Which of the following is true for $z = (3 + 2 i \sin θ) / (1 - 2 i \sin θ), where i = \sqrt{- 1}$

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z is purely real for $θ = n π, n \in Z$

none of these

z is purely real for $θ = n π \pm π / 3, n \in Z$

z is purely imaginary for $θ = n π \pm π / 2, n \in Z$

answer is C.

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Detailed Solution

let $Z = \frac{3 + 2 isin θ}{1 - 2 isin θ} = \frac{(3 + 2 isin θ) (1 + 2 isin θ)}{(1 - 2 isin θ) (1 + 2 isin θ)} = \frac{(3 - 4 \sin^{2} θ) + 8 isin θ}{1 + 4 \sin^{2} θ}$
Therefore, the real part of $Z = \frac{3 - 4 \sin^{2} θ}{1 + 4 \sin^{2} θ}$ and the imaginary part of $Z = \frac{8 \sin θ}{1 + 4 \sin^{2} θ}$
Z is real, if imaginary part $= \frac{8 \sin θ}{1 + 4 \sin^{2} θ} = 0$ or $\sin θ = 0 or θ = n π, \forall n \in I$
Z is purely imaginary, if real part $(3 - 4 \sin^{2} θ) / (1 + 4 \sin^{2} θ) = 0$
or $\sin^{2} θ = 3 / 4 = \sin^{2} (π / 3) or θ = n π \pm π / 3, \forall n \in I$