1

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 ?

Harry Peter
  • 8,193
math110
  • 94,932
  • 17
  • 148
  • 519
  • 1
    There is a general formula for cubic polynomials. Maybe there is an easier approach, I don't know. – Eff Jun 03 '16 at 11:58
  • You could try the Cardano method. It is a bit of an ugly formula but it should work just as well as the formula for the second degree polynomials –  Jun 03 '16 at 12:00
  • I have given the general solutions to both quartic and cubic here: https://math.stackexchange.com/a/4601016/151732 You can start lower down from the cubic solution. Please work through the algebra yourself, as it is not pleasant. According to Wolfram Alpha: https://www.wolframalpha.com/input?i=z%5E3-%282%2B4i%29z%5E2-3%281-3i%29z%2B14-2i%3D0 , there are no "nice" roots so it's unlikely any "clever tricks" apply here. – Deepak Dec 18 '22 at 06:54

1 Answers1

0

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.

boaz
  • 5,551