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

Solution:

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

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

Now, $\left(z\right)=0$                                        [ $\because z$ is purely imaginary]

$\begin{array}{l}\Rightarrow \frac{3-4{\sin }^2\theta }{1+4{\sin }^2\theta }=0\Rightarrow {\sin }^2\theta =\frac{3}{4}\\ \Rightarrow {\sin }^2\theta ={\sin }^2\frac{\pi }{3}\therefore \theta =n\pi \pm \frac{\pi }{3},n\in Z\end{array}$

$\theta =-\frac{\pi }{3},\frac{\pi }{3},\frac{2\pi }{3}$ $\left[\because \theta \in \left(\frac{-\pi }{2},\pi \right)\right]$

Hence, sum of elements of $f\left(A\right)=\frac{-\pi }{3}+\frac{\pi }{3}+\frac{2\pi }{3}=\frac{2\pi }{3}$