Cardinality of a set of complex numbers The question is basically to find the number of elements in the set $\{z \in \mathbb{C} : z^{60} = -1 , z^k \neq -1, 0<k<60 \}$.
As is quite obvious with the kind of question,I am a self-studying novice, but the only idea I could get was to just use $\theta = \dfrac{(2n+1)\pi}{60}$, which makes it a routine $n$-th root problem. But I am not sure how the condition on $k$ comes into this? A hint would suffice or a link if this is a duplicate(or close to being one).
 A: Taking modulus both sides: $|z|^{60} = |-1| = 1 \Rightarrow |z| = 1$ since $|z| \geq 0$. Thus $z = \cos \theta + i\sin \theta$, and using De Moivre's theorem: $z^{60} = \cos (60\theta) + i\sin(60\theta) = -1 \to \cos (60\theta) = -1, \sin (60\theta) = 0 \to 60\theta = (2n+1)\pi \to \theta = \dfrac{(2n+1)\pi}{60}$. You want those $n$ such that $\text{gcd}(2n+1,60) = 1$. Can you count them out?
A: You are right that all solutions of the equation $z^{60} = -1$ are of the form
$$z_n=e^{\frac{(2n + 1)\pi}{60}i}$$
There are $60$ values of $z$ here, but there is another condition you must fulfill! The value $z^k$ must not be equal to $-1$ for any other value $k<60$. If you take $n=1$ in your case, then
$z_1 = e^{\frac{3\pi}{60}i} = e^{\frac{\pi}{20}i}$, meaning that $z_1^{20} = e^{\pi i} = -1$. That means that $z_1$ is not in your set! There may be other candidates that are not in the set, and this is the part of the question you still need to explore.
A: Let
$$z=r\cos\theta + ir\sin\theta$$
Then
$$z^{60}=r^{60}\cos60\theta + ir^{60}\sin60\theta$$
Can you take it from there?
A: The set you find is a superset $\{z \in \mathbb{C}: z^{60} = -1\}$. You have to exclude those numbers which raised a smaller power than $60$ equals $-1$, i.e. $z^p = -1$ for some $p < 60$. But note that for such $z$ we have
$$-1 = z^{60} = (z^p)^{\frac{60}{p}} = (-1)^{\frac{60}{p}} \implies (-1)^{1-\frac{60}{p}} = 1$$
and therefore $1-\frac{60}{p} \neq 0$ is even and finally $\frac{60}{p} \neq 1$ is odd. This is only possible when $p$ is a factor of $60$. Of all factors of $60$, $\frac{60}{p} \neq 1$ is odd if and only if $p \in \{4,12,20\}$ if and only if $n \in \{1,2,7\}$.
A: It's almost a typical $n$-th root problem. The set of all 60th roots of $-1$ is the set
$$\left\{\text{exp}\left(\frac{(2n+1)\pi i}{60}\right) : n = 0, 1, 2, \ldots, 59\right\} = \text{exp}\left(\left\{\frac{\pi i}{60},\frac{\pi i}{20},\frac{\pi i}{12},\frac{7\pi i}{60}, \frac{3 \pi i}{20}, \cdots \right\}\right).$$
We need to find the subset of these numbers that aren't the $k$-th roots of $-1$ for any $k = 1, 2, \ldots 59$. But the 20th power of the second number is $-1$, and the 12th power of the third is $-1$, but for any power we put exp$\left(\frac{7 \pi i}{60} \right)$ to any power less than 60, we will not get $-1$ (why?). This limits us to only a few roots, namely those whose denominators in the exponent is not 60. 
