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Is there an existing method to solve the following equation: $z^4+z^3+z+1=0$?

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    $\begingroup$ $z=-1$ is one solution. It factors twice, so divide out $(z+1)^2$ and you get a quadratic for the other solutions. $\endgroup$ Commented Feb 2, 2016 at 19:00
  • $\begingroup$ Related: The roots of $t^5+1$. Note that $z^5 - 1 = (z-1)(z^4 + z^3 + z^2 + z + 1)$ and $z^5 + 1 = (z+1)(z^4 - z^3 + z^2 - z + 1).$ $\endgroup$ Commented Feb 2, 2016 at 19:05
  • $\begingroup$ Also, even without the nice factorizations, there's a method to solve this, as fourth-degree polynomials have exact closed-form solutions $\endgroup$
    – user304329
    Commented Feb 2, 2016 at 19:12

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Hint: $$z^4 + z^3 + z + 1 = z^4 + z + z^3 + 1 = z(z^3+1) + z^3 + 1 = (z+1)(z^3 + 1)$$

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  • $\begingroup$ Hah! Nice observation. +1. $\endgroup$ Commented Feb 2, 2016 at 19:08
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HINT: we have $$z^3(z+1)+z+1=(z+1)(z^3+1)$$

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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).

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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 $$\color{blue}{z+\frac{1}{z}=-2}\qquad \text{and}\qquad\color{blue}{z+\frac{1}{z}=1}$$

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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 $k:0-2$

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