Unit roots/Properties of unit roots

In this chapter, we will look at the basic properties of the root of unity.

An example

Example 1 It is given $a^2+a+1=0$, prove that

$a^{2012}+(\frac{1}{a}{)}^{2012}=a^{1987}+(\frac{1}{a}{)}^{1987}$.

Prove From the given equation, we can show that $a^3=1$:

$a^3=a^3-1+1=(a-1)(a^2+a+1)+1=1$.

Therefore,

$a^{2012}=a^{670\times 3+2}=(a^3{)}^{670}+a^2=a^2$

$(\frac{1}{a}{)}^{2012}=a^{-671\times 3+1}=(a^3{)}^{-671}+a=a$

$a^{1987}=a^{662\times 3+1}=(a^3{)}^{662}+a=a$

$(\frac{1}{a}{)}^{1987}=a^{-663\times 3+2}=(a^3{)}^{-663}+a^2=a^2$

So, both sides of the equation equal $a^2+a$. QED

Moreover, we can calculate the value of each side: $a^2+a=a^2+a+1-1=-1$.

In fact, we can obtain a more general result:

Example 2 Given $a^2+a+1=0$, and $n$ is a natural number. Evaluate $a^n+(\frac{1}{a}{)}^n$.

Solution

$a^n+(\frac{1}{a}{)}^n=\{\begin{array}{cc}-1& \text{if }n\text{ is not a multiple of 3}\\ 2& \text{if }n\text{ is a multiple of 3}\end{array}$

The roots of unity

We have make use of an important observation, namely $a^3=1$, in the examples above. Numbers that satisfy the equation:

$a^n=1,n\text{ is a natural number}$

are called the nth roots of unity or the unit roots. From the knowledge of algebra, the following formula:

${ϵ}_k=\cos \frac{2k\pi }{n}+i\sin \frac{2k\pi }{n},k\text{ is a natural number}$

always gives a root of unity. When $k=0,1,\dots ,n-1$, ${ϵ}_k$ takes distinct values, and when $k$ takes other values, ${ϵ}_k$ equals one of the values ${ϵ}_0,{ϵ}_1,\dots ,{ϵ}_{n-1}$. Moreover, as a polynomial equation of degree $n$, the equation has exactly $n$ roots. Therefore, ALL roots of unity are:

${ϵ}_0,{ϵ}_1,\dots ,{ϵ}_{n-1}$.

Note that ${ϵ}_0=1$.

On the other hand, the roots of unity are the solution of the equation:

$x^n-1=0$.

Moreover:

$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots +x+1)$.

Therefore, ${ϵ}_1,{ϵ}_2,\dots ,{ϵ}_{n-1}$ are all roots of the equation:

$x^{n-1}+x^{n-2}+\cdots +x+1=0$.

The cube roots of unity

The cube roots of unity is a good starting point in our study of the properties of unit roots.

Example 3 The cube roots of unity are:

${ϵ}_0=1$,

${ϵ}_1=\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$,

${ϵ}_2=\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$.

We usually write $\omega ={ϵ}_1$. Then:

${\omega }^2=(-\frac{1}{2}+\frac{\sqrt{3}}{2}i{)}^2=\frac{1}{4}-\frac{\sqrt{3}}{2}i-\frac{3}{4}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i={ϵ}_2$.

Therefore, the cube roots of unity can also be written as $1,\omega ,{\omega }^2$. The cube root of unity has the following properties:

1. They have a unit modulus: $|1|=|\omega |=|{\omega }^2|=1$.
2. $1,\omega ,{\omega }^2$ are the roots of the equation $x^3-1=0$.
3. $\omega ,{\omega }^2$ are the roots of the equation $x^2+x+1=0$.
4. ${ϵ}_2^2=(-\frac{1}{2}-\frac{\sqrt{3}}{2}i{)}^2=\frac{1}{4}+\frac{\sqrt{3}}{2}i-\frac{3}{4}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i={ϵ}_1$. So, the cube roots of unity still have the form of $1,\omega ,{\omega }^2$ if we let $\omega ={ϵ}_2$.
5. On the complex plane, the roots of unity are at the vertices of the regular triangle inscribed in the unit circle, with one vertex at 1.
6. $\bar{\omega }={\omega }^2$, $\overline{{\omega }^2}=\omega$.
7. $1^n+{\omega }^n+({\omega }^2{)}^n=\{\begin{array}{cc}0& \text{if }n\text{ is not a multiple of 3}\\ 3& \text{if }n\text{ is a multiple of 3}\end{array}$

General properties of roots of unity

After looking at the properties of the cube roots of unity, we are ready to study the general properties of the nth roots of unity.

Property 1 The nth roots of unity have a unit modulus, that is:

$|{ϵ}_k|=1k\text{ is an integer}$.

Proof It follows from the polar form of the unit roots.

Property 2 The product of two unit roots is also a unit root. Specifically, if $j$ and $k$ are integers, then:

${ϵ}_j\cdot {ϵ}_k={ϵ}_{j+k}$.

Proof From the multiplication rule of complex number:

${ϵ}_j\cdot {ϵ}_k=(\cos \frac{2j\pi }{n}+i\sin \frac{2j\pi }{n})\cdot (\cos \frac{2k\pi }{n}+i\sin \frac{2k\pi }{n})=(\cos \frac{2(j+k)\pi }{n}+i\sin \frac{2(j+k)\pi }{n})={ϵ}_{j+k}$.

This is a very important property of the roots of unity, from which a series of corollary can be derived:

Corollary 1 $({ϵ}_j{)}^{-1}={ϵ}_{-j}$.
Proof ${ϵ}_j\cdot {ϵ}_{-j}={ϵ}_{j+(-j)}={ϵ}_0=1$. Now, since ${ϵ}_j\ne 0$, multiplying its inverse on both sides yields $({ϵ}_j{)}^{-1}={ϵ}_{-j}$.

Corollary 2 For any integer $m$:

$({ϵ}_k{)}^m={ϵ}_{mk}$.

Proof When $m$ is positive, $({ϵ}_k{)}^m=\underset{m\text{ times}}{\underbrace{{ϵ}_k\cdot {ϵ}_k\cdot \cdots \cdot {ϵ}_k}}={ϵ}_{\underset{m\text{ times}}{\underbrace{{}_{k+k+\cdots +k}}}}={ϵ}_{mk}$.
When $m=0$, non-zero complex number raised to the power of 0 is 1, so $({ϵ}_k{)}^0=1={ϵ}_0$.
When $m$ is negative, $-m$ is positive, so $({ϵ}_j{)}^m=(({ϵ}_j{)}^{-m}{)}^{-1}=({ϵ}_{-mj}{)}^{-1}={ϵ}_{mj}$.

Corollary 3 If $r$ is the remainder when $k$ is divided by $n$, then ${ϵ}_k={ϵ}_r$.
Proof Let $k=nq+r$ where $q$ is an integer and $0\le r<n$, then:

${ϵ}_k={ϵ}_{nq+r}=({ϵ}_n{)}^q\cdot {ϵ}_r=1^n{ϵ}_r={ϵ}_r$.

Corollary 4 ${ϵ}_k=({ϵ}_1{)}^k$.
Any root of unity can be expressed as a power of ${ϵ}_1$.

We may ask the following question: is there any other root of unity ${ϵ}_{\ell }$ such that any root of unity can be expressed as a power of ${ϵ}_{\ell }$?

In fact we have seen such an example when we studied the cube root of unity. A unit root with such property is called a primitive root.

Corollary 5 The conjugate of a unit root is also a unit root.
Proof From the property of complex numbers $z\cdot \bar{z}=|z{|}^2$ and $|{ϵ}_k|=1$, $\overline{{ϵ}_k}=\frac{|{ϵ}_k{|}^2}{{ϵ}_k}=\frac{1}{{ϵ}_k}={ϵ}_{-k}={ϵ}_{n-k}$

Corollary 6 $({ϵ}_k{)}^j=({ϵ}_j{)}^k$.
Proof $({ϵ}_k{)}^j={ϵ}_{jk}=({ϵ}_j{)}^k$.

Property 3 Let $m$ be an integer, then:

$1+{ϵ}_1^m+{ϵ}_2^m+\cdots +{ϵ}_{n-1}^m=\{\begin{array}{cc}0& \text{if }m\text{ is not a multiple of }n\\ n& \text{if }m\text{ is a multiple of }n\end{array}$

Proof When $m$ is a multiple of $n$, $({ϵ}_k{)}^m=1$ for any integer $k$, so:

$1+{ϵ}_1^m+{ϵ}_2^m+\cdots +{ϵ}_{n-1}^m=\stackrel{n\text{ times}}{\overbrace{1+1+\cdots +1}}=n$

When $m$ is not a multiple of $n$, $({ϵ}_1{)}^m\ne 1$. Then:

$1+{ϵ}_1^m+{ϵ}_2^m+\cdots +{ϵ}_{n-1}^m=1+{ϵ}_m+({ϵ}_m{)}^2+\cdots +({ϵ}_m{)}^{n-1}=\frac{1-({ϵ}_m{)}^n}{1-{ϵ}_m}=\frac{1-({ϵ}_n{)}^m}{1-{ϵ}_m}=\frac{1-1}{1-{ϵ}_m}=0$.

Corollary 7 If $n>1$, the sum of all unit roots is zero: $1+{ϵ}_1+{ϵ}_2+\cdots +{ϵ}_{n-1}=0$.
Proof Take $m=1$. Alternatively, the sum of roots of the equation $x^n-1=0$ is zero.

Corollary 8 If $n>1$ and ${ϵ}_k\ne 1$, then $1+{ϵ}_k+({ϵ}_k{)}^2+\cdot +({ϵ}_k{)}^{n-1}=0$.
Proof Since ${ϵ}_k\ne 1={ϵ}_0$, $k$ is not a multiple of $n$. Then:

$1+{ϵ}_k+({ϵ}_k{)}^2+\cdots +({ϵ}_k{)}^{n-1}=1+({ϵ}_1{)}^k+({ϵ}_2{)}^k+\cdots +({ϵ}_{n-1}{)}^k=0$.

Therefore, if we exclude ${ϵ}_0=1$, the nth roots of unity ${ϵ}_1,{ϵ}_2,\dots ,{ϵ}_{n-1}$ are the roots of the equation:

$1+x+x^2+\cdots +x^{n-1}=0$.

Examples

Example 4 Find the fifth roots of unity.
Solution It can be proved that:

$\cos \frac{2\pi }{5}=\frac{\sqrt{5}-1}{4}$,

$\sin \frac{2\pi }{5}=\frac{\sqrt{10+2\sqrt{5}}}{4}$.

Therefore,

${ϵ}_0=1$,

${ϵ}_1=\frac{\sqrt{5}-1}{4}+\frac{\sqrt{10+2\sqrt{5}}}{4}i$,

by corollary 4 of property 2,

${ϵ}_2=(\frac{\sqrt{5}-1}{4}+\frac{\sqrt{10+2\sqrt{5}}}{4}i{)}^2=-\frac{\sqrt{5}+1}{4}+\frac{\sqrt{10-2\sqrt{5}}}{4}i$,

by corollary 5 of property 2,

${ϵ}_3=\overline{{ϵ}_2}=-\frac{\sqrt{5}+1}{4}-\frac{\sqrt{10-2\sqrt{5}}}{4}i$,

${ϵ}_4=\overline{{ϵ}_1}=\frac{\sqrt{5}-1}{4}-\frac{\sqrt{10+2\sqrt{5}}}{4}i$.

Example 5 Find the sixth roots of unity in terms of $\omega$.
Solution

${ϵ}_0=1$,

${ϵ}_1=\cos \frac{2\pi }{6}+i\sin \frac{2\pi }{6}=\frac{1}{2}+\frac{\sqrt{3}}{2}i=1+\omega$,

${ϵ}_2=\cos \frac{4\pi }{6}+i\sin \frac{4\pi }{6}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i=\omega$,

${ϵ}_3=\cos \frac{6\pi }{6}+i\sin \frac{6\pi }{6}=-1$,

${ϵ}_4=\cos \frac{8\pi }{6}+i\sin \frac{8\pi }{6}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i={\omega }^2$,

${ϵ}_5=\cos \frac{10\pi }{6}+i\sin \frac{10\pi }{6}=\frac{1}{2}-\frac{\sqrt{3}}{2}i=1+{\omega }^2$.

Example 6 Evaluate:

$(\genfrac{}{}{0ex}{}{0}{n})+(\genfrac{}{}{0ex}{}{3}{n})+(\genfrac{}{}{0ex}{}{6}{n})+(\genfrac{}{}{0ex}{}{9}{n})+\cdots +(\genfrac{}{}{0ex}{}{3\ell -3}{n})+(\genfrac{}{}{0ex}{}{3\ell }{n})$,

where $3\ell$ is the greatest multiple of 3 not exceeding $n$.
Analysis The expression is the sum of every first of three consecutive binomial coefficients:

$(\genfrac{}{}{0ex}{}{0}{n}),(\genfrac{}{}{0ex}{}{1}{n}),(\genfrac{}{}{0ex}{}{2}{n}),(\genfrac{}{}{0ex}{}{3}{n}),\dots ,(\genfrac{}{}{0ex}{}{n-1}{n}),(\genfrac{}{}{0ex}{}{n}{n})$.

A similar but more familiar sum is:

$(\genfrac{}{}{0ex}{}{0}{n})+(\genfrac{}{}{0ex}{}{2}{n})+(\genfrac{}{}{0ex}{}{4}{n})+\cdots$,

which can be computed by summing the binomial expansions:

$(1+x{)}^n=(\genfrac{}{}{0ex}{}{0}{n})+(\genfrac{}{}{0ex}{}{1}{n})x+(\genfrac{}{}{0ex}{}{2}{n})x^2+\cdots +(\genfrac{}{}{0ex}{}{n}{n})x^n$

for $x=+1,-1$ (note that these are the square root of unity). The sum is

$(1+1{)}^n+(1-1{)}^n=2(\genfrac{}{}{0ex}{}{0}{n})+(1+(-1))(\genfrac{}{}{0ex}{}{1}{n})+(1^2+(-1{)}^2))(\genfrac{}{}{0ex}{}{2}{n})+\cdots +(1^n+(-1{)}^n)(\genfrac{}{}{0ex}{}{n}{n})$.

The value $1^k+(-1{)}^k$(the coefficient of $(\genfrac{}{}{0ex}{}{k}{n})$) equals zero when $k$ is odd, but equals two when $k$ is even. (Note also that this follows from Property 3 for the square roots of unity.) Therefore,

$2(\genfrac{}{}{0ex}{}{0}{n})+2(\genfrac{}{}{0ex}{}{2}{n})+2(\genfrac{}{}{0ex}{}{4}{n})+\cdots =2^n$

$(\genfrac{}{}{0ex}{}{0}{n})+(\genfrac{}{}{0ex}{}{2}{n})+(\genfrac{}{}{0ex}{}{4}{n})+\cdots =2^{n-1}$

For the sum in this example, property 3 for the cube roots of unity may be useful.
Solution Summing the binomial expansions:

$(1+x{)}^n=(\genfrac{}{}{0ex}{}{0}{n})+(\genfrac{}{}{0ex}{}{1}{n})x+(\genfrac{}{}{0ex}{}{2}{n})x^2+\cdots +(\genfrac{}{}{0ex}{}{n}{n})x^n$

for $x=1,\omega ,{\omega }^2$ yields

$(1+1{)}^n+(1+\omega {)}^n+(1+{\omega }^2{)}^n=3(\genfrac{}{}{0ex}{}{0}{n})+(1+\omega +{\omega }^2)(\genfrac{}{}{0ex}{}{1}{n})+(1^2+{\omega }^2+({\omega }^2{)}^2)(\genfrac{}{}{0ex}{}{2}{n})+\cdots +(1^n+{\omega }^n+({\omega }^2{)}^n)(\genfrac{}{}{0ex}{}{n}{n})$.

By property 3, the coefficient of every first of three terms equals 3 and all other terms vanish. Therefore,

$3(\genfrac{}{}{0ex}{}{0}{n})+3(\genfrac{}{}{0ex}{}{3}{n})+3(\genfrac{}{}{0ex}{}{6}{n})+\cdots =2^n+(-{\omega }^2{)}^n+(-\omega {)}^n$

$=2^n+(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}{)}^n+(\cos \frac{-\pi }{3}+i\sin \frac{-\pi }{3}{)}^n$

$=2^n+2\cos \frac{n\pi }{3}$

$(\genfrac{}{}{0ex}{}{0}{n})+(\genfrac{}{}{0ex}{}{3}{n})+(\genfrac{}{}{0ex}{}{6}{n})+(\genfrac{}{}{0ex}{}{9}{n})+\cdots +(\genfrac{}{}{0ex}{}{3\ell -3}{n})+(\genfrac{}{}{0ex}{}{3\ell }{n})=\frac{1}{3}(2^n+2\cos \frac{n\pi }{3})$.