Complex Finite Product $\prod_{k=0}^{n-1} (1-\zeta^k z)$

Question

I am working on a review for a graduate level Complex Analysis course. The following problem is on the review:

Let $\zeta= e^{\frac{2\pi i}{n}}$ $(n\in \mathbb{N})$; show that

$\displaystyle{\prod_{k=0}^{n-1} (1-\zeta^k z)=1-z^n}$

$\displaystyle{\prod_{k=1}^{n-1} (1-\zeta^k z)=1+z\ +...+\ z^n}$

I have proved it to myself for $n=3$ and recognized a "telescoping" behavior if you will. I know I have seen this formula before but can't remember where. Could someone please either point me in the right direction or help prove it? Thank you for your help.

Answer 1

Hints: For your first formula, notice that $$(1-\zeta^kz) = (1-\overline{\zeta^{n-k}}z),$$ so that their product gives $1 - 2\Re(\zeta^k) + z^2$. Now you only have to separate the cases where $n$ is even or odd, and do some dirty work.

The second formula follows easily from the first by division of polynomials.

Answer 2

Just note this the first one nothing but the factorization of the of the polynomial

$$z^n-1=0\,. $$

Can you solve it? Here is how you find the roots

$$ z^n = e^{2\pi k i} \implies z = e^{2\pi k i/n}\, , \quad k=0,1,2,\dots,n-1$$

You should be able to figure out the other one!

Answer 3

For the first formula, notice that $p(z)=1-z^n$ and $q(z)=\prod_{k=0}^{n-1} (1-\zeta^k z)$ are both polynomials of degree $n$ with exactly the same roots because $z^n=1$ iff $z^{-n}=1$.

To conclude that $p(z)=q(z)$ we need to prove that they have the same leading coefficient, $-1$.

The leading coefficient $q_n$ of $q(z)$ is $(-1)^n \prod_{k=0}^{n-1} \zeta^k$.

But $\prod_{k=0}^{n-1} \zeta^k$ is the product of the roots of $p(z)$, and so is $(-1)^n\dfrac{1}{-1}=(-1)^n(-1)$ by Vieta's formulas.

Therefore, $q_n = (-1)^n (-1)^n (-1) = -1 = p_n$.

The second formula follows from the first: $$ 1-z^n =\prod_{k=0}^{n-1} (1-\zeta^k z) =(1-z)\prod_{k=1}^{n-1} (1-\zeta^k z) $$ and so $$ \prod_{k=1}^{n-1} (1-\zeta^k z) =\frac{1-z^n}{1-z}=1+z\ +\cdots+\ z^n $$

Answer 4

It was remarked that we can easily derive the second formula from the first by polynomial division, therefore we will prove the first formula here using what appears to be a somewhat exotic method, namely the Polya Enumeration Theorem (PET).

Given that $$\rho = \exp(2\pi i/n) \quad\text{we seek to show that}\quad f(z) = \prod_{k=0}^{n-1} (1-\rho^k z) = 1 - z^n.$$

Suppose $Z(P_n)$ is the cycle index of the set operator $\mathfrak{P}_{=n}$ given by the recurrence by Lovasz which is

$$Z(P_n) = \frac{1}{n} \sum_{l=1}^n (-1)^{l+1} a_l Z(P_{n-l}) \quad\text{where}\quad Z(P_0) = 1.$$

Now observe that $$[z^q] f(z) = (-1)^q \left. Z(P_q)(1+v+v^2+\cdots+v^{n-1}) \right|_{v=\rho}.$$

Next recall the OGF of the set operator $\mathfrak{P}_{=q}$ which is (this is the so-called exponential formula)

$$Z(P_q) = [w^q] \exp\left(a_1 w - a_2 \frac{w^2}{2} + a_3 \frac{w^3}{3} - a_4 \frac{w^4}{4} +\cdots \right).$$

or

$$Z(P_q) = [w^q] \exp\left(\sum_{r\ge 1} (-1)^{r+1} a_r \frac{w^r}{r}\right).$$

Nóte that $$\left.a_r\right|_{v=\rho} = \left.(1+v^r+v^{2r}+\cdots+v^{(n-1)r})\right|_{v=\rho} = n \times [[n|r]].$$

Therefore $$[z^q] f(z) = (-1)^q [w^q] \exp\left(\sum_{r\ge 1} (-1)^{nr+1} n \frac{w^{nr}}{nr}\right) \\ = (-1)^q [w^q] \exp\left(-\sum_{r\ge 1} \frac{(-w)^{nr}}{r}\right) \\ = (-1)^q [w^q] \exp\left(-\log\frac{1}{1-(-w)^n}\right) = (-1)^q [w^q] (1-(-w)^n).$$

It now follows that the constant coefficient (on $q=0$) is one, the coefficients on $[z^q]$ where $1\le q\lt n$ are zero, and the coefficient on $[z^n]$ i.e. $q=n$ is $$(-1)^n [w^n] (-(-w)^n) = - (-1)^{2n} = -1$$

and hence $$f(z) = 1 - z^n.$$

Commentary. The recurrence for the cycle index is included here for reference and can be used to prove the exponential formula and vice versa. We have also made use of the fact that for $r$ a positive integer

$$\sum_{q=0}^{n-1} \rho^{rq} = n \quad\text{when}\quad n|r.$$

On the other hand when $r$ is not a multiple of $n$ we get $$\frac{\rho^{rn}-1}{\rho^r-1} = 0.$$