TITLE: Number of $60$th primitive roots of $-1$
QUESTION [9 upvotes]: 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?

REPLY [3 votes]: 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 $\varphi(120)=32$
$r=1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73,77, 79, 83, 89, 91, 97, 101, 103,107,109,113, 119$