Question

If $|z-2+2i|=1,$ then

• A. maximum value of $|z|$ is $\sqrt{8}+1$
• B. maximum value of $|z|$ is $\sqrt{10}+1$
• C. minimum value of $|z|$ is $\sqrt{8}-1$
• D. minimum value of $|z|$ is $\sqrt{6}-1$

Answer 1

Answer: (A), (C)

minimum value of $|z|$ is $\sqrt{8}-1$

$|z-2+2i|=1$
$\Rightarrow |z-(2-2i)|=1$
$\because ||z_1|-|z_2||\le |z_1-z_2|\le |z_1|+|z_2|$
$\Rightarrow ||z|-\sqrt{8}|\le 1\le |z|+\sqrt{8}$
$||z|-\sqrt{8}|\le 1$ and $1\le |z|+\sqrt{8}$
$\Rightarrow -1\le |z|-\sqrt{8}\le 1$ and $|z|\ge -\sqrt{8}+1$
$\Rightarrow \sqrt{8}-1\le |z|\le \sqrt{8}+1$ and $|z|\ge 0$
$\therefore |z|\in [\sqrt{8}-1,\sqrt{8}+1]$
So, minimum value of $|z|$ is $\sqrt{8}-1$
and maximum value of $|z|$ is $\sqrt{8}+1$

Alternate solution :


$|z-(2-2i)|=1$
Let, $z=x+iy$
$\Rightarrow (x-2{)}^2+(y+2{)}^2=1$
It represents a circle with centre $(2,-2)$ and radius $1$ unit

$|z|=|z-0|=$ distance of point $P\left(z\right)$ from origin
$\therefore (OP{)}_{\text{min}}=OC-r=\sqrt{8}-1$
and $(OP{)}_{\text{max}}=OC+r=\sqrt{8}+1$