Question
Let $$A = \left\{ {\theta \in \left( { - \frac{\pi }{2},\pi } \right):\frac{{3 + 2i\sin \theta }}{{1 - 2i{{\sin }^2}\theta }}{\text{ is purely imaginary}}} \right\}$$
Then the sum of the elements in $$A$$ is:
A.
$$\frac{{5\pi }}{6}$$
B.
$$\pi $$
C.
$$\frac{{3\pi }}{4}$$
D.
$$\frac{{2\pi }}{3}$$
Answer :
$$\frac{{2\pi }}{3}$$
Solution :
Suppose $$z = \frac{{3 + 2i\sin \theta }}{{1 - 2i\,{{\sin }}\theta }}$$
Since, $$z$$ is purely imaginary, then $$z + \overline z = 0$$
$$\eqalign{
& \Rightarrow \,\,\frac{{3 + 2i\sin \theta }}{{1 - 2i\,{{\sin }}\theta }} + \frac{{3 - 2i\sin \theta }}{{1 + 2i\,{{\sin }}\theta }} = 0 \cr
& \Rightarrow \,\,\frac{{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right) + \left( {3 - 2i\sin \theta } \right)\left( {1 - 2i\sin \theta } \right)}}{{1 + 4\,{{\sin }^2}\theta }} = 0 \cr
& \Rightarrow \,\,{\sin ^2}\theta = \frac{3}{4}\sin \cr
& \Rightarrow \,\,\sin \theta = \frac{{\sqrt 3 }}{2} \cr
& \Rightarrow \,\,\theta = - \frac{\pi }{3},\frac{\pi }{3},\frac{{2\pi }}{3} \cr} $$
Now, the sum of elements in $$A$$
$$ = - \frac{\pi }{3} + \frac{\pi }{3} + \frac{{2\pi }}{3} = \frac{{2\pi }}{3}$$