$\left(\frac{1+\sqrt{3} i}{2}\right)$

$=\left[\frac 12+\frac{\sqrt 3}{2}\iota\right]$

$=[\cos \frac{\pi}{3}+\iota\sin \frac{\pi}{3}]$

$=[e^{\frac{\iota\pi}{3}}]$

Therefore $\left(\frac{1+\sqrt{3} i}{2}\right)^{n}=[e^{\frac{\iota\pi}{3}}]^n$

$=[e^{\frac{n\iota \pi}{3}}]$

Elements of $A$ are

Now for $n=1$

$a_1=e^{\frac{\iota \pi}{3}}$

for $n=2$

$a_2=e^{\frac{2\iota \pi}{3}}$

for $n=3$

$a_3=e^{\frac{3\iota \pi}{3}}=e^{\iota\pi}=-1$

for $n=4$

$a_4=e^{\frac{4\iota \pi}{3}}=e^{\iota\pi}\cdot e^{\frac{\iota \pi}{3}}= -e^{\frac{\iota \pi}{3}}$

for $n=5$

$a_5=-e^{\frac{2\iota \pi}{3}}$

for $n=6$

$a_6=e^{\frac{6\iota \pi}{3}}=(-1)^2=1$

Now for $n=7,8,\ldots$ the elements will repeat.

Therefore only $6$ distinct element in $A$.