The sum of all possible values of θ ∈[-π, 2 π] , for which (1+i θ)/(1-2 i θ) is purely imaginary, is equal

Correct answer: A. 3 π aligned & Z=(1+i θ)/(1-2 i θ) & Z=- Z (1+i θ)/(1-2 i θ)=-( 1+i θ 1-2 i θ ) & (1+i θ)( 1-2 i θ )=-(1-2 i θ)( 1+i θ ) & (1+i θ)(1+2 i θ)=-(1-2 i θ)(1-i θ) & 1+3 i θ-2 ^2 θ=-(1-3 i θ-2 ^2 θ) & 2-4 ^2 θ=0 & ^2 θ=(1)/(2) θ=-(π)/(4),-(3 π)/(4), (π)/(4), (3 π)/(4), (5 π)/(4), (7 π)/(4) & sum =3 π aligned

JEE Main 20248 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2024 · 8 April · Evening Shift · Mathematics · Complex Number

The sum of all possible values of θ ∈[-π, 2 π] , for which (1+i...

The sum of all possible values of

\theta \in[-\pi, 2 \pi]

, for which

\frac{1+i \cos \theta}{1-2 i \cos \theta}

is purely imaginary, is equal

Options

A $3 \pi$
B $2 \pi$
C $5 \pi$
D $4 \pi$

Correct answer

A. $3 \pi$

Step-by-step solution

\begin{aligned} & Z=\frac{1+i \cos \theta}{1-2 i \cos \theta} \\ & Z=-\bar{Z} \Rightarrow \frac{1+i \cos \theta}{1-2 i \cos \theta}=-\left(\frac{\overline{1+i \cos \theta}}{1-2 i \cos \theta}\right) \\ & (1+i \cos \theta)(\overline{1-2 i \cos \theta})=-(1-2 i \cos \theta)(\overline{1+i \cos \theta}) \\ & (1+i \cos \theta)(1+2 i \cos \theta)=-(1-2 i \cos \theta)(1-i \cos \theta) \\ & 1+3 i \cos \theta-2 \cos ^2 \theta=-\left(1-3 i \cos \theta-2 \cos ^2 \theta\right) \\ & 2-4 \cos ^2 \theta=0 \\ & \Rightarrow \cos ^2 \theta=\frac{1}{2} \Rightarrow \theta=-\frac{\pi}{4},-\frac{3 \pi}{4}, \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \\ & \text{ sum }=3 \pi\end{aligned}

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