Roots of $P_n(z)=1 + z + z^2 + ... + z^n$

Question

So I know that the factorization of $z^n-1$ is: $$z^{n}-1=\prod_{k=0}^{n-1}(z-e^{i(\frac{2\pi}{n}k)})$$ But I don't see how to factorize the monic polynomial $1+z+...+z^n$, where none of the coefficients is null.

Answer 1

As $$\displaystyle{z^{n+1}-1=\prod\limits_{k=0}^{n}(z-e^{i(\frac{2\pi}{n+1}k)})}$$ and $$\displaystyle{z^{n+1} -1 = (z-1)(1+\cdots+z^n)}$$ we get that : $$\displaystyle{(1+\cdots+z^n) = \prod\limits_{k=1}^{n}\left(z-e^{i(\frac{2\pi}{n+1}k)}\right)}$$

Answer 2

Hint:

$$P_n(z)=\frac{z^{n+1}-1}{z-1}$$

Answer 3

Notice that : $$ P_{n}\left(z\right)=\sum_{k=0}^{n}{z^{k}}=\frac{z^{n+1}-1}{z-1}=\prod_{k=1}^{n}{\left(z-\mathrm{e}^{\mathrm{i}\frac{2k\pi}{n+1}}\right)} $$