Let A = { ( 0 , 2 ) : 1 + 2 i sin 1 - i sin i s p u r e l y i m a g i n a r y } . Then the sum of the elements in A is [2023]

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Solution

Q.
Let $A = {θ \in (0, 2 π) : \frac{1 + 2 i \sin θ}{1 - i \sin θ} is purely imaginary} .$ Then the sum of the elements in A is [2023]

1 $2 π$

2 $4 π$

3 $π$

4 $3 π$

Ans.
(2)

Given, $\frac{1 + 2 i \sin θ}{1 - i \sin θ}$ is purely imaginary, then its real part must be zero.

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

$= \frac{1 - 2 \sin^{2} θ}{1 + \sin^{2} θ} + \frac{3 \sin θ}{1 + \sin^{2} θ} i$

Since real part is 0 $\Rightarrow \frac{1 - 2 \sin^{2} θ}{1 + \sin^{2} θ} = 0 \Rightarrow 2 \sin^{2} θ = 1$

$\Rightarrow \sin θ = \pm \frac{1}{\sqrt{2}} \Rightarrow θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$

Since, $θ \in (0, 2 π)$

Then sum of the elements in A is,

$\frac{π}{4} + \frac{3 π}{4} + \frac{5 π}{4} + \frac{7 π}{4} = \frac{16 π}{4} = 4 π$

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