Nth Roots of Unity Challenge Problem

[sbhatnagar]

Challenge Problem $1,a_1,a_2,a_3, \cdots ,a_{n-1}$ are the $n^{\text{th}}$ roots of unity.

Find the value of

i) $(1-a_1)(1-a_2)(1-a_3) \cdots (1-a_{n-1})$

ii)$\displaystyle \frac{1}{2-a_1}+\frac{1}{2-a_2}+\frac{1}{2-a_3}+\cdots +\frac{1}{2-a_{n-1}}$

 

[chisigma]

Challenge Problem $1,a_1,a_2,a_3, \cdots ,a_{n-1}$ are the $n^{\text{th}}$ roots of unity.

Find the value of

i) $(1-a_1)(1-a_2)(1-a_3) \cdots (1-a_{n-1})$

ii)$\displaystyle \frac{1}{2-a_1}+\frac{1}{2-a_2}+\frac{1}{2-a_3}+\cdots +\frac{1}{2-a_{n-1}}$

Is...

$\displaystyle f(x)=\frac{x^{n}-1}{x-1}= 1 + x + x^{2} + ... + x^{n-1} = \prod_{k=1}^{n-1} (x-a_{k})$ (1) ... so that the answer to i) is $f(1)=n$. The answer to ii) will be given in a successive post...

Kind regards

$\chi$ $\sigma$

 

[Opalg]

Challenge Problem $1,a_1,a_2,a_3, \cdots ,a_{n-1}$ are the $n^{\text{th}}$ roots of unity.

Find the value of

i) $(1-a_1)(1-a_2)(1-a_3) \cdots (1-a_{n-1})$

ii)$\displaystyle \frac{1}{2-a_1}+\frac{1}{2-a_2}+\frac{1}{2-a_3}+\cdots +\frac{1}{2-a_{n-1}}$

As chisigma has noted, the roots of $f(x)=\dfrac{x^{n}-1}{x-1}$ are $a_1,a_2,a_3, \ldots ,a_{n-1}$. The roots of $$f(2-x)=\frac{(2-x)^{n}-1}{(2-x)-1} = \bigl((2-x)^{n}-1\bigr)(1-x)^{-1} = \bigl((2-x)^{n}-1\bigr)(1+x+x^2+\ldots)$$ are $(2-a_1),(2-a_2),(2-a_3), \ldots ,(2-a_{n-1})$. The sum of the reciprocals of these roots will be the negative of the coefficient of $x$ in that last expression, divided by the constant term. Therefore $$ \frac{1}{2-a_1}+\frac{1}{2-a_2}+\frac{1}{2-a_3}+\ldots +\frac{1}{2-a_{n-1}} = \boxed{\dfrac{n2^{n-1} -2^n+1}{2^n-1}}.$$

You can check that result by working out the value of the sum when $n=4$. In that case, $a_1,a_2,a_3$ are $-1,i,-i$, and the sum is $\dfrac13+\dfrac1{2-i}+\dfrac1{2+i}$, which simplifies to $\dfrac{17}{15}$, in accordance with the boxed formula.

 

[sbhatnagar]

Here is another solution for 2!

We have
$$ \frac{x^n-1}{x-1}=(x-a_1)(x-a_2)(x-a_3)\cdots (x-a_{n-1}) $$
Let $f(x)=\log\frac{x^n -1}{x-1}$
$$f(x)=\log\dfrac{x^n-1}{x-1}=\log(x-a_1)+\log(x-a_2)+\log(x-a_3)+\cdots +\log(x-a_{n-1})$$
The derivative of $f(x)$ will be
$$f'(x)=\dfrac{ \left[n x^{-1+n}-\dfrac{-1+x^n}{(-1+x)}\right]}{-1+x^n}=\frac{1}{x-a_1}+\frac{1}{x-a_2}+\frac{1}{x-a_3}+\cdots +\frac{1}{x-a_{n-1}}$$

putting $x=2$

$$f'(2)=\frac{n 2^{n-1}-2^n+1}{2^n-1}=\frac{1}{2-a_1}+\frac{1}{2-a_2}+\frac{1}{2-a_3}+\cdots +\frac{1}{2-a_{n-1}}$$

 

[CellCoree]

I would approach this problem by first understanding what the nth roots of unity are. These are complex numbers that, when raised to the nth power, result in the value of 1. In other words, they are solutions to the equation $x^n = 1$.

With this knowledge, I can begin to tackle the challenge problem. For part i), I would start by expanding the expression using the distributive property. This would result in a sum of terms, each of which is a product of the form $(1-a_i)$. Since $a_i$ are the nth roots of unity, we know that $a_i^n = 1$. Using this fact, we can simplify each term to just $1-1=0$. Therefore, the entire expression simplifies to 0.

For part ii), I would start by finding a common denominator for all the fractions. Since each denominator is of the form $2-a_i$, we can multiply all the fractions by the product of these denominators. This results in a sum of terms, each of which is of the form $\frac{2-a_i}{(2-a_i)(2-a_j)...(2-a_{n-1})}$. Simplifying this, we get $\frac{1}{(2-a_j)...(2-a_{n-1})}$ for each term. Since the product in the denominator is just a constant, we can pull it out of the summation. This leaves us with $\frac{1}{(2-a_1)(2-a_2)...(2-a_{n-1})}$ as the final answer.

In conclusion, the value of $(1-a_1)(1-a_2)(1-a_3) \cdots (1-a_{n-1})$ is 0 and the value of $\displaystyle \frac{1}{2-a_1}+\frac{1}{2-a_2}+\frac{1}{2-a_3}+\cdots +\frac{1}{2-a_{n-1}}$ is $\frac{1}{(2-a_1)(2-a_2)...(2-a_{n-1})}$. These solutions highlight the unique properties of the nth roots of unity and demonstrate their usefulness in solving mathematical problems.