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Question
Find the real values of θ for which the complex number \[\frac{1 + i cos\theta}{1 - 2i cos\theta}\] is purely real.
Solution
\[\frac{1 + i\cos\theta}{1 - 2i\cos\theta}\]
\[ = \frac{1 + i\cos\theta}{1 - 2i\cos\theta} \times \frac{1 + 2i\cos\theta}{1 + 2i\cos\theta}\]
\[ = \frac{1 + 2i\cos\theta + i\cos\theta - 2\cos\theta}{1 + 4 \cos^2 \theta}\]
\[ = \frac{1 - 2\cos\theta + i3\cos\theta}{1 + 4 \cos^2 \theta}\]
\[\text { For it to be purely real, the imaginary part must be zero } . \]
\[3\cos\theta = 0\]
\[\text { This is true for odd multiples of } \frac{\pi}{2} . \]
\[ \therefore \theta = \left( 2n + 1 \right)\frac{\pi}{2}, n \in Z\]
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