Question

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

(a) $n\mathrm{\pi }+\frac{\mathrm{\pi }}{4}$

(b) $n\pi +{\left(-1\right)}^n\frac{\pi }{4}$

(c) $2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{2}$

(d) none of theses

Answer 1

Given $\frac{1+i\cos \theta }{1-2i\cos \theta }$ is a real number
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{1+i\cos \theta }{1-2i\cos \theta }\times \frac{1+2i\cos \theta }{1+2i\cos \theta } \\ =\frac{\left(1+i+\cos \theta \right)\left(1+2i\cos \theta \right)}{1-4i^2{\cos }^2\theta } \\ =\frac{1+2i\cos \theta +i\cos \theta +2i^2{\cos }^2\theta }{1+4{\cos }^2\theta } \\ =\frac{1-2{\cos }^2\theta +3i\cos \theta }{1+4{\cos }^2\theta } \\ =\frac{1-2{\cos }^2\theta }{1+4{\cos }^2\theta }+i\left(\frac{3\cos \theta }{1+4{\cos }^2\theta }\right) \end{gathered}$$

Since $\frac{1+i\cos \theta }{1-2i\cos \theta }$ is a real number

$$\begin{gathered} \Rightarrow \frac{3\cos \theta }{1+4{\cos }^2\theta }=0 \\ \mathrm{i}.\mathrm{e}\cos \theta =0 \\ \mathrm{i}.\mathrm{e}\theta =\frac{\left(2n\pm 1\right)\pi }{2}=n\pi \pm \frac{\pi }{2} \end{gathered}$$

Hence, the correct answer is option C.