Question

If `(3+2i sintheta)/(1-2 i sin theta)`is a real number and 0 < θ < 2π, then θ =

Options

• π

• `pi/2`

• `pi/3`

• `pi/6`

Answer 1

π
Given:

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

On rationalising, we get,

\[\frac{3 + 2i \sin \theta}{1 - 2i \sin \theta} \times \frac{1 + 2i \sin \theta}{1 + 2i \sin \theta} \]

\[ = \frac{(3 + 2i \sin \theta) (1 + 2i \sin \theta)}{(1 )^2 - (2i \sin \theta )^2}\]

\[ = \frac{3 + 2i \sin \theta + 6i \sin \theta + 4 i^2 \sin^2 \theta}{1 + 4 \sin^2 \theta}\]

\[ = \frac{3 - 4 \sin^2 \theta + 8i \sin \theta}{1 + 4 \sin^2 \theta} \left[ \because i^2 = - 1 \right]\]

\[ = \frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} + i\frac{8 \sin \theta}{1 + 4 \sin^2 \theta}\] For the above term to be real, the imaginary part has to be zero.

\[\therefore \frac{8\sin\theta}{1 + 4 \sin^2 \theta} = 0\]

\[ \Rightarrow 8\sin\theta = 0\]

For this to be zero,
sin\[\theta\]= 0

\[\Rightarrow\]\[\theta\]= 0,

\[\pi, 2\pi, 3\pi . . .\]

But

\[0 < \theta < 2\pi\]

Hence,

\[\theta = \pi\]