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If omega(!=1) is a cube root of unity, ...

If `omega(!=1)` is a cube root of unity, and `(1+omega)^7= A + B omega` . Then (A, B) equals

A

(0,1)

B

(1,1)

C

77

D

64

Text Solution

Verified by Experts

The correct Answer is:
B
We have,
`(1+omega)^(7)=A+Bomega`
`rArr (-omega)^(7)=A+Bomega`
`rArr -omega^(14)=A+Bomega`
`rArr -omega^(2)=A+Bomega rArr A=1,B=1`
Hence, `(A,B)=(1,1)`
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OBJECTIVE RD SHARMA ENGLISH-COMPLEX NUMBERS -Chapter Test
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  9. Find the sum 1xx(2-omega)xx(2-omega^(2))+2xx(-3-omega)xx(3-omega^(2))+...

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  10. The value of the expression (1+(1)/(omega))+(1+(1)/(omega^(2)))+(2+(1)...

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  11. The condition that x^(n+1)-x^(n)+1 shall be divisible by x^(2)-x+1 is ...

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  12. The expression (1+i)^(n1)+(1+i^(3))^(n2) is real iff

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  16. Sum of the series sum(r=0)^n (-1)^r ^nCr[i^(5r)+i^(6r)+i^(7r)+i^(8r)] ...

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  17. If az(1)+bz(2)+cz(3)=0 for complex numbers z(1),z(2),z(3) and real num...

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  18. If 2z1-3z2 + z3=0, then z1, z2 and z3 are represented by

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  19. If Re((z+4)/(2z-1)) = 1/2 then z is represented by a point lying on

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  20. The vertices of a square are z(1),z(2),z(3) and z(4) taken in the anti...

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  21. Let lambda in R . If the origin and the non-real roots of 2z^2+2z+lam...

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