- 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.
Best Answer
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 $z$.