# How many elements does the following set have? [duplicate]

$\{z\in \mathbb{C} \mid z^{60}=-1, z^k \not=-1 \text{ for } 0<k<60\}$ I tried to solve this but I've no idea on this type of questions, how to get start or in which way should I proceed. How can i do this?
That are just the primitive 120th roots, so we are looking for $$\varphi(120)=\varphi(8)\cdot\varphi(3)\cdot\varphi(5)=4\cdot 2\cdot 4=32.$$
Hint: First prove: \begin{align*} z^{60} = -1 \text{ and } z^k \ne -1 \text{ for } 0 < k < 60 &\iff z^{120} = 1 \text{ and } z^k \ne 1 \text{ for } 0 < k < 120. \end{align*} Then note that $$\{z^{120} = 1 \text{ and } z^k \ne 1 \text{ for } 0 < k < 120\}$$ is the set of primitive $120$th roots of unity.