Komplexa tal

$$\begin{gathered} (z+1-i)=6i-8 \\ (z+1-i)(z+1-i)=6i-8 \\ {a}^2-b^2+2a+2b+2abi-2ai+2bi=8i-8 \\ {\mathit{a}}^{\mathbf{2}}\mathbf{-}{\mathit{b}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{a}\mathbf{+}\mathbf{2}\mathit{b}\mathbf{=}\mathbf{-}\mathbf{8} \\ \mathbf{2}\mathit{a}\mathit{b}\mathbf{-}\mathbf{2}\mathit{a}\mathbf{+}\mathbf{2}\mathit{b}\mathbf{=}\mathbf{8} \\ {b}^4-4b^3-2b^2-16b-16=0 \\ Rätt?\hspace{0.17em}Hur löser jag denna isåfall? \end{gathered}$$