Question

If $z$ is any complex number satisfying $|z-3-2i|\le 2,$ then the minimum value of $|2z-6+5i|$ is

Answer 1

$|z-3-2i|\le 2$
It implies that $z$ lies on or inside the circle of radius $2$ and centre $(3,2).$



$|2z-6+5i{|}_{min}=2{|z-3+\left(\frac{5}{2}\right)i|}_{min}$
$=2\left(AB\right)=2(AC-BC)=2(\sqrt{(3-3{)}^2+{(2+\frac{5}{2})}^2}-2)$
$=2(2+\frac{5}{2}-2)$
$=5$

$\text{Alternate Solution}:$
Given $|z-3-2i|\le 2\dots \left(1\right)$
$|2z-6+5i|=|2(z-3-2i)+9i|$
$\ge ||2(z-3-2i)|-|9i||(\because |z_1+z_2|\ge ||z_1|-|z_2||)$
$\ge |2|z-3-2i|-|9i||$
$\Rightarrow |2z-6+5i|\ge |2|z-3-2i|-9|$
From equation $\left(1\right),$
$2|z-3-2i|\in [0,4]$
$\Rightarrow |2|z-3-2i|-9|\in [5,9]$
$\therefore |2z-6+5i|\ge 5$