The sum of all possible values of , for which 1 + i cos 1 - 2 i cos is purely imaginary, is equal to: [2024]

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Solution

Q.
The sum of all possible values of $θ \in [- π, 2 π]$ , for which $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is purely imaginary, is equal to: [2024]

1 $2 π$

2 $4 π$

3 $3 π$

4 $5 π$

Ans.
(3)

We have, $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is purely imaginary.

$∴ \frac{1 + i \cos θ}{1 - 2 i \cos θ} + \frac{1 - i \cos θ}{1 + 2 i \cos θ} = 0$ $(∵ z + \bar{z} = 0)$

$\Rightarrow \frac{1 + i \cos θ + 2 i \cos θ - 2 \cos^{2} θ + 1 - i \cos θ - 2 i \cos θ - 2 \cos^{2} θ}{1 + 4 \cos^{2} θ} = 0$

$\Rightarrow 2 - 4 \cos^{2} θ = 0 \Rightarrow \cos^{2} θ = \frac{1}{2}$

$θ = n π \pm \frac{π}{4}$

$θ$ can be $\frac{π}{4}, \frac{- π}{4}, \frac{3 π}{4}, \frac{- 3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ $(∵ θ \in [- π, 2 π])$

$∴$ Required $= \frac{π}{4} - \frac{π}{4} + \frac{3 π}{4} - \frac{3 π}{4} + \frac{5 π}{4} + \frac{7 π}{4} = 3 π$

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