Giải:
Đặt \(z=a+bi(a,b\in\mathbb{R})\)
Theo bài ra ta có:
\(\left\{\begin{matrix} |(a-2)+i(b-1)|=\sqrt{10}\\ z\overline{z}=|z|^2=a^2+b^2=25\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (a-2)^2+(b-1)^2=10\\ a^2+b^2=25\end{matrix}\right.\)
\(\left\{\begin{matrix} 2a+b=10\\ a^2+b^2=25\end{matrix}\right.\Rightarrow a^2+(10-2a)^2=25\rightarrow a=5\) hoặc \(a=3\)
\(\Rightarrow b=0;4\)
Vậy \(z\in \left \{5,3+4i\right\}\)