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$z^5+z^4+z^3+z^2+z+1 = 0$

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.

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    $\begingroup$ Doesn't the polynomial look a lot like a partial sum of a familiar series? $\endgroup$ Oct 18 '15 at 20:49
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    $\begingroup$ Whilst the given solutions are neater, you should also learn that the factoring you did gave a quadratic in $z^2$ that you know how to solve - so nothing more clever is actually required here. $\endgroup$
    – Keith
    Oct 19 '15 at 0:09
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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).

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  • $\begingroup$ this is clearly the best way :) $\endgroup$
    – Ant
    Oct 18 '15 at 20:57
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    $\begingroup$ Although I should have seen the solution the other ways, this will allow me to utilize the method I used previously. $\endgroup$ Oct 18 '15 at 21:01
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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: $$z^2+z+1=0\implies z_{1,2}=\frac{-1\pm\sqrt{1-4}}{2}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$$ $$z^2-z+1=0\implies z_{3,4}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$$

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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 $x\neq 1$.)

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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 $z^4 + z^2 + 1 = 0$.

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    $\begingroup$ Which is a quadratic equation in $x=z^2$. In fact, it's one of my favorite equations. $\endgroup$
    – Joel
    Oct 19 '15 at 10:15
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$$z^5+z^4+z^3+z^2+z+1 = 0 \Longleftrightarrow$$ $$(z+1)(z^2-z+1)(z^2+z+1) = 0 \Longleftrightarrow$$ $$z+1=0 \vee z^2-z+1=0 \vee z^2+z+1=0 \Longleftrightarrow$$ $$z=-1 \vee z=\frac{1\pm\sqrt{-3}}{2} \vee z=\frac{-1\pm\sqrt{-3}}{2} \Longleftrightarrow$$ $$z=-1 \vee z=\frac{1\pm i\sqrt{3}}{2} \vee z=\frac{-1\pm i\sqrt{3}}{2} $$

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$$\begin{align}z^6&=1\\ z^6-1&=0\\ (z-1)(\underbrace{z^5+z^4+z^3+z^2+z+1}_{f(z)})&=0 \end{align}$$ Hence the solutions to $f(z)$ are $$\begin{align}z^6&=1=e^{i2\pi} &&(z\neq1)\\ z&=e^{i\frac {2n\pi}6}=e^{i\frac {n\pi}3}&& (n=1,2,3,4,5)\quad\blacksquare\end{align}$$

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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.

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