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}` Wher

Question

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

Answer 1

Correct Answer - C::D
PLAN It is the simple representaion of points on Argand plane and to find the angle between the points.
Here, `P= W^n =(cos""pi/6+ I sin """"pi/6)^n= cos""(npi)/(6)+ isin((npi)/6)`
`H_={z inC:Re (z)gt 1/2}`
`therefore P capH_1` represents those points for which cod `(npi)/(6)` is + ve .
Hence , it belongs to I or IV quradrant .
`rArr z_1 = P capH_1 = cos ""(pi)/(6)+i sin ""(pi)/6 cos ""(11pi)/(6)+ i sin""(11 pi)/(6)`
`therefore z_1 = (sqrt(3))/2+i/2 or (sqrt(3))/(2)-i/2`
Similarly `z_2=P cap H_2` i.e those points for which
cos ` (npi)/(6)lt 0 `

`therefore z_2 = cos pi + i sin pi, cos (5 pi)/(6)+ i sin (5 pi)/6,(cos 7 pi)/6 + i sin (7 pi)/(6)`
`rArr z_1 = -1 ,(-sqrt(3))/2 + i/ 2,(-sqrt(3))/(2) - i/2`
Thus `angle z_1 Oz_2 = (2 pi)/(3),(5pi)/(6),pi`