Solve this equation $z^3-(2+4i)z^2-3(1-3i)z+14-2i=0,z\in C$

Question

Solve following equation $$z^3-(2+4i)z^2-3(1-3i)z+14-2i=0,z\in C$$

Try $z=a+bi$,then It's ugly can you more simple ?

Answer 1

Generally, you can try Cardano method, but sometimes it is worthwhile to search for "gaussian integers" solution first, that is, solutions of the form $a+bi$, where $a,b$ are integers. Of course, that method fails in case there are no "rational" solutions. If $z=a+bi$, is such a solution, then $$ a+bi\mid 14-2i=i(1+i)^3(1+2i)^2 $$ Therefore, $a+bi$ is of the form $a+bi=i^{\ell}(1+i)^m(1+2i)^n$, where $0\leqslant m\leqslant 3$, $0\leqslant n\leqslant 2$ and $0\leqslant \ell\leqslant 3$. In this case there are $48$ options to check.