Let z 1 and z 2 be n th roots of unity which subtend a right angle at the origin. Then n must be of the form [2001]

Question

Let $z_{1}$ and $z_{2}$ be $n^{th}$ roots of unity which subtend a right angle at the origin. Then $n$ must be of the form [2001]

Smart Achievers · Accepted Answer

(4)

$Let z = {(1)}^{1 / n} = {(\cos 2 k π + i \sin 2 k π)}^{1 / n}$

$\Rightarrow z = \cos \frac{2 k π}{n} + i \sin \frac{2 k π}{n}, k = 0, 1, 2, \dots, n - 1 .$

$Let z_{1} = \cos (\frac{2 k_{1} π}{n}) + i \sin (\frac{2 k_{1} π}{n})$

$and z_{2} = \cos (\frac{2 k_{2} π}{n}) + i \sin (\frac{2 k_{2} π}{n})$

$be the two values of z such that they subtend right angle at origin.$

$∴$ $\frac{2 k_{1} π}{n} - \frac{2 k_{2} π}{n} = \pm \frac{π}{2}$ $\Rightarrow 4 (k_{1} - k_{2}) = \pm n$

$As k_{1} and k_{2} are integers and k_{1} \neq k_{2},$

$∴$ $n = 4 k, k \in I .$

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