# Roots of Complex Numbers (proof)

1. Sep 10, 2009

### DEMJ

1. The problem statement, all variables and given/known data

If c is any nth root of unity other than 1, then

$$1 + c + c^2 + \cdots + c^{n-1} = 0$$

3. The attempt at a solution

This is what is done so far and I am at a dead stall for about 2 hours lol. Any ideas on what I should be thinking of next? Should I continue simplifying? I have tried to continue simplifying but it always leads to nothing relevant. Do I need to be using $$re^{i\theta}$$ ?

Proof:
Assume $$c^n = 1$$ and $$c \not = 1$$. Then $$c^n -1 = 0$$. Note that

$$(c-1)(1 + c + c^2 + \cdots + c^{n-1}) = (c)(1 + c + c^2 + \cdots c^{n-1}) + (-1)(1 + c + c^2 + \cdots + c^{n-1}) = c + c^2 + c^3 + \cdots + c^n - 1 - c - c^2 - \cdots - c^{n-1}$$

2. Sep 10, 2009

### lanedance

my browser doesn't show tex properly, but simplify your final equation by cancelling terms & i think you're pretty much there...

you have shown c-1 does not = 0, and you know c^n-1=0, so what does equating your start & final expressions tell you?

Last edited: Sep 10, 2009
3. Sep 10, 2009

### njama

$$S=1 + c + c^2 + \cdots + c^{n-1}$$

$$Sc=c+c^2+c^3 + \cdots + c^{n}$$

$$Sc-S=c^n-1$$

$$S(c-1)=c^n-1$$

$$S=\frac{c^n-1}{c-1}$$

So you got

$$\frac{c^n-1}{c-1}=0$$

I think you can solve it now.