Find the number of elements of the cyclic subgroup $\left\langle -\frac{1}{ 2}+\frac{\sqrt 3}{2} i \right\rangle$ of $\mathbb{C}^*$. Find the number of elements of the cyclic subgroup $\left\langle -\frac{1}{ 2}+\frac{\sqrt 3}{2} i \right\rangle$ of $\mathbb{C}^*$. 
Answer:
Since $\left\langle -\frac{1}{ 2}+\frac{\sqrt 3}{2} i \right\rangle=-\cos (2 \pi/6)+i \sin (2 \pi/6)$ is a generator of $U_6$ and $U_6$ has order $\phi(6)=2$, the cyclic subgroup generated by $\left\langle -\frac{1}{ 2}+\frac{\sqrt 3}{2} i \right\rangle$ has $2$ elements. 
Am I right ?
I think $\cos (2 \pi/n)+i \sin (2 \pi/n)$ has order $\phi(n)$ but in the above case it is like the form $-\cos (2 \pi/n)+i \sin (2 \pi/n)$. This confuses me
 A: Note that
$\cos \left ( \dfrac{2\pi}{3} \right ) = -\dfrac{1}{2}, \tag 1$
and
$\sin \left ( \dfrac{2\pi}{3} \right ) = \dfrac{\sqrt 3}{2}; \tag 2$
thus,
$-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} = \cos \left ( \dfrac{2\pi}{3} \right )  + i \sin \left ( \dfrac{2\pi}{3} \right ) = e^{2 \pi i /3}; \tag 3$
it follows that, for any $m \in \Bbb Z$,
$\left (-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} \right )^m = e^{2 m \pi i/3}; \tag 4$
in particular we have
$\left (-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} \right )^2 = e^{4\pi i/3}, \tag 5$
$\left (-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} \right )^3 = e^{6 \pi i/3} = e^{2\pi i} = 1; \tag 6$
at this point the powers if $e^{2\pi i/3}$ commence to cycle around so that
$\left (-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} \right )^4 = e^{8 \pi i/3} = e^{6\pi i / 3} e^{2 \pi i / 3}= e^{2 \pi i} e^{2 \pi i / 3}= e^{2 \pi i / 3}, \tag 7$
$\left (-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} \right )^5 = e^{10 \pi i/3} = e^{6\pi i / 3} e^{4 \pi i / 3}= e^{2 \pi i} e^{4 \pi i / 3}= e^{4 \pi i / 3}, \tag 8$
$\left (-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} \right )^6 = e^{12 \pi i/3} = e^{4 \pi i}= (e^{2 \pi i})^2 = 1^2 = 1, \tag 9$
and so forth.  Indeed, since the division algorithm yields
$m = 3k + j, \; j = 0, 1, 2,  \tag{10}$
(4) may be written
$\left (-\dfrac{1}{2} + i \dfrac{\sqrt 3}{2} \right )^m = e^{2 m \pi i/3} = e^{2 (3k + j) \pi i/3} = e^{6k\pi i / 3 + 2j \pi i / 3}$
$= e^{6k \pi i / 3}e^{2j\pi i / 3} = (e^{2\pi i})^k e^{2j \pi i /3} = e^{2j \pi i /3}, 0 \le j \le 2; \tag{11}$
which shows that every power of $e^{2 \pi i/ 3}$ lies in the set
$\{1, e^{2 \pi i / 3}, e^{4 \pi i / 3 } \}, \tag {12}$
which clearly is possessed of a group structure in light of the preceding calculations.  Since there exists precisely one group of order $3$,
$\Bbb Z_3 = \{0, 1, 2 \} \tag{13},$
and it is isomorphic to (12) via the map taking
$0 \mapsto 1, \; 1 \mapsto e^{2 \pi i / 3}, \; 2 \mapsto e^{4 \pi i / 3}, \tag{14}$
we conclude that the subgroup of $\Bbb C^\ast$ generated by $e^{2\pi i / 3}$ has exactly $3$ elements.
A: That is $\langle e^{\frac{2\pi i}3}\rangle $.  $\omega=e^{\frac{2\pi i}3}$ is a primitive $3$rd root of unity,  and has order $3$.  Thus we get the cyclic group of order $3$.  
Another description would be as the $3$-torsion subgroup of the circle group. 
$\phi (n)$ is the number of primitive $n$-th roots of unity.  Primitive roots generate the whole group $C_n$.
But an arbitrary element of $C_n$ is just an $n$-th root of unity (at least,  as one way of looking at it).   There are of course $n$ of them.
