Question

$\frac{3+2i\sin \theta }{1-2i\sin \theta }$ will be real, if $\theta =$

• A. $2n\pi$
• B. $n\pi +\frac{\pi }{2}$
• C. $n\pi$
• D. None of these

Answer 1

The correct option is C $n\pi$

Now,

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

$=\frac{(3+2i\sin \theta )(1+2i\sin \theta )}{1+4{\sin }^2\theta }$

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

Now the given complex number will be purely real if the imaginary part of the above complex number becomes zero.

That is $\sin \theta =0$

or, $\theta =n\pi$ for $n\in Z$ ( Set of integers)