Roots of Unity and Their Uses

Try this

1. $m=1$: the six roots of unity, head to tail, form a regular hexagon and return to the start.
2. Set $m$ to $2$ and then $3$: some points on the left are hit twice or three times. Does the chain still close?
3. Set $m$ to $6$ or $12$: what happens?
4. Tick “multiply everything by ${\epsilon }_1$”: every number turns by $\frac{360^{\circ }}{n}$, yet the set of points on the left does not change at all.

The $n$ roots of the equation $x^n=1$ are the $n$th roots of unity:

They are the vertices of a regular $n$-gon inscribed in the unit circle, and

• $|{\epsilon }_k|=1$, $\overline{{\epsilon }_k}={\epsilon }_k^{-1}={\epsilon }_{n-k}$, ${\epsilon }_j{\epsilon }_k={\epsilon }_{j+k}$ (indices taken mod $n$);
• $x^n-1=(x-{\epsilon }_0)(x-{\epsilon }_1)\cdots (x-{\epsilon }_{n-1})$;
• power sums: $\sum _{k=0}^{n-1}{\epsilon }_k^m=\{\begin{array}{ll}n,& n\mid m\\ 0,& n\nmid m\end{array}$

For $n=3$ we write $\omega ={\epsilon }_1=\frac{-1+\sqrt{3}i}{2}$; then ${\omega }^3=1$, $1+\omega +{\omega }^2=0$ and $\bar{\omega }={\omega }^2$.

Why

Application: filtering coefficients

The coefficients of $(1+x{)}^N$ are binomial coefficients. To find the sum of the binomial coefficients whose index is a multiple of $3$, substitute $x=1,\omega ,{\omega }^2$ in turn and add: $1+{\omega }^i+{\omega }^{2i}$ is $3$ when $3\mid i$ and $0$ otherwise. The power-sum formula “filters out” the terms we do not want.

Remove one condition

Practice