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My first post here

So I was doing some past year questions and this one popped out and I haven't been able to progress much upon it and I believe there's a trick which will get it done in no time.

If the roots of the equation $x^n-1=0$ are $1,\alpha,\beta,\gamma,\cdots$

we need to show that $$(1-\alpha)(1-\beta)(1-\gamma)\cdots= n$$

Can I be helped with some hints to push me in the right direction to solve it?

Thanks in advance.

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4 Answers 4

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Apparently

$ x^n-1 = (x-1)(x-\alpha)(x-\beta)(x-\gamma)... $

setting $P(x) = \tfrac{x^n-1}{x-1}$

$ P(x)= (x-\alpha)(x-\beta)(x-\gamma)... $

Now just evaluate at x = 1.

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  • $\begingroup$ This seems to be the 'trick' i was exactly looking for, thanks $\endgroup$
    – Anthony
    Mar 11, 2021 at 7:56
  • $\begingroup$ @Arthur Yes, it works out fine though, since $P(x)$ has a closed polynomial form. $\endgroup$
    – denklo
    Mar 11, 2021 at 8:01
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    $\begingroup$ @Arthur that does not require limits: $(x^n-1)/(x-1) = x^{n-1} + x^{n-2} + \cdots + x + 1$ as polynomials and the value of the right side at $x=1$ is $n$. Polynomials equal at infinitely many numbers are equal everywhere because a nonzero polynomial has only finitely many roots. $\endgroup$
    – KCd
    Mar 11, 2021 at 8:01
  • $\begingroup$ @KCd You're right. I missed that one. $\endgroup$
    – Arthur
    Mar 11, 2021 at 8:22
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Denote the roots by $x_0 = 1, x_1, x_2, x_3, \dots, x_{n-1}$
(roots of unity, usual indexing in counter-clockwise direction)

Then $x_k = x_1^k$

So this whole product is an expression involving just $x_1$

And from the roots of unity theory we know that

$$x_1 = \cos\frac{2\pi}{n} + i \cdot \sin\frac{2\pi}{n}$$

Proceeding from here is a matter of doing some simple algebra.

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  • $\begingroup$ $x_k = x_1^k$ really depends on which of the roots you call $\alpha$. It isn't necessarily the first counterclockwise root after $1$. I mean, having $\alpha$ be any other root would be confusing and obfuscating. But it should be clarified. $\endgroup$
    – Arthur
    Mar 11, 2021 at 7:54
  • $\begingroup$ @Arthur Yes. I call $x_1$ the usual one, the one from my formula. $\endgroup$ Mar 11, 2021 at 7:55
  • $\begingroup$ @peter.petrov My point is, you don't know that that's $\alpha$. And you say $x_1=\alpha$. It's a little pedantic, but I think it ought to be clarified. $\endgroup$
    – Arthur
    Mar 11, 2021 at 7:56
  • $\begingroup$ @Arthur Yeah, it's pedantic. I removed any references to $\alpha$ anyway. $\endgroup$ Mar 11, 2021 at 8:04
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Like Finding $\sum_{k=0}^{n-1}\frac{\alpha_k}{2-\alpha_k}$, where $\alpha_k$ are the $n$-th roots of unity OR Problem based on sum of reciprocal of $n^{th}$ roots of unity

using Transformation of equations, let $1-x=y$

$$\implies(1-y)^n=1\iff y^n-\binom n1y^{n-1}+\cdots+\binom n{n-1}(-1)^{n-1}y=0$$ whose roots are $y_k, 0\le k\le n-1$

Clearly, one of the them is $0,$ let $y_0=0$

So, the roots of $$y^{n-1}-\binom n1y^{n-2}+\cdots+\binom n{n-1}(-1)^{n-1}=0$$ will be $y_k, 1\le k\le n-1$

Finally use Vieta's formula

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If the roots are $r_i$, $i=1\ldots n$, with $r_1=1$, we have

$\prod\limits_{i=1}^{n}r_i= -(-1)^n$

$\sum\limits_{i=1}^{n}r_i = \sum\limits_{i<j}^{n}r_ir_j = \sum\limits_{i<j<k}^{n}r_ir_jr_k =\ldots= 0$ (till taking $n-1$ terms at a time)

Now, $\sum\limits_{i=2}^{n}r_i=\sum\limits_{i=1}^{n}r_i-1=0-1=-1$

Also, $\sum\limits_{i,j=2,i<j}^{n}r_ir_j=\sum\limits_{i,j=1,i<j}^{n}r_ir_j-\sum\limits_{i=2}^{n}r_i=0-(-1)=1$

We can show that the terms $\sum\limits_{i=2}^{n}r_i, \sum\limits_{i,j=2,i<j}^{n}r_ir_j,\sum\limits_{i,j,k=2,i<j<k}^{n}r_ir_jr_k, \ldots$ have alternating values $-1$ and $1$

$\therefore (1-\alpha)(1-\beta)(1-\gamma)\ldots$

$=\prod\limits_{i=2}^{n}(1-r_i)$

$=1-\sum\limits_{i=2}^{n}r_i+\sum\limits_{i,j=2}^{n}r_ir_j-\ldots$

$=1-(-1)+1-(-1)+\ldots=n$

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