Question

A value of $\theta$ for which $z=\frac{2+3isin\theta }{1-2isin\theta }$ is purely imaginary, is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $si{n}^{-1}\left(\frac{\sqrt{3}}{4}\right)$
• D. $si{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Answer 1

The correct option is D $si{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Let $z=\frac{2+3isin\theta }{1-2isin\theta }$ is purely imaginary. The, we have
Re(z) = 0
Now, consider $z=\frac{2+3isin\theta }{1-2isin\theta }$
$=\frac{(2+3isin\theta )(1+2isin\theta )}{(1-2isin\theta )(1+2isin\theta )}=\frac{(2+4isin\theta +3isin\theta +6i^{2}si{n}^2\theta )}{1^2-(2isin\theta {)}^2}=\frac{(2+7isin\theta -6si{n}^2\theta )}{1+4si{n}^2\theta }=\frac{2-6si{n}^2\theta }{1+4si{n}^2\theta }+i\frac{7sin\theta }{1+4si{n}^2\theta }$
$\because Re\left(z\right)=0$
$\therefore \frac{2-6si{n}^2\theta }{1+4si{n}^2\theta }=0\Rightarrow 2=6si{n}^2\theta$
$\Rightarrow si{n}^2\theta =\frac{1}{3}\Rightarrow sin\theta =\pm \frac{1}{3}\Rightarrow \theta =si{n}^{-1}(\pm \frac{1}{\sqrt{3}})=\pm si{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$