Solveeit Logo
Login

Question

Question: If \(1,{a_1},{a_2}.....,{a_{n - 1}}\) are the nth roots of unity then the value of \(\left( {1 - {a_...

If 1,a1,a2.....,an−11,{a_1},{a_2}.....,{a_{n - 1}} are the nth roots of unity then the value of (1−a1)(1−a2)......(1−an−1)\left( {1 - {a_1}} \right)\left( {1 - {a_2}} \right)......\left( {1 - {a_{n - 1}}} \right) is equal to
(A) 3\sqrt 3
(B) 12\dfrac{1}{2}
(C) nn
(D) 00

Explanation

Solution

If 1,a1,a2.....,an−11,{a_1},{a_2}.....,{a_{n - 1}} are the nth{n^{th}} roots of unity, that means they are the roots of the equation xn−1=0{x^n} - 1 = 0 . Express the roots of this equation in form of a product, i.e. xn−1=(x−1)(x−a1)(x−a2).....(x−an−2)(x−an−1){x^n} - 1 = \left( {x - 1} \right)\left( {x - {a_1}} \right)\left( {x - {a_2}} \right).....\left( {x - {a_{n - 2}}} \right)\left( {x - {a_{n - 1}}} \right) . Now use the expansion of (xn−1)\left( {{x^n} - 1} \right) in the left-hand side of the previous equation. Transpose the terms to obtain the required expression on one side. Put the value x=1x = 1 to find the desired answer.

Complete step-by-step answer:
Here in this problem, we are given that 1,a1,a2.....,an−11,{a_1},{a_2}.....,{a_{n - 1}} are the nth{n^{th}} roots of unity, i.e. nth{n^{th}} roots of 11 . And then we need to find the value of the expression (1−a1)(1−a2)......(1−an−1)\left( {1 - {a_1}} \right)\left( {1 - {a_2}} \right)......\left( {1 - {a_{n - 1}}} \right) . We need to find the correct answer among the four given options.
Let’s discuss what is the nth{n^{th}} roots of unity.
As we know that the equation having a power of ′n′'n' over the variable will have ′n′'n' number of roots. So when we consider the equation: xn=1{x^n} = 1 , here we know that the variable ‘x’ can obtain ‘n’ number of roots. And these are hence called the nth{n^{th}} roots of unity.
So, according to the question 1,a1,a2.....,an−11,{a_1},{a_2}.....,{a_{n - 1}} are the ‘n’ roots of the equation xn=1{x^n} = 1 , this can be represented by:
⇒xn=1⇒xn−1=0\Rightarrow {x^n} = 1 \Rightarrow {x^n} - 1 = 0
The product of difference of all the roots with the variable will be equal to zero. Now putting 1,a1,a2.....,an−11,{a_1},{a_2}.....,{a_{n - 1}} as the roots of this equation, we get:
⇒xn−1=(x−1)(x−a1)(x−a2).....(x−an−2)(x−an−1)\Rightarrow {x^n} - 1 = \left( {x - 1} \right)\left( {x - {a_1}} \right)\left( {x - {a_2}} \right).....\left( {x - {a_{n - 2}}} \right)\left( {x - {a_{n - 1}}} \right)
This can be further simplified by transposing the term (x−1)\left( {x - 1} \right) to the left side
⇒xn−1x−1=(x−a1)(x−a2).....(x−an−2)(x−an−1)\Rightarrow \dfrac{{{x^n} - 1}}{{x - 1}} = \left( {x - {a_1}} \right)\left( {x - {a_2}} \right).....\left( {x - {a_{n - 2}}} \right)\left( {x - {a_{n - 1}}} \right)
As we know the expansion from the binomial theorem, xn−1=(x−1)(xn−1+xn−2+......+x2+x+1){x^n} - 1 = \left( {x - 1} \right)\left( {{x^{n - 1}} + {x^{n - 2}} + ...... + {x^2} + x + 1} \right) . We can use this in the above equation:
⇒xn−1x−1=(xn−1+xn−2+......+x2+x+1)=(x−a1)(x−a2).....(x−an−2)(x−an−1)\Rightarrow \dfrac{{{x^n} - 1}}{{x - 1}} = \left( {{x^{n - 1}} + {x^{n - 2}} + ...... + {x^2} + x + 1} \right) = \left( {x - {a_1}} \right)\left( {x - {a_2}} \right).....\left( {x - {a_{n - 2}}} \right)\left( {x - {a_{n - 1}}} \right)
Therefore, we get:
⇒(xn−1+xn−2+......+x2+x+1)=(x−a1)(x−a2).....(x−an−2)(x−an−1)\Rightarrow \left( {{x^{n - 1}} + {x^{n - 2}} + ...... + {x^2} + x + 1} \right) = \left( {x - {a_1}} \right)\left( {x - {a_2}} \right).....\left( {x - {a_{n - 2}}} \right)\left( {x - {a_{n - 1}}} \right)
For obtaining the expression (1−a1)(1−a2)......(1−an−1)\left( {1 - {a_1}} \right)\left( {1 - {a_2}} \right)......\left( {1 - {a_{n - 1}}} \right)on the right side of the equation, we must substitute the value x=1x = 1 on both sides of the above equation.
For x=1x = 1, we get:
⇒(1n−1+1n−2+......+12+1+1)=(1−a1)(1−a2).....(1−an−2)(1−an−1)\Rightarrow \left( {{1^{n - 1}} + {1^{n - 2}} + ...... + {1^2} + 1 + 1} \right) = \left( {1 - {a_1}} \right)\left( {1 - {a_2}} \right).....\left( {1 - {a_{n - 2}}} \right)\left( {1 - {a_{n - 1}}} \right)
Since we know 1m=1{1^m} = 1 , so on further solving it, we have:
⇒(1−a1)(1−a2).....(1−an−2)(1−an−1)=(1n−1+1n−2+......+12+1+1)=(1+1+....n times)=1×n\Rightarrow \left( {1 - {a_1}} \right)\left( {1 - {a_2}} \right).....\left( {1 - {a_{n - 2}}} \right)\left( {1 - {a_{n - 1}}} \right) = \left( {{1^{n - 1}} + {1^{n - 2}} + ...... + {1^2} + 1 + 1} \right) = \left( {1 + 1 + ....n{\text{ }}times} \right) = 1 \times n
Therefore, we get the value of expression as:
⇒(1−a1)(1−a2).....(1−an−2)(1−an−1)=n\Rightarrow \left( {1 - {a_1}} \right)\left( {1 - {a_2}} \right).....\left( {1 - {a_{n - 2}}} \right)\left( {1 - {a_{n - 1}}} \right) = n
Hence, the option (C) is the correct answer.

Note: In questions like this, the use of the power expansions series plays an important role. The expansion used in the above solution i.e. xn−1=(x−1)(xn−1+xn−2+......+x2+x+1){x^n} - 1 = \left( {x - 1} \right)\left( {{x^{n - 1}} + {x^{n - 2}} + ...... + {x^2} + x + 1} \right) is derived using the theorem, i.e. xn−yn=(x−y)(xn−1+xn−2y+xn−3y2+.....+xyn−2+yn−1){x^n} - {y^n} = \left( {x - y} \right)\left( {{x^{n - 1}} + {x^{n - 2}}y + {x^{n - 3}}{y^2} + ..... + x{y^{n - 2}} + {y^{n - 1}}} \right) where we put y=1y = 1 .