$P(z)=(z^n-1)$ has roots $1, a_1,a_2,...,a_{n-1} $. Prove that $(1-a_1)(1-a_2)...(1-a_{n-1})=n$

Question

$P(z)=(z^n-1)$ has roots $1, a_1,a_2,...,a_{n-1} $. Prove that $(1-a_1)(1-a_2)...(1-a_{n-1})=n$

No idea how to prove this. Would really appreciate any help, hints...

Thanks in advance :)

Answer 1

Divide $z^n-1$ by $z-1$ using Ruffini. You will probably get $1+z+z^2+...+z^{n-1} $ if im not wrong (power series). This should factorize as $(z-a_1)... (z-a_{n-1}$ so you hace to evaluate on $z=1$. You will get $(1-a_1)... (1-a_{n-1})$ on one side and $1+1+1^2+...=n $ on the other side

Answer 2

Since

$P(z)= z^n - 1, \tag 1$

we have

$P(z) = (z - 1)(z^{n - 1} + z^{n - 2} + \ldots + z + 1) = (z - 1)Q(z), \tag 2$

where

$Q(z) = z^{n - 1} + z^{n - 2} + \ldots + z + 1 = \sum_0^{n - 1} z^i \tag 3$

it then follows that if $1 \ne b \in \Bbb C$ satisfies

$P(b) = (b - 1)(b^{n - 1} + b^{n - 2} + \ldots + b + 1) = (b - 1)Q(b) = 0, \tag 4$

we must have

$b^{n - 1} + b^{n - 2} + \ldots + b + 1 = \sum_0^{n - 1} b^i = Q(b) = 0; \tag 5$

thus any root $b$ of $P(z)$ other than $1$ must satisfy $Q(z)$; now $\deg Q(z) = n - 1$, so it has at most $n - 1$ zeroes, so if we assume the list of roots of $P(z)$,

$1, a_1, a_2, \ldots, a_{n - 1}, \tag 6$

is irredundant, that is, does not repeat itself (as seems to be the intention of our OP kjhg), then the zeroes of $Q(z)$ are the $a_i$, $1 \le i \le n - 1$. Thus we may write

$Q(z) = \displaystyle \prod_1^{n - 1} (z - a_i); \tag 7$

comparing (3) and (7), we see that

$\displaystyle \prod_1^{n - 1} (z - a_i) = z^{n - 1} + z^{n - 2} + \ldots + z + 1 = \sum_0^{n - 1} z^i;\tag 8$

taking $z = 1$ in (8), we reach

$\displaystyle \prod_1^{n - 1} (1 - a_i) = 1^{n - 1} + 1^{n - 2} + \ldots + 1^1 + 1 = n, \tag 8$

the desired result.