Proving sum of complex roots is zero

Question

Let $\alpha$ and $\beta$ be the complex roots of $z^3 =1$. Show that the sum of the first $n$ terms in the series $1 + \alpha + \alpha^2 +...$ is either $0, 1$ or $-\alpha^2$ depending on the remainder when $n$ is divided by $3$

For each case I have considered a GP sum:

$\text{If }n\equiv 0\:\left(\text{mod 3}\right)$, then $S_n=\frac{1-\alpha ^{3n}}{1-\alpha }$

$\text{If }n\equiv 1\:\left(\text{mod 3}\right)$, then $S_n=\frac{1-\alpha ^{3n+1}}{1-\alpha }$

$\text{If }n\equiv 2\:\left(\text{mod 3}\right)$, then $S_n=\frac{1-\alpha ^{3n+2}}{1-\alpha }$

I have tried algebraic manipulation but I can seem to show that they are equal to $0, 1 $ or $-\alpha^2$. I'm sure I would be able to algebraically manipulate the sums to a final term but I can't so I am thinking instead - are my GP sums correct?

Answer 1

The sum of the first $n$ terms, in any case, is given by $S_n=\frac{1-\alpha^n}{1-\alpha}$. Now we can evaluate $S_n$ in each of the three cases:

(a) If $n\equiv 0\, (\mathrm{mod}\, 3)$, then $n=3k$ for some integer $k$, hence $\alpha^n=\alpha^{3k}=(\alpha^3)^k=1^k=1$, so $S_n=0$;

(b) If $n\equiv 1\, (\mathrm{mod}\, 3)$, then $n=3k+1$ for some integer $k$, hence $\alpha^n=\alpha^{3k+1}=\alpha(\alpha^3)^k=\alpha$, so $S_n=1$;

(c) If $n\equiv 2\, (\mathrm{mod}\, 3)$, then $n=3k+2$ for some integer $k$, hence $\alpha^n=\alpha^{3k+2}=\alpha^2(\alpha^3)^k=\alpha^2$, so $S_n=\frac{1-\alpha^2}{1-\alpha}=1+\alpha=-\alpha^2$.

Answer 2

$\alpha$ is a root of $$0=z^3-1=(z-1)(z^2+z+1)$$ Since $\alpha\neq1$, we get that $\alpha$ is a root of $z^2+z+1=0$. Said differently, we have $$ \alpha^2+\alpha+1=0 $$ This immediately implies for any $n\geq0$ that $$ S_{n+3}=\alpha^{n}(\alpha^2+\alpha+1)+S_n=S_n $$ From here, it's not difficult to see that $S_0=0$, $S_1=1$ and $S_2=1+\alpha=-\alpha^2$ are the only three values that appear, in order, repeating indefinitely. An induction proof formalizes this neatly.

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Geometrically, we have this picture (remember that $\beta = \alpha^2$ and $1 = \alpha^3$, click for larger version):

Here you see $\alpha, \beta$ and the first four partial sums $S_0, S_1, S_2, S_3$ marked as points in the complex plane. The line from $S_1$ to $S_2$ is blue like the line from the origin to $\alpha$ to illustrate $S_1+\alpha = S_2$. And correspondingly for the two red lines with $\beta = \alpha^2$.