Sum of the n nth roots of unity 1) If $z=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$
2) Then apparently: $z^0+z^1+z^2...z^{n-1}=\frac{1-z^n}{1-z}=\frac{0}{1-z}=0$
Could someone please explain why: $z^0+z^1+z^2...z^{n-1}=\frac{1-z^n}{1-z}$?

Let $z^0+z^1+z^2...z^{n-1}= p(z)$. 
Then $z \ p(z)=z^1+z^2+z^3...z^{n}$. 
Hence $p(z)-zp(z)=1-z^n$. 
$p(z)(1-z)=1-z^n$. 
$p(z)=\frac{1-z^n}{1-z}$
If $z=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$. Then there is a formula (that is easy to prove) that states for any $i \in \mathbb{R}$, $z^i=\cos\frac{2i\pi}{n}+i\sin\frac{2i\pi}{n}$. Hence $z^n=\cos\frac{2n\pi}{n}+i\sin\frac{2n\pi}{n}=1$. Substitute that in the last expression of $p(z)$.
