Cube Roots of Unity: Definition, Derivation, and Formulas

Cube roots of Unity are the numbers which, when multiplied by itself three times or raised to the power three, gives the product as \({\rm{1}}{\rm{.}}\) In the lower classes, until complex numbers or imaginary numbers are introduced, we consider \({\rm{1}}\) as the only cube root of unity.

But, in higher classes, with the introduction of complex numbers imaginary number into mathematics, we come across a more comprehensive knowledge about the cube roots of unity, which includes the real cube root \({\rm{1}}\) and two more complex cube roots of unity represented by \(\omega \) and \({\omega ^2}.\) In this article, we shall discuss more about these three cube roots of unity. Continue reading to know more.

Define Cube Roots of Unity

Cube roots of Unity are the numbers which, when multiplied by themselves three times or raised to the power three, gives the product as \(1.\) In fact, the \({n^{{\rm{th}}}}\) root of unity is also one. The cube root of unity means cube root of one.

The cube root of the number is denoted by \(\sqrt[3]{{}}.\) It is also represented in the index form, as the number raised to the power of \(\frac{1}{3}.\)
\(\sqrt[3]{1} = {\left( 1 \right)^{\frac{1}{3}}} = 1\)

As we know, the cube root of unity is the number obtained by multiplying itself three times. The cube roots of the unit are the cube root of one. The cube root of one is always one.

There are three cubes of the unity, and they are \(1,\,\omega ,\,{\omega ^2}.\) Here, \(1\) is a real root of the unity and the numbers \(\omega ,\,{\omega ^2}\) are the complex roots of unity. The values of the complex numbers \(\omega ,\,{\omega ^2}\) are \(\frac{{ – 1 + i\sqrt 3 }}{2}\) and \(\frac{{ – 1 – i\sqrt 3 }}{2}.\)

Derivation of Cube Roots of Unity

As we know, the term unity means one. The derivation of the cube roots of the unity is explained below:

Step-1: Let assume the cube root of unity equals any variable, say \(x.\)
\(x = \sqrt[3]{1}\,\,….\left( 1 \right)\)

Step-2: Cubing on both sides of the equation \(\left( 1 \right),\)
\( \Rightarrow {x^3} = {\left( {\sqrt[3]{1}} \right)^3}\)
\( \Rightarrow {x^3} = 1\)

Step-3: Transpose the number one from R.H.S to L.H.S. Transposing the terms from one side of the equation to the other side will result in a reverse of Mathematical operations.

So, the sign of \(1\) becomes negative \(\left( – \right)\) from positive \(\left( + \right).\)
\( \Rightarrow {x^3} – 1 = 0\,\,…\left( 2 \right)\)

Step-4: Factorizing the above equation, by using an identity \({a^3} – {b^3} = \left( {a – b} \right)\left( {{a^2} + ab + {b^2}} \right),\) we get,
Thus, \(\left( {x – 1} \right)\left( {{x^2} + x + 1} \right) = 0\)

Step-5: Finding the factors of the above equation, such as
\(\left( {x – 1} \right) = 0\) or \(\left( {{x^2} + x + 1} \right) = 0\)
By solving, \(\left( {x – 1} \right) = 0,\) we get \(x = 1.\)

Step-6: Now, solving the equation \({x^2} + x + 1 = 0\) as follows:
Compare the above equation with quadratic equation \(a{x^2} + bx + c = 0,\) We get \(a = 1,\,b = 1,\,c = 1.\)
The quadratic formula to find the roots of quadratic equations is \(\frac{{ – b \pm \sqrt {{b^2} – 4\,ac} }}{{2\,a}}.\)
\( \Rightarrow x = \frac{{ – 1 \pm \sqrt {{1^2} – 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}}\)
\( \Rightarrow x = \frac{{ – 1 \pm \sqrt { – 3} }}{2}\)
Here, we know that, \(\sqrt { – 1} = i\)
\( \Rightarrow x = \frac{{ – 1 \pm i\sqrt 3 }}{2}\)
So, \(x = \frac{{ – 1 + i\sqrt 3 }}{2}\) or \(x = \frac{{ – 1 – i\sqrt 3 }}{2}\)

Step-7: The roots of the equation \(\left( 2 \right)\) are given as \(1,\,\frac{{ – 1 + i\sqrt 3 }}{2}\) and \(\frac{{ – 1 – i\sqrt 3 }}{2}.\)

Therefore, the cube roots of unity \(\left( {\sqrt[3]{1}} \right)\) are \(1,\,\frac{{ – 1 + i\sqrt 3 }}{2},\,\frac{{ – 1 – i\sqrt 3 }}{2}\)

Cube Roots of Unity: Formulas

Let a complex number ω is an imaginary number, then \({\omega ^2}\) is the other root. In general, \(\omega = \frac{{ – 1 + i\sqrt 3 }}{2};\,{\omega ^2} = \frac{{ – 1 – i\sqrt 3 }}{2}.\)

1.\(1 + \omega + {\omega ^2} = 0\)
2. \({\omega ^3} = 1\)
3. \({\omega ^2} = \frac{1}{\omega }\)
4. \(\omega = \frac{1}{{{\omega ^2}}}\)
5. \(\omega \times {\omega ^2} = {\omega ^3}\)

Properties of Cube Roots of Unity

We know that cube roots of unity are \(1,\,\frac{{ – 1 + i\sqrt 3 }}{2}\) and \(\frac{{ – 1 – i\sqrt 3 }}{2}.\) Its properties are explained below:

Property-1: In the three cube roots of unity, one root is real in nature, and the other two roots are complex in nature.

1. \(1\) is a real root.
2. \(\frac{{ – 1 + i\sqrt 3 }}{1}\) and \(\frac{{ – 1 – i\sqrt 3 }}{1}\) are the complex roots.

Property-2: The square of an imaginary root is equal to another imaginary root.

1. Square of \(\frac{{ – 1 + i\sqrt 3 }}{1}\) is equals to \(\frac{{ – 1 – i\sqrt 3 }}{1}\)
Proof:
\({\left( {\frac{{ – 1 + i\sqrt 3 }}{2}} \right)^2}\)
\( = \frac{{{{\left( { – 1} \right)}^2} + {{\left( {i\sqrt 3 } \right)}^2} + 2\left( { – 1} \right)\left( {i\sqrt 3 } \right)}}{{{2^2}}}\)
\( = \frac{{1 + {i^2} \times 3 – 2\,i\sqrt 3 }}{4}\)
\( = \frac{{1 – 3 – 2\,i\sqrt 3 }}{4}\)
\( = \frac{{ – 2 – 2\,i\sqrt 3 }}{4}\)
\( = \frac{{2\left( { – 1 – i\sqrt 3 } \right)}}{4}\)
\( = \frac{{\left( { – 1 – i\sqrt 3 } \right)}}{2}\)
\(\therefore {\left( {\frac{{ – 1 + i\sqrt 3 }}{2}} \right)^2} = \frac{{\left( { – 1 – i\sqrt 3 } \right)}}{2}\)

2. Square of \(\frac{{ – 1 – i\sqrt 3 }}{1}\) is equals to \(\frac{{ – 1 + i\sqrt 3 }}{1}\)
Proof:
\({\left( {\frac{{ – 1 – i\sqrt 3 }}{2}} \right)^2}\)
\( = \frac{{{{\left( { – 1} \right)}^2} + {{\left( {i\sqrt 3 } \right)}^2} + 2\left( { – 1} \right)\left( {i\sqrt 3 } \right)}}{{{2^2}}}\)
\( = \frac{{1 + {i^2} \times 3 + 2\,i\sqrt 3 }}{4}\)
\( = \frac{{1 – 3 +2\,i\sqrt 3 }}{4}\)
\( = \frac{{ – 2 + 2\,i\sqrt 3 }}{4}\)
\( = \frac{{2\left( { – 1 + i\sqrt 3 } \right)}}{4}\)
\( = \frac{{\left( { – 1 + i\sqrt 3 } \right)}}{2}\)
\(\therefore {\left( {\frac{{ – 1 – i\sqrt 3 }}{2}} \right)^2} = \frac{{\left( { – 1 + i\sqrt 3 } \right)}}{2}\)

Property-3: The product of three roots of the unity is one, or we can say that product of two imaginary roots is one.

\(1 \times \frac{{ – 1 + i\sqrt 3 }}{2} \times \frac{{ – 1 – i\sqrt 3 }}{2}\)
\( = \frac{{\left( { – 1 + i\sqrt 3 } \right) \times \left( { – 1 + i\sqrt 3 } \right)}}{{2 \times 2}}\)
\( = \frac{{\left[ {{{\left( { – 1} \right)}^2} – {{\left( {i\sqrt 3 } \right)}^2}} \right]}}{4}\)
\( = \frac{{1 – {i^2} \times 3}}{4}\)
\( = \frac{{1 – \left( { – 1} \right) \times 3}}{4}\)
\( = \frac{{1 + 3}}{4} = \frac{4}{4} = 1\)
\(1 \times \frac{{ – 1 + i\sqrt 3 }}{2} \times \frac{{ – 1 – i\sqrt 3 }}{2} = 1\)

Property-4: The sum of all three roots of unity is always zero.

\(1 + \frac{{ – 1 + i\sqrt 3 }}{2} + \frac{{ – 1 – i\sqrt 3 }}{2}\)
\( = 1 + \frac{{ – 1 + i\sqrt 3 – 1 – i\sqrt 3 }}{2}\)
\( = 1 + \frac{{ – 2}}{2}\)
\( = 1 – 1 = 0\)
\(1 + \frac{{ – 1 + i\sqrt 3 }}{2} + \frac{{ – 1 – i\sqrt 3 }}{2} = 0\)

Property-5: The reciprocal of one imaginary root is equals to another imaginary root.

1. The reciprocal of \(\frac{{ – 1 + i\sqrt 3 }}{2}\) is given by \(\frac{1}{{\frac{{ – 1 + i\sqrt 3 }}{2}}}\)
We have, reciprocal of \(\frac{{ – 1 + i\sqrt 3 }}{2} = \frac{2}{{\left( { – 1 + i\sqrt 3 } \right)}}\)
\( = \frac{2}{{\left( { – 1 + i\sqrt 3 } \right)}} \times \frac{{\left( { – 1 – i\sqrt 3 } \right)}}{{\left( { – 1 – i\sqrt 3 } \right)}}\)
\( = 2 \times \frac{{\left( { – 1 – i\sqrt 3 } \right)}}{{{{\left( { – 1} \right)}^2} – {{\left( {i\sqrt 3 } \right)}^2}}}\)
\( = 2 \times \frac{{\left( { – 1 – i\sqrt 3 } \right)}}{{{{\left( { – 1} \right)}^2} – {i^2} \times 3}}\)
\( = 2 \times \frac{{\left( { – 1 – i\sqrt 3 } \right)}}{{1 – \left( { – 1} \right) \times 3}}\)
\( = 2 \times \frac{{\left( { – 1 – i\sqrt 3 } \right)}}{{1 + 3}}\)
\( = 2 \times \frac{{\left( { – 1 – i\sqrt 3 } \right)}}{4}\)
\( = \frac{{\left( { – 1 – i\sqrt 3 } \right)}}{4}\)
So, the reciprocal of \(\frac{{ – 1 + i\sqrt 3 }}{2}\) is \(\frac{{ – 1 – i\sqrt 3 }}{2}\)

2. The reciprocal of \(\frac{{ – 1 – i\sqrt 3 }}{2}\) is given by \(\frac{1}{{\frac{{ – 1 – i\sqrt 3 }}{2}}}\)
We have, reciprocal of \(\frac{{ – 1 – i\sqrt 3 }}{2} = \frac{2}{{\left( { – 1 – i\sqrt 3 } \right)}}\)
\( = \frac{2}{{\left( { – 1 – i\sqrt 3 } \right)}} \times \frac{{\left( { – 1 + i\sqrt 3 } \right)}}{{\left( { – 1 + i\sqrt 3 } \right)}}\)
\( = 2 \times \frac{{\left( { – 1 + i\sqrt 3 } \right)}}{{{{\left( { – 1} \right)}^2} – {{\left( {i\sqrt 3 } \right)}^2}}}\)
\( = 2 \times \frac{{\left( { – 1 + i\sqrt 3 } \right)}}{{{{\left( { – 1} \right)}^2} – {i^2} \times 3}}\)
\( = 2 \times \frac{{\left( { – 1 + i\sqrt 3 } \right)}}{{1 – \left( { – 1} \right) \times 3}}\)
\( = 2 \times \frac{{\left( { – 1 + i\sqrt 3 } \right)}}{{1 + 3}}\)
\( = 2 \times \frac{{\left( { – 1 + i\sqrt 3 } \right)}}{4}\)
\( = \frac{{\left( { – 1 + i\sqrt 3 } \right)}}{2}\)
\( = \frac{{\left( { – 1 + i\sqrt 3 } \right)}}{2}\)
So, the reciprocal of \(\frac{{ – 1 – i\sqrt 3 }}{2}\) is \(\frac{{ – 1 + i\sqrt 3 }}{2}.\)

Property-6: The imaginary roots of \( – 1\) are \(\frac{{ – 1 + i\sqrt 3 }}{2}\) and \(\frac{{ – 1 – i\sqrt 3 }}{2}.\)

Cube Roots of Unity in Polar Form

Cube root of unity means the three roots of one. Cube roots of the unity are the solution of an equation for any variable, say \(x.\)
\({x^3} – 1 = 0\)
\( \Rightarrow {x^3} = 1\)
\( \Rightarrow x = \sqrt[3]{1}\)
\( \Rightarrow x = {\left( 1 \right)^{\frac{1}{3}}}\)

By using De Moivre’s theorem, the above equation can be written in the polar form as
\(x = \cos \left( {\frac{{2k\pi }}{3}} \right) + i\,\sin \left( {\frac{{2k\pi }}{3}} \right);\) where \(k = 0,\,1,2.\)

Therefore, the cube roots of the units in the polar form are
\(\cos \frac{{2\pi }}{3} + i\,\sin \frac{{2\pi }}{3};\,\cos \frac{{4\pi }}{3} + i\,\sin \frac{{4\pi }}{3}\) and \(1.\)

Cube Roots of Unity in Euler Form

Cube roots of the unit are the solution of an equation for any variable, say \(x.\)
\( \Rightarrow {x^3} – 1 = 0\)
\( \Rightarrow x = {\left( 1 \right)^{\frac{1}{3}}}\)

We know that the cube roots of the unity in the polar form are \(1,\,\cos \frac{{2\pi }}{3} + i\,\sin \frac{{2\pi }}{3}\) and \(\cos \frac{{4\pi }}{3} + i\,\sin \frac{{4\pi }}{3}.\) We know that the number, which is in the form of the polar form \(\left( {\cos \theta + i\,\sin \theta } \right)\) can be written polar form as \(\left( {\cos \theta + i\,\sin \theta } \right) = {e^{i\theta }}\)

So, the cube roots of units, which are given in the polar form, are \(1,\,\cos \frac{{2\pi }}{3} + i\,\sin \frac{{2\pi }}{3}\) and \(\cos \frac{{4\pi }}{3} + i\,\sin \frac{{4\pi }}{3}\) can be written in Euler form are \(1,\,{e^{\frac{{2\pi }}{3}}},\,{e^{\frac{{4\pi }}{3}}}.\)

Solved Examples on Cube Roots of Unity

Q.1. Evaluate \({\left( {1 + {\omega ^2}} \right)^3}.\)
Ans: Given \({\left( {1 + {\omega ^2}} \right)^3}\)
So, we have, \({\left( {1 + {\omega ^2}} \right)^3} = {\left( { – \omega } \right)^3}\left[ {1 + \omega + {\omega ^2} = 0} \right]\)
\(= {\left( { – 1} \right)^3}{\left( \omega \right)^3} = – 1 \times 1 = – 1\left[ {{\omega ^3} = 1} \right]\)

Q.2. Find the value of \({\left( {1 + \omega } \right)^3} – {\left( {1 + {\omega ^2}} \right)^3}.\)
Ans: Given \({\left( {1 + \omega } \right)^3} – {\left( {1 + {\omega ^2}} \right)^3}\)
So, we have, \({\left( {1 + \omega } \right)^3} – {\left( {1 + {\omega ^2}} \right)^3}\)
\( = {\left( { – {\omega ^2}} \right)^3} – {\left( { – \omega } \right)^3}\,\left[ {1 + \omega + {\omega ^2} = 0} \right]\)
\( = {\left( { – 1} \right)^3} \times {\left( {{\omega ^2}} \right)^3} – {\left( { – 1} \right)^3}{\left( \omega \right)^3}\)
\(= \, – 1 \times \,{\omega ^6} – \left( { – 1} \right) \times {\omega ^3}\)
\( = – 1 \times {\left( {{\omega ^3}} \right)^2} – \left( { – 1} \right) \times {\omega ^3}\)
\(= \, – 1 \times {\left( 1 \right)^2} – \left( { – 1} \right) \times 1{\mkern 1mu} \left[ {{\omega ^3} – 1} \right]\)
\(= \, – 1 \times 1 + 1\)
\(= \, – {\rm{ }}1{\rm{ }} + {\rm{ }}1\)
\( =0\)

Q.3. Find the factors of \({p^2} + pq + {q^2}\) by using the complex roots of the unity \(\left( {\omega ,\,{\omega ^2}} \right).\)
Ans: Given: \({p^2} + pq + {q^2}\)
It can be written as \({p^2} – \left( { – 1} \right)pq + \left( 1 \right){q^2}\)
From the properties of cube roots of unity, we know that sum of the roots of unity is zero, and also the product of roots is one.
\(1 + \omega + {\omega ^2} = 0\) and \({\omega ^3} = 1\)
So, the given equation can be rearranged as follows:
\({p^2} – \left( {\omega + {\omega ^2}} \right)pq + \left( {{\omega ^3}} \right){q^2}\)
\( \Rightarrow {p^2} – \omega pq – {\omega ^2}pq + {\omega ^3}{q^2}\)
\( \Rightarrow p\left( {p – \omega q} \right) – {\omega ^2}q\left( {p – \omega q} \right)\)
\( \Rightarrow \left( {p – {\omega ^2}q} \right)\left( {p – \omega q} \right)\)
The factors of \({p^2} + pq + {q^2}\) are \(\left( {p – {\omega ^2}q} \right)\left( {p – \omega q} \right).\)

Q.4. Find the value of \({\left( {2\,\omega } \right)^4} + {\left( {2\,{\omega ^2}} \right)^4}.\)
Ans: Given \({\left( {2\,\omega } \right)^4} + {\left( {2\,{\omega ^2}} \right)^4}\)
Thus,
\( = {2^4}{\omega ^4} + {2^4}{\left( {{\omega ^2}} \right)^4}\)
[By using the law, \({\left( {{a^m}} \right)^n} = {a^{mn}}\)]

\( = {2^4}{\omega ^4} + {2^4}{\omega ^8}\)
\( = {2^4}{\omega ^3} \times \omega + {2^4} \times {\omega ^6} \times {\omega ^2}\)
\( = {2^4} \times {\omega ^3} \times \omega + {2^4} \times {\left( {{\omega ^3}} \right)^2} \times {\omega ^2}\)
[We know that product of all three roots of unity is one.
\(1 \times \omega \times {\omega ^2} = {\omega ^3} = 1\)]

\( = {2^4} \times 1 \times \omega + {2^4} \times 1 \times {\omega ^2}\)
\( = {2^4}\left( {\omega + {\omega ^2}} \right)\)
[We know that sum of all the roots of unity is zero.
\(1 + \omega + {\omega ^2} = 0,\) then \(\omega + {\omega ^2} = – 1\)
\( = {2^4}\left( { – 1} \right) = – 16\)
Hence, the value of \({\left( {2\omega } \right)^4} + {\left( {2{\omega ^2}} \right)^4}\) is \(\left( { – 16} \right).\)

Q.5. Prove that, \(\left( {\omega – {\omega ^2}} \right) = \, – 3.\)
Ans: Let us consider L.H.S: \(\left( {\omega – {\omega ^2}} \right)\)
By using the identity: \({\left( {a – b} \right)^2} = {a^2} + {b^2} – 2\,ab.\)
Comparing the \(\left( {\omega – {\omega ^2}} \right)\) with \({\left( {a – b} \right)^2},\) we get \(a = \omega ,\,b = {\omega ^2}\)
Thus, L.H.S can be written as
\( = {\omega ^2} + {\left( {{\omega ^2}} \right)^2} – 2 \times \omega \times {\omega ^2}\)
\( = {\omega ^2} + {\omega ^4} – 2{\omega ^3}\)
\( = {\omega ^2} + {\omega ^3} \times \omega – 2{\omega ^3}\)
[We know that product of all three roots of unity is one.
\(1 \times \omega \times {\omega ^2} = {\omega ^3} = 1\)]
\( = {\omega ^2} + \omega – 2\left( 1 \right)\)
[We know that sum of all the roots of unity is zero.
\(1 + \omega + {\omega ^2} = 0,\) then \(\omega + {\omega ^2} = – 1\)]
\( = \, – {\rm{ }}1 – 2 = \, – {\rm{ }}3\)
Here, L.H.S = R.H.S.
Hence, proved.

Summary

In this article, we studied about Cube roots of Unity, which is obtained when a number multiplied itself three times or raised to the power three gives the product as \(1.\) In fact, the \({n^{{\rm{th}}}}\) root of unity is also one only. The cube roots of units mean cube root of one. Furthermore, Cube roots of unity have various applications in Mathematics. It is used in number systems, factorisation, complex numbers, etc.

It is also important to note that, in the three cube roots of unity, one root is real in nature, and the other two roots are complex in nature. Furthermore, the product of three roots of unity is one. In other words, we can state that the product of two imaginary roots is one.

FAQs on Cube Roots of Unity

Q.1. Explain cube roots of unity.
Ans: Cube roots of Unity are the number which, when multiplied itself three times or raised to the power three, gives the product as \(1.\) In fact, the \({n^{{\rm{th}}}}\) root of unity is also one only. The cube roots of unity mean the cube root of one.
There are three cubes of the unity \(1,\,\omega ,\,{\omega ^2}.\) Here, \(1\) is a real root of the unity and complex numbers \(\omega ,\,{\omega ^2}\) are the complex roots of unity.

Q.2. What are the properties of the cube roots of unity?
Ans: There are different properties of cube roots of unity, some of them listed below:
a. The sum of all three cube roots of the unity is zero.
b. The product of all three roots of unity is one.
c. One imaginary root of the cube roots of unity equals the square of another imaginary root.
d. Reciprocal of one imaginary root of the cube roots of unity is equal to the other imaginary root.

Q.3. What are the uses of cube roots of unity?
Ans: Cube roots of unity are used in the various branches in Mathematics like number systems, complex numbers, factorization etc. These are used to solve the \({n^{{\rm{th}}}}\) roots of the cubic polynomials.

Q.4. What is the value of \({\omega ^3}?\)
Ans: The three cubes of the unity are \(1,\,\omega ,\,{\omega ^2}.\) From the properties of cube roots of unity, we know that product of all three roots is one.
\(1 \times \omega \times {\omega ^2} = 1\)
By using the identity: \({a^m} \times {a^n} = {a^{m + n}}\)
\( \Rightarrow {\omega ^{1 + 2}} = 1\)
\( \Rightarrow {\omega ^3} = 1\)

Q.5. What are cube roots of unity, and how to implement cube roots of unity?
Ans: There are three cube roots of the unity \(1,\,\omega ,\,{\omega ^2}.\) Here, \(1\) is a real root of the unity and complex numbers \(\omega ,\,{\omega ^2}\) are the complex roots of unity.
The values of the complex numbers \(\omega ,\,{\omega ^2}\) are \(\frac{{ – 1 + i\sqrt 3 }}{2}\) and \(\frac{{ – 1 – i\sqrt 3 }}{2}.\)
The cube roots of the unity are implemented by using their properties.

We hope this article on cube roots of unity has provided significant value to your knowledge. If you have any queries or suggestions, feel to write them down in the comment section below. We will love to hear from you. Embibe wishes you all the best of luck!