Computing $\left|\{z \in \mathbb{C} | z^{60} =-1 , z^{k} \neq -1 \text{ for }0I am trying assignment question in algebra course and I was unable to solve this particular problem.

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

It's about $60^{th}$ roots of $-1$ and I think those roots will be $\cos(kπ/60) + i \sin(kπ/60)$,  $k$ belonging to $\{0,1,\ldots,59\}$ . But  I am not able to proceed further as I am a bit confused.
Can someone please give hint.
 A: OK, I'm going to answer this because one comment and one answer are wrong.
To answer this, it's better to look at 120th roots of $1$, rather than 60th roots of $-1$. So there are 120 roots of $1$, all powers $z^j$ of $z=\mathrm{e}^{2\pi\mathrm{i}/120}$. I claim that $w=z^j$ has your property ($w^a\neq -1$ for $1\leq a<60$) if and only if $\gcd(j,120)=1$. This yields $32$ options for $j$, not sixteen as mentioned above. Certainly if $j$ is even then $w^{60}=1$, not $-1$, and so throughout we assume that $j$ is odd.
To see this, suppose first that $\gcd(120,j)=b>1$, and note that $\gcd(60,j)=b>1$. Thus $w^{120/b}=z^{120j/b}$ is a power of $z^{120}$, so is $1$. If $w^{120/b}=1$ then  $w^{60/b}=\pm 1$. If $w^{60/b}=1$ then $w^{60}=1$, not allowed, or $w^{60/b}=-1$, and $j$ should be ignored.
On the other hand, suppose that $\gcd(120,j)=\gcd(60,j)=1$, but that $w^a=-1$ for some $a<60$, i.e., $z^{ja}=-1$ for some $a<60$. Since $z^{120}=1$ but no smaller power is unity, we obtain $2ja=120m$ for some integer $m$. Thus $ja=60m$, but $\gcd(120,j)=1$, so $60\mid a$.
Thus there are $\phi(120)=32$ solutions.
A: Hint:  Writing your roots closer to polar form, for example, something like $\mathrm{e}^{\pi\mathrm{i} + 2\pi \mathrm{i}k/60}$, may make testing the condition easier.
A: If you take $w_0=e^{i\pi/60}$ then your $z$'s will be $w_0^k$, where $k$ is an integer $7\le k\le 59$ and the greatest common divisor $(k,60)=1$, so the numbers are:
$$\{7,11,13,17,19,23,29,31, 37,41,43,47,49,53,57,59\}$$
A: The $60$-th roots of
$-1$ are:
$$
z_k=e^{\frac{(2k+1)}{60}i\pi}
$$
for $k=0,...,59$. If $\gcd(2k+1,60)=d>1$, then you can simplify:
$$
z_k=e^{\frac{\frac{(2k+1)}{d}}{\frac{60}{d}}i\pi}
$$
that is $z_k$ is a $\frac{60}{d}$-th root of $-1$ too. So you have to consider only those $z_k$'s, where $\gcd(60,2k+1)=1$. (Try to prove that all of them are good!)
We can generate the good $k$'s with a computer (or by hand)):
$\{
0, 3, 5, 6, 8, 9, 11, 14, 15, 18, 20, 21, 23, 24, 26, 29, 30, 33, 35, 36, 38, 39, 41, 44, 45, 48, 50, 51, 53, 54, 56, 59\}$
