Discussion Forum

A value of θ for which $\frac{2+3i\sin\theta}{1-2i\sin\theta}$ is purely imaginary , is

Answer:

Explanation:

Let $z= \frac{2+3i\sin\theta}{1-2i\sin\theta}$

is purely

 imaaginary Then , we have Re(z)=0

 NOw conside $z= \frac{2+3i\sin\theta}{1-2i\sin\theta}$

   $z= \frac{(2+3i\sin\theta)(1+2i\sin\theta)}{(1-2i\sin\theta)(1+2i\sin\theta)}$

 $z= \frac{(2+4i\sin\theta+3i\sin\theta+6i^{2}\sin^{2}\theta)}{(1^{2}-(2i\sin\theta)^{2}}$

   $z= \frac{(2+7i\sin\theta-6\sin^{2}\theta)}{(1+4\sin^{2}\theta)}$

  $z= \frac{(2-6\sin^{2}\theta)}{(1+4\sin^{2}\theta)}+i\frac{7\sin\theta}{(1+4\sin^{2}\theta)}$

 Re(z)=0

   $\frac{(2-6\sin^{2}\theta)}{(1+4\sin^{2}\theta)}=0$

   $2=6\sin^{2}\theta$

  $\Rightarrow sin^{2}\theta=\frac{1}{\sqrt{3}}$

$\Rightarrow sin^{2}\theta=\pm\frac{1}{\sqrt{3}}$

$\Rightarrow \theta= \sin^{-}(\pm \frac{1}{\sqrt{3}})=\pm sin^{-}(\frac{1}{\sqrt{3}})$