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Section 7.5 Roots of Unity

How do we find all solutions to \(z^{n}=1\text{?}\) We now have enough tools in place to answer this. We find them all in a few steps:

  • First, using polar coordinates, we can find one root rather easily: \(w=e^{\frac{2\pi i}{n}}\text{,}\) since then

    \begin{equation*} w^{n} = e^{\frac{2\pi i}{n}\cdot n}=e^{2\pi i}=1\text{.} \end{equation*}

  • In particular, this means that any power of \(w\) is also a root, since if \(j\in\mathbb{N}\text{,}\)

    \begin{equation*} (w^{j} )^{n} = (w^{n})^{j}=1\text{.} \end{equation*}

  • Finally, we will show later that for any degree \(n\) polynomial there are at most \(n\) distinct roots. So if we show that the numbers

    \begin{equation*} 1,w,w^{2},\ldots,w^{n-1} \end{equation*}

    are all distinct, then we will have all the roots.

  • Suppose for the sake of a contradiction that there are distinct integers \(j\) and \(k\) with \(0\leq j,k\lt n\) so that \(w^{j}=w^{k}\text{.}\) Then

    \begin{equation*} 1=w^{j-k}=e^{(j-k)\frac{2\pi i}{n}} = \cos \left((j-k)\frac{2\pi }{n}\right)+i\sin \left((j-k)\frac{2\pi }{n}\right)\text{.} \end{equation*}

    The only way this can be \(1\) is if the cosine is 1 and the sine is zero, so which only happens if their arguments are multiples of \(2\pi\text{,}\) that is, we must have

    \begin{equation*} (j-k)\frac{2\pi }{n} = 2\pi \ell \text{ for some integer } \ell \end{equation*}

    which implies \(j-k=n\ell\) for some \(\ell\in\mathbb{Z}\text{,}\) but this is impossible since \(0\leq j,k\lt n\text{,}\) and hence \(-n\lt j-k\lt n\text{,}\) so \(j-k\) must be zero, and so \(j\) and \(k\) cannot be distinct. Thus, the numbers \(1,w,\ldots,w^{n-1}\) are distinct and form all the roots.

We have thus shown the following.

The third roots of unity are

\begin{align*} z_1=\amp 1\\ z_2=\amp e^{2\pi i/3} = \cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3} = -\frac{1}{2}+i\frac{\sqrt{3}}{2}\\ z_3= \amp e^{4\pi i/3} = \cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3} = -\frac{1}{2}-i\frac{\sqrt{3}}{2} \end{align*}

Note: the cube roots of unity are used so frequently that writers often define \(\omega := e^{2\pi i/3}\text{.}\) Then we have \(\omega^2=\bar{\omega}\text{,}\) and lots of other interesting and useful algebraic relationships.

How do we find all solutions to \(z^{n}=a\) where \(a\in\CC\text{?}\)

Again, we can use \(n\)th roots of unity. We showcase the method in an example:

Suppose we set \(a=16i\) and wish to find the fourth roots of \(a\) (i.e., solve \(z^4=16i\)). We can find one quick root using polar coordinates: note that \(a=16e^{i\frac{\pi}{2}}\text{,}\) so we can spot one root as

\begin{equation*} z=16^{1/4}e^{i\frac{\pi}{2}\frac{1}{4}} = 2e^{i\frac{\pi}{8}}\text{,} \end{equation*}

since then \(z^{4} = a\text{.}\)

Recall that there are at most \(4\) distinct roots for a degree \(4\) polynomial. Note that since \(z\) is a solution, if \(1,w,w^2,w^3\) (that is, \(1,e^{i2pi/4}=e^{i\pi/2}, e^{i4\pi/4}=e^{i\pi}\text{,}\) and \(e^{i6\pi/4}=e^{i3\pi/2}\)) are the \(4\)th roots of unity, then for \(k=0,1,2,3\text{,}\)

\begin{equation*} (zw^{k})^{4}=z^{4}w^{4k}=a\cdot 1=a\text{.} \end{equation*}

Thus, the other solutions are \(z,zw,zw^2,zw^3\text{.}\) For \(k \in \{0,1,2,3\}\) this gives the following (in exponential form)

\begin{equation*} 2e^{i\frac{\pi}{8}},2e^{i\frac{5\pi}{8}},2e^{i\frac{9\pi}{8}},2e^{i\frac{13\pi}{8}} \end{equation*}

or equivalently

\begin{equation*} 2e^{i\frac{\pi}{8}},2e^{i\frac{5\pi}{8}},2e^{i\frac{-7\pi}{8}},2e^{i\frac{-3\pi}{8}} \end{equation*}