Let z 1 and z 2 be the n t h 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

Step-by-step solution

1 1 n = cos ⁡ 2 r π n + i sin ⁡ 2 r π n , r = 0 , 1 , … … , n - 1 Let, z 1 = cos ⁡ 2 r 1 π n + i sin ⁡ 2 r 1 π n And, z 2 = cos ⁡ 2 r 2 π n + i sin ⁡ 2 r 2 π n . Then, ∠ z 1 O z 2 = a m p z 1 z 2 = a m p z 1 − a m p z 2 = 2 r 1 - r 2 π n = π 2 (Given) ∴ n = 4 r 1 - r 2 = 4 × integer, so n is of the form 4 k .