Question

Let \[A=\left\{ \theta \in \left( \dfrac{-\pi }{2},\pi \right):\dfrac{3+2i\sin \theta }{1-2i\sin \theta }\text{ is purely imaginary} \right\}\]. Then the sum of elements of A is
\[\begin{align}
  & A)\dfrac{5\pi }{6} \\
 & B)\dfrac{2\pi }{3} \\
 & C)\dfrac{3\pi }{4} \\
 & D)\pi \\
\end{align}\]

Answer 1

Hint: We know that a complex number is said to be purely imaginary if the real part of the complex number is equal to zero. So, we can say that the value of a in \[a+ib\] is equal to zero. By using this concept, we can find the value of \[\sin \theta \] where the given complex number is imaginary. Now we should find the values of \[\theta \] in the range where \[\theta \in \left( \dfrac{-\pi }{2},\pi \right)\]. Now we should find the sum of values of \[\theta \].

Complete step by step answer:
From the question, it is clear that the value of \[\dfrac{3+2i\sin \theta }{1-2i\sin \theta }\] is purely imaginary. Before solving the question, we should know that a complex number is said to be purely imaginary if the real part of the complex number is equal to zero.
Let us assume the complex number \[\dfrac{3+2i\sin \theta }{1-2i\sin \theta }\] is equal to \[a+ib\].
\[\Rightarrow \dfrac{3+2i\sin \theta }{1-2i\sin \theta }=a+ib.......(1)\]
Now let us multiply and divide with \[1+2i\sin \theta \] on L.H.S of equation (1).
 \[\begin{align}
  & \Rightarrow \dfrac{3+2i\sin \theta }{1-2i\sin \theta }\times \dfrac{1+2i\sin \theta }{1+2i\sin \theta }=a+ib \\
 & \Rightarrow \dfrac{\left( 3+2i\sin \theta \right)\left( 1+2i\sin \theta \right)}{\left( 1-2i\sin \theta \right)\left( 1+2i\sin \theta \right)}=a+ib \\
\end{align}\]
We know that \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\].
\[\Rightarrow \dfrac{\left( 3+2i\sin \theta \right)\left( 1+2i\sin \theta \right)}{1-4{{i}^{2}}{{\sin }^{2}}\theta }=a+ib\]
We know that the value of \[{{i}^{2}}\] is equal to -1.
\[\begin{align}
  & \Rightarrow \dfrac{\left( 3+2i\sin \theta \right)\left( 1+2i\sin \theta \right)}{1+4{{\sin }^{2}}\theta }=a+ib \\
 & \Rightarrow \dfrac{\left( 3+2i\sin \theta \right)\left( 1 \right)+\left( 3+2i\sin \theta \right)\left( 2i\sin \theta \right)}{1+4{{\sin }^{2}}\theta }=a+ib \\
 & \Rightarrow \dfrac{3+2i\sin \theta +\left( 3 \right)\left( 2i\sin \theta \right)+\left( 2i\sin \theta \right)\left( 2i\sin \theta \right)}{1+4{{\sin }^{2}}\theta }=a+ib \\
\end{align}\]
\[\Rightarrow \dfrac{3+2i\sin \theta +6i\sin \theta +4{{i}^{2}}{{\sin }^{2}}\theta }{1+4{{\sin }^{2}}\theta }=a+ib\]
We know that the value of \[{{i}^{2}}\] is equal to -1.
\[\begin{align}
  & \Rightarrow \dfrac{3+2i\sin \theta +6i\sin \theta -4{{\sin }^{2}}\theta }{1+4{{\sin }^{2}}\theta }=a+ib \\
 & \Rightarrow \dfrac{3-4{{\sin }^{2}}\theta +8i\sin \theta }{1+4{{\sin }^{2}}\theta }=a+ib \\
\end{align}\]
\[\Rightarrow \dfrac{3-4{{\sin }^{2}}\theta }{1+4{{\sin }^{2}}\theta }+\dfrac{8i\sin \theta }{1+4{{\sin }^{2}}\theta }=a+ib\]
\[\begin{align}
  & \Rightarrow \dfrac{3-4{{\sin }^{2}}\theta }{1+4{{\sin }^{2}}\theta }=a.....(2) \\
 & \Rightarrow \dfrac{8\sin \theta }{1+4{{\sin }^{2}}\theta }=b.....(3) \\
\end{align}\]
We know that a complex number is said to be purely imaginary if the real part of the complex number is equal to zero. So, we can say that the value of a in \[a+ib\] is equal to zero.
Then from equation (2), we can write that
\[\Rightarrow \dfrac{3-4{{\sin }^{2}}\theta }{1+4{{\sin }^{2}}\theta }=0\]
Now by using cross multiplication, we get
\[\begin{align}
  & \Rightarrow 3-4{{\sin }^{2}}\theta =0 \\
 & \Rightarrow 4{{\sin }^{2}}\theta =3 \\
\end{align}\]
Now by using cross multiplication, we get
\[\begin{align}
  & \Rightarrow {{\sin }^{2}}\theta =\dfrac{3}{4} \\
 & \Rightarrow \sin \theta =\pm \dfrac{\sqrt{3}}{2} \\
\end{align}\]
From the question, we were given that \[\theta \in \left( \dfrac{-\pi }{2},\pi \right)\].
We know that the values of \[\theta \] if \[\theta \in \left( \dfrac{-\pi }{2},\pi \right)\] for \[\sin \theta =\pm \dfrac{\sqrt{3}}{2}\] are equal to \[\dfrac{-\pi }{3},\dfrac{\pi }{3},\dfrac{2\pi }{3}\].
Now we should find the sum of values of all possible values of \[\theta \].
Let us assume the sum of values of \[\theta \] is equal to \[\sum{\theta }\].
\[\begin{align}
  & \Rightarrow \sum{\theta }=\dfrac{-\pi }{3}+\dfrac{\pi }{3}+\dfrac{2\pi }{3} \\
 & \Rightarrow \sum{\theta }=\dfrac{2\pi }{3}....(4) \\
\end{align}\]
From equation (4), it is clear that the sum of all possible values of \[\theta \] are equal to \[\dfrac{2\pi }{3}\].

So, the correct answer is “Option B”.

Note: Some students may have a misconception that a complex number is said to be purely imaginary if the imaginary part of the complex number is equal to zero. Then, we will say that the value of b in \[a+ib\] is equal to zero. If this misconception is followed, then we cannot get the correct answer. The final answer will definitely get interrupted. So, students should have a clear view of the concept.