A value of θ for which is purely imaginary, is

Engineering

Mathematics

Introduction to Complex Number

Question

A value of θ for which $\frac{2 + 3 i sinθ}{1 - 2 isinθ}$ is purely imaginary, is

$\sin^{- 1} (\frac{\sqrt{3}}{4})$

$\frac{π}{6}$

$\frac{π}{3}$

$\sin^{- 1} (\frac{1}{\sqrt{3}})$

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$\frac{2 + 3 i sinθ}{1 - 2 isinθ} \times \frac{1 + 2 i sinθ}{1 + 2 i sinθ} = \frac{2 - 6 \sin^{2} θ}{1 + 4 \sin^{2} θ} + i \frac{7 sinθ}{1 + 4 \sin^{2} θ}$

For purely imaginary,

$2 - 6 \sin^{2} θ = 0 \Rightarrow \sin^{2} θ = \frac{1}{3} \Rightarrow θ = \sin^{- 1} (\frac{1}{\sqrt{3}}) .$