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?
 A: 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. 
A: $z = \exp(\pi i j/60)$ 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 $C_{120}(z)$.
A: 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 $\varphi(120)=32$
$r=1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73,77, 79, 83, 89, 91, 97, 101, 103,107,109,113, 119$
