Problem: Let $\mathbb{C}^\star$ denotes the set of all complex number without zero. Determine the number of elements in the set \[ \left\{ z\in \mathbb{C} ^\star :~z^{60}=-1,\ z^k\neq -1\ \text{for } 0 \lt k \lt 60 \right\}. \]
Solution: We want to find the number of elements in the set \[ H = \left\{ z\in \mathbb{C} ^{\star} :~z^{60}=-1,\ z^k\neq -1\ \text{for } 0 \lt k \lt 60 \right\}. \] The set $H$ is equal to the set \[ H^\prime = \left\{ z\in \mathbb{C} ^{\star} :~z^{120}=1,\ z^k\neq 1\ \text{for } 0\lt k \lt 120 \right\}. \] Thus the give set $H$ is the set of all elements of $\mathbb{C} ^{\star} $ whose order is $120$. Therefore the order of $H$ will be \[ o(H) = \phi (120) = \phi (8\times 3\times 5) = \phi (8) \times \phi (3) \times \phi (5) = 32, \] where $\phi(n) $ is the set of all positive numbers less than $n$ which is coprime to $n$.