How do I solve the following equation: $z^4+z^3+z+1=0$ Is there an existing method to solve the following equation:
$z^4+z^3+z+1=0$?
 A: Hint: 
$$z^4 + z^3 + z + 1 = z^4 + z + z^3 + 1 = z(z^3+1) + z^3 + 1 = (z+1)(z^3 + 1)$$
A: HINT: we have $$z^3(z+1)+z+1=(z+1)(z^3+1)$$
A: As @David observed, since this polynomial is front-to-back symmetrical (the $z^2$ coefficient is in the middle, so its presence or absence doesn't matter), and a general method applies to cut the degree in half, as follows. Divide through by $z^2$ to get $z^2+z+{1\over z}+{1\over z^2}=0$, and let $w=z+{1\over z}$. Since $w^2=z^2+2+{1\over z^2}$, the equation becomes $w^2-2+w=0$ or $w^2+w-2=0$. Since this is quadratic, we solve it easily... $w=-2, 1$. Then $z+{1\over z}=-2$ and $z+{1\over z}=1$ give subsequent quadratics in $z$ (in this case).
A: Another approach:
We see that $0$ does not satisfy the equation, so we can divide by $z^2$ to get the equivalent equation
$$z^2+z+\frac{1}{z}+\frac{1}{z^2}=0\tag{1}$$
By making $u=z+\frac{1}{z}$ we get
$$u^2+u-2=0$$
The roots of this are $-2$ and $1$. Now we only need to solve
$$\color{blue}{z+\frac{1}{z}=-2}\qquad \text{and}\qquad\color{blue}{z+\frac{1}{z}=1}$$
A: When $z\in\mathbb{C}$:
$$z^4+z^3+z+1=0\Longleftrightarrow$$
$$(z+1)(z^3+1)=0$$

So, we get the following equations:


*

*$$z+1=0\Longleftrightarrow z=-1$$


Or:


*

*$$z^3+1=0\Longleftrightarrow$$
$$z^3=-1\Longleftrightarrow$$
$$z^3=e^{\pi i}\Longleftrightarrow$$
$$z=\left(e^{\left(2\pi k+\pi\right)i}\right)^{\frac{1}{3}}\Longleftrightarrow$$
$$z=e^{\frac{1}{3}\left(2\pi k+\pi\right)i}$$


With $k\in\mathbb{Z}$ and  $k:0-2$
