compute $(\sqrt 3 + i)^{14} + ( \sqrt 3 - i)^{14}$

Question

I want to compute $(\sqrt 3 + i)^{14} + ( \sqrt 3 - i)^{14}$.

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My attempts: I was thinking about De Moivre's theorem

$$(\cos\theta + i \sin\theta)^n= \cos(n\theta) + i \sin(n\theta)$$

but I don't know how it can be applied here. Any hints/solution will be appreciated.

Thank you.

Answer 1

You're right but you have have to express it in exponential (polar) form, $$\sqrt{3} + i = 2 e^{i \frac{\pi}{6}}$$ and $$\sqrt{3} - i = 2 e^{-i \frac{\pi}{6}}$$ So that $$(\sqrt{3} + i)^{14} = 2^{14} e^{i \frac{14\pi}{6}}$$ and $$(\sqrt{3} - i)^{14} = 2^{14} e^{-i \frac{14\pi}{6}}$$ Since $(\sqrt{3} + i)^{14}$ and $(\sqrt{3} - i)^{14}$ are complex conjugates, then $(\sqrt{3} + i)^{14} + (\sqrt{3} - i)^{14} = 2\Re\{(\sqrt{3} + i)^{14}\} = 2(2^{14} \cdot\Re\{e^{i \frac{\pi}{6}}\}) $

Answer 2

HINT: $\sqrt{3}+i=2(\cos(\pi/6)+i\sin(\pi/6))$, $\sqrt{3}-i=2(\cos(-\pi/6)+i\sin(-\pi/6))$.