SATHEE: Complex Numbers 1 Question 5

Complex Numbers 1 Question 5

5. Let $A = θ \in - \frac{π}{2}, π : \frac{3 + 2 i \sin θ}{1 - 2 i \sin θ}$ is purely imaginary Then, the sum of the elements in $A$ is (2019 Main, 9 Jan I)

(a) $\frac{3 π}{4}$

(b) $\frac{5 π}{6}$

(d) $\frac{2 π}{3}$

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Answer:

Correct Answer: 5. (d)

Solution:

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

$\begin{matrix} (rationalising the denominator) \\ = \frac{3 - 4 \sin^{2} θ + 8 i \sin θ}{1 + 4 \sin^{2} θ} \\ [∵ a^{2} - b^{2} = (a + b) (a - b) and i^{2} = - 1] \\ = \frac{3 - 4 \sin^{2} θ}{1 + 4 \sin^{2} θ} + \frac{8 \sin θ}{1 + 4 \sin^{2} θ} i \end{matrix}$

As $z$ is purely imaginary, so real part of $z = 0$

$∴ \frac{3 - 4 \sin^{2} θ}{1 + 4 \sin^{2} θ} = 0 \Rightarrow 3 - 4 \sin^{2} θ = 0$

$\Rightarrow \sin^{2} θ = \frac{3}{4} \Rightarrow \sin θ = \pm \frac{\sqrt{3}}{2}$

$\Rightarrow θ \in - \frac{π}{3}, \frac{π}{3}, \frac{2 π}{3}$

Sum of values of $θ = \frac{2 π}{3}$ .

Complex Numbers 1 Question 5

5. Let $A = θ \in - \frac{π}{2}, π : \frac{3 + 2 i \sin θ}{1 - 2 i \sin θ}$ is purely imaginary Then, the sum of the elements in $A$ is (2019 Main, 9 Jan I)

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Complex Numbers 1 Question 5

5. Let A=θ∈−π2,π:3+2isin⁡θ1−2isin⁡θ is purely imaginary Then, the sum of the elements in A is (2019 Main, 9 Jan I)

Answer:

Engineering Preparation

Medical Preparation

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Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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5. Let $A = θ \in - \frac{π}{2}, π : \frac{3 + 2 i \sin θ}{1 - 2 i \sin θ}$ is purely imaginary Then, the sum of the elements in $A$ is (2019 Main, 9 Jan I)