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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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$$z^n - 1 = (z-1)(z^{n-1} + \cdots + z + 1) = (z-1)\prod_{i=2}^n(z-a_i)$$

$$\Rightarrow z^{n-1} + \cdots + z + 1 = \prod_{i=2}^n(z-a_i) \stackrel{z=1}{\Rightarrow} n = \prod_{i=2}^n(1-a_i)$$

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  • $\begingroup$ In the first equation in the second $=$ sign, why does $z^{n-1}+...+z+1=(z-a_2)\cdot\dots\cdot (z-a_n)$? $\endgroup$ – J. Doe Mar 20 at 15:29
  • $\begingroup$ @J.Doe : The equality in the first line holds because of the fundamental theorem of algebra: 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 $z_0, \ldots , z_m$) $\endgroup$ – trancelocation Mar 20 at 16:33
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L'hopital's (sp?) rule applied to $\lim_{x \to 1} \dfrac{x^n-1}{x-1}$.

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