Question

$LetA=\{\theta \in (-\frac{\pi }{2},\pi ):\frac{3+2i\sin \theta }{1-2i\sin \theta }\text{is purely imaginary}\}$. The sum of all the elements in $A$ is :

• A. $\pi$
• B. $\frac{2\pi }{3}$
• C. $\frac{3\pi }{4}$
• D. $\frac{5\pi }{6}$

Answer 1

The correct option is B $\frac{2\pi }{3}$

Given :
$\theta \in (-\frac{\pi }{2},\pi )\text{and}\frac{3+2i\sin \theta }{1-2\iota \sin \theta }\text{is purely imaginary}$

Let us assume $Z=\frac{3+2i\sin \theta }{1-2i\sin \theta }$,

Multiplying and dividing $Z$ with $1+2i\sin \theta$, we will get:

$Z=\frac{(3+2i\sin \theta )(1+2i\sin \theta )}{1^2-(2i\sin \theta {)}^2}$
$\Rightarrow Z=\frac{3+6i\sin \theta +2i\sin \theta +4i^2{\sin }^2\theta }{1-4i^2{\sin }^2\theta }$
$\Rightarrow Z=\frac{3+8i\sin \theta -4{\sin }^2\theta }{1+4{\sin }^2\theta }$
$\Rightarrow Z=\frac{3-4{\sin }^2\theta }{1+4{\sin }^2\theta }+i\left(\frac{8\sin \theta }{1+4{\sin }^2\theta }\right)$

Since $Z$ is purely imaginary,
$\therefore \frac{3-4{\sin }^2\theta }{1+4{\sin }^2\theta }=0$
$\Rightarrow 3-4{\sin }^2\theta =0$
$\Rightarrow {\sin }^2\theta =\frac{3}{4}$
$\Rightarrow \sin \theta =\sqrt{\frac{3}{4}}=\pm \frac{\sqrt{3}}{2}$
$\Rightarrow \theta =-\frac{\pi }{3},\frac{\pi }{3},\frac{2\pi }{3}$ $[\because \theta \in (-\frac{\pi }{2},\pi )]$

$\therefore$ $A=\{-\frac{\pi }{3},\frac{\pi }{3},\frac{2\pi }{3}\}$
Hence the sum of all the values of A $=-\frac{\pi }{3}+\frac{\pi }{3}+\frac{2\pi }{3}=\frac{2\pi }{3}$