Complex Numbers 1 Question 6
6. A value of $\theta$ for which $\frac{2+3 i \sin \theta}{1-2 i \sin \theta}$ is purely imaginary, is
(2016 Main)
(a) $\frac{\pi}{3}$
(b) $\frac{\pi}{6}$
(c) $\sin ^{-1} \frac{\sqrt{3}}{4}$
(d) $\sin ^{-1} \frac{1}{\sqrt{3}}$
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Answer:
Correct Answer: 6. (d)
Solution:
- Let $z=\frac{2+3 i \sin \theta}{1-2 i \sin \theta}$ is purely imaginary. Then, we have $\operatorname{Re}(z)=0$
Now, consider $z=\frac{2+3 i \sin \theta}{1-2 i \sin \theta}$
$$ \begin{aligned} & =\frac{(2+3 i \sin \theta)(1+2 i \sin \theta)}{(1-2 i \sin \theta)(1+2 i \sin \theta)} \\ & =\frac{2+4 i \sin \theta+3 i \sin \theta+6 i^{2} \sin ^{2} \theta}{1^{2}-(2 i \sin \theta)^{2}} \\ & =\frac{2+7 i \sin \theta-6 \sin ^{2} \theta}{1+4 \sin ^{2} \theta} \\ & =\frac{2-6 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}+i \frac{7 \sin \theta}{1+4 \sin ^{2} \theta} \end{aligned} $$
$\because \quad \operatorname{Re}(z)=0$
$\therefore \quad \frac{2-6 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}=0 \Rightarrow 2=6 \sin ^{2} \theta$
$\Rightarrow \quad \sin ^{2} \theta=\frac{1}{3}$
$\Rightarrow \quad \sin \theta= \pm \frac{1}{\sqrt{3}}$
$\Rightarrow \quad \theta=\sin ^{-1} \pm \frac{1}{\sqrt{3}}= \pm \sin ^{-1} \frac{1}{\sqrt{3}}$
