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JEE Mains · Maths · STD 11 - 4.1 complex nubers

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

  1. A \(\pi\)
  2. B \(2 \pi\)
  3. C \(4 \pi\)
  4. D \(3 \pi\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4 \pi\)

Step-by-step Solution

Detailed explanation

\(z=\frac{1+2 i \sin \theta}{1-i \sin \theta} \times \frac{1+i \sin \theta}{1+i \sin \theta}\) \(z=\frac{1-2 \sin ^2 \theta+i(3 \sin \theta)}{1+\sin ^2 \theta}\) \(\operatorname{Re}(z)=0\) \(\frac{1-2 \sin ^2 \theta}{1+\sin ^2 \theta}=0\) \(\sin \theta=\frac{ \pm 1}{\sqrt{2}}\)…
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