Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$ $z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution. 
I solved for all roots of $z^4 = -4$ but the structure for this example was more simple.
 A: Hint:
Since $$z^4+z^2+1=(z^2+1)^2-z^2=(z^2+z+1)(z^2-z+1)$$
Solve $z^2+z+1=0$ and $z^2-z+1=0$ by using the quadratic formula:
$$z^2+z+1=0\implies z_{1,2}=\frac{-1\pm\sqrt{1-4}}{2}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$$
$$z^2-z+1=0\implies z_{3,4}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$$
A: Here is another way to pursue this (just for fun!).
Recall that for $|x| < 1$ we have:
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$$
Multiplying both sides by $x^6$ yields:
$$\frac{x^6}{1-x} = x^6 + x^7 + x^8 + x^9 + \cdots$$
Now we subtract the former line from the latter:
$$\frac{x^6 - 1}{1-x} = -(1 + x + x^2 + x^3 + x^4 + x^5)$$
We now multiply both sides by $-1$ to obtain your original expression:
$$\frac{x^6 - 1}{x-1} = 1 + x + x^2 + x^3 + x^4 + x^5$$
In one way, this is no different than the suggestion in your accepted answer. But I think (hope) seeing the connection arise in a different manner is of interest. Anyway, at this point you seek to find the roots of $x^6 - 1 = (x^3 + 1)(x^3 - 1)$, and you can now factor the latter two parenthetical expressions as the sum and difference of cubes, respectively, to find the roots. 
(Noting that/why $x\neq 1$.)
A: Note that $(z-1)(1+z+z^2+z^3+z^4+z^5) = z^6-1$. Solve $z^6-1=0$ and discard the $z=1$ solution (which comes from the $z-1$ factor).
A: 
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution. 

False.
It also shows the remaining four solutions are roots of $z^4 + z^2 + 1 = 0$.
A: $$z^5+z^4+z^3+z^2+z+1 = 0 \Longleftrightarrow$$
$$(z+1)(z^2-z+1)(z^2+z+1) = 0 \Longleftrightarrow$$
$$z+1=0 \vee z^2-z+1=0 \vee z^2+z+1=0 \Longleftrightarrow$$
$$z=-1 \vee z=\frac{1\pm\sqrt{-3}}{2} \vee z=\frac{-1\pm\sqrt{-3}}{2} \Longleftrightarrow$$
$$z=-1  \vee z=\frac{1\pm i\sqrt{3}}{2}  \vee z=\frac{-1\pm i\sqrt{3}}{2} $$
A: $$\begin{align}z^6&=1\\
z^6-1&=0\\
(z-1)(\underbrace{z^5+z^4+z^3+z^2+z+1}_{f(z)})&=0
\end{align}$$
Hence the solutions to $f(z)$ are
$$\begin{align}z^6&=1=e^{i2\pi} &&(z\neq1)\\
z&=e^{i\frac {2n\pi}6}=e^{i\frac {n\pi}3}&& (n=1,2,3,4,5)\quad\blacksquare\end{align}$$
A: I quote: Factoring it into (z+1)(z4+z2+1)(z+1)(z4+z2+1) didn't do anything but show -1 is one solution. Put $z^2=u$ then you will find the solutions.
