Roots of unity: Can somebody explain the following

Question

The n-th roots of unity are the solutions to the equation $z^n-1=0$

The coefficient of $z^{n-1}$ being zero implies that the sum of the n-th roots of unity equals zero for the following reason: The coefficients of the $z^{n-1}$ term is the sum of the roots since the polynomial can be factored into linear factors.

Can somebody explain it in a different way, because I don't really get it.

Answer 1

Let's take a look at how polynomials expand.

$$(x-a)(x-b) = x^2 - (a+b)x +ab$$

$$(x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab + ac + bc)x - abc$$

In general, the second term of this polynomial will be the (negative of the) sum of the roots. It's not relevant to this question, but it's worth knowing that the constant term will always be the product of the roots, up to sign (you might show this by induction, but other more sophisticated tools are available).

So what do we have in your case? We have

$$z^n - 1 = (z - \zeta^0)(z - \zeta^1)(z-\zeta^2)\cdots(z-\zeta^{n-1})$$

for $\zeta$ some primitive $n$th root of unity.

But now we can look at the $z^{n-1}$ coefficient of both sides! On the left we have $0$, and on the right we have $-\sum \zeta^i$. Which is exactly what you're trying to show.

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I hope this helps ^_^

Answer 2

For a generic polynomial over the complex numbers, you can factor it into linear factors: $$p(z) = a(z-r_1)(z-r_2)...(z-r_n)$$ If you multiply out that polynomial, the coefficient of $z^n$ is $1$ and the coefficient of $z^{n-1}$ is $-r_1-r_2-...-r_n$, the negative of the sum of the roots (times $a$), just like the constant term is $(-1)^n\cdot$(product of roots times $a$), and so forth for all the other coefficients.

The geometric explanation for why the sum of the $n$-th roots of unity is zero is that they form a regular $n$-gon around the origin when plotted in the complex plane.