If i^2 = -1,then (1 + sqrt{3} i)^{2022} - (sq | vedclass

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(A) Let $z_1 = 1 + \sqrt{3} i = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 2e^{i\pi/3}$.Then $z_1^{2022} = 2^{2022} e^{i(2022\pi/3)} = 2^{2022} e^

Vedclass · Accepted Answer

$2^{2023}$

Vedclass · Answer

(A) Let $z_1 = 1 + \sqrt{3} i = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 2e^{i\pi/3}$.Then $z_1^{2022} = 2^{2022} e^{i(2022\pi/3)} = 2^{2022} e^{i(674\pi)} = 2^{2022} 	imes 1 = 2^{2022}$.Let $z_2 = \sqrt{3} - i = 2(\cos \frac{\pi}{6} - i \sin \frac{\pi}{6}) = 2e^{-i\pi/6}$.Then $z_2^{2022} = 2^{2022} e^{-i(2022\pi/6)} = 2^{2022} e^{-i(337\pi)} = 2^{2022} 	imes (-1) = -2^{2022}$.Therefore,$z_1^{2022} - z_2^{2022} = 2^{2022} - (-2^{2022}) = 2^{2022} + 2^{2022} = 2 	imes 2^{2022} = 2^{2023}$.

If $i^2 = -1$,then $(1 + \sqrt{3} i)^{2022} - (\sqrt{3} - i)^{2022} = $

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