Question

Consider complex number $z=\frac{1-i\sin \theta }{1+i\cos \theta }$
The value of $\theta$ for which $arg\left(z\right)=\frac{\pi }{4}$ is given by

• A. $n\pi \pm \pi /6$
• B. $n\pi \pm \frac{\pi }{3}$
• C. $n\pi \pm \frac{\pi }{4}$
• D. $(2n+1)\pi$ or $2n\pi -\frac{\pi }{2},nϵI$

Answer 1

The correct option is A $n\pi \pm \pi /6$

Given:$\arg \left(z\right)=\frac{\pi }{4}\Rightarrow \tan \frac{\pi }{4}=1$

On multiplying numerator and denominator by conjugate of denominator

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

$=\frac{1-{\sin }^2\theta -2i\sin \theta }{1+{\sin }^2\theta }$

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

$\Rightarrow \arg z=\frac{\frac{-2\sin \theta }{1+{\sin }^2\theta }}{\frac{{\cos }^2\theta }{1+{\sin }^2\theta }}=1$

$\Rightarrow \frac{-2\sin \theta }{1+{\sin }^2\theta }=\frac{{\cos }^2\theta }{1+{\sin }^2\theta }$

$\Rightarrow -2\sin \theta ={\cos }^2\theta$

$\Rightarrow -2\sin \theta -{\cos }^2\theta =0$

$\Rightarrow {\cos }^2\theta +2\sin \theta =0$

$\Rightarrow 1-{\sin }^2\theta +2\sin \theta =0$

$\Rightarrow -{\sin }^2\theta +2\sin \theta +1=0$

$\Rightarrow {\sin }^2\theta -2\sin \theta -1=0$ which is a quadratic equation

$\sin \theta =\frac{2\pm \sqrt{4+4}}{2}$

$=\frac{2\pm \sqrt{8}}{2}=\frac{2\pm 2\sqrt{2}}{2}=1\pm \sqrt{2}$

$\therefore \sin \theta =\sqrt{2}+1$ is not valid

Hence $\sin \theta =1-\sqrt{2}$

$\theta =n\pi \pm \frac{\pi }{6}$