A value of θ for which 2 + 3 i sin θ 1 - 2 i sin θ is purely imaginary, is

Let, z = 2 + 3 i sin θ 1 - 2 i sin θ Multiplying the denominator by its conjugate, we get, z = 2 + 3 i sin θ 1 + 2 i sin θ 1 + 4 sin 2 θ = 2 - 6 sin 2 θ + i 7 sin θ 1 + 4 sin 2 θ To make it purely imaginary. Its real part should be 0 . Hence, 2 = 6 sin 2 ⁡ θ ⇒ sin ⁡ θ = ± 1 3

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Mathematics - Complex Number Question with Solution | TestHub

MathematicsComplex NumberGeneral(Modulus,Argument,Conjugate)Medium2 minPYQ_2016

MathematicsMediumsingle choice

A value of $θ$ for which $\frac{2 + 3 i \sin θ}{1 - 2 i \sin θ}$ is purely imaginary, is

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Answer:

Solution:

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

Multiplying the denominator by its conjugate, we get,

$z = \frac{(2 + 3 i \sin θ) (1 + 2 i \sin θ)}{1 + 4 \sin^{2} θ}$

$= \frac{(2 - 6 \sin^{2} θ) + i (7 \sin θ)}{1 + 4 \sin^{2} θ}$

To make it purely imaginary. Its real part should be $0$ .

Hence, $2 = 6 \sin^{2} θ$

$\Rightarrow \sin θ = \pm \frac{1}{\sqrt{3}}$

Stream:JEESubject:MathematicsTopic:Complex NumberSubtopic:General(Modulus,Argument,Conjugate)

⏱ 2mℹ️ Source: PYQ_2016

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