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Question: If w is a complex cube root of unity then show that \[\left( {2 - w} \right)\left( {2 - {w^2}} \righ...

If w is a complex cube root of unity then show that (2−w)(2−w2)(2−w10)(2−w11)=49\left( {2 - w} \right)\left( {2 - {w^2}} \right)\left( {2 - {w^{10}}} \right)\left( {2 - {w^{11}}} \right) = 49?

Explanation

Solution

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., 13\sqrt[3]{1}. To show that (2−w)(2−w2)(2−w10)(2−w11)=49\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,
(2−w)(2−w2)(2−w10)(2−w11)=49\left( {2 - w} \right)\left( {2 - {w^2}} \right)\left( {2 - {w^{10}}} \right)\left( {2 - {w^{11}}} \right) = 49
Since, w∈Cw \in Ci.e., w is a complex cube root of unity, hence we know that:
⇒w=13\Rightarrow w = \sqrt[3]{1}
⇒w3=1\Rightarrow {w^3} = 1 ……………. 1
And according to the properties of the cube root of 1, the sum of its root is zero. So,
1+w+w2=01 + w + {w^2} = 0
⇒w2+w=−1\Rightarrow {w^2} + w = - 1 …………. 2
Now, 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\\} …………….. 3
Now, substitute the value of w2+w{w^2} + w,w3{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\\} ………………. 4
As, we have simplified the terms of 2(w11+w10)2\left( {{w^{11}} + {w^{10}}} \right), w21{w^{21}}as:
2(w11+w10)=2w9(w+w2)2\left( {{w^{11}} + {w^{10}}} \right) = 2{w^9}\left( {w + {w^2}} \right)and w21=(w3)7{w^{21}} = {\left( {{w^3}} \right)^7}, hence now, substitute the value of w2+w{w^2} + w,w3{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\\}
=(7)(7)= \left( 7 \right)\left( 7 \right)
Multiplying both the terms, we get:
=49= 49
Hence, (2−w)(2−w2)(2−w10)(2−w11)=49\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≠1w \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+w2=01 + 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, −12+i32- \dfrac{1}{2} + i\sqrt {\dfrac{3}{2}} and −12−i32- \dfrac{1}{2} - i\sqrt {\dfrac{3}{2}}.