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I have done the first part of this question and proved the statement is true by induction. But I am not sure about the second part.

Use induction on $n$ to erify that $$1 + z + \cdots + z^n = \frac{1-z^{n+1}}{1-z}\quad\text{(for }z\neq 1\text{)}$$ Use this to show that if $c$ is an $n$th root of $1$ and $c\neq 1$, then $1+c+\cdots + c^n = 0$.

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You probably mean $1 + c + \cdots + c^{n-1} = 0$. –  mrf Feb 12 '12 at 23:02
    
As mrf says, it isn't true as stated. E.g. take $c=-1$ and $n=2$. The corrected version would follow immediately from the first part. –  Jonas Meyer Feb 12 '12 at 23:03
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up vote 4 down vote accepted

Well if $c$ is the $n$th root of 1, then $c^{n + 1} = 1\cdot c = c$. Thus $$1 + c +\cdots + c^n = \frac{1 - c^{n + 1}}{1 - c} = \frac{1 - c}{1 - c} = 1$$

So you would either you mean $$1 + c + \cdots + c^n = 1$$ or $$1 + c + \cdots + c^{n - 1} = 0.$$ Which follows immediately by subtracting $c^n = 1$ from both sides of the first equation.

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Hint: a root $c$ of $fg$ not a root of $f$ is a root of $g$ since $\:f(c)\:g(c) = 0,\ f(c)\ne 0\: \Rightarrow\: g(c) = 0\:$ in $\mathbb C$

Your exercise is the special case $\:f = z-1,\ \ g = \dfrac{z^n-1}{z-1}\: =\: z^{n-1}+\:\cdots\:+z+1$

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