Show that if $r$ is an nth root of $1$ and $r\ne1$, then $1 + r + r^2 + ... + r^{n-1} = 0$. Show that if $r$ is an nth root of $1$ and $r\ne1$, then $1 + r + r^2 + ... + r^{n-1} = 0$.
I think I can represent all the roots of 1 as follows:
$r = 1^{\frac{1}{n}} ( \frac{\cos{2\pi k}}{n} +  i\frac{\sin{2\pi k}}{n} )$
From there I am not sure how to get to $1 + r + r^2 + ... + r^{n-1} = 0$.
 A: We know that $r^n - 1 = 0$ but we can factorise the LHS as $(r-1)(r^{n-1}+r^{n-2}+...+r+1)=0$.
However $r\neq 1$ so the other bracket must be $0$.
A: Just multiply your sum by $r$.  You get the same thing you started with. Since $r \neq 1$, this is only possible if your sum is $0$.
A: Remember that the equation $r^{n}=1$ has $n$ roots in the complex plane (because $r^n$-1 is an nth degree polynomial).  You know one of the roots is 1, so the polynomial $r-1$ must be a factor of this nth degree polynomial.  Now what happens if you divide by $r-1$ (Assuming $r \neq 1$)?
$\frac{r^{n}-1}{r-1}=1+r+r^2+...+r^{n-1}$ by long division.
So this is a true equation for $r \neq 1$, and the left-hand-side will be 0 for any other root of unity.
A: Multiply top and bottom by $(1-r)$.  
$$1+r+r^2+\cdots+r^{n-1}=\frac{(1+r+r^2+\cdots+r^{n-1})(1-r)}{(1-r)}=\frac{1-r^n}{1-r}=\frac{0}{1-r}=0$$
Note: $r\neq 1$ is necessary, else the statement is false.
A: $1 + r + r^2 + r^3 + ... + r^{n-1}$ is the sum of the first $n$ terms of a geometric series.  The formula for this summation, as long as $r \ne 1$, is
$$\frac{r^n - 1}{r - 1}$$
Since $r$ is an nth root of unity, $r^n - 1 = 0$.  Since the denominator is nonzero, this quantity is equal to zero.
Hope this helps!
A: The sum of the roots of any polynomial $\sum a_{n}x^{n}$ is $-\frac{a_{n-1}}{a_{n}}$. In the case of the polynomial $x^{n}-1$ this expression is zero.  
As Carsten Schultz points out, this argument only proves the case when $r$ is a primitive $n^{th}$ root of unity. This does not exclude any cases, though, as every root of unity is primitive for some $n$.  It is also important to see that this excludes the case of $r=1$, since then $-a_{n-1}/a_{n}$ is non-zero.
A: Since no one has posted this - geometrically the roots of unity point to edges of a regular polygon from centre (0,0). Clearly by symmetry, they must vectorially add to zero. 
