# Number of $60$th primitive roots of $-1$

How many elements does the set $$\{z\in \mathbb{C}\mid z^{60}= -1; z^k \neq -1\text{ for } 0\lt k< 60\}$$ have?

$1.\quad24$

$2.\quad30$

$3.\quad32$

$4.\quad45$

I assumed that set consists of elements of order $120$ (as $(-1)^{2}=1$) i.e no lesser number than 120 can take them to $1$ because if a number lesser than 60 can take them to $-1$ then a lesser than 120 number can take them to $1$ also. So the number of primitive $60$th roots of $1$ , that happens to be 32. But the answer is given 30. What am I missing here?

• I get 32, I think the given answer is wrong. By the way, your text should read "So the number of primitive $120^{th}$ roots of 1".
– lulu
Jul 20, 2015 at 20:56
• The question was given to me in the above form and I was not sure whether I was thinking right. Jul 20, 2015 at 21:04
• Well, my thinking was identical to yours. And I see the posted answers all got 32 as well so I doubt that we have it wrong.
– lulu
Jul 20, 2015 at 21:11

If you look at it in trigonometric form, you're looking to solve:

$(\cos \phi + i.\sin \phi)^{60} = \cos \pi + i.\sin \pi$

It's well-known that the solutions are 60, here they are:

$\phi = \phi(k) = (\pi + 2k\pi)/60$ , $k=0,1,2,...,59$.

Let's denote these solutions by $z_k$ i.e. $z_k = cos \phi(k) + i.sin \phi(k)$.

i.e.

$\phi = (2k+1)\pi/60$ , $k=0,1,2,...,59$.

For this root to be primitive (2k+1) has to be relatively prime with 60 (otherwise there exists a degree m smaller than 60 such that $z_k^m=-1$). There're 32 such values for k (and 28 values for which it is not relatively prime). One can check this directly. Hence the answer is 32.

• I get $32$ such values: $2k+1 = 1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 77, 79, 83, 89, 91, 97, 101, 103, 107, 109, 113, 119$. Which two are you missing? Jul 20, 2015 at 20:36
• You are right, I also get 32 such values: 1 7 11 13 17 19 23 29 31 37 41 43 47 49 53 59 61 67 71 73 77 79 83 89 91 97 101 103 107 109 113 119. So 32 is the correct answer. Jul 20, 2015 at 21:03

$z = \exp(\pi i j/60)$ satisfies the condition if $j$ is an odd integer in $\{1,2, \ldots, 119\}$ coprime to $60$. The number of these is the number of integers in $\{1,2,\ldots, 119\}$ coprime to $120$, which is the Euler totient $\varphi(120) = 32$. The answer given is wrong.

These are also the roots of the cyclotomic polynomial $C_{120}(z)$.

suppose $$z=e^{\frac{2\pi i}{120}}$$ then $z$ is a $120^{th}$($60^{th}$ of $-1$) primitive root of $1$. $z^r$ is also a $120^{th}$ primitive root of $-1$ if $gcd(r,120)=1$

No. of primitive roots is $\varphi(120)=32$

$r=1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73,77, 79, 83, 89, 91, 97, 101, 103,107,109,113, 119$