If z = {left( {frac{{sqrt 3 }}{2} + frac{i}{2 | vedclass

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(B) Given $z = {\left( {\frac{{\sqrt 3 }}{2} + i\frac{1}{2}} \right)^5} + {\left( {\frac{{\sqrt 3 }}{2} - i\frac{1}{2}} \right)^5}$.We know that $\fra

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$	ext{Im}(z) = 0$

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(B) Given $z = {\left( {\frac{{\sqrt 3 }}{2} + i\frac{1}{2}} ight)^5} + {\left( {\frac{{\sqrt 3 }}{2} - i\frac{1}{2}} ight)^5}$.We know that $\frac{{\sqrt 3 }}{2} + i\frac{1}{2} = \cos \left( {\frac{\pi }{6}} ight) + i\sin \left( {\frac{\pi }{6}} ight) = e^{i\pi/6}$.Similarly,$\frac{{\sqrt 3 }}{2} - i\frac{1}{2} = \cos \left( {\frac{\pi }{6}} ight) - i\sin \left( {\frac{\pi }{6}} ight) = e^{-i\pi/6}$.Using De Moivre's Theorem,$z = (e^{i\pi/6})^5 + (e^{-i\pi/6})^5 = e^{i5\pi/6} + e^{-i5\pi/6}$.This simplifies to $z = 2\cos \left( {\frac{{5\pi }}{6}} ight)$.Since $\cos \left( {\frac{{5\pi }}{6}} ight) = -\frac{{\sqrt 3 }}{2}$,we have $z = 2 	imes \left( -\frac{{\sqrt 3 }}{2} ight) = -\sqrt 3$.Since $z$ is a real number,its imaginary part $	ext{Im}(z) = 0$.

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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