The value of left(frac{1+sqrt{3} i}{1-sqrt{3} | vedclass

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(B) Let $\omega = \frac{-1+i\sqrt{3}}{2}$ be the cube root of unity. Then $1+i\sqrt{3} = 2\omega^2$ and $1-i\sqrt{3} = 2\omega$.Substituting these int

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$-1$

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(B) Let $\omega = \frac{-1+i\sqrt{3}}{2}$ be the cube root of unity. Then $1+i\sqrt{3} = 2\omega^2$ and $1-i\sqrt{3} = 2\omega$.Substituting these into the expression:$\left(\frac{2\omega^2}{2\omega}\right)^{64} + \left(\frac{2\omega}{2\omega^2}\right)^{64} = (\omega)^{64} + \left(\frac{1}{\omega}\right)^{64}$Since $\omega^3 = 1$,we have $\omega^{64} = (\omega^3)^{21} \cdot \omega = \omega$ and $\frac{1}{\omega^{64}} = \omega^{64} = \omega^2$ (since $\frac{1}{\omega} = \omega^2$).Thus,the expression becomes $\omega + \omega^2$.Using the identity $1 + \omega + \omega^2 = 0$,we get $\omega + \omega^2 = -1$.

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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