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Question: The real value of \(\theta\) for which the expression \(\frac{1 + i\cos\theta}{1 - 2i\cos\theta}\)is...

The real value of θ\theta for which the expression 1+icos⁡θ1−2icos⁡θ\frac{1 + i\cos\theta}{1 - 2i\cos\theta}is a real number, is

A

nπ+π/2n\pi + \pi/2

B

nπ−π/2n\pi - \pi/2

C

nπ±π/2n\pi \pm \pi/2

D

None of these

Answer

nπ±π/2n\pi \pm \pi/2

Explanation

Solution

Sol. Given that 1+icos⁡θ1−2icos⁡θ\frac{1 + i\cos\theta}{1 - 2i\cos\theta}

=(1+icos⁡θ)(1+2icos⁡θ)(1−2icos⁡θ)(1+2icos⁡θ)= \frac{(1 + i\cos\theta)(1 + 2i\cos\theta)}{(1 - 2i\cos\theta)(1 + 2i\cos\theta)}

=[(1−2cos⁡2θ)(1+4cos⁡2θ)]+i[3cos⁡θ1+4cos⁡2θ]= \left\lbrack \frac{(1 - 2\cos^{2}\theta)}{(1 + 4\cos^{2}\theta)} \right\rbrack + i\left\lbrack \frac{3\cos\theta}{1 + 4\cos^{2}\theta} \right\rbrack

Since Im(z)=0{Im}(z) = 0, then 3cos⁡θ=03\cos\theta = 0 ⇒ θ=nπ±π/2.\theta = n\pi \pm \pi/2.