Question

MODERATOR CHALLENGE:


don't try to spam it will be useless as your answer will be reported and your points will be deducted ​

Answer 1

$\huge\boxed{\underline{\underline{\green{\tt Solution}}}} </p><p>$

Let

$\displaystyle \: \: z \: = \frac{1 - i \: cos \theta}{1 + 2 i \: cos \theta}$

$\implies \: \displaystyle \: \: z \: = \frac{(1 - i \: cos \theta)(1 - 2 i \: cos \theta)}{(1 - 2 i \: cos \theta)(1 + 2 i \: cos \theta)}$

$\implies \: \displaystyle \: \: z \: = \frac{(1 - 2 i \: cos \theta -i \: cos \theta + 2 {i}^{2} {cos}^{2} \theta)}{(1 - 4 {i}^{2} \: {cos}^{2} \theta)}$

$\implies \: \displaystyle \: \: z \: = \frac{(1 - 3 i \: cos \theta - 2 {cos}^{2} \theta)}{(1 + 4 \: {cos}^{2} \theta)}$

$\implies \: \displaystyle \: \: z \: = \frac{(1 - 2 {cos}^{2} \theta)}{(1 + 4 \: {cos}^{2} \theta)} - \frac{ 3 i \: cos \theta }{(1 + 4 \: {cos}^{2} \theta)}$

Now z is a real number if Imaginary part = 0

So

$\displaystyle \: \frac{ - 3 \: cos \theta }{(1 + 4 \: {cos}^{2} \theta)} = 0$

$\displaystyle \: \: cos \theta \: = 0$

Hence

$\theta \: = (2n + 1) \frac{\pi}{2} \: \: \: , \: \: n \: \in I$