# 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 109109 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 110110 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 111111 $$e^{\theta i} = \cos (\theta ) + i \sin (\theta )$$, 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

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*}