Verified Solution • 2025–26 Session moderate

QJEEM19Ma7904: JEE Main 2019 Mathematics - Class 11 JEEM19Ma7904 Solution: Let A = . Then the sum of the elements in A is :

Let
A = {θ∈(−π2,π):3+2isin⁡θ1−2isin⁡θis purely imaginary}\left\{ {\theta \in \left( { - {\pi \over 2},\pi } \right):{{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}is\,purely\,imaginary} \right\}
. Then the sum of the elements in A is :

Multiple Choice Options

  • A 5π6{5\pi \over 6}
  • B π\pi
  • C 3π4{3\pi \over 4}
  • D 2π3{{2\pi } \over 3}

Step-by-Step Verified Solution

Given complex number,

3+2isin⁡θ1−2isin⁡θ{{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}

=(3+2isin⁡θ)(1+2isin⁡θ)1+4sin⁡2θ= {{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}

=3+6isin⁡θ+2isin⁡θ−4sin⁡2θ1+4sin⁡2θ= {{3 + 6i\sin \theta + 2i\sin \theta - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}

=(3−4sin⁡2θ)+i(8sin⁡θ)1+4sin⁡2θ= {{\left( {3 - 4{{\sin }^2}\theta } \right) + i\left( {8\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}

As complex number is purely imaginary, So real part of this complex number is zero.

∴\therefore     3−4sin⁡2θ1+4sin⁡2θ{{3 - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }} = 0

⇒\Rightarrow     3−4sin⁡2θ=03 - 4{\sin ^2}\theta = 0

⇒\Rightarrow     sin⁡θ=±32\sin \theta = \pm {{\sqrt 3 } \over 2}

as    θ\theta ∈\in (−π2,π)\left( { - {\pi \over 2},\pi } \right)

∴\therefore     θ\theta == −- π3,π3,2π3{\pi \over 3},{\pi \over 3},{{2\pi } \over 3}

∴\therefore    Sum of those values of A is

=−π3+π3+2π3= - {\pi \over 3} + {\pi \over 3} + {{2\pi } \over 3}

=2π3= {{2\pi } \over 3}

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