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Question

3+2i sin θ12i sin θ will be real, if θ =
[IIT 1976; EAMCET 2002]

A
2nπ
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B
nπ+π2
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C
nπ
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D
None of these
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Solution

The correct option is C nπ
(3+2i sin θ)(1+2i sinθ)(12i sin θ)(1+2i sinθ)=(34 sin2 θ1+4 sin2 θ)+i(8 sin θ1+4 sin2 θ)
Now, since it is real, therefore Im (z) = 0
8 sin θ1+4 sin2 θ=0 sinθ=0, θ=nπ

where n = 0, 1, 2, 3, ......

Trick : Check for (a), if the given number is absolutely real but (c) also satisfies this condition and in (a) and (c), (c) is most general value of θ.


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