left(frac{sqrt{6}-sqrt{2}}{4}+frac{sqrt{6}+sq | vedclass

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(D) Given expression: $z = \left(\frac{\sqrt{6}-\sqrt{2}}{4}+i \frac{\sqrt{6}+\sqrt{2}}{4}\right)^{2020}$ We know that $\cos \frac{5 \pi}{12} = \frac{

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

Vedclass · Answer

(D) Given expression: $z = \left(\frac{\sqrt{6}-\sqrt{2}}{4}+i \frac{\sqrt{6}+\sqrt{2}}{4}ight)^{2020}$ We know that $\cos \frac{5 \pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}$ and $\sin \frac{5 \pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4}$. So,$z = \left(\cos \frac{5 \pi}{12} + i \sin \frac{5 \pi}{12}ight)^{2020}$. Using De Moivre's theorem,$(\cos 	heta + i \sin 	heta)^n = \cos(n 	heta) + i \sin(n 	heta)$: $z = \cos \left(2020 	imes \frac{5 \pi}{12}ight) + i \sin \left(2020 	imes \frac{5 \pi}{12}ight)$ $z = \cos \left(\frac{505 	imes 5 \pi}{3}ight) + i \sin \left(\frac{505 	imes 5 \pi}{3}ight)$ $z = \cos \left(\frac{2525 \pi}{3}ight) + i \sin \left(\frac{2525 \pi}{3}ight)$ Since $\frac{2525 \pi}{3} = 841 \pi + \frac{2 \pi}{3} = 842 \pi - \frac{\pi}{3}$,$z = \cos \left(-\frac{\pi}{3}ight) + i \sin \left(-\frac{\pi}{3}ight)$ $z = \cos \frac{\pi}{3} - i \sin \frac{\pi}{3} = \frac{1}{2} - \frac{\sqrt{3}}{2} i$.

$\left(\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4} i\right)^{2020} =$

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