Can we use Cardano's method if the coefficients of the equation is not real?

Question

Can we use Cardano's method if the coefficients of the equation is not real?

If yes, is it done the same way? If no, how can we solve such equations.

Answer 1

Yes. Actually, Cardano's formula works on any field whose characteristic is neither $2$ nor $3$ and on which every number has a square root and a cube root.

Here is an example (over $\mathbb C$). Consider the equation$$x^3-13ix-15+15i=0.\tag1$$Then$$\frac{(-15+15i)^2}4+\frac{(-13i)^3}{27}=-\frac{1\,681}{54}i=\left(\frac{41}{6\sqrt3}-\frac{41}{6\sqrt3}i\right)^2.$$Now, consider the numbers$$-\frac{-15+15i}2+\left(\frac{41}{6\sqrt3}-\frac{41}{6\sqrt3}i\right)\tag2$$and$$-\frac{-15+15i}2-\left(\frac{41}{6\sqrt3}-\frac{41}{6\sqrt3}i\right).\tag3$$It turns out that a cube root of $(2)$ is$$1+\frac{5\sqrt3}6i+\left(\frac12-\frac{2\sqrt3}3\right)i\tag4$$and that a cube root of $(3)$ is$$1-\frac{5\sqrt3}6i+\left(\frac12+\frac{2\sqrt3}3\right)i;\tag5$$besides, the product of these cube roots is $\frac{13i}3$. So, by Cardano's formula, $(4)+(5)$ should be a root of $(1)$. And it is: $(4)+(5)=2+i$ and you can check that $2+i$ is indeed a root of $(1)$.