Let w = (`sqrt 3 + iota/2)` and `P = { w^n : n = 1,2,3, ..... },` Further `H_1 = { z in C: Re(z) > 1/2} and H_2 = { z in c : Re(z) < -1/2}` Where C is set of all complex numbers. If `z_1 in P nn H_1 , z_2 in P nn H_2` and O represent the origin, then `/_Z_1OZ_2` =

Options

A. `pi/2, (5pi)/6`
B. `pi, (2pi)/3`
C. `(2pi)/3, (5pi)/3`
D. `(5pi)/3, (7pi)/3`

Answer

B

Explanation

We have, `omega=sqrt(3)/2+1/2i=cospi/6+isinpi/6`
`rArr omega^(n)=cos(npi)/6+isin(npi)/6, n=1,2,3,…………`
`therefore P={omega^(n), n=1,2,3,…..} = {cos(npi)/6+ isin(npi)/6, n=1,2,3,….}`

Now, `z_(1) in P frown H_(1)`
`rArr z_(1) in P` and `z_(1) in H_(1)`
`rArr z_(1)cos(npi)/6+isin(npi)/6` for some `n in Z` and Re `(z_(1)) gt 1/2`
`rArr z_(1)=sqrt(2)/2 +1/2i, sqrt(3)/2-1/2i`
`z_(2) in P frown H_(2)`
`rArr z_(2) in P` and `z_(2) in H_(2)`
`rArr z_(2) = cos(npi)/6+isin(npi)/6` for some `n in N` and Re `z_(2) lt -1/2`
`rArr z_(2) = -sqrt(3)/2+1/2i, -sqrt(3)/2-1/2i`
If `z_(1)=sqrt(3)/2+1/2i` and `z_(2)=-sqrt(3)/2+1/2i`, then
`anglez_(1)Oz_(2)=angleAOC=pi-(pi/6+pi/6)=(2pi)/3`
If `z_(1)=sqrt(3)/2+1/2i` and `z_(2)=-sqrt(3)/2-1/2i`, then
`anglez_(1)Oz_(2)=angleAOD=pi`
If `z_(1)=sqrt(3)/2+1/2i` and `z_(2)=-sqrt(3)/2-1/2i`, then
If `z_(1)=sqrt(3)/2-1/2i` and `z_(2)=-sqrt(3)/2-1/2i`, then
`anglez_(1)Oz_(2)=angleBOD=(2pi)/3`
`If z_(1)=sqrt(3)/2-1/2i` and `z_(2)=-sqrt(3)/2+1/2i`, then
`anglez_(1)Oz_(2)=angleBOC=pi`
Hence, `anglez_(1)Oz_(2)=pi` or `(2pi)/3`