PT⇔z3−z2−iz2+3z+iz−3i=0⇔z3−z2+3z=iz2−iz+3i⇔z(z2−z+3)=i(z2−z+3)⇔(z−i)(z2−z+3)=0$\Leftrightarrow \left[ {\begin{matrix} z=i (1)\\ z^2-z+3=0 (2) \end{matrix}} \right.$(2)⇔[z=12(1−i√11)z=12(1+i√11)
PT
⇔z3−z2−iz2+3z+iz−3i=0⇔z3−z2+3z=iz2−iz+3i⇔z(z2−z+3)=i(z2−z+3)⇔(z−i)(z2−z+3)=0$\Leftrightarrow \left[ {\begin{matrix} z=i
\\ z^2-z+3=0 (
1) \end{matrix}} \right.$(
1)
⇔[z=12(1−i√11)z=12(1+i√11)