Question

If $\frac{3+2i\mathrm{sin\theta }}{1-2i\mathrm{sin\theta }}$ is a real number and 0 < θ < 2π, then θ =
(a) π
(b) $\frac{\mathrm{\pi }}{2}$
(c) $\frac{\mathrm{\pi }}{3}$
(d) $\frac{\mathrm{\pi }}{6}$

Answer 1

(a) π

Given:

$\frac{3+2i\sin \theta }{1-2i\sin \theta }$ is a real number

On rationalising, we get,

$$\begin{gathered} \frac{3+2i\sin \theta }{1-2i\sin \theta }\times \frac{1+2i\sin \theta }{1+2i\sin \theta } \\ =\frac{(3+2i\sin \theta )(1+2i\sin \theta )}{(1{)}^2-(2i\sin \theta {)}^2} \\ =\frac{3+2i\sin \theta +6i\sin \theta +4i^2{\sin }^2\theta }{1+4{\sin }^2\theta } \\ =\frac{3-4{\sin }^2\theta +8i\sin \theta }{1+4{\sin }^2\theta }\left[\because i^2=-1\right] \\ =\frac{3-4{\sin }^2\theta }{1+4{\sin }^2\theta }+i\frac{8\sin \theta }{1+4{\sin }^2\theta } \end{gathered}$$

For the above term to be real, the imaginary part has to be zero.

$$\begin{gathered} \therefore \frac{8\sin \theta }{1+4{\sin }^2\theta }=0 \\ \Rightarrow 8\sin \theta =0 \end{gathered}$$

For this to be zero,
sin $\theta$= 0
$\Rightarrow$ $\theta$ = 0, $\mathrm{\pi },2\mathrm{\pi },3\mathrm{\pi }...$
But $0<\theta <2\pi$
Hence, $\theta =\pi$