Then the sum of the elements in A is:
Question:

Let $\mathrm{A}=\left\{\theta \in\left(-\frac{\pi}{2}, \pi\right): \frac{3+2 \mathrm{i} \sin \theta}{1-2 \mathrm{i} \sin \theta}\right.$ is purely imaginary $\}$.

Then the sum of the elements in $A$ is:

  1. (1) $\frac{5 \pi}{6}$

  2. (2) $\pi$

  3. (3) $\frac{3 \pi}{4}$

  4. (4) $\frac{2 \pi}{3}$


Correct Option: , 4

Solution:

Suppose $z=\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$

Since, $z$ is purely imaginary, then $z+\bar{z}=0$

$\Rightarrow \frac{3+2 i \sin \theta}{1-2 i \sin \theta}+\frac{3-2 i \sin \theta}{1+2 i \sin \theta}=0$

$\Rightarrow \frac{(3+2 i \sin \theta)(1+2 i \sin \theta)+(3-2 i \sin \theta)(1-2 i \sin \theta)}{1+4 \sin ^{2} \theta}=0$

$\Rightarrow \sin ^{2} \theta=\frac{3}{4} \Rightarrow \sin \theta=\frac{\sqrt{3}}{2}$

$\Rightarrow \theta=-\frac{\pi}{3}, \frac{\pi}{3}, \frac{2 \pi}{3}$

Now, the sum of elements in $A=-\frac{\pi}{3}+\frac{\pi}{3}+\frac{2 \pi}{3}=\frac{2 \pi}{3}$

 

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