Question

Let $A=\{\theta ϵ(-\frac{\pi }{2},\pi ):\frac{3+2i\sin \theta }{1-2i\sin \theta }\text{is purely imaginary}\}$

Then the sum of the elements in $A$ is

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

Answer 1

Answer: (B)

$\frac{2\pi }{3}$

Given: $z=\frac{3+2i\sin \theta }{1-2i\sin \theta }$ is purely img so the real part becomes zero.

$\Rightarrow z=\left(\frac{3+2i\sin \theta }{1-2i\sin \theta }\right)\times \left(\frac{1+2i\sin \theta }{1+2i\sin \theta }\right)$

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

Now $Re\left(z\right)=0$

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

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

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

$\because \theta ϵ(-\frac{\pi }{2},\pi )$

then sum of the elements in $A$ is

$\Rightarrow -\frac{\pi }{3}+\frac{\pi }{3}+\frac{2\pi }{3}=\frac{2\pi }{3}$.