Question

Consider the complex numbaers $z=\frac{(1-isin\theta )}{(1+icos\theta )}$. If argument of z is $\pi /4$, then

• A. $\theta =n\pi ,nϵI$ only
• B. $\theta =(2n+1),nϵI$ only
• C. Both $\theta =n\pi$ and $\theta =(2n+1)\frac{\pi }{2},nϵI$
• D. None of these

Answer 1

Answer: (D)

None of these

$z=\frac{1-i\sin \theta }{1+i\cos \theta }=\frac{(1-i\sin \theta )(1-i\cos \theta )}{(1+i\cos \theta )(1-i\cos \theta )}$
$=\frac{\left(10\sin \theta \cos \theta \right)-i(\cos \theta +\sin \theta )}{(1+{\cos }^2\theta )}$
If $z$ is purely real, then
$\cos \theta +\sin \theta =0$
or $\tan \theta =-1$\Rightarrow \theta=n\pi -\frac {\pi}{4}, n\epsilon I$If$z$ispurelyimaginary.$1- \sin\theta \cos\theta =0$or$\sin \theta \cos\theta =1$,whichisnotpossible.$|z|=\left |\dfrac {1-i \sin\theta}{1+ i \cos\theta}\right |=\dfrac {\sqrt {1+ \sin^2\theta}}{\sqrt {1+\cos^2\theta}}$If$|z|=1$,then$\cos^2\theta=\sin^2\theta \Rightarrow \tan^2\theta 1\Rightarrow \theta=n\pi \pm \frac {\pi}{4}, n\epsilon I$Wehave,$arg (z)=\tan^{-1}\left (\frac {-(\cos\theta + \sin\theta)}{(1-\sin\theta \cos\theta)}\right )$Now,$arg(z)=\pi / 4\Rightarrow \dfrac {-(\cos\theta+\sin\theta)}{(1-\sin\theta \cos\theta)}=1\Rightarrow \cos^2\theta+ \sin^2\theta+ 2 \sin\theta \cos\theta=1+\sin^2\theta \cos^2\theta-2 \sin\theta \cos\theta\Rightarrow 1+4 \sin\theta \cos\theta =1+\sin^2\theta \cos^2\theta\Rightarrow \sin^2\theta \cos^2\theta -4 \sin\theta \cos \theta =0\Rightarrow \sin\theta \cos\theta (\sin\theta \cos\theta -4)=0\Rightarrow \sin\theta \cos\theta =0 (\because \sin\theta \cos\theta =4$isnotpossible)$\Rightarrow \theta=(2n+1)\pi$or$\theta=(4n-1)\pi/2, n\epsilon 1 (\because -cos\theta - sin\theta > 0)$.