Find roots of cubic polynomial with complex coefficient

[BearY]

Homework Statement

Find roots of
$$
-\lambda ^3 +(2+2i)\lambda^2-3i\lambda-(1-i) = 0
$$

Homework Equations

The Attempt at a Solution

I tried my old trick
I tried to separating the 4 terms into 2 pairs and try to find a common factor in the form of $\lambda + z$ between them,
$$
-\lambda ^2 (\lambda -2-2i) - 3i\lambda -1+i
$$
It doesn't seem to work.
$$
-\lambda (\lambda^2 +3i) - (2+2i)\lambda -1+i
$$
Nope.
And I am out of tricks.
Should I keep trying to factorize it or there is something else I should do about polynomial with complex coefficient?

 

[Ray Vickson]

Homework Statement

Find roots of
$$
-\lambda ^3 +(2+2i)\lambda^2-3i\lambda-(1-i) = 0
$$

Homework Equations

The Attempt at a Solution

I tried my old trick
I tried to separating the 4 terms into 2 pairs and try to find a common factor in the form of $\lambda + z$ between them,
$$
-\lambda ^2 (\lambda -2-2i) - 3i\lambda -1+i
$$
It doesn't seem to work.
$$
-\lambda (\lambda^2 +3i) - (2+2i)\lambda -1+i
$$
Nope.
And I am out of tricks.
Should I keep trying to factorize it or there is something else I should do about polynomial with complex coefficient?

You can use Cardano's formula for the solution of a cubic. It works for any coefficients, not just for real ones.

 

[LCKurtz]

The roots are $1,~i,~1+i$.

 

[fresh_42]

The shortest way is usually to guess one root, divide the polynomial by the corresponding linear factor and solve the quadratic rest polynomial. That $1$ is a root helps a lot.