solve the equation $1+2z+2z^2+\ldots +2z^{n-1}+z^n=0$ im trying to solve the equation  $$(E)\quad 1+2z+2z^2+\ldots +2z^{n-1}+z^n=0$$
attempt : because $1$ isnt a solution we have  $$\begin{aligned}
1+2 z+2 z^{2}+\cdots+2 z^{n-1}+z^{n} &=2\left(z^{0}+z+z^{2}+\cdots+z^{n-1}\right)-1+z^{n} \\
&=2 \frac{1-z^{n}}{1-z}-1+z^{n} \\
&=\frac{2-2 z^{n}+z^{n}-z^{n+1}}{1-z}-1 \\
&=\frac{2-z^{n}-z^{n+1}}{1-z}-1
\end{aligned}$$ then (E) becomes $$z^{n+1}+z^{n}-z-1=0$$ but i dont know how to get any further.

$$\begin{align}
1 + 2z + 2z^2 + \cdots + 2z^{n-1} + z^n 
&= (1 + z + \cdots + z^{n-1}) + (z + z^2 + \cdots + z^n) \\
&= (1+z)(1 + z + z^2 + \cdots + z^{n-1}) \\
&= \frac{(z+1)(z^n - 1)}{z - 1}.
\end{align}$$
Thus the roots are $$z \in \{ e^{2\pi i k/n} : k \in \{1, 2, \ldots, n-1\} \} \cup \{ -1 \}.$$  When $n$ is even, $-1$ is a double root; when $n$ is odd, it is a single root.
