Q. Let `z_1` and `z_2` be nth roots of unity which subtend a right angle at the origin, then n must be the form `4k`.

Options

A. `4k +1`
B. `4 k +2`
C. `4 k +3`
D. 4k

Answer

D

Explanation

Since arg`z_1/z_2=pi/2`
`rArr " " z_1/z_2=cos""lpi/2+I sin ""pi/2=i`
`therefore z_1/z_2=cos""pi/2+ I sin ""pi/2=i`
`therefore" "z_1^n=(i)^n rArr i^n =1" "[because |z_2|=|z_1|=1]`
`rArr n=4k `
Alternate Solution
Since arg `z_2/z_1 = pi/2`
`therefore z_2/z_1=|z_2/z_1|e^(ipi/2)`
`rArr z_2/z_1 =i " "[ because |z_1|=|z_2|=1]`
`rArr ((z_2)/(z_1))^n=i^n`
`therefore z_1 " and " z_2` are nth roots of unity
`rArr ((z_2)/(z_1))^(z_1^n=z_2^n=1)`
`rArr i^n=1`
`rArr n =4k` where k is an integer.