Let {z_1} and {z_2} be n^{th} roots of unity | vedclass

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(D) The $n^{th}$ roots of unity are given by ${z_r} = \cos \frac{2r\pi}{n} + i\sin \frac{2r\pi}{n}$ for $r = 0, 1, \dots, n-1$.Let ${z_1} = \cos \frac

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$4k$

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(D) The $n^{th}$ roots of unity are given by ${z_r} = \cos \frac{2r\pi}{n} + i\sin \frac{2r\pi}{n}$ for $r = 0, 1, \dots, n-1$.Let ${z_1} = \cos \frac{2r_1\pi}{n} + i\sin \frac{2r_1\pi}{n}$ and ${z_2} = \cos \frac{2r_2\pi}{n} + i\sin \frac{2r_2\pi}{n}$.The angle subtended by the segment joining ${z_1}$ and ${z_2}$ at the origin is given by the argument of the ratio $\frac{z_1}{z_2}$.$	ext{arg}\left(\frac{z_1}{z_2}ight) = 	ext{arg}(z_1) - 	ext{arg}(z_2) = \frac{2(r_1 - r_2)\pi}{n}$.Since the angle is a right angle,we have $\frac{2(r_1 - r_2)\pi}{n} = \pm \frac{\pi}{2}$.This implies $\frac{2(r_1 - r_2)}{n} = \pm \frac{1}{2}$,which simplifies to $n = \pm 4(r_1 - r_2)$.Since $r_1$ and $r_2$ are integers,$n$ must be a multiple of $4$,i.e.,$n = 4k$ for some integer $k$.

Let ${z_1}$ and ${z_2}$ be $n^{th}$ roots of unity which are ends of a line segment that subtend a right angle at the origin. Then $n$ must be of the form

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