If 3+2i 1-2i is a real number and 0 < < 2 , then = - Vidyadip

MCQ

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

✓
$\pi$
B
$\frac{\pi}{2}$
C
$\frac{\pi}{3}$
D
$\frac{\pi}{6}$

✓

Answer

Correct option: A.

$\pi$

Given:
$\frac{3+2\text{i}\sin\theta}{1-2\text{i}\sin\theta}$ is a real number
On rationalising, we get,
$\frac{3+2\text{i}\sin\theta}{1-2\text{i}\sin\theta}\times\frac{{1+2\text{i}\sin\theta}}{{1+2\text{i}\sin\theta}}$
$=\frac{(3+2\text{i}\sin\theta)({1+2\text{i}\sin\theta})}{(1)^2-(2\text{i}\sin\theta)^2}$
$=\frac{3+2\text{i}\sin\theta+{6\text{i}\sin\theta}+4\text{i}^2\sin^2\theta}{1+4\sin^2\theta}$
$=\frac{3-4\text{i}\sin^2\theta+{8\text{i}\sin\theta}}{1+4\sin^2\theta} \ [\because\text{i}^2=-1]$
$=\frac{3-4\text{i}\sin^2\theta}{1+4\sin^2\theta}+\text{i}\frac{8\sin\theta}{1+4\sin^2\theta}$
For the above term to be real, the imaginary part has to be zero.
$\therefore\frac{8\sin\theta}{1+4\sin^2\theta}=0$
$\Rightarrow8\sin\theta=0$
For this to be zero,
$\sin\theta=0$
$\Rightarrow\theta=0,\pi,2\pi,3\pi...$
But $0<\theta<2\pi$
Hence, $\theta=\pi$

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