Question

Find real $\theta$ such that
$\frac{3+2isin\theta }{1-2isin\theta }$ is purely real.

Answer 1

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

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

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

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

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

$8\sin \theta =0\Rightarrow \sin \theta =0$

$\theta =n\pi$