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If $z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5$ then
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Verified Answer
The correct answer is:
$\operatorname{Im}(z)=0$
Given that $z=\left(\frac{\sqrt{3}}{2}+i \frac{1}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-i \frac{1}{2}\right)^5$
$\begin{aligned} & =\left[\cos \left(\frac{\pi}{6}\right)+i \sin \left(\frac{\pi}{6}\right)\right]^5+\left[\cos \left(\frac{\pi}{6}\right)-i \sin \left(\frac{\pi}{6}\right)\right]^5 \\ & =\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}+\cos \frac{5 \pi}{6}-i \sin \frac{5 \pi}{6}\end{aligned}$
$\begin{aligned} & =\left[\cos \left(\frac{\pi}{6}\right)+i \sin \left(\frac{\pi}{6}\right)\right]^5+\left[\cos \left(\frac{\pi}{6}\right)-i \sin \left(\frac{\pi}{6}\right)\right]^5 \\ & =\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}+\cos \frac{5 \pi}{6}-i \sin \frac{5 \pi}{6}\end{aligned}$
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