Question

If $\tan \theta -i,$ where $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$
can be written as $r{e}^{i\beta }$, then $(r,\beta )$ is

• A. $(\sec \theta ,\theta -\frac{\pi }{2})$
• B. $(\sec \theta ,\theta +\frac{\pi }{2})$
• C. $(\text{cosec}\theta ,\theta -\frac{\pi }{2})$
• D. $(\text{cosec}\theta ,\frac{\pi }{2}-\theta )$

Answer 1

The correct option is A $(\sec \theta ,\theta -\frac{\pi }{2})$

$z=\tan \theta -i,\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$
$|z|=\sqrt{1+{\tan }^2\theta }=\sqrt{{\sec }^2\theta }=|\sec \theta |=\sec \theta$
Now, case $-1:$ If $\theta \in (0,\frac{\pi }{2})$
Then $Re\left(z\right)>0$ and $z$ lies in $1^{st}$ quadrant.
$\tan \beta =\left|\frac{-1}{\tan \theta }\right|=\cot \theta =\tan (\frac{\pi }{2}-\theta )$
$\Rightarrow \beta =\frac{\pi }{2}-\theta$
$\therefore \text{arg}\left(z\right)=\frac{\pi }{2}-\theta$

Case $-2:$ If $\theta \in (-\frac{\pi }{2},0)$
Then $Re\left(z\right)<0$ and $z$ lies in $4^{th}$ quadrant.
$\tan \beta =\left|\frac{-1}{\tan \theta }\right|=\frac{1}{-\tan \theta }=-\tan (\frac{\pi }{2}-\theta )=\tan (\theta -\frac{\pi }{2})$
$\Rightarrow \beta =\theta -\frac{\pi }{2}$
$\therefore \text{arg}\left(z\right)=-\beta \therefore \text{arg}\left(z\right)=\frac{\pi }{2}-\theta$