$\mathbf{29.}$ The subgroup of $U_6$ generated by $\cos\frac{2\pi}3+i\sin\frac{2\pi}3.$
$\mathbf{30.}$ The subgroup of $U_5$ generated by $\cos\frac{4\pi}5+i\sin\frac{4\pi}5.$
$\mathbf{31.}$ The subgroup of $U_8$ generated by $\cos\frac{3\pi}2+i\sin\frac{3\pi}2.$
where $U_n = \{z \in \mathbb{C} : z^n = 1 \}$ (nth roots of unity)
For example in (29), I have to multiply $e^{\frac{2\pi i}{3}}$ three times to get back 1. So the order is 1. However I don't see how the number "6" under the U plays a role here.
Also for instance, for the subgroup of $U_8$ generated by $e^{\frac{5\pi i}{4}}$, the order is 8 because $(e^{\frac{5\pi i}{4}})^8 = 1$. The answer given was
The 1st number which makes $ e^{\frac{5\pi i}{4}} = 1$ is 8 because the number must be either 2,4, or 8. 2 and 4 are not working so $|\langle e^{\frac{5\pi i}{4}}\rangle| = 8$
I don't understand how 2 and 4 come up here or even related here?
Also, sorry for the inconvenience I caused