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This question already has an answer here:

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.......

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marked as duplicate by Community Oct 9 '15 at 7:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I think this question already have been discused some days ago $\endgroup$ – Chiranjeev_Kumar Aug 26 '15 at 4:28
  • $\begingroup$ @ – Chiranjeev ) then please give the link... $\endgroup$ – Empty Aug 26 '15 at 5:09
  • $\begingroup$ I saw this question here some months ago. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 26 '15 at 6:09
  • $\begingroup$ @ Michael Hardy) then please give the link...Your ' here ' does not contain any link $\endgroup$ – Empty Aug 26 '15 at 6:59
  • $\begingroup$ I was searching this but unable to find that.. Sorry $\endgroup$ – Chiranjeev_Kumar Aug 26 '15 at 14:02
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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 \ $.

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  • $\begingroup$ 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.. $\endgroup$ – Empty Sep 4 '15 at 16:26
  • $\begingroup$ 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) $\endgroup$ – colormegone Sep 4 '15 at 21:42
  • $\begingroup$ 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.) $\endgroup$ – colormegone Sep 4 '15 at 21:45
  • $\begingroup$ 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 ..... $\endgroup$ – Empty Sep 5 '15 at 3:03
  • $\begingroup$ Please see my updated answer...There are four options ... $\endgroup$ – Empty Sep 5 '15 at 3:08

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