Roots of Complex Numbers (proof)

[DEMJ]

Homework Statement

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

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

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}

 

[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?

 

[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.