<s>[INST] Help please on complex polynomials I wanted to know if there's any good approaches to these questions
a)By considering $z^9-1$ as a difference of two cubes, write $1+z+z^2+z^3+z^4+z^5+z^6+z^7+z^8$
as a product of two real factors one of which is a quadratic.
b) Solve $z^9-1=0$ and hence write down the 6 solutions of $z^6+z^3+1=0$
c)By letting $y=z+\frac{1}{z}$ and dividing $z^6+z^3+1=0$ by $z^3$, deduce that:
   $cos\frac{2\pi}{9}+cos\frac{4\pi}{9}+cos\frac{8\pi}{9}=0$
Any help would be greatly appreciated.
 [/INST] (a)
\begin{align*}
(z - 1)(z^8 + z^7 + \ldots + 1) = z^9 - 1 & = (z^3 - 1)(z^6 + z^3 + 1) \\
& = (z - 1)(z^2 + z + 1)(z^6 + z^3 + 1) \\
\therefore \qquad z^8 + z^7 + \ldots + 1 & = (z^2 + z + 1)(z^6 + z^3 + 1).
\end{align*}
(b)
The roots of $z^9 - 1$ are $r_n = e^{\frac{2\pi}9ni}$, for $n = 0, \ldots, 8$. Since $r_0 = 1$ is the root of $z - 1$, the other 8 roots are precisely the roots of $z^8 + z^7 + \ldots + 1 = (z^2 + z + 1)(z^6 + z^3 + 1)$. The quadratic $z^2 + z + 1$ has roots $\frac{-1 \pm \sqrt{3}i}{2}$, which are $r_3$ and $r_6$. Therefore, the 6 roots of $z^6 + z^3 + 1$ are $r_1, r_2, r_4, r_5, r_7, r_8$.
(c)
\begin{align*}
y^3
& = z^3 + 3z + 3\frac 1z + \frac 1{z^3} \\
y^3 - 3y + 1 & = z^3 + 1 + \frac 1{z^3} = \frac{1}{z^3}(z^6 + z^3 + 1) \\
& = \frac{1}{z^3}(z - r_1)(z - r_2)(z - r_4)(z - r_5)(z - r_7)(z - r_8).
\end{align*}
The roots of $y^3 - 3y + 1$ are $r_i + \frac{1}{r_i}$, but some $r_i$ give the same root because $r_i = \frac{1}{r_{9-i}}$. More precisely, the three roots are $y_1 = r_1 + \frac{1}{r_1} = r_8 + \frac{1}{r_8}$, $y_2 = r_2 + \frac{1}{r_2} = r_7 + \frac{1}{r_7}$, and $y_3 = r_4 + \frac{1}{r_4} = r_5 + \frac{1}{r_5}$.
We also know that the sum of the roots of $y^3 - 3y + 1$ is $0$ because the coefficient in front of $y^2$ is $0$. This means
\begin{align*}
y_1 + y_2 + y_3 = r_1 + \frac 1{r_1} + r_2 + \frac 1{r_2} + r_4 + \frac 1{r_4} & = 0 \\
2\left(\cos\frac{2\pi}9 + \cos\frac{4\pi}9 + \cos\frac{8\pi}9\right) & = 0
\end{align*}
</s>