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Let w = (sqrt 3 + iota/2) and P = { w^n ...

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

`5pi//6`

Text Solution

Verified by Experts

The correct Answer is:
C, D
`w=(sqrt(3) +i)/(2) = e^((ipi)/(6))`,so`w^(n) = e^(i((npi)/( 6))), n =0,1,2,3,.......,12`

Now, for `z_(1) ,cos.(npi)/(6)gt(1)/(2)` and for `z_(2),cos.(npi)/(6)lt -(1)/(2)`
Possible position of `z_(1)` are `A_(1),A_(2),A_(3)` whereas of `z_(2)` are `B_(1),B_(2),B_(3)` (as show in the figure).
So possible value of `/_z_(1)Oz_(2)` according to the given options is `(2pi)/(3) or (5pi)/(6)`
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