If $\{ z_k\}_{0\leq k < n}$ are the $n$-th roots of a unity with $z_0 = 1$ prove that $n = \prod (z_0-z_k)$

Question

Let $z_k$ be a $n$-th root of the unity, i.e.

$$z_k = e^{2\pi i\frac{k}{n}}\\ k\in \{0,1,\cdots,n-1\}$$

Prove that $$(z_0 - z_1)(z_0 - z_2)\ldots (z_0 - z_{n-1})= n$$

This problem was at the end of the introduction chapter of a book about complex numbers, way before talking about poles and residues. I thought of doing it by induction but for $n+1$ we don't have the same $z_k$ as for $n$.

What surprises me the most is the fact that $z_k = \bar{z}_{n-k}$ So:

$$(z_0 - z_k)(z_0 - z_{n-k}) = 2 - 2\Re(z_k)$$

Thus the total product becomes:

$$2^{\lceil\frac{n}{2}\rceil}\prod_{k=1}^{\lceil\frac{n}{2}\rceil}(1-\Re(z_k))\stackrel{?}{=}n$$ Any hints?

Chen Jiang · Accepted Answer · 2014-08-04T15:17:14.187

6

We know that $$ (x-1)(x-z_1)\ldots(x-z_{n-1})=x^n-1. $$ Hence $$ (x-z_1)\ldots(x-z_{n-1})=\frac{x^n-1}{x-1}=x^{n-1}+x^{n-2}+\ldots+x+1. $$ If take $x=1$, then $$ (1-z_1)\ldots(1-z_{n-1})=n. $$

edited Aug 04 '14 at 15:17

answered Aug 04 '14 at 14:51

Chen Jiang

1,393

1

maybe add that $\frac{x^n-1}{x-1}=x^{n-1}+x^{n-2}+ \cdots +x +1$. – Nicky Hekster Aug 04 '14 at 14:56
I don't know where you got that from, and if $x=z_0=1$ to get my product, then I get $\frac{0}{0}$! – Darth Geek Aug 04 '14 at 14:57
@DarthGeek see the comment above you. – Chen Jiang Aug 04 '14 at 15:02
@ChenJiang I see it, I don't see the relation, tho. Care to elaborate a bit more? – Darth Geek Aug 04 '14 at 15:10
What I didn't get was that $z_k$ are the roots of the polynomial $x^n - 1$. I get it now, thanks. – Darth Geek Aug 04 '14 at 15:24
@DarthGeek you are welcome :) – Chen Jiang Aug 04 '14 at 15:25

If $\{ z_k\}_{0\leq k < n}$ are the $n$-th roots of a unity with $z_0 = 1$ prove that $n = \prod (z_0-z_k)$

1 Answers1