Question:

Starting with $a = \frac{2-\sqrt{-3}}{3} = -1 + i$ and $\bar{a} = \sqrt{2}\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = \sqrt{2}\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$. We compute $a^{10} + \bar{a}^{10} = \left[\sqrt{2}\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)\right]^{10} + \left[\sqrt{2}\left(\cos\frac{2\pi}{3} - i\sin\frac{2\pi}{3}\right)\right]^{10}$. Using De Moivre's theorem: $= 2^5\left(\cos\frac{20\pi}{3} + i\sin\frac{20\pi}{3}\right) + 2^5\left(\cos\frac{20\pi}{3} - i\sin\frac{20\pi}{3}\right) = 2^5 \cdot 2\cos\frac{20\pi}{3}$. Since $\frac{20\pi}{3} = 6\pi + \frac{2\pi}{3}$, we have $\cos\frac{20\pi}{3} = \cos\frac{2\pi}{3} = -\frac{1}{2}$. Therefore $a^{10} + \bar{a}^{10} = 32 \cdot 2 \cdot (-\frac{1}{2}) = -32$. However, the solution shows this equals $-256$.