Let `Z_0` is the root of equation `x^2+x+1=0` and `Z=3+6i(Z_0)^(81)-3i(Z_0)^(93)` Then arg `(Z)` is equal to (a) `(pi)/(4)` (b) `(pi)/(3)` (c) `pi` (d) `(pi)/(6)`

Options

A. `pi/4`
B. `pi/6`
C. 0
D. `pi/3`

Answer

A

Explanation

Given `x^2 + x +1 =0 `
`rArr x- (-1 pm sqrt(3)i)/2`
[`because` Roots of quadratic equation `alphax^2+bx+c=0 `are given by
are given by x ` x =(-b pm sqrt(b^2-4 ac ))/(2a)`
`rArr z_0 = omega ,omega^2 [" where" omega =(-1 + sqrt(3)i)/2 " and " omega^2=(-1 - sqrt(3) i )/(2)`
are the cube roots of unity and `omega , omega ^ 2 ne 1)`
Now consider `z= 3+ 6i z_0^81- 3 i zz_0^93 `
`= 3+ 6i - 3i " " ( because omega ^(3n) = ( omega^(2))^(3n)=1)`
`= 3+ 3 i =3( 1+ i)`
If `theta` is the argument of z, then
`tan theta =(Im(z))/(Re (z)) " " [ becaues ` z is in the first quadrant ]
`=3/3 = 1 rArr theta- pi/4`