Question

If complex number $z$ satisfies the inequality $|z-1-2i|\le 1$, then which of the following option(s) is (are) CORRECT ?

• A. maximum value of $|z|$ is $\sqrt{5}+1$
• B. minimum value of $|z|$ is $\sqrt{5}$
• C. maximum value of $\arg \left(z\right)$ is $\frac{\pi }{2}$
• D. minimum value of $\arg \left(z\right)$ is ${\tan }^{-1}\left(\frac{3}{4}\right)$

Answer 1

Answer: (A), (C), (D)

minimum value of $\arg \left(z\right)$ is ${\tan }^{-1}\left(\frac{3}{4}\right)$

$|z-1-2i|\le 1$ represents the points inside and on the circle with centre $C(1,2)$ and radius $r=1$


$\therefore |z{|}_{min}=OC-r=\sqrt{1^2+2^2}-1=\sqrt{5}-1$
and $|z{|}_{max}=OC+r=\sqrt{1^2+2^2}+1=\sqrt{5}+1$

Clearly, $\arg \left(z\right)$ is maximum for $M\left(z\right)$.
$max\left\{\arg \right(z\left)\right\}=\frac{\pi }{2}$
$\arg \left(z\right)$ is minimum for $N\left(z\right)$.
$min\left\{\arg \right(z\left)\right\}=\theta$
Now, $\tan \theta =\tan (\frac{\pi }{2}-2\alpha )$
$=\cot 2\alpha$
$=\frac{1}{\tan 2\alpha }$
$=\frac{1-{\tan }^2\alpha }{2\tan \alpha }$
$=\frac{1-{\left(\frac{1}{2}\right)}^2}{2\times \frac{1}{2}}=\frac{3}{4}$
$\therefore \theta ={\tan }^{-1}\frac{3}{4}$