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If $\alpha=\frac{1+\mathrm{i} \sqrt{3}}{2}$, then what is the value of $1+\alpha^{8}+\alpha^{16}+$ $\alpha^{24}+\alpha^{32} ?$
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The correct answer is:
$-\omega^{2}$
$\frac{-1+\mathrm{i} \sqrt{3}}{2}$ and $\frac{-1-\mathrm{i} \sqrt{3}}{2}$
are complex cube roots of unity $\omega$ and $\omega^{2}$
$\begin{aligned} & \alpha=\frac{1+\mathrm{i} \sqrt{3}}{2}=-\omega^{2} \\ \therefore \quad & 1+\alpha^{8}+\alpha^{16}+\alpha^{24}+\alpha^{32} \\=& 1+\omega^{16}+\omega^{32}+\omega^{48}+\omega^{64} \\=& 1+\omega+\omega^{2}+1+\omega=0+1+\omega=-\omega^{2} \end{aligned}$
are complex cube roots of unity $\omega$ and $\omega^{2}$
$\begin{aligned} & \alpha=\frac{1+\mathrm{i} \sqrt{3}}{2}=-\omega^{2} \\ \therefore \quad & 1+\alpha^{8}+\alpha^{16}+\alpha^{24}+\alpha^{32} \\=& 1+\omega^{16}+\omega^{32}+\omega^{48}+\omega^{64} \\=& 1+\omega+\omega^{2}+1+\omega=0+1+\omega=-\omega^{2} \end{aligned}$
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