Complex Numbers 1 Question 5
5. Let $A=\theta \in-\frac{\pi}{2}, \pi: \frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is purely imaginary Then, the sum of the elements in $A$ is (2019 Main, 9 Jan I)
(a) $\frac{3 \pi}{4}$
(b) $\frac{5 \pi}{6}$
(c) $\pi$
(d) $\frac{2 \pi}{3}$
Show Answer
Answer:
Correct Answer: 5. (d)
Solution:
- Let $z=\frac{3+2 i \sin \theta}{1-2 i \sin \theta} \times \frac{1+2 i \sin \theta}{1+2 i \sin \theta}$
$$ \begin{gathered} \text { (rationalising the denominator) } \\ =\frac{3-4 \sin ^{2} \theta+8 i \sin \theta}{1+4 \sin ^{2} \theta} \\ {\left[\because a^{2}-b^{2}=(a+b)(a-b) \text { and } i^{2}=-1\right]} \\ =\frac{3-4 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}+\frac{8 \sin \theta}{1+4 \sin ^{2} \theta} i \end{gathered} $$
As $z$ is purely imaginary, so real part of $z=0$
$\therefore \frac{3-4 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}=0 \Rightarrow 3-4 \sin ^{2} \theta=0$
$\Rightarrow \quad \sin ^{2} \theta=\frac{3}{4} \Rightarrow \sin \theta= \pm \frac{\sqrt{3}}{2}$
$\Rightarrow \quad \theta \in-\frac{\pi}{3}, \frac{\pi}{3}, \frac{2 \pi}{3}$
Sum of values of $\theta=\frac{2 \pi}{3}$.
