\(z=x+yi\Rightarrow\left|x+\left(y-5\right)i\right|\le3\)
\(\Rightarrow x^2+\left(y-5\right)^2\le9\)
\(\Rightarrow\left(y-5\right)^2\le9\Rightarrow-3\le y-5\le3\)
\(\Rightarrow2\le y\le8\Rightarrow4\le y^2\le64\)
\(\left|z\right|=\sqrt{x^2+y^2}\ge\sqrt{y^2}\ge\sqrt{4}=2\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\) hay \(z=2i\)