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Let z(k) = cos((2kpi)/(10)) -isin ((2kp...

Let `z_(k) = cos((2kpi)/(10)) -isin ((2kpi)/(10)), k = 1,2,…..,9`

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The correct Answer is:
1
`z_(k)` is `10^(th)` root of unity.
So, `barz_(k)` will also be `10^(th)` root of unity.
Take `barz(j)` as `barz_(k)`.
b `z_(1) ne 0 ` take `z = (z_(k))/(z_(1))` we can always find z.
c `z^(10)-1= (z-1)(z-z_(1))….(z-z_(9))`
`rArr (z-z_(1)) (z-z_(2))....(z-z_(9))`
`= 1+ z+ z^(2) + .....+ ^(9) AA z in ` complex number
Put z = 1
`rArr (1-z_(1)) (1-z_(2)) .....(1-z_(9)) = 10`
`rArr (|1-z_(1)||1-Z_(2)|......|1-z_(9)|)/(10) = 1`
d `1+z_(1)+z_(2)+.......+z_(9) = 0`
`rArr Re (1) + Re(z_(1)) + .....+Re(z_(9)) = 0`
`rArr Re(z_(1)) + Re(z_(2)) + .......+Re(z_(9)) = -1`
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