For $1<i\leq n$, let $a_i$ be the n-th roots of $1\in\mathbb{C}$, why does $(1-a_2)\cdot...\cdot(1-a_n)=n$?
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1https://math.stackexchange.com/questions/3038472/if-x-1-x-2-ldots-x-n-are-the-roots-for-1xx2-ldotsxn-0-find-the-value/3038481#3038481 – lab bhattacharjee Mar 20 '19 at 14:22
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You can use the derivative of $x^n-1$ to show the slightly more general rule that for any $j-1,2,\dots,n:$ $$na_j^{-1}=\prod_{i\neq j} (a_j-a_i).$$ – Thomas Andrews Mar 20 '19 at 14:33
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$$z^n - 1 = (z-1)(z^{n-1} + \cdots + z + 1) = (z-1)\prod_{i=2}^n(z-a_i)$$
$$\Rightarrow z^{n-1} + \cdots + z + 1 = \prod_{i=2}^n(z-a_i) \stackrel{z=1}{\Rightarrow} n = \prod_{i=2}^n(1-a_i)$$
trancelocation
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In the first equation in the second $=$ sign, why does $z^{n-1}+...+z+1=(z-a_2)\cdot\dots\cdot (z-a_n)$? – J. Doe Mar 20 '19 at 15:29
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@J.Doe : The equality in the first line holds because of the fundamental theorem of algebra: https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra . The second line follows for $z \neq 1$. But as we have polynomials on both sides which are equal for all $z \neq 1$, then they are also equal for $z = 1$. (Note that two polynomials of degree $m$ are identical iff they are equal at $m+1$ points $z_0, \ldots , z_m$) – trancelocation Mar 20 '19 at 16:33
