Question

If $\frac{3+2i\sin x}{1-2i\sin x}$ is purely imaginary then $x=$ ?

• A. $n\pi \pm \frac{\pi }{6}$
• B. $n\pi \pm \frac{\pi }{3}$
• C. $2n\pi \pm \frac{\pi }{3}$
• D. $2n\pi \pm \frac{\pi }{6}$

Answer 1

The correct option is B $n\pi \pm \frac{\pi }{3}$

A complex number is said to be purely imaginary if $z+\overline{z}=0$

If $z=\frac{3+2isin\theta }{1-2isin\theta }$

then $\overline{z}=\overline{\frac{3+2isin\theta }{1-2isin\theta }}=\frac{3-2isin\theta }{1+2isin\theta }$

$z+\overline{z}=\frac{3+2isin\theta }{1-2isin\theta }+\frac{3-2isin\theta }{1+2isin\theta }$

So,$\frac{3+2isin\theta }{1-2isin\theta }+\frac{3-2isin\theta }{1+2isin\theta }=0$

$\frac{(3+2isin\theta )(1+2isin\theta )+(3-2isin\theta )(1-2isin\theta )}{(1-2isin\theta )(1+2isin\theta )}=0$

$3+6isin\theta +2isin\theta -4{sin}^2\theta +3-6isin\theta -2isin\theta -4{sin}^2\theta =0$

$6-{8sin}^2\theta =0$

${sin}^2\theta =\frac{3}{4}$

$sin\theta =\frac{\sqrt{3}}{2}=sin\frac{\pi }{3}$

$\theta =n\pi +{(-1)}^n\left(\frac{\pi }{3}\right)$

$sin\theta =\frac{\sqrt{3}}{2}=sin-\frac{\pi }{3}$

$\theta =n\pi +{(-1)}^n\left(\frac{-\pi }{3}\right)=n\pi +{(-1)}^{n+1}\left(\frac{\pi }{3}\right)$