Developing a polynomial.

Question

Let $P(x)=\prod_{k=0}^{n-1}(1-w_kX)$ where $w_k=e^{\frac{2k\pi}{n}}$

Attempt to solve:

• for $n =1$ : $P(x)=1-X$

• for $n = 2$ : $P(x)=(1-X)(1+X) = 1-X^{2}$

• for $n = 3$ : $P(x)=(1-X)(1-jX)(1-j^2X) = 1-X^{3}$

By induction:

Assume $P(x)=\prod_{k=0}^{n-1}(1-w_kX)= (1-X^n)$ And prove $\prod_{k=0}^{n}(1-w_kX)= (1-X^{n+1})$

$\prod_{k=0}^{n}(1-w_kX)=(1-X^n)(1-e^{\frac{2n\pi}{n+1}}) = 1 -X^n +e^{\frac{2n\pi}{n+1}}X^{n+1}-e^{\frac{2n\pi}{n+1}}X$

And i have no idea how to continue from here nor i am sure if my assumption is even correct.

Thanks in advance.

Answer 1

You must be careful as $w_k$ depends of $k$ and $n$. So in fact when you write $$\prod_{k=0}^n(1-w_k X)$$ you mean: $$\prod_{k=0}^n(1-e^\frac{2ik \pi}{\color{red}{n+1}} X) = \prod_{k=0}^{n-1}(1-e^\frac{2ik \pi}{\color{red}{n+1}} X) (1-e^\frac{2in \pi}{n+1} X) \neq \prod_{k=0}^n(1-e^\frac{2ik \pi}{\color{red}{n}} X) (1-e^\frac{2in \pi}{n+1} X)$$

for the original question you can notice that $P_n(X)$ and $1-X^n$ have the same degree, the sames roots and the same leading coeeficient.

Answer 2

Just note that for fixed $n$ and variable $k$, $\omega_k^n = 1$, so $X^n - 1$ has all the $\omega_k$ ($n$ distinct quantities) as roots. $X^n - 1$ can have at most $n$ roots because it has degree $n$, and you're done.