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If z=sqrt(3)/2+i/2(i=sqrt(-1)) then (1+ ...

If `z=sqrt(3)/2+i/2(i=sqrt(-1))` then `(1+ iz+z^5+iz^8)^9` is equal to

A

1

B

`(-1 + 2i )^9`

C

`-1`

D

0

Text Solution

Verified by Experts

The correct Answer is:
C
Given `z = sqrt(3)/2 + (1/2)i= cos""pi/6 + i sin ""pi/6 = e^(ipi/6)`
so, `(1 + iz + z^2 + iz ^8 )^9`
`= (1+ ie^(ipi/6)+e^(i(5pi)/6)+ie^(i(8pi)/6))^9`
`= (1+ e^(ipi/2)+e^(i(pi)/6)+e^(i(5pi)/6)+e^(i(pi)/2)+e^(i(4pi)/3))^9" "[becausei=e^(ipi/2)]`
`= (1+e^(i(2pi)/3)+e^(i(5pi)/(2))+e^(i(11pi)/6))^9`
`=[1+(cos""(2pi)/(3)+ i sin""(2pi)/3)+(cos""(5pi)/6+i sin ""(5pi)/6)+(cos""(11pi)/6+isin""(11pi)/(6))]^9`
`= (1-1/2+ (isqrt(3))/2-sqrt(3)/2+1/2i+sqrt(3)/2-i/2)^9`
`=(1/2+sqrt(3i)/2)^9=(cos ""(pi)/3+i sin""pi/3)^9`
`= cos 3pi + i sin 3 pi ` [for any natural number 'n' `(cos theta + i sin theta)^n= cos (n theta)+ i sin (n theta)]` =-1
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