If z_1 and z_2 are two of the n^{text{th}} ro | vedclass

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(A) The $n^{	ext{th}}$ roots of unity are given by $z_r = e^{i \frac{2\pi r}{n}}$ for $r = 0, 1, \dots, n-1$.Let $z_1 = e^{i \frac{2\pi r_1}{n}}$ and

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$4k$

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(A) The $n^{	ext{th}}$ roots of unity are given by $z_r = e^{i \frac{2\pi r}{n}}$ for $r = 0, 1, \dots, n-1$.Let $z_1 = e^{i \frac{2\pi r_1}{n}}$ and $z_2 = e^{i \frac{2\pi r_2}{n}}$.The angle subtended by the line segment joining $z_1$ and $z_2$ at the origin is the difference of their arguments: $	heta = |\arg(z_1) - \arg(z_2)| = |\frac{2\pi r_1}{n} - \frac{2\pi r_2}{n}| = \frac{2\pi}{n} |r_1 - r_2|$.Given that this angle is a right angle,we have $\frac{2\pi}{n} |r_1 - r_2| = \frac{\pi}{2}$.This simplifies to $\frac{2}{n} |r_1 - r_2| = \frac{1}{2}$,which implies $n = 4 |r_1 - r_2|$.Since $|r_1 - r_2|$ is an integer,let $|r_1 - r_2| = k$,where $k$ is a positive integer.Therefore,$n$ must be of the form $4k$.

If $z_1$ and $z_2$ are two of the $n^{\text{th}}$ roots of unity such that the line segment joining them subtends a right angle at the origin,then for a positive integer $k$,$n$ takes the form

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