Question

If principal argument of $z_0$ satisfying $|z-3|\le \sqrt{2}$ and $\arg (z-5i)=\frac{-\pi }{4}$ simultaneously is $\theta$ then identify the correct statement(s)?

• A. $\left|z_0\right|=17$
• B. $\tan 2\theta =\frac{8}{15}$
• C. $\tan \theta =\frac{-1}{4}$
• D. $\left|z_0-5i\right|=4\sqrt{2}$

Answer 1

Answer: (B), (D)

$\arg (z-5i)=\frac{-\pi }{4}$ is a ray emanating from $(0,5)$ and making an angle $\frac{\pi }{4}$ from the $x$-axis in clockwise direction.
$\Rightarrow \arg (x+i(y-5))=\frac{-\pi }{4}\Rightarrow {\tan }^{-1}\left(\frac{y-5}{x}\right)=\frac{-\pi }{4}\Rightarrow y-5=-x\Rightarrow x+y=5$
Also, $|z-3|\le \sqrt{2}\Rightarrow |(x-3)+iy|\le \sqrt{2}\Rightarrow (x-3{)}^2+y^2\le 2$
$\Rightarrow x+y=5$ touches the circle. Let $z_0$ be the complex number corresponding to point of contact.
Let $z_0=x_0+i{y}_0$

$x_0=3+\sqrt{2}\cos \frac{\pi }{4}=4$
$y_0=\sqrt{2}\sin \frac{\pi }{4}=1$
$\therefore z_0=4+i$
$\text{ Now, }\tan \theta =\frac{1}{4}$
$\Rightarrow \tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{2\cdot \frac{1}{4}}{1-\frac{1}{16}}=\frac{8}{15}$
Also, $\left|z_0\right|=\sqrt{17}$
and $\left|z_0-5i\right|=|4+i-5i|=4|1-i|=4\sqrt{2}$.