Write in polynomial in factored form in complex number Write the following polynomial in factored form(in complex number):
$$1+z+z^2+z^3+z^4+z^5+z^6$$
Also, is there general solution of factoring for $1+z+z^2...z^n$ types of polynomial?
 A: $$1+z+z^2+z^3+z^4+z^5+z^6=\frac{1-z^7}{1-z}$$
$$1+z+z^2+z^3+z^4+z^5+...+z^n=\frac{1-z^{n+1}}{1-z}$$
Denote $$1+z+z^2+...+z^n=S$$ then multiply by z we get
$$z+z^2+...+z^{n+1}=Sz$$ 
from firs equation subtract the second then solve by $S$
A: I'd extend Adi Dani's answer a bit
$$
1 + z^2 + \ldots + z^6 = \frac {1-z^7}{1-z}
$$
Now decompose $1-z^7$ in factors. In order to do that, find all roots of
$$
1-z^7 = 0 \\
z = \sqrt [7]1 = \sqrt[7]{e^{0\cdot pi}} = \{ e^{\frac {2\pi ki}7}\}, k = 0,1,\ldots,6
$$
So
$$
1-z^7 = \prod_{k = 0}^6 \left (z - e^{\frac {2\pi ki}7}\right)
$$
To remove singularity at $z = 1$ we can cancel first $z - e^{\frac 07} = z-1$ term with denominator. So
$$
\sum_{k=0}^6 z^k = \prod_{k=1}^6 \left(z-e^{\frac {2\pi ki}7} \right) 
$$
Update
I overlooked second part of the problem, but it's pretty much the same as Adi Dani's answer
$$
\sum_{k=0}^n z^k = \frac {1-z^{n+1}}{1-z} = \frac {\displaystyle \prod_{k = 0}^n \left(z-e^{\frac {2\pi ki} n} \right)}{1-z} = \prod_{k = 1}^n \left(z-e^{\frac {2\pi ki}n} \right)
$$
