Question

Consider the complex numbers $z=\frac{(1-i\sin \theta )}{(1+i\cos \theta )},$
The value of $\theta$ for which z is purely imaginary real are,

• A. $n\pi -\frac{\pi }{4},n\in I$
• B. $n\pi +\frac{\pi }{4},n\in I$
• C. $n\pi ,n\in I$
• D. none of these

Answer 1

The correct option is A $n\pi -\frac{\pi }{4},n\in I$

$z=\frac{(1-i\sin \theta )}{(1+i\cos \theta )}$

$z=\frac{(1-i\sin \theta )}{(1+i\cos \theta )}\times \frac{(1-i\cos \theta )}{(1-i\cos \theta )}$

$z=\frac{1-i(\sin \theta +\cos \theta )-\sin \theta \cos \theta }{1+{\cos \theta }^2}$

$Forztobepurelyreal,$

$\sin \theta =-\cos \theta$

$\Rightarrow \theta =n\pi -\frac{\pi }{4}$

Hence option ${}^{\prime }{A}^{\prime }$ is the answer.