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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.

Donnie Darko
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2 Answers2

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$$\begin{align} 1 + 2z + 2z^2 + \cdots + 2z^{n-1} + z^n &= (1 + z + \cdots + z^{n-1}) + (z + z^2 + \cdots + z^n) \\ &= (1+z)(1 + z + z^2 + \cdots + z^{n-1}) \\ &= \frac{(z+1)(z^n - 1)}{z - 1}. \end{align}$$

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.

heropup
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Hint: \begin{align} z^{n+1}+z^n-z-1&=z^n(z+1)-(z+1)\\ &=(z^n-1)(z+1) \end{align}

Also, I think you should justify why the second equality holds, i.e. explain why you can assume $1-z\neq 0$.

tca
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