Question

Consider the complex numbers $z=\frac{(1-i\sin \theta )}{(1+i\cos \theta )},$ then the value of $\theta$ for z is purely imaginary 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. no real value of $\theta$

Answer 1

The correct option is D no real value of $\theta$

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

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

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

For $z$ to be purely imaginary,

$1=\sin \theta \cos \theta$ which is not possible for any value of $\theta$

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