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If `z_(1)` and `z_(2)` are the complex roots of the equation `(x-3)^(3) + 1=0`, then `z_(1) +z_(2)` equal to

A

1

B

3

C

5

D

7

Text Solution

Verified by Experts

The correct Answer is:
D

`(x-3)^(2)+1=0`
`rArr((x-3)/(-1))^(3)=1`
`rArr (x-3)/(-1)=1,omega,omega^(2)`
`rArr x = 2,3 - omega,3-omega^(2)`
Hence, the sum of complex roots is `6-(omega-omega^(2))=6+1=7`.
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