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The value of $\left(\frac{1+\sqrt{3} i}{1-\sqrt{3 i}}\right)^{64}+\left(\frac{1-\sqrt{3} i}{1+\sqrt{3} i}\right)^{64}$ is
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The correct answer is:
-1
We know that, $\omega=\frac{-1+\sqrt{3} i}{2} 1-\sqrt{3} i=-2 \omega$
and $\omega^{2}=\frac{-1-\sqrt{3} i}{2} 1+\sqrt{3} i=-2 \omega^{2}$
Now, $\begin{aligned}\left(\frac{1+\sqrt{3} i}{1-\sqrt{3 i}}\right)^{64}+\left(\frac{1-\sqrt{3} i}{1+\sqrt{3} i}\right)^{64} \\ &=\left(\frac{-2 \omega^{2}}{-2 \omega}\right)^{64}+\left(\frac{-2 \omega}{-2 \omega^{2}}\right)^{64} \end{aligned}$
$=\omega^{64}+\frac{1}{\omega^{64}}=\omega+\omega^{2}$
$$
\left[\because \cdot \omega^{3}=1\right]
$$
$$
\left.=-1 \quad \mid \because 1+\omega+\omega^{2}=0\right]
$$
and $\omega^{2}=\frac{-1-\sqrt{3} i}{2} 1+\sqrt{3} i=-2 \omega^{2}$
Now, $\begin{aligned}\left(\frac{1+\sqrt{3} i}{1-\sqrt{3 i}}\right)^{64}+\left(\frac{1-\sqrt{3} i}{1+\sqrt{3} i}\right)^{64} \\ &=\left(\frac{-2 \omega^{2}}{-2 \omega}\right)^{64}+\left(\frac{-2 \omega}{-2 \omega^{2}}\right)^{64} \end{aligned}$
$=\omega^{64}+\frac{1}{\omega^{64}}=\omega+\omega^{2}$
$$
\left[\because \cdot \omega^{3}=1\right]
$$
$$
\left.=-1 \quad \mid \because 1+\omega+\omega^{2}=0\right]
$$
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