If w is complex root of unity, then what is the value of 1+w+w2

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Hint: The cube roots of unity can be defined as the numbers which when raised to the power of 3 gives the result as 1 i.e., in simple words, the cube root of unity is the cube root of 1 i.e., \[\sqrt[3]{1}\]. To show that \[\left( {2 - w} \right)\left( {2 - {w^2}} \right)\left( {2 - {w^{10}}} \right)\left( {2 - {w^{11}}} \right) = 49\] hence, consider the properties of the cube root to obtain the given equation.

Complete step by step answer:

Given,\[\left( {2 - w} \right)\left( {2 - {w^2}} \right)\left( {2 - {w^{10}}} \right)\left( {2 - {w^{11}}} \right) = 49\]Since, \[w \in C\]i.e., w is a complex cube root of unity, hence we know that:\[ \Rightarrow w = \sqrt[3]{1}\]\[ \Rightarrow {w^3} = 1\] ……………. 1And according to the properties of the cube root of 1, the sum of its root is zero. So, \[1 + w + {w^2} = 0\]\[ \Rightarrow {w^2} + w = - 1\] …………. 2Now, the given expression is: \[ = \left\{ {\left( {2 - w} \right)\left( {2 - {w^2}} \right)} \right\}\left\{ {\left( {2 - {w^{10}}} \right)\left( {2 - {w^{11}}} \right)} \right\}\]Multiply the terms each as directed:\[ = \left\{ {\left( {2\left( {2 - {w^2}} \right) - w\left( {2 - {w^2}} \right)} \right)} \right\}\left\{ {\left( {2\left( {2 - {w^{11}}} \right) - {w^{10}}\left( {2 - {w^{11}}} \right)} \right)} \right\}\]Multiplying the terms, we get:\[ = \left\{ {\left( {4 - 2{w^2} - 2w + {w^3}} \right)} \right\}\left\{ {\left( {4 - 2{w^{11}} - 2{w^{10}} + {w^{21}}} \right)} \right\}\]Now, combine the common terms i.e., 2 is common in both the terms, hence we get:\[ = \left\{ {4 - 2\left( {{w^2} + w} \right) + {w^3}} \right\}\left\{ {4 - 2\left( {{w^{11}} + {w^{10}}} \right) + {w^{21}}} \right\}\] …………….. 3Now, substitute the value of \[{w^2} + w\],\[{w^3}\]from equation 1 and 2 in equation 3:\[ = \left\{ {4 - 2\left( { - 1} \right) + 1} \right\}\left\{ {4 - 2{w^9}\left( {w + {w^2}} \right) + {{\left( {{w^3}} \right)}^7}} \right\}\] ………………. 4As, we have simplified the terms of \[2\left( {{w^{11}} + {w^{10}}} \right)\], \[{w^{21}}\]as:\[2\left( {{w^{11}} + {w^{10}}} \right) = 2{w^9}\left( {w + {w^2}} \right)\]and \[{w^{21}} = {\left( {{w^3}} \right)^7}\], hence now, substitute the value of \[{w^2} + w\],\[{w^3}\]in equation 4 as:\[ = \left\{ {4 + 2 + 1} \right\}\left\{ {4 - 2{{\left( 1 \right)}^3}\left( { - 1} \right) + {{\left( 1 \right)}^7}} \right\}\]\[ = \left\{ 7 \right\}\left\{ {4 - 2\left( 1 \right)\left( { - 1} \right) + \left( 1 \right)} \right\}\]Evaluating the terms, we get:\[ = \left\{ 7 \right\}\left\{ {4 + 2 + 1} \right\}\]\[ = \left( 7 \right)\left( 7 \right)\]Multiplying both the terms, we get:\[ = 49\]Hence, \[\left( {2 - w} \right)\left( {2 - {w^2}} \right)\left( {2 - {w^{10}}} \right)\left( {2 - {w^{11}}} \right) = 49\]

Note: Since w is a non-real complex root of unity, then \[w \ne 1\]. We must note that one imaginary cube root of unity is the square of the other and if two imaginary cube roots are multiplied then the product we get is equal to 1. And as \[1 + w + {w^2} = 0\], it can be said that the cube root of unity is collinear and the cube root of unity values are 1, \[ - \dfrac{1}{2} + i\sqrt {\dfrac{3}{2}} \] and \[ - \dfrac{1}{2} - i\sqrt {\dfrac{3}{2}} \].

If ω is a complex cube root of unity, find the value of (1 + ω2)3

ω is a complex cube root of unity
∴ ω3 = 1 and 1 + ω + ω2 = 0

Also, 1 + ω2 = - ω, 1 + ω = - ω2 and ω + ω2 = – 1

(1 + ω2)3 = (- ω)3 = - ω3 = - 1

Concept: Cube Root of Unity

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Page 2

If ω is a complex cube root of unity, find the value of (1 - ω - ω2)3 + (1 - ω + ω2)3

ω is a complex cube root of unity
∴ ω3 = 1 and 1 + ω + ω2 = 0

Also, 1 + ω2 = - ω, 1 + ω = - ω2 and ω + ω2 = – 1

(1 - ω - ω2)3 + (1 - ω + ω2)3

= [1 - (ω + ω2)]3 + [(1 + ω2) - ω]3

= [1 - (-1)]3 + (- ω - ω)3

= 23 + (- 2ω)3

= 8 - 8ω3 = 8 – 8(1) = 0

Concept: Cube Root of Unity

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Page 3

If `omega` is a complex cube root of unity, find the value of `(1 + omega)(1 + omega^2)(1 + omega^4)(1 + omega^8)`

ω is a complex cube root of unity
∴ ω3 = 1 and 1 + ω + ω2 = 0

Also, 1 + ω2 = -ω, 1 + ω = - ω2 and ω + ω2 = – 1

(1 + ω)(1 + ω2)(1 + ω4)(1 + ω8)

= (1 + ω)(1 + ω2)(1 + ω)(1 + ω2) ...[∵ ω3 = 1, therefore ω4 = ω]

= (- ω2)(- ω)(- ω2)(- ω) = ω6 = (ω3)2 = (1)2 =1.

Concept: Cube Root of Unity

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Page 4

If α and β are the complex cube roots of unity, show that α2 + β2 + αβ = 0.

α and β are the complex cube roots of unity.

∴ α = `(-1 + "i"sqrt(3))/2 and beta = (-1 - "i"sqrt(3))/2`

∴ αβ = `((- 1 + "i"sqrt(3))/2)((-1 - "i"sqrt(3))/2)`

= `((-1)^2 - (isqrt(3))^2)/4`

= `(1 - (-1)(3))/4` ...[∵ i2 = – 1]

= `(1 + 3)/4`

∴ αβ = 1

Also, α + β = `(-1 + "i"sqrt(3))/2 + (-1 - "i"sqrt(3))/2`

= `(-1 + "i"sqrt(3) - 1 - "i" sqrt(3))/2`

= `(-2)/2`

∴ α + β = – 1

L.H.S. = α2 + β2 + αβ
= α2 + 2αβ + β2 + αβ – 2αβ ...[Adding and subtracting 2αβ]
= (α2 + 2αβ + β2) – αβ
= (α + β)2 – αβ
= (– 1)2 – 1 = 1 – 1

= 0 = R.H.S.

Concept: Cube Root of Unity

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Page 5

x = a + b, y = αa + βb and z = aβ + bα
α and β are the complex cube roots of unity.

∴ α = `(-1 + "i"sqrt(3))/2 and beta = (-1 - "i"sqrt(3))/2`

∴ αβ = `((-1 + "i"sqrt(3))/2)((-1 - "i"sqrt(3))/2)`

= `((-1)^2 - ("i"sqrt(3))^2)/4`

= `(1 - (-1)(3))/4` ...[∵ i2 = – 1]

= `(1 + 3)/4`

∴ αβ = 1

Also, α + β = `(-1 + "i"sqrt(3))/2 + (-1 - "i"sqrt(3))/2`

= `(-1 + "i"sqrt(3) - 1 - "i"sqrt(3))/2`

= `(-2)/2`

∴ α + β = -1L.H.S. = xyz = (a + b)(αa + βb)(aβ + bα)

= (a + b)(αβa2 + α2ab + β2ab + αβb2)

= (a + b)[1. (a2) + (α2 + β2)ab + 1. (b2)]
= (a + b) {a2 + [(α + β)2 – 2αβ ]ab + b2}
= (a + b) {a2 + [(– 1)2 – 2(1)]ab + b2}
= (a + b) [a2 + (1 – 2)ab + b2]
= (a + b)(a2 – ab + b2)
= a3 +b3
= R.H.S.