Is $(2+i)^n + (2-i)^n $ a real number ($\in \Bbb R$)? The question:
$$\forall n \in \Bbb N$$
is the number
$$(2+i)^n+(2-i)^n$$
in the real numbers ($\Bbb R$)?

My try for solution using Newton binom:
$$(2+i)^n = \sum_{k=0}^{n}\binom{n}{k}2^{n-k}i^{k}$$
$$(2-i)^n = \sum_{k=0}^{n}\binom{n}{k}2^{n-k}(-i)^{k}$$
$$(2+i)^n + (2-i)^n$$
$$\Downarrow$$
$$\forall k - odd: \sum_{k=0}^{n}\binom{n}{k}2^{n-k}i^{k} + \sum_{k=0}^{n}\binom{n}{k}2^{n-k}(-i)^{k} = 0$$
$$\forall k - even: \sum_{k=0}^{n}\binom{n}{k}2^{n-k}i^{k} + \sum_{k=0}^{n}\binom{n}{k}2^{n-k}(-i)^{k} = Real \ number$$
So the answer is YES, the sum $(2+i)^n + (2-i)^n \in \Bbb R.$
Correct? any better/smarter/more efficient way? 

*I tried to solve it trigonometricaly, didnt realy work, stuck...:
$$(2+i) \Rightarrow r = \sqrt{2^2+1^2} =\sqrt{5}, \tan \theta = \frac{1}{2} \Rightarrow \theta = 26.565$$
$$(2+i) = \sqrt{5}(\cos 26.565 + i \sin 26.565)$$
$$(2+i)^n = \sqrt{5}^n(\cos 26.565n + i \sin 26.565n)$$
$$(2-i)^n = \sqrt{5}^n(\cos 26.565n - i \sin 26.565n)$$
$$(2+i)^n + (2-i)^n = ?$$
 A: More efficient way:
A complex number plus its complex conjugate is real, so
$(2+i)^n+(2-i)^n=(2+i)^n+(\overline{2+i})^n=(2+i)^n+\overline{(2+i)^n}\in\mathbb R.$
A: $2+i$ and $2-i$ are conjugates. Then $(2+i)^n$ and $(2-i)^n$ are conjugates for all natural numbers $n$. Hence their sum is real.
A: A good method to try to prove that a complex number is real is to use the complex conjugate, which takes a complex number $z=a+bi$ to $\bar z = a-bi$. A complex number is real if and only if it equals its conjugate; similar tricks prove helpful in some more general contexts (e.g. to show that Binet's formula for Fibonnaci numbers yields rational numbers or to show all sorts of things in Galois theory).
If you prove two relationships about conjugation, you're pretty much good to solve your problem:
$$\overline{z_1 \cdot z_2} = \overline{z_1}\cdot \overline{z_2}$$
$$\overline{z_1 + z_2} = \overline{z_1}+ \overline{z_2}$$
which say that complex conjugation is a symmetry of arithmetic over the complex numbers; there are also rules such as $\overline{e^z} = e^{\overline{z}}$, though those aren't necessary here. If you repeatedly apply the first rule, you get $\overline{z^n}=\overline{z}^n$ for $n\in \mathbb N$, and then you can see
$$\overline{(2+i)^n+(2-i)^n}=\overline{(2+i)}^n+\overline{(2-i)}^n=(2-i)^n+(2+i)^n$$
which shows that your expression is equal to its conjugate, hence real.

Somewhat unrelatedly, your trigonometric approach can work; you can observe that $2+i=\sqrt{5}\cdot e^{i\tan^{-1}(1/2)}$ by working in polar coordinates (note $e^{ix}=\cos(x)+i\sin(x)$) and $2-i=\sqrt{5}\cdot e^{-i\tan^{-1}(1/2)}$. Then,
$$(2+i)^n + (2-i)^n = \sqrt{5}^n \cdot \left(e^{ni\tan^{-1}(1/2)} + e^{-ni\tan^{-1}(1/2)}\right).$$
Observing that $e^{ix}+e^{-ix}=2\cos(x)$ we get
$$(2+i)^n + (2-i)^n = 2\sqrt{5}^n \cos(n\tan^{-1}(1/2)).$$
