If $1,\omega,\omega^2,...\omega^{n-1}$ are complex numbers such that $\omega^n-1=0$, then
- they are the $n^{th}$ roots of unity
- they lie on the vertices of a regular polygon on the unit circle with center at the origin
For $n\geq2,$ $1+\omega+\omega^2+...+\omega^{n-1}=0$ can be proved using:
- Consider the LHS as a geometric series, and use the formula for the sum of that.
- Use vector addition to the equally spaced vertices, and note that their sum is invariant under rotation, and hence their sum is the zero vector.
Consider a third proof: Since it is a regular polygon with centroid at the origin:
- the sides are congruent
- when we consider the sides as position vectors they have angle $\theta, 2\theta, ...,n\theta.$ Hence, they also form a set of vectors that are the position vectors of a regular polygon, and in the second proof above, we showed that they add up to zero. So this reduces to the second case, and is now proved.
Is this correct?
Edit: Added that the polygon has centroid at the origin based on yoyo's answer.
The oriented edges of any polygon will sum to zero (as vectors in $\mathbb{R}^2$ or complex numbers). This says nothing about the sum of the vertices (as complex numbers).
For instance, the points $$ \{0, 1, 1+i, i,\} $$ sum to $2+2i$ but the edges $$ \{1-0, 1+i-1, i-(1+i), 0-i\}=\{1,i,-1,-i\} $$ sum to $0$.
[A square is a regular polygon; I don' know what a "cyclic" polygon is (the four vertices lie on a circle). In any case, your "third proof" doesn't make sense to me, although after your edits, it seems you're saying that the oriented edges, translated and thought of as vertices, reproduce the same geometric figure (the roots of unity polygon).]