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If z=omega,omega^2w h e r eomega is a no...

If `z=omega,omega^2w h e r eomega` is a non-real complex cube root of unity, are two vertices of an equilateral triangle in the Argand plane, then the third vertex may be represented by `z=1` b. `z=0` c. `z=-2` d. `z=-1`

A

`z=1`

B

`z=0`

C

`z= -2`

D

`z = - 1`

Text Solution

Verified by Experts

The correct Answer is:
A, C
Clearly, we have to find it for real z . Let z = x Then.
`|x-w|= |x=w^(2)| = |w- w^(2)|`
`rArr (x+ (1)/(2))^(2) + (3)/(4)=|(-1+sqrt(3i))/(2) -(-1-sqrt(3i))/(2)|= 3 rArr x +(1)/(2) = pm (3)/(2)`
`rArr x = 1, -2`
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