Roots of unity.

Definition 1.5.4   Let $ n$ be a natural number such that $ n \geq 2$ . A complex number $ u$ such that $ u^n=1$ is called an $ n^{th}-$ root of unity.

Example 1.5.5       

  1. $ i$ is a $ 4^{th}$ root of unity, as $ i^4=1$ .
  2. $ -1/2 +i\sqrt{3}/2$ is a $ 3^{rd}$ root of unity. Please check this.

By the method of subsection 5.1, we compute all the $ n^{th}-$ roots of unity:

$\displaystyle z^n=1 \Leftrightarrow \begin{cases}\vert z^n\vert=1 \ \operatorn...
...ert z\vert=1 \ \operatorname{arg}\nolimits (z)= \frac {2k \pi}{n} \end{cases}.$    

For $ k=0, \dots n-1$ , we have the following roots:

$\displaystyle z_0$ $\displaystyle = \cos 0 + i \; sin 0 =1,$    
$\displaystyle z_1$ $\displaystyle =\cos \frac {2\pi}{n}+ i \sin \frac {2 \pi}{n},$    
$\displaystyle z_2$ $\displaystyle =\cos \frac {4 \pi}{n}+ i \sin \frac {4 \pi}{n},$    
$\displaystyle \dots$ $\displaystyle \dots$    
$\displaystyle z_{n-1}$ $\displaystyle = \cos \frac {2(n-1) \pi}{n} + i \; \sin \frac {2(n-1) \pi}{n}.$    

The images in Gauss-Argand plane of the $ n^{th}$ roots of unity are the vertices of a regular polygone inscribed in the unit circle.

Example 1.5.6       

(i)
The square roots of unity ($ n=2$ ) are $ 1$ and $ -1$ (see Figure 9(a).
(ii)
For $ n=3$ , the roots of unity are $ 1$ , $ -1/2+i \; \sqrt{3}/2$ and $ -1/2-i \; \sqrt{3}/2$ (see Figure 9(b).
(iii)
For $ n=4$ , the roots of unity are $ 1$ , $ i$ , $ -1$ and $ -i$ (see Figure 9(c).

Figure 9: The images of the roots of unity.
\begin{figure}\mbox{
\subfigure[$n=2$]{\epsfig{file=2RootsUnity.eps,height=4cm}...
...ad
\subfigure[$n=4$]{\epsfig{file=4RootsUnity.eps,height=4cm}}
}\end{figure}

By De Moivre's formula De Moivre's formula, the following holds:

$\displaystyle z_2$ $\displaystyle = z_0^2,$    
$\displaystyle z_3$ $\displaystyle = z_0^3,$    
$\displaystyle \dots$ $\displaystyle \dots$    
$\displaystyle z_{n-1}$ $\displaystyle = z_0^{n-1},$    
$\displaystyle z_0$ $\displaystyle = z_0^n.$    

An immediate consequence is as follows:

Proposition 1.5.7   Let $ n$ be a natural number such that $ n \geq 2$ . The sum of all $ n^{th}$ roots of unity is equal to 0.

Proof. We have:

$\displaystyle z_0+z_1+z_2+ \dots +z_{n-1}$ $\displaystyle = 1+ z_1 + z_1^2 + z_1^3 + \dots + z_1^{n-1}$    
$\displaystyle \quad$ $\displaystyle = \frac {1 -z_1^n}{1-z_1}$    
$\displaystyle \quad$ $\displaystyle =0.$    

$ \qedsymbol$

Now, let us see a nice application of these roots of unity.

Proposition 1.5.8   Let $ n$ be a natural number such that $ n \geq 2$ . Suppose that $ u$ and $ v$ are two complex numbers such that $ v^n=u$ . Then we obtain all the $ n^{th}$ roots of $ v$ by separate multiplication of $ u$ by all the $ n^{th}$ roots of unity.

Example 1.5.9   Take the number

$\displaystyle v = \sqrt{2+\sqrt{2}} + i \; \sqrt{2-\sqrt{2}}.$    

Squarring twice, we obtain

$\displaystyle v^4=16i.$    

By Proposition 5.8, the $ 4^{th}$ roots of $ 16i$ are given by:

$\displaystyle 1 \cdot v$ $\displaystyle = \sqrt{2+\sqrt{2}} + i \; \sqrt{2-\sqrt{2}}$    
$\displaystyle i \cdot v$ $\displaystyle = -\sqrt{2-\sqrt{2}} + i \; \sqrt{2+\sqrt{2}}$    
$\displaystyle (-1) \cdot v$ $\displaystyle = -\sqrt{2+\sqrt{2}} - i \; \sqrt{2-\sqrt{2}}$    
$\displaystyle (-i) \cdot v$ $\displaystyle = \sqrt{2-\sqrt{2}} - i \; \sqrt{2+\sqrt{2}}$    

An additional result of this example is the following: compute polar forms for the $ 4^{th}$ roots of $ 16i$ . Theses roots have arguments $ \pi /8$ , $ 5\pi /8$ , $ 9\pi /8$ and $ 13\pi /8$ . They fit exactly the order of the roots given above.

Noah Dana-Picard 2007-12-24