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

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(D) We know that $\frac{\sqrt{3}}{2} + \frac{i}{2} = e^{i\pi/6}$ and $\frac{\sqrt{3}}{2} - \frac{i}{2} = e^{-i\pi/6}$.Thus,$z = (e^{i\pi/6})^5 + (e^{-

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$I(z) = 0$

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(D) We know that $\frac{\sqrt{3}}{2} + \frac{i}{2} = e^{i\pi/6}$ and $\frac{\sqrt{3}}{2} - \frac{i}{2} = e^{-i\pi/6}$.Thus,$z = (e^{i\pi/6})^5 + (e^{-i\pi/6})^5 = e^{i5\pi/6} + e^{-i5\pi/6}$.Using Euler's formula,$e^{i	heta} + e^{-i	heta} = 2\cos(	heta)$,we get $z = 2\cos(5\pi/6)$.Since $\cos(5\pi/6) = -\frac{\sqrt{3}}{2}$,we have $z = 2(-\frac{\sqrt{3}}{2}) = -\sqrt{3}$.Here,$R(z) = -\sqrt{3}$ and $I(z) = 0$.Therefore,$I(z) = 0$ is the correct statement.

Let $z = \left( \frac{\sqrt{3}}{2} + \frac{i}{2} \right)^5 + \left( \frac{\sqrt{3}}{2} - \frac{i}{2} \right)^5$. If $R(z)$ and $I(z)$ respectively denote the real and imaginary parts of $z$,then:

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