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If $\mathrm{z}=\left(\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2}\right)^{107}+\left(\frac{\sqrt{3}}{2}-\frac{\mathrm{i}}{2}\right)^{107}$, then what is the
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$z=\left[\frac{\sqrt{3}}{2}+\frac{i}{2}\right]^{107}+\left[\frac{\sqrt{3}}{2}-\frac{i}{2}\right]^{107} .$
$\because \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2} \& \sin \frac{\pi}{6}=\frac{1}{2}$
$\Rightarrow \quad \mathrm{z}=\left[\cos \frac{\pi}{6}+\mathrm{i} \sin \frac{\pi}{6}\right]^{107}+\left[\cos \frac{\pi}{6}-\mathrm{i} \sin \frac{\pi}{6}\right]^{107}$
$\mathrm{Also},(\cos \theta+\mathrm{i} \sin \theta)^{\mathrm{n}}=\cos n \theta+\mathrm{i} \sin \mathrm{n} \theta$
$\Rightarrow \quad \mathrm{z}=\cos \frac{107 \pi}{6}+\mathrm{i} \sin \frac{107 \pi}{6}+\cos \frac{107 \pi}{6}-\mathrm{i} \sin \frac{107 \mathrm{t}}{6} .$
$\Rightarrow \quad \operatorname{Im}(\mathrm{z})=0$
$\because \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2} \& \sin \frac{\pi}{6}=\frac{1}{2}$
$\Rightarrow \quad \mathrm{z}=\left[\cos \frac{\pi}{6}+\mathrm{i} \sin \frac{\pi}{6}\right]^{107}+\left[\cos \frac{\pi}{6}-\mathrm{i} \sin \frac{\pi}{6}\right]^{107}$
$\mathrm{Also},(\cos \theta+\mathrm{i} \sin \theta)^{\mathrm{n}}=\cos n \theta+\mathrm{i} \sin \mathrm{n} \theta$
$\Rightarrow \quad \mathrm{z}=\cos \frac{107 \pi}{6}+\mathrm{i} \sin \frac{107 \pi}{6}+\cos \frac{107 \pi}{6}-\mathrm{i} \sin \frac{107 \mathrm{t}}{6} .$
$\Rightarrow \quad \operatorname{Im}(\mathrm{z})=0$
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