Show that $\frac{1-x^n}{1-x}=0\,$ has a real solution at $x$ equals $-1$

Question

I'm trying to solve this question:

$$S_n = 1+x+x^2+x^3...+x^{2021}$$ $$xS_n = x+x^2+x^3+x^4...+x^{2022}$$ Subtracting the bottom equation from the top.. $$S_n(1-x)=1-x^{2022}$$ $$S_n=\dfrac{1-x^{2022}}{1-x}$$

Giving that $S_n = 0$, I ended up with this equation: $\dfrac{1-x^{2022}}{1-x}=0$

When graphed, there is only one real solution, which is $-1$, as long as $x$ is raised to an even power. How can I solve that equation to get $-1$? That's my question.

Answer 1

The solutions are the $n-1$ $n$-th roots of unity other than $1$. Those are $e^{2\pi ik/n},\,k=1,\dots n-1$. For $n$ even, we have $e^{\pi i}=-1$ as the only real one.

That's other than $k=n/2$, they all have nonzero imaginary part $\sin(2\pi k/n)$.

(The $n$-th roots of unity are evenly spaced on the unit circle, if you know anything about the circle group.)

Answer 2

Hint Factor the sum as $$S = (1 + x)(1 + x^2 + \cdots + x^{2020}) .$$