Question

$\frac{3+2isin\theta }{1-2isin\theta }$ will be purely imaginary, if $\theta$ =

• A. $2n\pi -\frac{\pi }{3}$
• B. $n\pi +\frac{\pi }{3}$
• C. $n\pi -\frac{\pi }{3}$
• D. None of these

Answer 1

The correct option is C $n\pi -\frac{\pi }{3}$

$\frac{3+2i\sin \theta }{1-2i\sin \theta }$ will be purely imaginary, if the real part vanishes, i.e.,
$\frac{3-4{\sin }^2\theta }{1+4{\sin }^2\theta }=0\Rightarrow 3-4{\sin }^2\theta =0$( Only if $\theta$ be real)
$\Rightarrow \sin \theta =-\frac{\sqrt{3}}{2}=\sin [-\frac{\pi }{3}]\Rightarrow \theta =n\pi +(-1{)}^n[-\frac{\pi }{3}]=n\pi -\frac{\pi }{3}$