Product of one minus the tenth roots of unity [closed]

If $1$, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots \alpha_9$ are the $10$th roots of unity, then what is the value of $$ (1 - \alpha_1)(1 - \alpha_2)(1 - \alpha_3) \cdots (1 - \alpha_9)? $$

I am not being able to solve this. Please help!

Solution 1:

Let $$f(x)=x^{10}-1=(x-1)(x-\alpha_1)\cdots(x-\alpha_9).$$ Then $$ f'(x)=10x^9=\sum_{i=0}^9\prod_{0\le j\le9, j\neq i}(x-\alpha_j) $$ (where I denote $\alpha_0=1$).

Can you calculate $f'(1)$? Do you see why it answers your question?

Solution 2:

Hint: Let $ \mu_n $ be the set of $ n $th roots of unity. Then, we have

$$ \prod_{\zeta \in \mu_n} (x - \zeta) = x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \ldots + x + 1) $$

and therefore

$$ \prod_{\zeta \in \mu_n - \{1\}} (x - \zeta) = x^{n-1} + x^{n-2} + \ldots + x + 1 $$

Solution 3:

Let the roots of $x^n-1=0$ be $$1,\alpha_1,\alpha_2,\cdots,\alpha_{n-1}$$

If $1-x=y\iff x=1-y$

So, the equation whose roots are $$1-1,1-\alpha_1,1-\alpha_2,\cdots,1-\alpha_{n-1}$$ will be $$(1-y)^n-1=0\iff(-1)^ny^n+(-1)^{n-1}\binom n1y^{n-1}\cdots-\binom n1y=0$$

So, the roots of $$y^{n-1}-\binom n1y^{n-1}\cdots+(-1)^{n-1}\binom n1=0$$ will be $$1-\alpha_1,1-\alpha_2,\cdots,1-\alpha_{n-1}$$

Can you use Vieta's formula now?

Solution 4:

Hint: What is the product of the roots of the polynomial $(1-x)^{10} - 1$, other than $0$?