Question

For all complex numbers $z_1,z_2$ satisfying $|z_1|=12$ and $|z_2-3-4i|=5$ respectively, the minimum values of $|z_1-z_2|$ is

Answer 1

We have, $|z_1|=12$
and $|z_2-(3+4i)|=5$
$\Rightarrow ||z_2|-5|\le 5\le |z_2|+5$
$\Rightarrow -5\le |z_2|-5\le 5\text{and}|z_2|\ge 0$
$\Rightarrow 0\le |z_2|\le 10\text{and}|z_2|\ge 0$
$\therefore |z_2|\in [0,10]$

Now, $||z_1|-|z_2||\le |z_1-z_2|$
$\therefore |z_1|-|z_2|\in [2,12]$
$\Rightarrow$ The minimum value of $|z_1-z_2|$ is $2$ which occurs when $|z_2|=10$

$\text{Alternate Solution}:$


$(0,0)$ lies on $|z_2-3-4i|=5.$
Clearly, from the diagram, we can conclude that the minimum value of $|z_1-z_2|$ occurs at the common diameter.
Hence, the minimum value of $|z_1-z_2|$ is $12-10=2$