# Product of roots of a polynomial decreased from one

I got a question recently, and have been unable to solve it.

If $1,\alpha_1,\alpha_2,\dots,\alpha_{n-1}$ are the roots of $X^n-1=0$, then find the value of $\prod{(1-\alpha_i)}$.

My solution:

The product will expand to $$1-\sum_{cyc}\alpha_i+\sum_{cyc}\alpha_i\alpha_{i+1}-\dots$$

This means I can use the relations of roots of a polynomial to its coefficients, thus getting this equal to

$1+0+0+\dots+0+0\pm1$, depending on whether $n$ is even or odd.

But the book gives the answer as $n$.

How does this happen? Please help.

## 2 Answers

The main flaw in your argument is that $\sum_{i=1}^{n-1} \alpha_i$ is not the sum of the roots of the polynomial: you are missing the root 1. Similarly $\sum_{i=1}^{n-1}\sum_{j=1}^{n-1}\alpha_i\alpha_j \neq 0.$

Here's a hint: given that the $\alpha_i$ are roots of $x^n-1$, can you write down a polynomial whose roots are $1-\alpha_i$?

• Oh. I get it. I missed adding the $1$. Now, that's embarrassing. Thanks! – MalayTheDynamo Jul 28 '18 at 8:17

Notice that

$$\prod_{i=1}^{n-1} (X - \alpha_i) = \frac{X^n-1}{X-1} = X^{n-1} + X^{n-2} + \cdots + X + 1$$

so for $X = 1$ we get

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

Your approach can also be used but notice that $\alpha_1, \ldots, \alpha_{n-1}$ are roots of $X^{n-1} + X^{n-2} + \cdots + X + 1$, and not $X^{n}-1$ because the root $1$ is missing.

So

$$\prod_{i=1}^{n-1}(1-\alpha_i) = 1-\sum_{cyc}\alpha_i+\sum_{cyc}\alpha_i\alpha_{i+1}-\cdots = 1 - (-1) + 1 - \cdots = n$$

• What do you mean by $X-\alpha_n$? Is it $\alpha_i$? And is this true for all $X,\alpha$? – MalayTheDynamo Jul 28 '18 at 8:19
• @MalayTheDynamo Yes, $\alpha_i$ of course. – mechanodroid Jul 28 '18 at 8:19