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How many elements does the set $\{z\in \mathbb C:z^{60}=-1,z^k\not=-1\text{ for } 0<k<60\}$ have ?

(A) $24$ (B) $30$ (C) $32$ (D) $45$. Which is correct ?

$z^{60}=-1=\cos(2k\pi+\pi)+i\sin(2k\pi+\pi)$. Then , $z=\cos\left(\frac{2k+1}{60}\pi\right)+i\sin\left(\frac{2k+1}{60}\pi\right)$ , for $k=0,1,\ldots,59$. But I am unable to use the second condition.

Please anyone help me.......

Community
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Empty
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1 Answers1

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Hint: How much integers from $ \ 1 \ $ to $ \ 119 \ $ are relatively prime to $ \ 60 \ $ ?

Your sixty complex roots of $ \ -1 \ $ are $ \ cis(3º) \ , \ cis(9º) \ , \ cis(15º) \ , \ \ldots \ , \ cis(357º) \ $ . Consider which ones will give you $ \ cis(180º) \ $ if you raise them to any power other than $ \ 60 \ $ (you want to discard those), and which ones can't.

EDIT (9/5) : I was just not seeing that I had stopped with the first thirty roots. I have corrected the upper end of the range from $ \ 59 \ $ to $ \ 119 \ $.

colormegone
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  • I found that the number of integers from $1$ to $59$ are relatively prime to $60$ is 16...which differs from the answer ...I am unable to understand your 2nd argument.. – Empty Sep 04 '15 at 16:26
  • Is $ \ k = 0 \ $ to be included or not? The question says no, but your listing says yes. The first root that satisfies the second condition $ \ z^k \ \neq \ -1 \ $ is $ \ cis \ \frac{\pi}{60} \ $ . The others, $ \ cis \ \frac{m \ \pi}{60} \ $ that work are integer $ \ m \ $ being any prime $ \ 7 \ \le \ m \ \le \ 59 \ $ and $ \ m = 49 \ $ . In my second statement, I am saying that the second condition means we don't want any power other than 60 to give -1 . So $ \ ( e^{\frac{3 \pi}{60}})^{20} \ = \ -1 \ $ requires us to reject that root, (continued) – colormegone Sep 04 '15 at 21:42
  • but $ \ ( e^{\frac{7 \pi}{60}})^{k} \ = \ -1 \ $ is only true for $ \ k = 60 \ $ and no lower positive integer. I count 16 roots if we include $ \ cis \ \frac{\pi}{60} \ $ and 15 if we don't. (Also, your use of $ \ k \ $ differs from that in the question.) – colormegone Sep 04 '15 at 21:45
  • Integers from $1$ to $59$ which are relatively prime are : $1,7,11,13,17,19,23,29,31,37,41,43,47,49,53,59$.....In my list excluding $0$ there are $16$ integers ..... – Empty Sep 05 '15 at 03:03
  • Please see my updated answer...There are four options ... – Empty Sep 05 '15 at 03:08
  • What is the source of this question? I've read it a number of times and it seems like it is asking for the "primitive roots" of $ \ -1 \ $ , but the size of the choices makes me think the poser meant to ask about the complement of that set. – colormegone Sep 05 '15 at 04:05
  • It is a question of a competitive exam. Yes..Number of Primitive n-th roots of $1$ is $\phi(n)$...But is it same for $-1$ ? I think NOT. – Empty Sep 05 '15 at 05:04
  • I think I realized what the issue is: I'm not going high enough. (I've only been going through thirty of the roots. I should have been counting up to $ \ e^{\frac{119 \ \pi}{60}} \ $ . ) We need the numbers from 1 to 119 that are relatively prime to 60. So that's 1, all the primes through 113, and the composite integers which do not have factors of 2, 3, or 5 . (The composites are 49, 77, 91, and 119.) Now I count 32. [Sorry about that -- I was just not seeing that I stopped too soon.] – colormegone Sep 05 '15 at 06:28
  • Check this https://math.stackexchange.com/questions/1368003/number-of-60th-primitive-roots-of-1 – Chiranjeev_Kumar Sep 12 '15 at 17:31
  • I had tried a search on primitive roots early on, but hadn't turned up that page. (Unfortunately, as this site continues to expand, it is becoming more difficult to hunt for duplications that are well-answered.) This same question has come up at least once again in the past week or so. – colormegone Sep 12 '15 at 20:22