how to solve a third degree equation of complex roots and coefficients

Question

It's not a homework it came in one of our exams and I didn't find anything on the internet that is a high-school level. please give me any hint or answer to solve this in a noncomplicated way. solve the equation: $$z^3+(2-i)z^2+(5-2i)z-5i=0.$$ Where the unknown z belongs to the set of complex numbers.

Answer 1

$$z^3+(2-i)z^2+(5-2i)z-5i=0$$

You can read the relevant section in Wikipedia on roots of a cubic function to proceed programmatically, or you can scan the equation to test for roots with some informed "guesses", which in this case, yields one root to be $z_1 = i$.

Then, just as, given a root $z_1 = c$, divide by the factor $(z - c)$, so using the root $z_1 = i$, divide your cubic by the factor $(z-i)$.

Now you're down to a degree two equation, and finding the other two roots should proceed smoothly for you.

Answer 2

One way to see what's going on here is to rewrite the equation with all the $i$'s on one side:

$$z^3+2z^2+5z=i(z^2+2z+5)$$

Factoring the left hand side as $z(z^2+2z+5)$, it's clear you can move things back to get

$$(z-i)(z^2+2z+5)=0$$

By happenstance (or design), the other two roots, $-1+2i$ and $-1-2i$, are also Gaussian integers.