Xét các số phức ${{z}_{1}}=x-2+\left( y+2 \right)i$ và...

$\left| {{z}_{1}} \right|=1\Leftrightarrow {{\left( x-2 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=1 (1)$
Đặt $\left\{ \begin{aligned}
& x-2=\cos t \\
& y+2=\sin t \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=2+\cos t \\
& y=-2+\sin t \\
\end{aligned} \right.$
Đặt $r=\left| z \right|,r\ge 0$
${{r}^{2}}={{\left| {{z}_{2}} \right|}^{2}}={{x}^{2}}+{{y}^{2}}={{\left( 2+\cos t \right)}^{2}}+{{\left( -2+\sin t \right)}^{2}}=9+4\left( \cos t-\sin t \right)$
${{r}^{2}}=9+4\sqrt{2}\cos \left( x+\dfrac{\pi }{4} \right)$
Do $-1\le \cos \left( x+\dfrac{\pi }{4} \right)\le 1\Leftrightarrow 1\Leftrightarrow 9-4\sqrt{2}\le {{r}^{2}}=9+4\sqrt{2}\cos \left( x+\dfrac{\pi }{4} \right)\le 9+4\sqrt{2}$
$\Leftrightarrow {{\left( 2\sqrt{2}-1 \right)}^{2}}\le {{r}^{2}}\le {{\left( 1+2\sqrt{2} \right)}^{2}}\Leftrightarrow 2\sqrt{2}-1\le r\le 2\sqrt{2}+1$
$\Rightarrow \left\{ \begin{aligned}
& M=2\sqrt{2}+1 \\
& m=2\sqrt{2}-1 \\
\end{aligned} \right.\Rightarrow M.m=7$