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?
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 $z \neq 1$.
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 $k = 1, 2, ..., n-1$.
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.
