Question

Let \(A = \left\{\theta \in \left(-\frac\pi2, \pi\right) : \frac{3 + 2i\sin\theta}{1 - 2i\sin\theta}\ \text{is purely imaginary}\right\}\). Then the sum of the elements in A is :

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

Hint:

A complex number is purely imaginary when its real part is \(0\); rationalise the denominator and set the real part to zero to get \(\sin^2\theta\).

Answer 1

The Correct Option is D

Let \(s = \sin\theta\). \(\frac{3 + 2is}{1 - 2is} = \frac{(3 + 2is)(1 + 2is)}{1 + 4s^2} = \frac{(3 - 4s^2) + 8is}{1 + 4s^2}\).
Purely imaginary \(\Rightarrow\) real part \(= 0 \Rightarrow 3 - 4s^2 = 0 \Rightarrow \sin\theta = \pm\frac{\sqrt3}{2}\) (the imaginary part \(8s \ne 0\), fine).
In \(\left(-\frac\pi2, \pi\right)\): \(\sin\theta = \frac{\sqrt3}{2} \Rightarrow \theta = \frac\pi3, \frac{2\pi}{3}\); \(\sin\theta = -\frac{\sqrt3}{2} \Rightarrow \theta = -\frac\pi3\) (\(-\frac{2\pi}{3}\) lies outside the interval).
Sum \(= \frac\pi3 + \frac{2\pi}{3} - \frac\pi3 = \frac{2\pi}{3}\).
Answer: (4) \(\frac{2\pi}{3}\)