Question

The real value of $\theta$ for which the expression, $\frac{1+i\cos \theta }{1-2i\cos \theta }$ is real number is

• A. $n\pi \pm \frac{\pi }{2}$
• B. $n\pi -\frac{\pi }{2}$
• C. $n\pi +\frac{\pi }{2}$
• D. $2n\pi \pm \frac{\pi }{2}$

Answer 1

The correct option is A $n\pi \pm \frac{\pi }{2}$

Let $z=\frac{1+i\cos \theta }{1-2i\cos \theta }$
$=\frac{(1+i\cos \theta )(1+2i\cos \theta )}{(1-2i\cos \theta )(1+2i\cos \theta )}$
$=\frac{1-2{\cos }^2\theta +i3\cos \theta }{1+4{\cos }^2\theta }$
$=\frac{1-2{\cos }^2\theta }{1+4{\cos }^2\theta }+i\left(\frac{3\cos \theta }{1+4{\cos }^2\theta }\right)$
If the above complex number is purely real, then
$Im\left(z\right)=0$
$3\cos \theta =0$
$\theta =\frac{(2n\pm 1)\pi }{2}$, where $nϵN$