If z = left(frac{sqrt{3}+i}{2}right)^5 + left | vedclass

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(D) Let $\omega = \frac{\sqrt{3}+i}{2} = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.Then the given expression is $

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$\operatorname{Im}(z) = 0$

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(D) Let $\omega = \frac{\sqrt{3}+i}{2} = \cos\left(\frac{\pi}{6}ight) + i\sin\left(\frac{\pi}{6}ight) = e^{i\pi/6}$.Then the given expression is $z = (e^{i\pi/6})^5 + (e^{-i\pi/6})^5$.$z = e^{i5\pi/6} + e^{-i5\pi/6}$.Using the identity $e^{i	heta} + e^{-i	heta} = 2\cos(	heta)$, we get $z = 2\cos\left(\frac{5\pi}{6}ight)$.Since $\cos\left(\frac{5\pi}{6}ight) = -\frac{\sqrt{3}}{2}$, we have $z = 2\left(-\frac{\sqrt{3}}{2}ight) = -\sqrt{3}$.Since $z = -\sqrt{3} + 0i$, the imaginary part $\operatorname{Im}(z) = 0$.

If $z = \left(\frac{\sqrt{3}+i}{2}\right)^5 + \left(\frac{\sqrt{3}-i}{2}\right)^5$, then

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