Gọi $z= a +bi \Rightarrow \overline{z}=a-bi$
Vậy $a+bi -(1+i)(a-bi)=(\overline{1+2i})^2=(1-2i)^2 =1-4i+4i^2=-3-4i $
$\Leftrightarrow a+bi -a+bi-ai +bi^2=-3-4i$
$\Leftrightarrow -b +(2b-a)i =-3-4i$
$\Leftrightarrow b= 3;\ 2b-a=-4 \Rightarrow b=3;\ a= 10$
Vậy $z=10 +3i$