Question

Consider the complex numbers $z=\frac{(1-isin\theta )}{(1+icos\theta )}$. The value of $\theta$ for which z is purely real are

• A. $n\pi -\frac{\pi }{4},nϵI$
• B. $n\pi +\frac{\pi }{4},nϵI$
• C. $n\pi ,nϵI$
• D. None of these

Answer 1

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

Given, $z=\frac{1-i\sin \theta }{1+i\cos \theta }=\frac{(1-i\sin \theta )(1-i\cos \theta )}{(1+i\cos \theta )(1-i\cos \theta )}$
$=\frac{\left(10\sin \theta \cos \theta \right)-i(\cos \theta +\sin \theta )}{(1+{\cos }^2\theta )}$
If $z$ is purely real, then $\cos \theta +\sin \theta =0$
$\tan \theta =-1$
$\Rightarrow \theta =n\pi -\frac{\pi }{4},nϵI$