# Prove that $z^{2n} - 2 a^n z^n \cos (nθ) + a^{2n}=\prod_{k=0}^{n-1}\left[z^2-2az\cos\left(\theta+\frac{2\pi k}{n}\right)+a\right]$

Prove that $$z^{2n} - 2 a^n z^n \cos (nθ) + a^{2n}= \prod_{k=0}^{n-1}\left[z^2-2az\cos\left(\theta+\frac{2\pi k}{n}\right)+a\right].$$

I personally know a similar question stating that $$z^{2n}+1=\prod_{k=0}^{n-1}\left[z^2-2z\cos\left(\frac{(2k+1)\pi}{2n}\right)+1\right].$$

And we can show this by using the roots of $$z^{2n}+1=0$$, i.e., multiplying each $$(z-z_{n})$$ and correspond each term to its conjugate.

However, I found this approach not that practical to be used in the above more generalised situation. I also tried Induction, and it turns out that it will be very tedious to right out all the terms.

First, it suffices to prove this identity for $$a=1$$ (why?)
Then, we need to prove the following polynomial identity: $$z^{2n}-2\cos nθ\cdot z^n+1=\prod_{k=0}^{n-1}(z^2-2\cos(\theta+2\pi k/n)\cdot z+1).$$ Now note that $$z^2-2\cos\varphi\cdot z+1=(z-e^{i\varphi})(z-e^{-i\varphi})$$ Can you continue now?
Here is an answer that I have attained eventually. Apply the factorization $$z^n-b^n=\prod_{k=0}^{n-1}(z- b e^{i\frac{2\pi k}n})$$ to \begin{align} z^{2n} - 2 a^n z^n \cos nθ+ a^{2n} =& \left[z^n - \left(a e^{i\theta}\right)^n \right]\left[z^n - \left(a e^{-i\theta}\right)^n \right]\\ =& \prod_{k=0}^{n-1}\left[z- a e^{i(\theta+\frac{2\pi k}n)}\right]\prod_{k=0}^{n-1}\left[z- a e^{-i(\theta+\frac{2\pi k}n)}\right] \\ =& \prod_{k=0}^{n-1}\left[z^2-2az\cos\left(\theta+\frac{2\pi k}{n}\right)+a\right] \end{align}