How do I solve this : $z^4+z^3+z^2+1=0$ in $\mathbb{C}$?

Question

let z be a complex variable , How do I solve this : $z^4+z^3+z^2+1=0$ in $\mathbb{C}$ without using numerical method ?

Note: Wolfram alpha show this result

Answer 1

You could use Ferrari's method. Setting $w=1/z$ gives $$w^4+w^2+w+1=0.$$ Then, for a new unknown $t$, one has $$(w^2+t)^2=(2t-1)w^2-w+t^2-1.$$ If the quadratic on the RHS has equal roots we can take a square root. This is the case iff the discriminant $$1-4(2t-1)(t^2-1)=0.$$ This is a cubic equation that can be solved by Cardan's method.

Answer 2

Write $$z^4+z^3+z^2+1=\left(z^2+\frac{1}{2}z+k\right)^2-\left(\left(2k-\frac{3}{4}\right)z^2+kz+k^2-1\right).$$ Now, choose $k$ such that $k^2-4\left(2k-\frac{3}{4}\right)(k^2-1)=0$ and $2k-\frac{3}{4}>0$.

I got $$k=\frac{1}{6}+\frac{\sqrt{13}}{3}\cos\left(\frac{1}{3}\arccos\left(-\frac{43}{26\sqrt{13}}\right)\right)=1.0993...$$ and solve two quadratic equations.