Question

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

Answer 1

$\frac{1+icos\theta }{1-2icos\theta }$

Rationalizing denominator

$=\frac{(1+icos\theta )(1+2icos\theta )}{(1-2icos\theta )(1+2icos\theta )}$

$=\frac{1+2icos\theta +icos\theta -co{s}^2\theta }{1+4co{s}^2\theta }(\because i^{-2}-1)$

$=\frac{1+3icos\theta -2co{s}^2\theta }{1+4co{s}^2\theta }$

Separating real and imaginary parts.

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

for purely real number imaginary part = 0

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

$\Rightarrow cos\theta =0$

Thus

$\theta =\frac{n\pi }{2}$

where n = 1,3,5,7.....