Question

Consider the complex numbers $z=\frac{(1-i\sin \theta )}{(1+i\cos \theta )},$ then the value of $\theta$ for which $z$ is unimodular is given by,

• A. $n\pi \pm \frac{\pi }{6},n\in I$
• B. $n\pi \pm \frac{\pi }{3},n\in I$
• C. $n\pi \pm \frac{\pi }{4},n\in I$
• D. no real values of $\theta$

Answer 1

Answer: (C)

$n\pi \pm \frac{\pi }{4},n\in I$

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

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

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

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

$\Rightarrow 1+{{\sin }^2\theta }^{}{{\cos }^2\theta }^{}-2\sin \theta \cos \theta +1+2\sin \theta \cos \theta =1+{{\cos }^4\theta }^{}+2{{\cos }^2\theta }^{}$

$\Rightarrow 1+{{\cos }^2\theta }^{}-{{\cos }^4\theta }^{}={{\cos }^4\theta }^{}+2{{\cos }^2\theta }^{}$

$\Rightarrow 2{{\cos }^4\theta }^{}+{{\cos }^2\theta }^{}-1=0$

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

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

$\Rightarrow \cos \theta =\pm \frac{1}{\sqrt{2}}$

$\Rightarrow \theta =n\pi \pm \frac{\pi }{4},n\in I$

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