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 

Hint: if $\,\omega=\cfrac{1 + i \sqrt{3}}{2}\,$ then $\omega^2=-\overline{\omega}\,$, $\omega^3 = -1\,$ and $\omega^6=1\,$, so the sequence $\omega^n$ is periodic and non-constant, therefore the limit doesn't exist.
