# Complex numbers and Roots of unity

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 2:58
• – lhf Jan 22 '14 at 12:51

## 3 Answers

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 $210$.

• 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 0:58
• @Lucian: I meant to say *12 positive integers. – Mathy Person Apr 22 '15 at 0:23
• @MathyPerson: I suggest that you revise your computations. – Lucian Apr 22 '15 at 6:43

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.

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

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