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# The value of $\theta$ for which$\dfrac{{2 + 3i\sin \theta }}{{1 - 2i\sin \theta }}$ is purely imaginary isA. $\dfrac{\pi }{3}$ B. $\dfrac{\pi }{6}$ C. ${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{4}} \right)$ D. ${\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)$

Last updated date: 04th Aug 2024
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Hint: In this question, first we will simplify the expression by making the denominator a real number by rationalization. After this we will separate the real and imaginary terms and then equate the real terms to zero. Finally solve the trigonometric equation to get the answer.

We have to find the value of if the given complex number$\dfrac{{2 + 3i\sin \theta }}{{1 - 2i\sin \theta }}$ is purely imaginary.
For the given expression to be purely imaginary, its real part must be zero.
We will first simplify the expression and make the denominator a real number.
So we will rationalize the denominator part.
$\dfrac{{2 + 3i\sin \theta }}{{1 - 2i\sin \theta }} \times \dfrac{{1 + 2i\sin \theta }}{{1 + 2i\sin \theta }}$
Now let’s simply the numerator and denominator part:
$\dfrac{{(2 + 3i\sin \theta )(1 + 2i\sin \theta )}}{{(1 - 2i\sin \theta )(1 + 2i\sin \theta )}} = \dfrac{{2 + 7i\sin \theta - 6{{\sin }^2}\theta }}{{(1 - 2i\sin \theta )(1 + 2i\sin \theta )}}$
We know that ${a^2} - {b^2} = (a + b)(a - b)$ and also ${i^2} = - 1$
$\therefore$ The above expression can be written as:
$\dfrac{{2 + 7i\sin \theta - 6{{\sin }^2}\theta }}{{{1^2} - {{(2i\sin \theta )}^2}}} = \dfrac{{2 + 7i\sin \theta - 6{{\sin }^2}\theta }}{{1 + 4{{\sin }^2}\theta }}$.
On separating the real part and imaginary part, we have:
$\dfrac{{2 + 7i\sin \theta - 6{{\sin }^2}\theta }}{{1 + 4{{\sin }^2}\theta }} = \dfrac{{2 - 6{{\sin }^2}\theta }}{{1 + 4{{\sin }^2}}} + i\dfrac{{7\sin \theta }}{{1 + 4{{\sin }^2}}}$
On equating the real part to zero, we get:
$\dfrac{{2 - 6{{\sin }^2}\theta }}{{1 + 4{{\sin }^2}}}$=0
$\Rightarrow 2 - 6{\sin ^2}\theta = 0$
On solving, we get:
$\Rightarrow {\sin ^2}\theta = \dfrac{2}{6}$
On taking square root on both sides, we have:
$\Rightarrow \sin \theta = \pm \sqrt {\dfrac{1}{3}}$ .
$\Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right){\text{ or }}{\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 3 }}} \right)$

So, the correct answer is “Option D”.

Note: In such a type of question to make purely imaginary means we have to make the real part zero and if it asks for purely real then we will have to make the imaginary part zero. The purely imaginary term lies on the $j\omega$ axis or imaginary axis of the complex plane and the purely real term lies on the real axis.