The Minimum Complex ?

$\begin{aligned} |z-3-2i| & \le 2 & \small {\color{#3D99F6}\text{Let }z = x+yi} \\ |x-3+(y-2)i| & \le 2 \\ (x-3)^2 + (y-2)^2 & \le 4 \end{aligned}$

We note that $y$ is maximum when $x=3$ . That is $(y-2)^2 = 4$ $\implies y_{max} = 4$ .

Now, consider

$\begin{aligned} |2z-6+5i| & = 2|z-3+2.5i| \\ & = 2\sqrt{(x-3)^2 + (y+2.5)^2} \\ & = 2\sqrt{(x-3)^2 + (y-2)^2 - (y-2)^2 + (y+2.5)^2} \\ & \le 2\sqrt{4 - (y-2)^2 + (y+2.5)^2} \\ & \le 2\sqrt{4 - y^2+4y-4 + y^2+5y+6.25} \\ & \le 2\sqrt{9y+6.25} \\ & \le 2\sqrt{9y_{max}+6.25} \\ & \le 2\sqrt{42.25} \\ & \le \boxed{13} \end{aligned}$

$|2z-6+5i|=|2z-6-4i+9i| \leq |2z-6-4i| + |9i|= 2|z-3-2i| +|9i| \leq 2.2+9=13$

For two complex numbers $z_1,z_2$ we know that $|z_1|+|z_2| \geq|z_1+z_2|$

The minimum value of $|z-3-2i|=2$ which is given