Question

Let $A=\left\{0\in \left(-\frac{\pi }{2},\pi \right):\frac{3+2i\sin \theta }{1-2i\sin \theta }\text{is}\text{purely}\text{imaginary}\right\}$ 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)

Given $z=\frac{3+2i\sin \theta }{1-2i\sin \theta }$ is purely img
so real part becomes zero.
$z=\left(\frac{3+2i\sin \theta }{1-2i\sin \theta }\right)\times \left(\frac{1+2i\sin \theta }{1+2i\sin \theta }\right)$
$z=\frac{\left(3-4{\sin }^2\theta \right)+i\left(8\sin \theta \right)}{i+4{\sin }^2\theta }$
Now Re(z) = 0
$\frac{3-4{\sin }^2\theta }{1+4{\sin }^2\theta }=0$
${\sin }^2\theta =\frac{3}{4}\sin \theta =\pm \frac{\sqrt{3}}{2}\Rightarrow \theta =-\frac{\pi }{3},\frac{\pi }{3},\frac{2\pi }{3}$
$\because \theta \in \left(-\frac{\pi }{2},\pi \right)$
then sum of the elements in A is
$-\frac{\pi }{3}+\frac{\pi }{3}+\frac{2\pi }{3}=\frac{2\pi }{3}$