I need to find the real and imaginary part of this $Z_{n}=\left (\frac{ \sqrt{3} + i }{2}\right )^{n} + \left (\frac{ \sqrt{3} - i }{2}\right )^{n}$ I have a test tomorrow and i have some troubles understanding this kind of problems, would really appreciate some help with this
$$ Z_{n}=\left (\frac{ \sqrt{3} + i }{2}\right )^{n} + \left (\frac{ \sqrt{3} - i }{2}\right )^{n}
$$
$$ 
Z_{n}\epsilon  \mathbb{C}
$$

The above two can be written using binomial theorem
$$\sum_{k=0}^n \binom{n}{k}\left(\frac{3}{4}\right)^\frac{n-k}{2}\left(\frac{i}{2}\right)^k$$
And
$$\sum_{k=0}^n \binom{n}{k}\left(\frac{3}{4}\right)^\frac{n-k}{2}\left(\frac{-i}{2}\right)^k$$
We rewrite the second as
$$\sum_{k=0}^n \binom{n}{k}\left(\frac{3}{4}\right)^\frac{n-k}{2}\left(\frac{i}{2}\right)^k(-1)^k$$
Such that even k terms are the same and odd k terms are opposite signs 
Combining the sums we have 
$$\sum_{k=0}^{\frac{n}{2}} 2\binom{n}{2k}\left(\frac{3}{4}\right)^\frac{n-2k}{2}\left(\frac{i}{2}\right)^{2k}$$
$$\sum_{k=0}^{\frac{n}{2}} 2\binom{n}{2k}\left(\frac{3}{4}\right)^{\frac{n}{2}-k}\left(\frac{-1}{4}\right)^{k}$$
$$\sum_{k=0}^{\frac{n}{2}} 2\binom{n}{2k}\left(\frac{3}{4}\right)^{\frac{n}{2}}\left(\frac{-1}{3}\right)^{k}$$
Which calculating the sum from wolfram alpha gives, $$Re(Z_n)=2cos(\frac{n\pi}{6})$$ 
And
$$Im(Z_n)=0$$
The phase would flip between $\pi$ and $0$ and the function is always real. (Which I should have realized by seeing that this clearly resembles the cos function)
