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Mathematics · Complex Numbers

JEE Main 2024 — 8 April, Shift 2 — Question 6

The sum of all possible values of θ∈[−π,2π]\theta \in[-\pi, 2 \pi], for which 1+icos⁡θ1−2icos⁡θ\frac{1+i \cos \theta}{1-2 i \cos \theta} is purely imaginary, is equal to

  1. Option A:

    2π2 \pi

  2. Option B:

    3π3 \pi

    Correct
  3. Option C:

    5π5 \pi

  4. Option D:

    4π4\pi

Answer: B

Step-by-step solution

Z=1+icos⁡θ1−2icos⁡θZ=\frac{1+i \cos \theta}{1-2 i \cos \theta}

Z=Z‾⇒1+icos⁡θ1−2icos⁡θ=−(1+icos⁡θ‾1−2icos⁡θ)\mathrm{Z}= \overline{\mathrm{Z}} \Rightarrow \frac{1+\mathrm{i} \cos \theta}{1-2 \mathrm{i} \cos \theta}=-\left(\frac{\overline{1+\mathrm{i} \cos \theta}}{1-2 \mathrm{i} \cos \theta}\right)

(1+icos⁡θ)(1−2icos⁡θ‾)=−(1−2icos⁡θ)(1+icos⁡θ‾)(1+\mathrm{i} \cos \theta)(\overline{1-2 \mathrm{i} \cos \theta})=-(1-2 \mathrm{i} \cos \theta)(\overline{1+\mathrm{i} \cos \theta})

(1+icos⁡θ)(1+2icos⁡θ)=−(1−2icos⁡θ)(1−icos⁡θ)(1+\mathrm{i} \cos \theta)(1+2 \mathrm{i} \cos \theta)=-(1-2 \mathrm{i} \cos \theta)(1-\mathrm{i} \cos \theta)

1+3icos⁡θ−2cos⁡2θ=−(1−3icos⁡θ−2cos⁡2θ)1+3 \mathrm{i} \cos \theta-2 \cos ^{2} \theta=-\left(1-3 \mathrm{i} \cos \theta-2 \cos ^{2} \theta\right)

2−4cos⁡2θ=02-4 \cos ^{2} \theta=0

⇒cos⁡2θ=12⇒θ=−π4,−3π4,π4,3π4,5π4,7π4\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}

Sum =3π=3 \pi

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
8 April, Shift 2
Subject
Mathematics
Chapter
Complex Numbers
Topic
Introduction to Complex Numbers