If omega neq 1 is a cube root of unity,then o | vedclass

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(A) We know that $1+\omega+\omega^2 = 0$,so $1+\omega = -\omega^2$.We are looking for the $7^{	ext{th}}$ roots of $-\omega^2$.Note that $-\omega^2 =

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$1+\omega$

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(A) We know that $1+\omega+\omega^2 = 0$,so $1+\omega = -\omega^2$.We are looking for the $7^{	ext{th}}$ roots of $-\omega^2$.Note that $-\omega^2 = e^{i\pi} \cdot e^{i4\pi/3} = e^{i(7\pi/3)} = e^{i\pi/3} = -\omega^2$ is not quite right,let's use polar form.$-\omega^2 = -(\cos(4\pi/3) + i\sin(4\pi/3)) = \cos(\pi/3) + i\sin(\pi/3) = e^{i\pi/3}$.The $7^{	ext{th}}$ roots are given by $z_k = e^{i(\pi/3 + 2k\pi)/7}$ for $k=0, 1, ..., 6$.For $k=1$,$z_1 = e^{i(7\pi/3)/7} = e^{i\pi/3} = \cos(\pi/3) + i\sin(\pi/3) = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.Alternatively,note that $(-\omega^2) = (-\omega^2)^8$ is not helpful. Let's check the options.If $z = -\omega^2$,then $z^7 = (-\omega^2)^7 = -\omega^{14} = -\omega^2 = 1+\omega$.Thus,$z = -\omega^2$ is a $7^{	ext{th}}$ root of $1+\omega$.Since $-\omega^2 = 1+\omega$,the correct option is $A$.

If $\omega \neq 1$ is a cube root of unity,then one root among the $7^{\text{th}}$ roots of $(1+\omega)$ is

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