If i is the root of the equation x^2+1=0,then | vedclass

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(B) Let $z_1 = 1+\sqrt{3}i = 2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 2e^{i\pi/3}$.Similarly,$z_2 = 1-\sqr

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$2^{2023}$

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(B) Let $z_1 = 1+\sqrt{3}i = 2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 2e^{i\pi/3}$.Similarly,$z_2 = 1-\sqrt{3}i = 2(\cos \frac{\pi}{3} - i\sin \frac{\pi}{3}) = 2e^{-i\pi/3}$.Then,$(1+\sqrt{3}i)^{2023} + (1-\sqrt{3}i)^{2023} = (2e^{i\pi/3})^{2023} + (2e^{-i\pi/3})^{2023}$.$= 2^{2023} (e^{i2023\pi/3} + e^{-i2023\pi/3})$.Since $2023 = 3 	imes 674 + 1$,we have $\frac{2023\pi}{3} = 674\pi + \frac{\pi}{3}$.Thus,$e^{i2023\pi/3} = e^{i(674\pi + \pi/3)} = e^{i\pi/3} = \cos \frac{\pi}{3} + i\sin \frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.Similarly,$e^{-i2023\pi/3} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.Sum $= 2^{2023} (\frac{1}{2} + i\frac{\sqrt{3}}{2} + \frac{1}{2} - i\frac{\sqrt{3}}{2}) = 2^{2023}(1) = 2^{2023}$.

If $i$ is the root of the equation $x^2+1=0$,then $(1+\sqrt{3}i)^{2023}+(1-\sqrt{3}i)^{2023}=$

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