2. If the complex number $z=\frac{\sqrt{3}}{2}+\frac{1}{2} \mathrm{i}$, then $z^{2016}=(\quad)$.
(A) -1
(B) $-\mathrm{i}$
(C) $\mathrm{i}$
(D) 1
2. D.

It is easy to get $z^{3}=\mathrm{i}$. Then
$$
z^{2016}=\left(z^{3}\right)^{672}=i^{672}=\left(i^{4}\right)^{168}=1 .
$$