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JEE Mains · Maths · STD 11 - 4.1 complex nubers

Let \(A = \left\{ {0 \in \left( { - \frac{\pi }{2},\pi } \right):\frac{{3 + 2i\,\sin \,\theta }}{{1 - 2i\,\sin \,\theta }}\,{\rm{purely \,imaginary}}} \right\}\). Then the sum of the elements in \(A\) is

  1. A \(\frac{{5\pi }}{6}\)
  2. B \(\pi\)
  3. C \(\frac{{3\pi }}{4}\)
  4. D \(\frac{{2\pi }}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{{2\pi }}{3}\)

Step-by-step Solution

Detailed explanation

\(z=\frac{3+2 i \sin \theta}{1-2 i \sin \theta} \) \(\times \frac{1+2 i \sin \theta}{1+2 i \sin \theta}\) \(z=\frac{\left(3-4 \sin ^{2} \theta\right)+8 i \sin \theta}{1+4 \sin ^{2} \theta}\) For purely imaginary real part should be zero. i.e. \(3-4 \sin ^{2} \theta=0\) ie.…