# Find the Real Values of θ for Which the Complex Number 1 + I C O S θ 1 − 2 I C O S θ is Purely Real. - Mathematics

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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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 13 Complex Numbers
Exercise 13.2 | Q 10 | Page 32