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 ,
$ where n is integer