Question

Find the real values of $\theta$ for which the complex number $\frac{1+icos\theta }{1-2icos\theta }$ is purely real.

Answer 1

$Letz=\frac{1+icos\theta }{1-2icos\theta }=\frac{1+icos\theta }{1-2icos\theta }\times \frac{1+2icos\theta }{1+2icos\theta }=\frac{1+2icos\theta +icos\theta (1+2icos\theta )}{1^2+(2cos\theta {)}^2}=\frac{1+2icos\theta +icos\theta -2co{s}^2\theta }{1+4co{s}^2\theta }=\frac{1-2co{s}^2\theta +3icos\theta }{1+4co{s}^2\theta }=\frac{1-2co{s}^2\theta }{1+4co{s}^2\theta }+\frac{3cos\theta }{1+4co{s}^2\theta }$

We know that z is purely real if and only if Imz = 0

$\therefore \frac{3cos\theta }{1+4co{s}^2\theta }=0$

( $\because$ z is given to be purely real)

$\Rightarrow 3cos\theta =0\Rightarrow cos\theta =0\Rightarrow cos\theta =cos\frac{\pi }{2}$

$\therefore$ The general solution is given by

$\theta =2n\pi \pm \frac{\pi }{2},nϵZ$