Number of $60$th primitive roots of $-1$

If you look at it in trigonometric form, you're looking to solve:

$(\cos \phi + i.\sin \phi)^{60} = \cos \pi + i.\sin \pi$

It's well-known that the solutions are 60, here they are:

$\phi = \phi(k) = (\pi + 2k\pi)/60$ , $k=0,1,2,...,59$.

Let's denote these solutions by $z_k$ i.e. $z_k = cos \phi(k) + i.sin \phi(k)$.

i.e.

$\phi = (2k+1)\pi/60$ , $k=0,1,2,...,59$.

For this root to be primitive (2k+1) has to be relatively prime with 60 (otherwise there exists a degree m smaller than 60 such that $z_k^m=-1$). There're 32 such values for k (and 28 values for which it is not relatively prime). One can check this directly. Hence the answer is 32.

$z = \exp(\pi i j/60)$ satisfies the condition if $j$ is an odd integer in $\{1,2, \ldots, 119\}$ coprime to $60$. The number of these is the number of integers in $\{1,2,\ldots, 119\}$ coprime to $120$, which is the Euler totient $\varphi(120) = 32$. The answer given is wrong.

These are also the roots of the cyclotomic polynomial $C_{120}(z)$.

suppose $$z=e^{\frac{2\pi i}{120}}$$ then $z$ is a $120^{th}$($60^{th}$ of $-1$) primitive root of $1$. $z^r$ is also a $120^{th}$ primitive root of $-1$ if $gcd(r,120)=1$

No. of primitive roots is $\varphi(120)=32$

$r=1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73,77, 79, 83, 89, 91, 97, 101, 103,107,109,113, 119$