Gọi $z=a+bi=>\overline{z}=a-bi(a,b\in R)$

Ta có:

$\begin{align} & (1+i)z+3i\overline{z}={{(\frac{2i}{i-1})}^{2}} \\ & <=>(1+i)(a+bi)+3i(a-bi)=-2i \\ & <=>a+2b+(4a+b)i=-2i \\ \end{align}$

$<=>\left\{ \begin{align} & a+2b=0 \\ & 4a+b=-2 \\ \end{align} \right.$

$<=>\left\{ \begin{align} & a=-\frac{4}{7} \\ & b=\frac{2}{7} \\ \end{align} \right.$

$<=>z=\frac{-4}{7}+\frac{2}{7}i=>\text{w}=-6+2i=>\overline{\text{w}}=-6-2i$