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$\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ will be purely imaginary, if $\theta$ is equal to
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$n \pi \pm \frac{\pi}{3}$
$z=\frac{3+2 i \cos \theta}{1-2 i \sin \theta}$ is purely imaginary
$\begin{aligned} & z=\frac{3+2 i \cos \theta}{1-2 i \sin \theta} \times \frac{1+2 i \sin \theta}{1+2 i \sin \theta} \\ & =\frac{(3-4 \sin \theta \cos \theta)+i(6 \sin \theta+2 \cos \theta)}{1+4 \sin ^2 \theta}\end{aligned}$
for purely imaginary $\operatorname{Re}(Z)=0$
$\begin{aligned} & \therefore \frac{3-4 \sin \theta \cos \theta}{1+4 \sin ^2 \theta}=0 \\ & \therefore 3-4 \sin \theta \cos \theta=0\end{aligned}$
$\sin \theta \cos \theta=\frac{3}{4} \Rightarrow \sin 2 \theta=\frac{3}{2}>1$
which is not possible
$\begin{aligned} & z=\frac{3+2 i \cos \theta}{1-2 i \sin \theta} \times \frac{1+2 i \sin \theta}{1+2 i \sin \theta} \\ & =\frac{(3-4 \sin \theta \cos \theta)+i(6 \sin \theta+2 \cos \theta)}{1+4 \sin ^2 \theta}\end{aligned}$
for purely imaginary $\operatorname{Re}(Z)=0$
$\begin{aligned} & \therefore \frac{3-4 \sin \theta \cos \theta}{1+4 \sin ^2 \theta}=0 \\ & \therefore 3-4 \sin \theta \cos \theta=0\end{aligned}$
$\sin \theta \cos \theta=\frac{3}{4} \Rightarrow \sin 2 \theta=\frac{3}{2}>1$
which is not possible
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