Since complex number field $\mathbb{C}$ is algebraically closed, every polynomials with complex coefficients have linear polynomial decomposition. In this case, it's $$ z^3 - 3z^2 + 6z - 4 = (z - 1)(z - 1 + \sqrt{3}i)(z - 1 - \sqrt{3}i). $$ So you can see the solution of the equation easily from this representation.

One way to find out such decomposition is simply put $$ z^3 - 3z^2 + 6z - 4 = c(z - \alpha)(z - \beta)(z - \gamma) $$ to find out the coefficients by equating coefficients of power of $z$.

If you substitute in $z=x+iy$ the real and imaginary parts are each cubics-you just sort out the terms that have a factor $i$ for the imaginary part and you have a real equation. But as Daniel Fischer says, there is an easier approach here. To see what happens, we have $$(x+iy)^3-3(x+iy)^2+6(x+iy)-4=0\\x^3+3ix^2y-3xy^2-iy^3-3x^2-6ixy+3y^2+6x+6iy-4=0\\x^3-3xy^2-3x^2+3y^2+6x-4=0\\3x^2y-y^3-6xy+6y=0$$ where the second came from the imaginary part. It is no worse than the real part, which is the next to last line.

$z=x+1$ $\Longrightarrow$ $x^{3}+3x=0$

$\therefore$ $x=0$ , $\pm\sqrt{3}i$

$\therefore$ $z=1$ , $1\pm\sqrt{3}i$

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 $$z_{1} = 1, \hspace{10pt}z_{2} = 1 + i\sqrt{3}, \hspace{10pt}z_{3} = 1 - i\sqrt{3}.$$

@Petch Puttichai , after we've guessed the root say $p(1)$, how do we find the other factors , is it that we divide the polynomial with first factor , are there any other methods ?

For example if I have a polynomial of $z^4+z^3+... $ after we find the first root by guessing, how to get the other polynomial which I think should be of the form $z^3+z^2+...$.

Thanks for all your help.

Arif