Question-4 \frac{3+2i\sin\theta}{1-2i\sin\theta} will be purely imaginary, if \theta equals __ (where i=\sqrt{-1})
A)
2 n \pi \pm \frac{\pi}{3}B)
n \pi+\frac{\pi}{3}C)
n \pi\pm\frac{\pi}{3}D) None
Answer: C)
Explanation
\frac{3+2 i \sin \theta}{1-2 i \sin \theta} will be purely imaginary, if the real part is zero and imaginary part is non-zero.
Now,
\left(\frac{3+2 i \sin \theta}{1-2 i \sin \theta}\right) \times\left(\frac{1+2 i \sin \theta}{1+2 i \sin \theta}\right)\Rightarrow\left(\frac{3-4 \sin ^{2} \theta+8 i \sin \theta}{1+4 \sin ^{2} \theta}\right).
Real part of the above expression is given by
\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 ^{2} \theta=\frac{3}{4} \Rightarrow \sin ^{2} \theta=\sin ^{2} \frac{\pi}{3}\Rightarrow \boldsymbol{\theta}=\boldsymbol{n} \pi \pm \frac{\pi}{3}, \boldsymbol{n} \in \boldsymbol{I}.