{left( frac{-1 + isqrt{3}}{2} right)^{20}} + | vedclass

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Question

(D) We know that the cube roots of unity are $1, \omega, \omega^2$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(D) We know that the cube roots of unity are $1, \omega, \omega^2$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.Given expression is $\omega^{20} + (\omega^2)^{20} = \omega^{20} + \omega^{40}$.Since $\omega^3 = 1$,we have $\omega^{20} = (\omega^3)^6 \cdot \omega^2 = 1^6 \cdot \omega^2 = \omega^2$.Similarly,$\omega^{40} = (\omega^3)^{13} \cdot \omega = 1^{13} \cdot \omega = \omega$.Thus,the expression becomes $\omega^2 + \omega$.Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 + \omega = -1$.

${\left( \frac{-1 + i\sqrt{3}}{2} \right)^{20}} + {\left( \frac{-1 - i\sqrt{3}}{2} \right)^{20}} = $

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