Question

Find $\theta$ such that $\frac{3+2i\sin \theta }{1-2i\sin \theta }$ is real

• A. $\theta =\frac{n\pi }{2}$ where n is an integer.
• B. $\theta =n\pi$ where n is an integer.
• C. $\theta =2n\pi$ where n is an integer.
• D. $\theta =\frac{n\pi }{4}$ where n is an integer.

Answer 1

The correct option is B $\theta =n\pi$ where n is an 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 $n\in Z$