Geometric series with complex numbers In a geometric series
$a_{1}=1-\sqrt{3}i$  and $a_{2}=2$
Prove that for every natural number $n$, the numbers at the $3n+2$ location in the series are real numbers.
I have started by finding the ratio of the series, which is:
$q=1-\sqrt{3}i$
Now I am stuck. The elements of the series are $5,8,11,14,\ldots$, i.e., jumps of $3$. I tried looking at the first one, using De Moivre's formula, and got that it is indeed real. But how do I prove it for every $n$?
Thank you

The geometric progression has ratio
$$\frac{a_2}{a_1}=\frac{2}{1-i \sqrt{3}}=\frac{1}{2} \left(1+i \sqrt{3}\right)$$
Therefore the sequence can be written as
$$\{a_n\}=\left\{\left(1-i \sqrt{3}\right) \left(\frac{1}{2} \left(1+i \sqrt{3}\right)\right)^{n-1}\right\};\;n\geq 1$$
Observing that $\frac{1}{2} \left(1+i \sqrt{3}\right)$ is a cubic root of $-1$ we have
$$a_{3n+2}=\left(1-i \sqrt{3}\right) \left(\frac{1}{2} \left(1+i \sqrt{3}\right)\right)^{3n+1}=$$
$$=\left(1-i \sqrt{3}\right) \left(\frac{1}{2} \left(1+i \sqrt{3}\right)\right)^{3n}\left(\frac{1}{2} \left(1+i \sqrt{3}\right)\right)=\left(1-i \sqrt{3}\right) (-1)^{n}\left(\frac{1}{2} \left(1+i \sqrt{3}\right)\right)=$$
$$=\frac{1}{2}(-1)^n(1-i\sqrt{3})(1+i\sqrt{3})=(-1)^n \cdot 2$$
Therefore the terms indexed $(3n+2)$ are $2$ or $-2$.
