Question

What will be the values of real $\theta$ such that $f\left(\theta \right)=\left(\frac{3+2i\sin \theta }{1-2i\sin \theta }\right)$ is real

• A. $n\pi /2$ (n integer)
• B. $n\pi$ (n integer)
• C. $n\pi /3$ (n integer)
• D. $n\pi /4$ (n integer)

Answer 1

Answer: (B)

$n\pi$ (n integer)

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

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

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

$=\frac{3+8i\sin \theta -4{\sin }^2\theta }{1+4{\sin }^2\theta },[\because i^2=-1]$

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

Now for $z$ to be real, we must have,

$\frac{8\sin \theta }{1+4{\sin }^2\theta }=0\Rightarrow \sin \theta =0$

$\Rightarrow \theta =n\pi$, where $n\in Z$