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If a=cos((4pi)/3)+isin((4pi)/3) then ...

If `a=cos((4pi)/3)+isin((4pi)/3)` then `|(1,1,1),(1,a,a^2),(1,a^2,a)|`

A

purely real

B

purely imaginary

C

`0`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
`(b)` `a=cos'(4pi)/(3)+isin"(4pi)/(3)=omega^(2)`
`:.``|(1,1,1),(1, omega^(2),omega),(1,omega,omega^(2))|`=`3(omega-omega^(2))` purely imaginery
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