Question

If $1,{a_1},{a_2}.....,{a_{n - 1}}$ are the nth roots of unity then the value of $\left( {1 - {a_1}} \right)\left( {1 - {a_2}} \right)......\left( {1 - {a_{n - 1}}} \right)$ is equal to
(A) $\sqrt 3 $
(B) $\dfrac{1}{2}$
(C) $n$
(D) $0$

Answer 1

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

Complete step-by-step answer:
Here in this problem, we are given that $1,{a_1},{a_2}.....,{a_{n - 1}}$ are the ${n^{th}}$ roots of unity, i.e. ${n^{th}}$ roots of $1$ . And then we need to find the value of the expression $\left( {1 - {a_1}} \right)\left( {1 - {a_2}} \right)......\left( {1 - {a_{n - 1}}} \right)$ . We need to find the correct answer among the four given options.
Let’s discuss what is the ${n^{th}}$ roots of unity.
As we know that the equation having a power of $'n'$ over the variable will have $'n'$ number of roots. So when we consider the equation: ${x^n} = 1$ , here we know that the variable ‘x’ can obtain ‘n’ number of roots. And these are hence called the ${n^{th}}$ roots of unity.
So, according to the question $1,{a_1},{a_2}.....,{a_{n - 1}}$ are the ‘n’ roots of the equation ${x^n} = 1$ , this can be represented by:
$ \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 $1,{a_1},{a_2}.....,{a_{n - 1}}$ as the roots of this equation, we get:
$ \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 $\left( {x - 1} \right)$ to the left side
$ \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, ${x^n} - 1 = \left( {x - 1} \right)\left( {{x^{n - 1}} + {x^{n - 2}} + ...... + {x^2} + x + 1} \right)$ . We can use this in the above equation:
$ \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:
$ \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 $\left( {1 - {a_1}} \right)\left( {1 - {a_2}} \right)......\left( {1 - {a_{n - 1}}} \right)$on the right side of the equation, we must substitute the value $x = 1$ on both sides of the above equation.
For $x = 1$, we get:
$ \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 ${1^m} = 1$ , so on further solving it, we have:
$ \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:
$ \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. ${x^n} - 1 = \left( {x - 1} \right)\left( {{x^{n - 1}} + {x^{n - 2}} + ...... + {x^2} + x + 1} \right)$ is derived using the theorem, i.e. ${x^n} - {y^n} = \left( {x - y} \right)\left( {{x^{n - 1}} + {x^{n - 2}}y + {x^{n - 3}}{y^2} + ..... + x{y^{n - 2}} + {y^{n - 1}}} \right)$ where we put $y = 1$ .