Q: Sixth root of $-1$ We know that $\frac{\sqrt3}{2}+ \frac{i}{2} = \cos(\frac{\pi}{6})+i \sin (\frac{\pi}{6})=e^\frac{i \pi}{6}$. Therefore we have $(e^\frac{i \pi}{6})^6 =e^{i\pi}= \cos(\pi)+i \sin (\pi)=-1$. (please correct me if this calculation is wrong)
Therefore $\frac{\sqrt3}{2}+ \frac{i}{2}$ is one complex sixth root of $-1$.
However when I manually computed $(\frac{\sqrt3}{2}+ \frac{i}{2})^6$ by computing its square $3$ times I end up having the expression $\frac{1+ \sqrt3.i}{2}$.
why are these two expressions are not matching?
I have one further question : $(1)$ If we have  $\frac{\sqrt3}{2}+ \frac{i}{2}$ is a complex $6$ th root of $-1$, then can explicitly find other $6$th root of $-1$ distinct from the previous one?
Any remark from anyone is welcome.
