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Complex Numbers — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSComplex Numbers3 Marks
Question
Find the real values of $\theta$ for which the complex number $\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}$ is purely real.
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Answer

Let $\text{z}=\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}$ $=\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}\times\frac{1+2\text{i}\cos\theta}{1+2\text{i}\cos\theta}$ $=\frac{1+2\text{i}\cos\theta+\text{i}\cos\theta(1+2\text{i}\cos\theta)}{1^2+(2\cos\theta)^2}$ $=\frac{1+2\text{i}\cos\theta+\text{i}\cos\theta-2\cos^2\theta)}{1+4\cos^2\theta}$ $=\frac{1-2\cos^2\theta+3\text{i}\cos\theta}{1+4\cos^2\theta}$ $=\frac{1-2\cos^2\theta}{1+4\cos^2\theta}+\frac{3\cos\theta}{1+4\cos^2\theta}\text{i}$ We know that z is purely real if and on;y if Im z=0 $\therefore \ \frac{3\cos\theta}{1+4\cos^2\theta}=0$ $\big(\because$ z is given to be purely real $\big)$ $\Rightarrow3\cos\theta=0$ $\Rightarrow\cos\theta=0$ $\Rightarrow\cos\theta=\cos\frac{\pi}{2}$ $\therefore$ The general solution is given by $\theta=2\text{n}\pi\pm\frac{\pi}{2},\text{n}\in\text{z}$

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