SATHEE: Complex Numbers 1 Question 6

Complex Numbers 1 Question 6

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

(2016 Main)

(a) $\frac{π}{3}$

(b) $\frac{π}{6}$

(d) $\sin^{- 1} \frac{1}{\sqrt{3}}$

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Answer:

Correct Answer: 6. (d)

Solution:

Let $z = \frac{2 + 3 i \sin θ}{1 - 2 i \sin θ}$ is purely imaginary. Then, we have $Re (z) = 0$

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

$\begin{aligned} = \frac{(2 + 3 i \sin θ) (1 + 2 i \sin θ)}{(1 - 2 i \sin θ) (1 + 2 i \sin θ)} \\ = \frac{2 + 4 i \sin θ + 3 i \sin θ + 6 i^{2} \sin^{2} θ}{1^{2} - (2 i \sin θ)^{2}} \\ = \frac{2 + 7 i \sin θ - 6 \sin^{2} θ}{1 + 4 \sin^{2} θ} \\ = \frac{2 - 6 \sin^{2} θ}{1 + 4 \sin^{2} θ} + i \frac{7 \sin θ}{1 + 4 \sin^{2} θ} \end{aligned}$

$∵ Re (z) = 0$

$∴ \frac{2 - 6 \sin^{2} θ}{1 + 4 \sin^{2} θ} = 0 \Rightarrow 2 = 6 \sin^{2} θ$

$\Rightarrow \sin^{2} θ = \frac{1}{3}$

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

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

Complex Numbers 1 Question 6

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

Answer:

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Complex Numbers 1 Question 6

6. A value of θ for which 2+3isin⁡θ1−2isin⁡θ is purely imaginary, is

Answer:

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