1−2isinθ3+2isinθ will be purely imaginary, if θ =

1−2isinθ3+2isinθ will be purely imaginary, if the real part vanishes, i.e.,1+4sin2θ3−4sin2θ=0 ⇒3−4sin2θ=0( Only if θ be real) ⇒sinθ=−23 =sin[−3π ]⇒θ=nπ+(−1)n[−3π ]=nπ−3π

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{3+2i sin }{1-2 i sin } will be purely imaginary, if =

Question

$\frac{3 + 2 i s in θ}{1 - 2 i s in θ}$ will be purely imaginary, if $θ$ =

2 nπ - \frac{π}{3}

nπ + \frac{π}{3}

nπ - \frac{π}{3}

None of these

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\frac{3 + 2 i s in θ}{1 - 2 i s in θ}

will be purely imaginary, if

θ

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Text solutionVerified

\frac{3 + 2 i sin θ}{1 - 2 i sin θ}

will be purely imaginary, if the real part vanishes, i.e.,

\frac{3 - 4 sin ^{2} θ}{1 + 4 sin ^{2} θ} = 0 \Rightarrow 3 - 4 sin^{2} θ = 0

( Only if

θ

be real)

\Rightarrow sin θ = - \frac{3}{2} = sin [- \frac{π}{3}] \Rightarrow θ = nπ + (- 1)^{n} [- \frac{π}{3}] = nπ - \frac{π}{3}

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Question Text	$\frac{3 + 2 i s in θ}{1 - 2 i s in θ}$ will be purely imaginary, if $θ$ =
Answer Type	Text solution:1
Upvotes	150

1−2isinθ3+2isinθ​ will be purely imaginary, if θ =

Text solutionVerified

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$\frac{3 + 2 i s in θ}{1 - 2 i s in θ}$ will be purely imaginary, if $θ$ =