# Find the number of elements of a complex subset [duplicate]

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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.......

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

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

## 1 Answer

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 \$.

• 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 4 '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 4 '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 4 '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 5 '15 at 3:03
• Please see my updated answer...There are four options ... – Empty Sep 5 '15 at 3:08