what are the properties of cube root
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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, -12 + i√32 and -12 - i√32.
Hence, we conclude that from the cube roots of unity we get 1 is real and the other two i.e., 12 + i√32 and -12 - i√32 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.
(−1+3√i2)2 = 14[(- 1)^2 - 2 ∙ 1 ∙ √3i + (√3i)2]
= 14[1 - 2√3i - 3]
= −1−3√i2,
And (−1−3√i2)2 = 14[(1^2 + 2 ∙ 1 ∙ √3i + (√3i)2]
= 14[1 + 2√3 i - 3]
= −1+3√i2,
Hence, we conclude that square of any cube root of unity is equal to the other.
Therefore, suppose ω2 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 = −1−3√i2; then, ω2 = −1+3√i2
Therefore, the product of the two imaginary or complex cube roots = ω ∙ ω2 = −1−3√i2 × −1+3√i2
Or, ω3 = 14[(-1)2 - (√3i)2] = 14[1 - 3i2] = 14[1 + 3] = 14 × 4 = 1.
Again, the cube roots of unity are 1, ω, ω2. So, product of cube roots of unity = 1 ∙ ω ∙ ω2 = ω3 = 1.
Therefore, product of the three cube roots of unity is 1.
Property IV: ω3 = 1
We know that ω is a root of the equation z3 - 1 = 0. Therefore, ω satisfies the equation z3 - 1 = 0.
Consequently, ω3 - 1 = 0
or, ω = 1.
Note: Since ω3 = 1, hence, ωn = ωm, where m is the least non-negative remainder obtained by dividing n by 3.
Property V: The sum of the three cube roots of unity is zero i.e., 1 + ω + ω2 = 0.
We know that, the sum of the three cube roots of unity = 1 + −1−3√i2 + −1+3√i2
Or, 1 + ω + ω2 = 1 - 12 + √32i - 12 - √32i = 0.
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Answer:
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, -12 + i√32 and -12 - i√32.
Hence, we conclude that from the cube roots of unity we get 1 is real and the other two i.e., 12 + i√32 and -12 - i√32 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.
(−1+3√i2)2 = 14[(- 1)^2 - 2 ∙ 1 ∙ √3i + (√3i)2]
= 14[1 - 2√3i - 3]
= −1−3√i2,
And (−1−3√i2)2 = 14[(1^2 + 2 ∙ 1 ∙ √3i + (√3i)2]
= 14[1 + 2√3 i - 3]
= −1+3√i2,
Hence, we conclude that square of any cube root of unity is equal to the other.
Therefore, suppose ω2 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 = −1−3√i2; then, ω2 = −1+3√i2
Therefore, the product of the two imaginary or complex cube roots = ω ∙ ω2 = −1−3√i2 × −1+3√i2
Or, ω3 = 14[(-1)2 - (√3i)2] = 14[1 - 3i2] = 14[1 + 3] = 14 × 4 = 1.
Again, the cube roots of unity are 1, ω, ω2. So, product of cube roots of unity = 1 ∙ ω ∙ ω2 = ω3 = 1.
Therefore, product of the three cube roots of unity is 1.
Property IV: ω3 = 1
We know that ω is a root of the equation z3 - 1 = 0. Therefore, ω satisfies the equation z3 - 1 = 0.
Consequently, ω3 - 1 = 0
or, ω = 1.
Note: Since ω3 = 1, hence, ωn = ωm, where m is the least non-negative remainder obtained by dividing n by 3.
Property V: The sum of the three cube roots of unity is zero i.e., 1 + ω + ω2 = 0.
We know that, the sum of the three cube roots of unity = 1 + −1−3√i2 + −1+3√i2
Or, 1 + ω + ω2 = 1 - 12 + √32i - 12 - √32i = 0.
Notes:
(i) The cube roots of 1 are 1, ω, ω2 where, ω = −1−3√i2 or, −1+3√i2
(ii) 1 + ω + ω2 = 0 ⇒ 1 + ω = - ω2, 1 + ω2