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Let $z = \left (\frac{\sqrt3}{2} + \frac{i}{2} \right )^5 + \left (\frac{\sqrt3}{2} - \frac{i}{2} \right )^5$ . If $R(z)$ and $I(z)$ respectively denote the real and imaginary parts of $z$ , then :
Option 1)
$R(z) < 0$ and $I(z) > 0$
Option 2)
$R(z) > 0$ and $I(z) >0$
Option 3)
$R(z) = -3$
Option 4)

$I(z) =0$

Answers (1)

best_answer

Cube roots of unity -

$z=\left ( 1 \right )^{\frac{1}{3}}\Rightarrow z=\cos \frac{2k\pi }{3}+i\sin \frac{2k\pi }{3}$

k=0,1,2 so z gives three roots

$\Rightarrow 1,\frac{-1}{2}+i\frac{\sqrt{3}}{2}\left ( \omega \right ),\frac{-1}{2}-i\frac{\sqrt{3}}{2}\left ( \omega^{2} \right )$

- wherein

$\omega=\frac{-1}{2} +\frac{i\sqrt{3}}{2},\omega^{2}=\frac{-1}{2} -\frac{i\sqrt{3}}{2},\omega^{3}=1, 1+\omega+\omega^{2}=0$

$1,\omega,\omega^{2}$ are cube roots of unity.

Given

$z=(\frac{\sqrt 3}{2}+\frac{i}{2})^{5}+(\frac{\sqrt 3}{2}-\frac{i}{2})^{5}$

$z=(e^{i\frac{\pi}{6}})^{5}+(e^{-i\frac{\pi}{6}})^{5}$

$=\cos \frac{5\pi}{6}+i\sin \frac{5\pi}{6}+\cos \frac{5\pi}{6}-i\sin \frac{5\pi}{6}$

$=2\cos \frac{5\pi}{6}<0$

$Img(z)=0$

Option 1)

$R(z) < 0$ and $I(z) > 0$

Option 2)

$R(z) > 0$ and $I(z) >0$

Option 3)

$R(z) = -3$

Option 4)

$I(z) =0$

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