Question

Find real values of $\theta$ for which$\left(\frac{4+3i\sin \theta }{1-2i\sin \theta }\right)$ is purely real.

Answer 1

Given:

$\Rightarrow \frac{4+3i\sin \left(\theta \right)}{1-2i\sin \theta }$ .......(i)

As the function is purely real, the imaginary part of eqn.(i) would be 0

rationalizing eqn.(i) we get,

$\Rightarrow \frac{4+3i\sin \left(\theta \right)}{1-2i\sin \theta }\times \frac{1+2i\sin \theta }{1+2i\sin \theta }$

$\Rightarrow \frac{(4+3i\sin (\theta \left)\right)\times (1+2i\sin (\theta \left)\right)}{1^2-(2i\sin \theta {)}^2}$

.......{$since,(a+b)(a-b)=a^2-b^2$ }

$\Rightarrow \frac{4+8i\sin \theta +3i\sin \theta +6i^2{\sin }^2\theta }{1-4i^2{\sin }^2\theta }$

$\Rightarrow \frac{4+11i\sin \theta -6{\sin }^2\theta }{1+4{\sin }^2\theta }$

.......{$since,i^2=-1$ }

$\Rightarrow \frac{1}{1+4\sin \theta }\times \left[\right(4-6{\sin }^2\theta )+i(11\sin \theta \left)\right]$

the imainary part is equal to zero for eqn. (i) to be purely real.

$\therefore$

$\Rightarrow \frac{11\sin \theta }{1+4{\sin }^2\theta }=0$

$\Rightarrow \sin \theta =0$

the general solution for $\sin \theta =0$ is $\theta \in n\pi where,n\in Z$

Hence, for $\theta \in n\pi where,n\in Z$

$\Rightarrow \frac{4+3i\sin \left(\theta \right)}{1-2i\sin \theta }$ is purely real.