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
$$ Z_{n}=\left (\frac{ \sqrt{3} + i }{2}\right )^{n} + \left (\frac{ \sqrt{3} - i }{2}\right )^{n}
$$
$$ 
Z_{n}\epsilon  \mathbb{C}
$$
 A: Note that $z=\left (\frac{ \sqrt{3} + i }{2}\right )^{n}=(a+bi)^n$ we have 
$$|z|=\sqrt{(\sqrt{3}/2)^2+(1/2)^2}=1, \quad \theta_z=\arctan(\tfrac{b}{a})=\arctan(\tfrac{1}{\sqrt{3}})=\tfrac{\pi}{6}$$
and for $w=\left (\frac{ \sqrt{3} - i }{2}\right )^{n}=(a-bi)^n$ we have 
$$|w|=\sqrt{(\sqrt{3}/2)^2+(1/2)^2}=1, \quad \theta_w=\arctan(-\tfrac{b}{a})=\arctan(-\tfrac{1}{\sqrt{3}})=-\tfrac{\pi}{6}$$
Then, 
$$z=|z|^n(\cos\theta_z+i\sin\theta_z)^n=\left[\cos\left(\tfrac{\pi}{6}\right)+i\sin\left(\tfrac{\pi}{6}\right)\right]^n$$
$$w=|w|^n(\cos\theta_w+i\sin\theta_w)^n=\left[\cos\left(-\tfrac{\pi}{6}\right)+i\sin\left(-\tfrac{\pi}{6}\right)\right]^n$$
Using de Moivre's formula
$$z^n=|z|^n(\cos\theta+i\sin\theta)^n=|z|^n(\cos(n\theta)+i\sin(n\theta))^n$$
we get 
$$z=\left[\cos\left(\tfrac{\pi}{6}\right)+i\sin\left(\tfrac{\pi}{6}\right)\right]^n=\left[\cos\left(\tfrac{n\pi}{6}\right)+i\sin\left(\tfrac{n\pi}{6}\right)\right]$$
$$w=\left[\cos\left(-\tfrac{\pi}{6}\right)+i\sin\left(-\tfrac{\pi}{6}\right)\right]^n=\left[\cos\left(-\tfrac{n\pi}{6}\right)+i\sin\left(-\tfrac{n\pi}{6}\right)\right]=\left[\cos\left(\tfrac{n\pi}{6}\right)-i\sin\left(\tfrac{n\pi}{6}\right)\right]$$
Finally, 
$$Z_n=z+w=2\cos\left(\tfrac{n\pi}{6}\right)$$
A: 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)
A: Let $\omega=\frac{\sqrt{3}+i}{2}$. Then $Z_n$ is twice the real part of $\omega^n$.
Now, $\omega$ is a primitive $12$-th root of unity and so $\omega = \exp(\frac{2\pi i}{12})$.
Therefore, $\omega^n=\exp(n\frac{2\pi i}{12})$, whose real part is $\cos(n\frac{2\pi}{12})=\cos(\frac{n\pi}{6})$.
A: Obviously $\dfrac{\sqrt 3+i}2=\mathrm e^{\tfrac{i\pi}6}$ and   $\dfrac{\sqrt 3-i}2$ is its conjugate $\mathrm e^{-\tfrac{i\pi}6}$. Thus 
$ Z_n= \mathrm e^{\tfrac{in\pi}6}+\mathrm e^{-\tfrac{in\pi}6}$ is the sum ot two conjugate numbers, so that
$$\operatorname{Im}(Z_n)=0\quad \text{and}\quad\operatorname{Re}(Z_n)=2\cos \frac{n\pi}6.$$
Let's give explicit formulæ depending on the values of $n$:
$$\operatorname{Re}(Z_n)=\begin{cases}
2&\text{if }n\equiv 0 \mod 12, \\\sqrt 3&\text{if }n\equiv 1\;\text{ or }\;11,\\1&\text{if }n\equiv 2\;\text{ or }\;10, \\
0&\text{if }n\equiv 3\;\text{ or }\;9, \\-1&\text{if }n\equiv 4\;\text{ or }\;8, \\ -\sqrt3 &\text{if }n\equiv 5\;\text{ or }\;7, \\
-2 &\text{if }n\equiv 6.
\end{cases}$$
