If n is a positive integer, then (1 + isqrt{3 | vedclass

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

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$2^{2n+1} \cos(\frac{2n\pi}{3})$

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(A) Let $z_1 = 1 + i\sqrt{3} = 2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 2e^{i\pi/3}$.Similarly, $z_2 = 1 - i\sqrt{3} = 2(\cos \frac{\pi}{3} - i \sin \frac{\pi}{3}) = 2e^{-i\pi/3}$.Then, $(z_1)^{2n} + (z_2)^{2n} = (2e^{i\pi/3})^{2n} + (2e^{-i\pi/3})^{2n}$.$= 2^{2n} e^{i2n\pi/3} + 2^{2n} e^{-i2n\pi/3}$.$= 2^{2n} (e^{i2n\pi/3} + e^{-i2n\pi/3})$.Using Euler's formula, $\cos 	heta = \frac{e^{i	heta} + e^{-i	heta}}{2}$, so $e^{i	heta} + e^{-i	heta} = 2 \cos 	heta$.$= 2^{2n} \cdot 2 \cos(\frac{2n\pi}{3}) = 2^{2n+1} \cos(\frac{2n\pi}{3})$.

If $n$ is a positive integer, then $(1 + i\sqrt{3})^{2n} + (1 - i\sqrt{3})^{2n}$ is equal to .........

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