Question

If $\frac{3+2isin\theta }{1-2isin\theta }$ is a real number and 0 < $\theta <2<2\pi \text{then}\theta$

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is A

$\pi$

Given :

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

On retionalising, we get

$\frac{3+2isin\theta }{1-2isin\theta }\times \frac{1+2isin\theta }{1+2isin\theta }=\frac{(3+2isin\theta )(1+2isin\theta )}{(1{)}^2-(2isin\theta {)}^2}=\frac{3+2isin\theta +6isin\theta +4i^{2}si{n}^2\theta }{1+4isi{n}^2\theta }=\frac{3-4si{n}^2\theta +8isin\theta }{1+4isi{n}^2\theta }[\because i^2=-1]=\frac{3-4si{n}^2\theta }{1+4si{n}^2\theta }+i\frac{8sin\theta }{1+4si{n}^2\theta }$

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

$\therefore \frac{8sin\theta }{1+4si{n}^2\theta }=0\Rightarrow 8sin\theta =0\text{For this to be zero,}sin\theta =0\Rightarrow \theta =0,\pi ,2\pi ,3\pi ...\text{But}0<\theta <2\pi \text{Hence,}\theta =\pi$