We will discuss here about the cube roots of unity and their properties.
Suppose let us assume that the cube root of 1 is z i.e., ∛1 = z.
Then, cubing both sides we get, z\(^{3}\) = 1
or, z\(^{3}\)  1 = 0
or, (z  1)(z\(^{2}\) + z + 1) = 0
Therefore, either z  1 = 0 i.e., z = 1 or, z\(^{2}\) + z + 1 = 0
Therefore, z = \(\frac{1\pm \sqrt{1^{2}  4\cdot 1\cdot 1}}{2\cdot 1}\) = \(\frac{1\pm \sqrt{ 3}}{2}\) = \(\frac{1}{2}\) ± i\(\frac{√3}{2}\)
Therefore, the three cube roots of unity are
1, \(\frac{1}{2}\) + i\(\frac{√3}{2}\) and \(\frac{1}{2}\)  i\(\frac{√3}{2}\)
among them 1 is real number and the other two are conjugate complex numbers and they are also known as imaginary cube roots of unity.
Properties of the cube roots of unity:
Property I: Among the three cube roots of unity one of the cube roots is real and the other two are conjugate complex numbers.
The three cube roots of unity are 1, \(\frac{1}{2}\) + i\(\frac{√3}{2}\) and \(\frac{1}{2}\)  i\(\frac{√3}{2}\).
Hence, we conclude that from the cube roots of unity we get 1 is real and the other two i.e., \(\frac{1}{2}\) + i\(\frac{√3}{2}\) and \(\frac{1}{2}\)  i\(\frac{√3}{2}\) are conjugate complex numbers.
Property II: Square of any one imaginary cube root of unity is equal to the other imaginary cube root of unity.
\((\frac{1 + \sqrt{3}i}{2})^{2}\) = \(\frac{1}{4}\)[( 1)^2  2 ∙ 1 ∙ √3i + (√3i)\(^{2}\)]
= \(\frac{1}{4}\)[1  2√3i  3]
= \(\frac{1  \sqrt{3}i}{2}\),
And \((\frac{1  \sqrt{3}i}{2})^{2}\) = \(\frac{1}{4}\)[(1^2 + 2 ∙ 1 ∙ √3i + (√3i)\(^{2}\)]
= \(\frac{1}{4}\)[1 + 2√3 i  3]
= \(\frac{1 + \sqrt{3}i}{2}\),
Hence, we conclude that square of any cube root of unity is equal to the other.
Therefore, suppose ω\(^{2}\) is one imaginary cube root of unity then the other would be ω.
Property III: The product of the two imaginary cube roots is 1 or, the product of three cube roots of unity is 1.
Let us assume that, ω = \(\frac{1  \sqrt{3}i}{2}\); then, ω\(^{2}\) = \(\frac{1 + \sqrt{3}i}{2}\)
Therefore, the product of the two imaginary or complex cube roots = ω ∙ ω\(^{2}\) = \(\frac{1  \sqrt{3}i}{2}\) × \(\frac{1 + \sqrt{3}i}{2}\)
Or, ω\(^{3}\) = \(\frac{1}{4}\)[(1)\(^{2}\)  (√3i)\(^{2}\)] = \(\frac{1}{4}\)[1  3i\(^{2}\)] = \(\frac{1}{4}\)[1 + 3] = \(\frac{1}{4}\) × 4 = 1.
Again, the cube roots of unity are 1, ω, ω\(^{2}\). So, product of cube roots of unity = 1 ∙ ω ∙ ω\(^{2}\) = ω\(^{3}\) = 1.
Therefore, product of the three cube roots of unity is 1.
Property IV: ω\(^{3}\) = 1
We know that ω is a root of the equation z\(^{3}\)  1 = 0. Therefore, ω satisfies the equation z\(^{3}\)  1 = 0.
Consequently, ω\(^{3}\)  1 = 0
or, ω = 1.
Note: Since ω\(^{3}\) = 1, hence, ω\(^{n}\) = ω\(^{m}\), where m is the least nonnegative remainder obtained by dividing n by 3.
Property V: The sum of the three cube roots of unity is zero i.e., 1 + ω + ω\(^{2}\) = 0.
We know that, the sum of the three cube roots of unity = 1 + \(\frac{1  \sqrt{3}i}{2}\) + \(\frac{1 + \sqrt{3}i}{2}\)
Or, 1 + ω + ω\(^{2}\) = 1  \(\frac{1}{2}\) + \(\frac{√3}{2}\)i  \(\frac{1}{2}\)  \(\frac{√3}{2}\)i = 0.
Notes:
(i) The cube roots of 1 are 1, ω, ω\(^{2}\) where, ω = \(\frac{1  \sqrt{3}i}{2}\) or, \(\frac{1 + \sqrt{3}i}{2}\)
(ii) 1 + ω + ω\(^{2}\) = 0 ⇒ 1 + ω =  ω\(^{2}\), 1 + ω\(^{2}\) =  ω and ω + ω\(^{2}\) = 1
(iii) ω\(^{4}\) = ω\(^{3}\) ∙ ω = 1 ∙ ω = ω;
ω\(^{5}\) = ω\(^{3}\) ∙ ω\(^{2}\) = 1 ∙ ω\(^{2}\) = ω\(^{2}\);
ω\(^{6}\) = (ω\(^{3}\))\(^{2}\) = (1)\(^{2}\) = 1.
In general, if n be a positive integer then,
ω\(^{3n}\) = (ω\(^{3}\))\(^{n}\) = 1\(^{n}\) = 1;
ω\(^{3n + 1}\) = ω\(^{3n}\) ∙ ω = 1 ∙ ω = ω;
ω\(^{3n + 2}\) = ω\(^{3n}\) ∙ ω\(^{2}\) = 1 ∙ ω\(^{2}\) = ω\(^{2}\).
Property VI: The reciprocal of each imaginary cube roots of unity is the other.
The imaginary cube roots of unity are ω and ω\(^{2}\), where ω = \(\frac{1 + \sqrt{3}i}{2}\).
Therefore, ω ∙ ω\(^{2}\) = ω\(^{3}\) = 1
⇒ ω = \(\frac{1}{ω^{2}}\) and ω\(^{2}\) = \(\frac{1}{ω}\)
Hence, we conclude that the reciprocal of each imaginary cube roots of unity is the other.
Property VII: If ω and ω\(^{2}\) are the roots of the equation z\(^{2}\) + z + 1 = 0 then  ω and  ω\(^{2}\) are the roots of the equation z\(^{2}\)  z + 1 = 0.
Property VIII: Cube roots of 1 are 1,  ω and  ω\(^{2}\).
11 and 12 Grade Math
From The Cube Roots of Unity to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.