Verified Solution • 2025–26 Session moderate

QJEEM23Macb52: JEE Main 2023 Mathematics - Class 11 JEEM23Macb52 Solution: Let is purely imaginary . Then the sum of the elements in is :

Let A={θ∈(0,2π):1+2isin⁡θ1−isin⁡θA=\left\{\theta \in(0,2 \pi): \frac{1+2 i \sin \theta}{1-i \sin \theta}\right. is purely imaginary }\} . Then the sum of the elements in A\mathrm{A} is :

Multiple Choice Options

  • A 3π3 \pi
  • B π\pi
  • C 2π2 \pi
  • D 4π4 \pi

Step-by-Step Verified Solution

amp; Here, z=1+2isin⁡θ1−isin⁡θ×1+isin⁡θ1+isin⁡θamp;1+isin⁡θ+2isin⁡θ−2sin⁡2θ1−i2sin⁡2θamp;=(1−2sin⁡2θ)+i(3sin⁡θ)1+sin⁡2θ\begin{aligned}& \text { Here, } z=\frac{1+2 i \sin \theta}{1-i \sin \theta} \times \frac{1+i \sin \theta}{1+i \sin \theta} \\\\& \frac{1+i \sin \theta+2 i \sin \theta-2 \sin ^2 \theta}{1-i^2 \sin ^2 \theta} \\\\& =\frac{\left(1-2 \sin ^2 \theta\right)+i(3 \sin \theta)}{1+\sin ^2 \theta}\end{aligned}

∵z\because z is purely imaginary, so Re⁡z=0\operatorname{Re} z=0

amp;⇒1−2sin⁡2θ1+sin⁡2θ=0amp;⇒2sin⁡2θ=1⇒sin⁡2θ=12amp;⇒sin⁡θ=±12\begin{aligned}& \Rightarrow \frac{1-2 \sin ^2 \theta}{1+\sin ^2 \theta}=0 \\\\& \Rightarrow 2 \sin ^2 \theta=1 \Rightarrow \sin ^2 \theta=\frac{1}{2} \\\\& \Rightarrow \sin \theta= \pm \frac{1}{\sqrt{2}}\end{aligned}

amp;∴A=[π4,3π4,5π4,7π4]∵θ∈(0,2π)amp;∴ Sum =π+3π+5π+7π4=16π4=4π\begin{aligned}& \therefore A=\left[\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\right] \because \theta \in(0,2 \pi)\\\\& \therefore \text { Sum }=\frac{\pi+3 \pi+5 \pi+7 \pi}{4}=\frac{16 \pi}{4}=4 \pi\end{aligned}

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