A calculus of the absurd

14.8 The roots of unity

The nth roots of unity are the nth roots of one. How can there be more than one (\(1\)) nth root of one? Well, some of them are complex, of course!

Note that 111111 In this document the natural numbers include zero!

\begin{equation*} e^{\pm 2 n \pi i} = 1 \text { where } n \in \mathbb {N} \end{equation*}

This is because of Euler’s formula.

Therefore, if we take the \(n\)th roots of both sides, we get that

\begin{equation*} e^{\pm \frac {2 \pi }{n} i} = 1^{\frac {1}{n}} \text { where } n \in \mathbb {N} \end{equation*}

Which we can use to compute the \(n\)th roots of unity. Note that by the fundamental theorem of algebra for the \(nth\) roots of unity, there are \(n\) different values.

  • Example 14.8.1 By considering the ninth roots of unity, show that 112112 I believe this question comes from the textbook "Further Pure Mathematics" [?]

    \begin{equation*} \cos \left (\frac {2\pi }{9} \right ) + \cos \left ( \frac {4\pi }{9} \right ) + \cos \left (\frac {6\pi }{9} \right ) + \cos \left ( \frac {8\pi }{9} \right ) = -\frac {1}{2} \end{equation*}

Solution

Using Euler’s formula 113113 \(e^{\theta i} = \cos (\theta ) + i \sin (\theta )\), see above for more we can write the sum of \(\cos (\frac {2\pi }{9}) + \cos (\frac {4\pi }{9}) + \cos (\frac {6\pi }{9}) + \cos (\frac {8\pi }{9})\) as

\begin{equation*} \frac {1}{2} \left [ e^{\frac {2\pi }{9}i} + e^{-\frac {2\pi }{9}i} + e^{\frac {4\pi }{9}i} + e^{-\frac {4\pi }{9}i} + e^{\frac {6\pi }{9}i} + e^{-\frac {6\pi }{9}i} + e^{\frac {8\pi }{9}i} + e^{-\frac {8\pi }{9}i} \right ] \end{equation*}

This is actually the sum of eight of the nine roots of unity in disguise! Note that if we add \(2\pi \) to anything in the form \(e^{ai}\), this has no effect (again, Euler’s formula and the fact that \(2\pi \) radians is a full rotation). Therefore, the previous expression is the same as

\begin{equation*} \frac {1}{2} \left [ e^{\frac {2\pi }{9}i} + e^{\frac {16\pi }{9}i} + e^{\frac {4\pi }{9}i} + e^{\frac {14\pi }{9}i} + e^{\frac {6\pi }{9}i} + e^{\frac {12\pi }{9}i} + e^{\frac {8\pi }{9}i} + e^{\frac {10\pi }{9}i} \right ] \end{equation*}

which can be re-ordered as

\begin{equation*} \frac {1}{2} \left [ e^{\frac {2\pi }{9}i} + e^{\frac {4\pi }{9}i} + e^{\frac {6\pi }{9}i} + e^{\frac {8\pi }{9}i} + e^{\frac {10\pi }{9}i} + e^{\frac {12\pi }{9}i} + e^{\frac {14\pi }{9}i} + e^{\frac {16\pi }{9}i} \right ] \end{equation*}

which are all the ninth roots of unity (except \(1\)).

Thus we can write that

\begin{align*} \frac {1}{2} \left [ \cos \left (\frac {2\pi }{9} \right ) + \cos \left ( \frac {4\pi }{9} \right ) + \cos \left (\frac {6\pi }{9} \right ) + \cos \left ( \frac {8\pi }{9} \right ) \right ] &= -1 + 1 + \omega + \omega ^2 + ... + \omega ^ 8 \\ &= -1 + \frac {1 - \omega ^ 9}{\omega - 1} \\ &= -1 + \frac {1 - 1}{\omega - 1 } \text { ($\omega ^9=1$ as $\omega $ is a 9th root of 1)} \\ &= -1 \end{align*}

which means that

\begin{equation*} \cos \left (\frac {2\pi }{9} \right ) + \cos \left ( \frac {4\pi }{9} \right ) + \cos \rbrackets { \frac {6\pi }{9} } + \cos \left ( \frac {8\pi }{9} \right ) = -\frac {1}{2} \end{equation*}