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Let omega ne 1 be a complex cube root of...

Let `omega ne 1` be a complex cube root of unity. If `(3-3omega+2omega^(2))^(4n+3) + (2+3omega-3omega^(2))^(4n+3)+(-3+2omega+2omega^(2))^(4n+3)=0`, then the set of possible value(s) of n is are

A

N

B

`{3k: k in N}`

C

`N-(3k:k in N)`

D

`{6k:k in N}`

Text Solution

Verified by Experts

The correct Answer is:
C

We have,
`(3-3omega+2omega^(2))+(2+3omega-3omega^(2))^(4n+3) +(-3+2omega+3omega^(2))^(4n+3)=0`
`rArr (3-3omega+2omega^(2))^(4n+3)(1+omega^(4n+3))(1+omega^(n)+omega^(2n))=0`
`rArr 1+omega^(n)+omega^(2n)=0`
`rArr` n is not a multiple of 3
Hence, the set of values of n is `N-(n:n=3k, k in N)`
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OBJECTIVE RD SHARMA-COMPLEX NUMBERS -Chapter Test
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  12. The expression (1+i)^(n1)+(1+i^(3))^(n(2)) is real iff

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