# 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 $$(1 - \alpha_1)(1 - \alpha_2)(1 - \alpha_3) \cdots (1 - \alpha_9)?$$

• Hint: How does $x^{10}+1=0$ factor over $\mathbb{C}$? – Justin Benfield Aug 31 '16 at 5:29
• – lhf Aug 31 '16 at 14:50

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?

• This is a very nifty solution, +1. – 6005 Aug 31 '16 at 6:11
• Glad you liked it @6005. There are (at least) three approaches to this question. I recall thinking about each and every one of them as "nifty" the first time I saw them :-) – Jyrki Lahtonen Aug 31 '16 at 8:41
• As instructive as pedagogically valuable! Nice little gem! (+1) – Markus Scheuer Aug 31 '16 at 18:00

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

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

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?

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

Here is an other approach. You have this result:

Roots to any polynomial with real coefficients are either real or come in complex conjugate pairs.

We have one real root, -1, the contribution from -1 would be multiplication by $(1-(-1))=2$.

Then let us consider the product of any complex conjugate pair $$(1-z)(1-z^*) = (1+R(z)^2+I(z)^2-2R(z))=2(1-R(z))$$ Where in the last step we used the $R(z)^2+I(z)^2 = 1$ which is true on the unit circle.

Now consider the distribution of these $R(z)$ on our circle. Maybe there is some way we can arrange them to make the pairwise product simpler. We had 8 roots, that means 4 complex conjugate pairs.

But we can also pair these pairs up so we have 2 pairs of pairs of opposite real signs. The real parts are cosine values. So the pairwise products become $1-\cos^2$ which should ring a trig-one bell and then we are rather close to a solution!