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Let A=θπ2,π:3+2isinθ12isinθ is purely imaginary . Then the sum of the elements in A .is

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a
3π/4
b
2π/3
c
π
d
5π/6

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detailed solution

Correct option is B

Let z=3+2isinθ12isinθ

=3+2isinθ12isinθ×1+2isinθ1+2isinθ=34sin2θ+i(8sinθ)1+4sin2θ

Now, (z)=0                                        [ z is purely imaginary]

34sin2θ1+4sin2θ=0sin2θ=34sin2θ=sin2π3 θ=nπ±π3,nZ

θ=π3,π3,2π3 θπ2,π

Hence, sum of elements of fA=π3+π3+2π3=2π3

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