Question

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

• A. This gives $\theta =n\pi \pm \frac{\pi }{3},$ where n is an integer.
• B. This gives $\theta =n\pi \pm \frac{\pi }{6},$ where n is an integer.
• C. This gives $\theta =\frac{n\pi }{2}\pm \frac{\pi }{3},$ where n is an integer.
• D. This gives $\theta =\frac{n\pi }{2}\pm \frac{\pi }{6},$ where n is an integer.

Answer 1

The correct option is A This gives $\theta =n\pi \pm \frac{\pi }{3},$ 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 purely imaginary, we must have,
$\frac{3-4{\sin }^2\theta }{1+4{\sin }^2\theta }=0\Rightarrow {\sin }^2\theta =\frac{3}{4}$
$\Rightarrow \theta =n\pi \pm \frac{\pi }{3}$, where $n\in Z$