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

& A)\dfrac{5\pi }{6} \\\ & B)\dfrac{2\pi }{3} \\\ & C)\dfrac{3\pi }{4} \\\ & D)\pi \\\ \end{aligned}$$

Answer 1

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).

& \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{aligned}$$ We know that $$\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{aligned} & \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{aligned}$$ $$\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{aligned} & \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{aligned}$$ $$\Rightarrow \dfrac{3-4{{\sin }^{2}}\theta }{1+4{{\sin }^{2}}\theta }+\dfrac{8i\sin \theta }{1+4{{\sin }^{2}}\theta }=a+ib$$ $$\begin{aligned} & \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{aligned}$$ We know that a complex number is said to be purely imaginary if the real part of the complex number is equal to zero. So, we can say that the value of a in $$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{aligned} & \Rightarrow 3-4{{\sin }^{2}}\theta =0 \\\ & \Rightarrow 4{{\sin }^{2}}\theta =3 \\\ \end{aligned}$$ Now by using cross multiplication, we get $$\begin{aligned} & \Rightarrow {{\sin }^{2}}\theta =\dfrac{3}{4} \\\ & \Rightarrow \sin \theta =\pm \dfrac{\sqrt{3}}{2} \\\ \end{aligned}$$ From the question, we were given that $$\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{aligned} & \Rightarrow \sum{\theta }=\dfrac{-\pi }{3}+\dfrac{\pi }{3}+\dfrac{2\pi }{3} \\\ & \Rightarrow \sum{\theta }=\dfrac{2\pi }{3}....(4) \\\ \end{aligned}$$ From equation (4), it is clear that the sum of all possible values of $$\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.