How to solve a complex polynomial? 
*

*Solve:
$$ z^3 - 3z^2 + 6z - 4 = 0$$


How do I solve this?
Can I do it by basically letting $ z = x + iy$ such that $ i = \sqrt{-1}$ and $ x, y \in \mathbf R $ and then substitute that into the equation and get a crazy long equation? If I did that I suspect I wouldn't be able to decipher the imaginary part of the equation.
Or should I change it to one of the forms below:
$$ z^n = r^n \mathbf{cis} n \theta $$
$$ z^n = r^n e^{n\theta i} $$
And then plug that into the equation? I did that. But it looked unsolvable. I'm so confused.

The easiest thing is just try to guest a root of the polynomial first. In this case, for
$$p(z) = z^3 - 3z^2 + 6z - 4,$$
we have that $p(1) = 0$.
Therefore, you can factorize it further and get
$$z^3 - 3z^2 + 6z - 4 = (z-1)(z^2 - 2z + 4)$$
$$= (z-1)((z-1)^2 + 3).$$
Their roots are just $$z_{1} = 1, \hspace{10pt}z_{2} = 1 + i\sqrt{3}, \hspace{10pt}z_{3} = 1 - i\sqrt{3}.$$
