Find the roots of the given equation (complex numbers).

Question

How can I find the roots of the given equation in $\mathbb{C}$?

$$z^4 + 4z^3 + 7z^2 + 6z + 3 = 0 $$

Answer 1

Since $(1,4,7,6,3)=(1,4,6,4,1)+(0,0,1,2,1)+(0,0,0,0,1)$, $$ p(z)=z^4+4z^3+7z^2+6z+3 = (z+1)^4+(z+1)^2+1 $$ so in order to find the roots of $p(z)$ it is enough to find the roots of $$ q(z) = z^4+z^2+1 = \color{red}{\frac{z^6-1}{z^2-1}} $$ that clearly lie on the unit circle.

Answer 2

We have \begin{align} z^4+4z^3+7z^2+6z+3 & = (z^4+z^3+z^2) + (3z^3+3z^2+3z) + (3z^2+3z+3)\\ & = z^2(z^2+z+1) + 3z(z^2+z+1) + 3(z^2+z+1)\\ & = (z^2+3z+3)(z^2+z+1) \end{align}