6-24+6+24i2020 3-122+i3+1222020 cos5π12+i sin 5π122020 cos 2020 5π12+i sin 2020 5π12 cos 2525π3+i sin2525π3 =cos 5π3+i sin5π3=12-i32

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${[\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{6} + \sqrt{2}}{4} i]}^{2020} =$

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$\frac{1}{2} + \frac{\sqrt{3}}{2} i$

$\frac{- 1}{2} + \frac{\sqrt{3}}{2} i$

$\frac{- 1}{2} - \frac{\sqrt{3}}{2} i$

$\frac{1}{2} - \frac{\sqrt{3}}{2} i$

answer is D.

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Detailed Solution

$\begin{array}{rcl} {(\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{6} + \sqrt{2}}{4} i)}^{2020} \\ {(\frac{\sqrt{3} - 1}{2 \sqrt{2}} + i \frac{\sqrt{3} + 1}{2 \sqrt{2}})}^{2020} \\ {(\cos \frac{5 π}{12} + i \sin \frac{5 π}{12})}^{2020} \\ \cos 2020 (\frac{5 π}{12}) + i \sin 2020 (\frac{5 π}{12}) \\ \cos \frac{2525 π}{3} + i \sin \frac{2525 π}{3} \\ = & \cos \frac{5 π}{3} + i \sin \frac{5 π}{3} = \frac{1}{2} - \frac{i \sqrt{3}}{2} \end{array}$