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Question

Let A={θ(π2,π):3+2isinθ12isinθ is purely imaginary }. The sum of all the elements in A is :

A
π
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B
2π3
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C
3π4
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D
5π6
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Solution

The correct option is B 2π3
Given :
θ(π2,π) and 3 + 2isinθ1 2ιsinθ is purely imaginary

Let us assume Z=3 + 2isinθ1 2isinθ,

Multiplying and dividing Z with 1+2isinθ, we will get:

Z=(3 + 2isinθ) (1 + 2isinθ)12 (2isinθ)2
Z=3 + 6isinθ + 2isinθ + 4i2sin2θ1 4i2sin2θ
Z=3 + 8isinθ 4sin2θ1 + 4sin2θ
Z=3 4sin2θ1 + 4sin2θ +i(8sinθ1 + 4sin2θ)

Since Z is purely imaginary,
3 4sin2θ1 + 4sin2θ = 0
34sin2θ=0
sin2θ=34
sinθ=34=±32
θ=π3,π3,2π3 [θ(π2,π)]

A={π3,π3,2π3}
Hence the sum of all the values of A =π3+π3+2π3=2π3

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