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Let Z0 is the root of equation x^2+x+1=0...

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)`

A

`pi/4`

B

`pi/6`

C

`0`

D

`pi/3`

Text Solution

Verified by Experts

The correct Answer is:
A
`Z_(0)" is a root of "x^(2)+x+1=0," "Z_(2)" is either " omega or omega^(2)`
`Z=3+6i(omega)^(81)-3i(omega)^(93)," "=3+6i-3i=3+3i`
`"arg "(z)=(pi)/(4)`
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