Question

The real value of $\alpha$ for which the expression
$\frac{1-i\sin \alpha }{1+2i\sin \alpha }$ is purely real, is:

• A. $(2n+1)\frac{\pi }{2}$
• B. $(n+1)\frac{\pi }{2}$
• C. $n\pi$
• D. none of these

Answer 1

Answer: (C)

If $z$ is purely real, then the coefficient of imaginary part will be zero. Let
$z=\frac{1-i\sin \alpha }{1+2i\sin \alpha }$
$=\frac{1-i\sin \alpha }{1+2i\sin \alpha }\times \frac{(1-2i\sin \alpha )}{(1-2i\sin \alpha )}$
$=\frac{1-2{\sin }^2\alpha -3i\sin \alpha }{1+4{\sin }^2\alpha }$
$=\frac{(1-2{\sin }^2\alpha )-3i\sin \alpha }{1+4{\sin }^2\alpha }$
Since, z is real, therefore the coefficient of imaginary part will be zero.
$\Rightarrow$ $3\sin \alpha =0$
$\Rightarrow$ $\alpha =n\pi ,<br/><br/>$ where n is integer