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Câu 2:Đặt: $z=x+yi,x,y\in\mathbb{R}$.Hệ trở thành: $\left\{\begin{array}{l}|x+(y-2)i|=|x+yi|\\|x+(y-1)i|=|(x-1)+yi|\end{array}\right.$$\Leftrightarrow \left\{\begin{array}{l}x^2+(y-2)^2=x^2+y^2\\x+(y-1)^2=(x-1)^2+y^2\end{array}\right.$$\Leftrightarrow \left\{\begin{array}{l}4-4y=0\\-2y=-2x\end{array}\right.\Leftrightarrow x=y=1$Vậy $z=1+i$.

Câu 2:Đặt: $z=x+yi,x,y\in\mathbb{R}$.Hệ trở thành: $\left\{\begin{array}{l}|x+(y-2)i|=|x+yi|\\|x+(y-1)i|=|(x-1)+yi|\end{array}\right.$$\Leftrightarrow \left\{\begin{array}{l}x^2+(y-2)^2=x^2+y^2\\x^2+(y-1)^2=(x-1)^2+y^2\end{array}\right.$$\Leftrightarrow \left\{\begin{array}{l}4-4y=0\\-2y=-2x\end{array}\right.\Leftrightarrow x=y=1$Vậy $z=1+i$.