 · Answer 3 · edited Apr 13 '17 at 12:20

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So, we have $$x^{72}-1=0\text{ and }x^{70}-1=0$$

From Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$, $$(x^{72}-1=0,x^{70}-1)=x^{(72,70)-1}=x^2-1$$

edited Apr 13 '17 at 12:20

Community

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answered Jan 22 '14 at 03:33

lab bhattacharjee

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Complex numbers and Roots of unity

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