Finding the roots of $z^4+z^3+z^2+z+1$ I need to find the roots of $z^4+z^3+z^2+z+1$. One approach I have is to guess the first linear solution, then use polynomial long division to find a 3rd degree polynomial, guess another from that, long division again, then use the quadratic formula to solve the quadratic. 
I think this will work but seems a bit work-heavy. Is there a simpler way?
 A: Observe that this is a geometric sum.
$$\sum_{i=0}^4 z^i = \frac{z^5-1}{z-1}=0$$
Solve for $z^5=1$ and $z \neq 1$.
A: And, of course,
the obvious generalization:
If
$f(z)
=1+z+z^2+...+z^{n-1}
=\sum_{k=0}^{n-1} z^k
$,
since
$f(z)(z-1)
=z^{n}-1
$,
the roots of $f(x)$
are the $n$th 
roots of unity except for $1$,
or
$\exp(2\pi ik/n)$
for $k = 1, 2, ..., n-1$.
A: In addition to the multiply-by-$\left(z - 1\right)$ trick, there is also the "symmetric" method to solve for the roots directly:
$$z^4 + z^3 + z^2 + z + 1 = z^2 \left(z^2 + z + 1 + \frac{1}{z} + \frac{1}{z^2}\right) = 0 $$
Since $z^2 \ne 0$, we can cancel it out, and we recognize the almost-square $z^2 + \frac{1}{z^2}$:
$$z^2 + z + 1 + \frac{1}{z} + \frac{1}{z^2} = \left(z + \frac{1}{z}\right)^2 - 2 + \left(z + \frac{1}{z}\right) + 1 = 0 $$
Now we can substitute $y = z + \frac{1}{z}$:
$$y^2 + y - 1 = 0 \to y = \frac{1\pm\sqrt{5}}{2}$$
We also have:
$$z + \frac{1}{z} = y \to z^2 + 1 = yz \to z=\frac{y\pm\sqrt{y^2 - 4}}{2}$$
We plug in $ y = \frac{1\pm\sqrt{5}}{2} $ into the equation above, and the reader verifies that the answers are equivalent to:
$$e^{i\frac{2\pi k}{5}}=\cos{\frac{2\pi k}{5}} + i\sin{\frac{2\pi k}{5}}, 0\le k\le 4$$
as obtained from the multiply-by-$\left(z-1\right)$ method.
