Without using the binomial expansion. Show that $(\sqrt 3 +i)^n + (\sqrt 3 -i)^n$ is real for any positive integer $n$. I have to show that $(\sqrt 3 +i)^n + (\sqrt 3 -i)^n$ is REAL for any positive integer $n$. My initial thought was to use trial and error using values $1,2,3,\ldots,n$ but that does not seem like a thorough proof. Many thanks in advance!
 A: Hint A complex number $z$ is real if and only if $z=\overline{z}.$
(I'm not sure for the positive part)
A: I think can rewrite $C=|r|e^{i\theta},C^*=|r|e^{-i\theta}$, and use $C^n+C^{*n}=|r|^n (e^{ni\theta}+e^{-ni\theta})$ and use $e^{ia}=\cos (a)+i\sin (a)$
A: $z^n + \bar z ^n = z^n + \overline{z^n} = 2\operatorname{Re}z^n\in\mathbb R$
A: Well, if $x_n=(\sqrt 3 +i)^n + (\sqrt 3 -i)^n$, then one can show that $x_{n}=2\sqrt3 x_{n-1}-4x_{n=2}$. 
$x_0=2, x_1=2\sqrt3$ are both real, the reals are closed under $\pm$, so by induction, $x_n$ is real for all $n$.
A: If $a$ and $b$ are real then the complex conjugate of $a+bi$ is $a-bi$.
If you add a complex number to its complex conjugate, you get a real  number: $$(a+bi) + (a-bi)= 2a.$$
So if you can show that $(\sqrt 3 -i)^n$ is the complex conjugate of $(\sqrt 3 +i)^n$, then you've got it.
You get $(\sqrt3+i)^n$ by multiplying:
$$
\underbrace{(\sqrt3+i)\cdots\cdots(\sqrt3+i)}_\text{$n$ factors}. \tag 1
$$
If you take the complex conjugate of each factor separately you get
$$
\underbrace{(\sqrt3-i)\cdots\cdots(\sqrt3-i)}_\text{$n$ factors} = (\sqrt3-i)^n. \tag 2
$$
So the question is: If you conjugate each factor separately, do you get the same complex number as if you conjugate the product?
\begin{align}
(a-bi)(c-di) & = (ac-bd) - i(ad+bc) \\[10pt]
= {} &  \Big( \text{the conjugate of } (ac-bd) + i (ad+bc) \Big) = (a+bi)(c+di).
\end{align}
So the answer to that question is "yes".  Therefore $(2)$ is the conjugate of $(1)$.
