Evaluate $(\sqrt{3}+i)^{14}+(\sqrt{3}-i)^{14}$ Evaluate $(\sqrt{3}+i)^{14}+(\sqrt{3}-i)^{14}$
I tried by using the De'moivers theorem but I didn't get proper value I get a mess value 4..but I am not sure about the answer can anyone please tell me

First of all, try to factorise by $2^{14}$ :).
$(\sqrt{3} + i)^{14} + (\sqrt{3} - i)^{14} = 2^{14}(\dfrac{\sqrt{3}}{2} + \dfrac{i}{2})^{14} + 2^{14}(\dfrac{\sqrt{3}}{2} - \dfrac{i}{2})^{14} = 2^{14}\left((\dfrac{\sqrt{3}}{2} + \dfrac{i}{2})^{14} + (\dfrac{\sqrt{3}}{2} - \dfrac{i}{2})^{14}\right)$
Now, use the fact that: $\dfrac{\sqrt{3}}{2} + \dfrac{i}{2} = \cos(\pi/6) + i\sin(\pi/6) = \exp(i\pi/6)$.
And similarly: $\dfrac{\sqrt{3}}{2} - \dfrac{i}{2} = \cos(-\pi/6) + i\sin(-\pi/6) = \exp(-i\pi/6)$.
I think you'll be able to finish from there, by using De Moivre's theorem :)
