# Roots of unity?

The $n$th roots of unity are the complex numbers: $1,w,w^2,...,w^{n-1}$, where $w = e^{\frac{2\pi i} {n}}$. If $n$ is even:

1. The $n$th roots are plus-minus paired, $w^{\frac{n}{2}+j} = -w^j$.

2. Squaring them produces the $\frac{n}{2}$nd roots of of unity.

1. Could someone explain the first statement? I understand why the $n$ roots are plus-minus paired (if $n$ is even), but what does the equation mean? An explanation of the equation will be appreciated.

2. The statement states $w^0$ (or $1$) is a root of unity. Which root is it? Aside from the obvious fact that $1$ is a always a root of $1$, doesn't $w...w^{n-1}$ cover every root? What specific root is $w^0$ referring to?

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I think it's a good idea to draw these on the complex plane for small $n$. –  Dylan Moreland May 17 '12 at 0:34

Since $\omega=\exp\left(\frac{2\pi i}{n}\right)$, we have $\omega^{n/2}=\exp\left(\frac{2\pi i}{n}\frac{n}{2}\right)=\exp(\pi i)=-1$. This and the law $\omega^{a+b}=\omega^a\omega^b$ give the formula you ask about in your first question. In words: multiplying a complex number by $\omega$ rotates the vector in the complex plane corresponding to that number by $360^\circ/n$. Repeating $n$ times brings the vector back to its starting position; repeating only $n/2$ times rotates by $180^\circ$, which is the same as multiplying by $-1$.
As to your second question, when $n$ is odd, $\omega^0=1$ is the only real root; it's the root we already knew about before we learned about complex numbers; all other roots have an imaginary part. When $n$ is even, there are two real roots, $1$ and $-1=\omega^{n/2}$. Again, we knew both of these roots before we learned complex numbers. (E.g. the two real sixth roots of 1 are $1$ and $-1$.) All other roots have an imaginary part.
1. Think of the roots as points on a circle equally spaced apart. The negative of a root is the root on the entire opposite side of the circle, which means you have to traverse half the roots to get from any particular rotos to its negative. Traversing half the circle means cycling forward (or backwards, if you wanted) $n/2$ roots, which means (if we're considering the root $\omega^j$), we have that $-\omega^j=\omega^{j+n/2}$. This can also be seen by simply noticing that $-1$ must be the $n/2$th root on the cirlce, being halfway around, and computing $-\omega^j=\omega^{n/2}\omega^j=\omega^{n/2+j}$. Ultimately this tells you not only that the roots are plus-minus pairs, but it explicitly tells you which root is the negative of any given one, where the given one is in the form $\omega^j$.
2. Of course $\omega^0=1$. None of the other powers of $\omega$ - that is, none of $\omega^1,\omega^2,\cdots,\omega^{n-1}$ - can also be the number $1$. For if, say, $\omega^k=1$ with $k\in\{1,2,\cdots,n-1\}$, then we have $e^{2\pi i (k/n)}=1$, or in other words that $1$ on the unit circle has argument $2\pi\cdot k/n$, which is not an integer multiple of the full circle, $2\pi$, and hence is geometrically impossible. This should be rather obvious if you actually draw roots on a circle: the root $\omega$ is the first counterclockwise from $1$ and $\omega^{n-1}=\omega^{-1}$, being the reflection of $\omega$ across the real axis, must be the last one right before the root $1$.