Number of elements in $\{z \in \Bbb{C}: z^{60} = -1, z^k \neq -1 \:\text{ for }\: 0 < k < 60\}$ How many elements does the set $S = \{z \in \Bbb{C}: z^{60} = -1, z^k \neq -1 \:\text{ for }\: 0 < k < 60\}$ have?
If $z \in S$ then $z^{60}$ = $-1$ and hence $z^{120}$ = $1$.
How can I find any condition on $z$ from here?

We know $z=e^{\dfrac{in \pi}{60}}$ for some odd $n$.
Assume $\gcd(n, 60)=1$.  Then if $\frac{nk}{60}$ is an odd integer (as is required for $z^k=-1$ to hold), then we know $60 \mid nk$.  But since $n$ and $60$ are relatively prime, then $60 \mid k$, so in particular $k \geq 60$.  Thus, $\{ e^{\dfrac{in \pi}{60}} \mid \gcd (n, 60)=1 \} \subseteq S$.
Conversely, suppose $d \gt 1$ is a common factor of $n$ and $60$.  Then for some $a, b \in \Bbb Z, n=da$ and $60=db.$  Moreover, because $n$ is odd, $a$ must also be odd.  Thus, $z=e^{\dfrac{ia \pi}{b}}$ and $z^b=-1$ where $b \lt 60$.  Thus, $S=\{ e^{\dfrac{in \pi}{60}} \mid \gcd (n, 60)=1 \}$.
There are $16$ integers less than $60$ that are relatively prime to $60$, so $\vert S \vert = 16$.
