Proving a complex identity Can someone please help me prove that 
$(\sqrt{3}+i)^{n} + (\sqrt{3}-i)^n$ is purely real
I know it's true because the imaginary parts will be the negative of each other but it came at my test and i don't think i had it correct and I won't like to also fail it at the exam.
Help
 A: I am not sure what tools you have at your disposal.
but...any complex number $z$ can be represented as an exponential so if
if $\sqrt 3 + i =  {\rho e^i\theta})$ then $\sqrt 3 - i =  \rho e^{-i\theta})$ 
$(\sqrt 3 + i)^n +  (\sqrt 3 - i)^n = \rho ^n e^{in\theta} + \rho ^n e^{-in\theta}$
and $\rho ^n e^{in\theta} + \rho ^n e^{-in\theta}$ are conjugates
you could take that down a level...
$\sqrt 3 + i = 2 (\cos \frac {\pi}{6} + i\sin \frac {\pi}{6})\\
\sqrt 3 - i = 2 (\cos -\frac {\pi}{6} + i\sin -\frac {\pi}{6})\\
(\sqrt 3 + i)^n =  2^n (\cos \frac {n\pi}{6} + i\sin \frac {n\pi}{6})\\
(\sqrt 3 - i)^n = 2^n (\cos \frac {-n\pi}{6} + i\sin \frac {-n\pi}{6})$
From DeMoivre's theorem, and again, conjugates.
I suppose taking down to a primative level
if $z$ and $\bar z$ are conjugates
$(z+\bar z)$ is a real number and $z\bar z$ is a real number
$(z+\bar z)^2 - 2 z\bar z= z^2 + \bar z^2$ is real
proof by induction 
Propostion $z^n + \bar z^n$ is real
we have covered the base case
suppose the proposition is true, for all $k \le n$  We must show that
$z^{n+1} + \bar z^{n+1}$ is real
$(z^n + \bar z^n)(z + \bar z)\\
z^{n+1} + z^n\bar z + z\bar z^n + \bar z^{n+1}\\
z^{n+1} +  \bar z^{n+1}+ |z|^2 (z^{n-1}+ z^{n-1})$
$(z^{n-1}+ z^{n-1})$ is real based on the inductive hypothesis
$|z|^2 (z^{n-1}+ z^{n-1})$ is real
$z^{n+1} +  \bar z^{n+1}$ is real 
QED
A: Notice
$$z=|z|e^{i\theta}=|z|(\cos\theta + i \sin\theta)$$
Thus
$$(\sqrt{3}+i)^{n} + (\sqrt{3}-i)^n=2^n\bigg((e^{\frac{\pi}{6}})^n + (e^{-\frac{\pi}{6}})^n\bigg)$$
$$=2^n(e^{\frac{n\pi}{6}}+e^{-\frac{n\pi}{6}})=2^n\bigg(\cos(\frac{n\pi}{6}) + i\sin(\frac{n\pi}{6})+\cos(-\frac{n\pi}{6})+i\sin(-\frac{n\pi}{6})\bigg)$$
$$=2^{n+1}\cos(\frac{n\pi}{6}) \in \mathbb R$$
And the above process could be generalized to give easier proof about $z^{n} +  \bar z^{n}$ is real 
A: Hint: note that $\sqrt{3}-i = \overline{\sqrt{3}+i}$, then using the properties of the complex conjugate $\overline{uv} = \bar u \bar v\,$ (which implies $\overline{u^n} = \overline{u}^{\,n}$), $\,\overline{u+v} = \bar u + \bar v\,$, and $\,u+ \bar u = 2 \operatorname{Re}(u)\in \mathbb{R}\,$:
$$
\left(\sqrt{3}+i\right)^n + \left(\sqrt{3} - i\right)^n = \left(\sqrt{3}+i\right)^n + \left(\overline{\sqrt{3} + i}\right)^n = \left(\sqrt{3}+i\right)^n + \overline{\left(\sqrt{3} + i\right)^n} \;\in\; \mathbb{R}
$$
