evaluate $lim_{n \to \infty}\left (\frac{1+\sqrt{3}i}{2}\right)^n$ 
$$\lim_{n \to \infty} \left(\frac{1+\sqrt{3}i}{2}\right)^n$$

In general we look at $\lim(x+yi)$ as $\lim(x,y)$ which we obviously can not here.
So I looked at $$\lim_{n \to \infty} \frac{\left(1+\sqrt{3}i\right)^n}{2^n}$$
$1+\sqrt{3}i=2e^{\frac{\pi i}{3}}$
So $$\lim_{n \to \infty}  \frac{2e^{\frac{\pi i n}{3}}}{2^n}=\lim_{n \to \infty}  \frac{e^{\frac{\pi i n}{3}}}{2^{n-1}}$$
But I still can not get rid of the imaginary part 

Let’s call your number $\frac{1+\sqrt3i}2=z$. You want to notice that $|z|=1$, since $(\frac12)^2+(\frac{\sqrt3}2)^2=1$. This means that the powers of $z$ bounce around on the unit circle, never changing their distance from the origin. So no limit! A closer examination shows that $z$ is a sixth root of $1$, so that its powers actually take on only six different values.
