Question

If the expression $\frac{1-i\sin \alpha }{1+2i\sin \alpha }$ is purely real, which of following is/are correct?

• A. $Re\left(z\right)=-\frac{1}{5}$
• B. $Re\left(z\right)=1$
• C. $\alpha =n\pi ,n\in Z$
• D. $\alpha =\frac{n\pi }{2},n\in Z$

Answer 1

Answer: (B), (C)

The correct options are
B $Re\left(z\right)=1$
C $\alpha =n\pi ,n\in Z$

$\frac{1-i\sin \alpha }{1+2i\sin \alpha }$
$=\frac{1-i\sin \alpha }{1+2i\sin \alpha }.\frac{1-2i\sin \alpha }{1-2i\sin \alpha }$
$=\frac{(1-2{\sin }^2\alpha )-3i\sin \alpha }{1+4{\sin }^2\alpha }$
$=\frac{1-2{\sin }^2\alpha }{1+4{\sin }^2\alpha }-\frac{3\sin \alpha }{1+4{\sin }^2\alpha }i$
Which is purely real if
$\frac{3\sin \alpha }{1+4{\sin }^2\alpha }=0$
$\Rightarrow \sin \alpha =0$
$\Rightarrow \alpha =n\pi ,n\in Z$

$\therefore Re\left(z\right)=\frac{1-2{\sin }^2\alpha }{1+4{\sin }^2\alpha }=1$