Prove that$(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$

Question

Prove that $(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$ where $\omega$ is the cube root of unity

My attempt:

$(1+\omega^n+\omega^{2n})=0$ $\Rightarrow (1-\omega^n+\omega^{2n})=-2\omega^n$ $\Rightarrow \prod_{n=1 \to 2n}(-2\omega^n) =2^{2n}\omega^{1+2+3…2n}$ $\Rightarrow \prod_{n=1 \to 2n}(-2\omega^n) =2^{2n}\omega^{n(2n+1)}$

If $n$ is $3k$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{3m}=2^{2n}$

If $n$ is $3k+1$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{3m}=2^{2n}$

If $n$ is $3k+2$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{(3k+2)(6k+5)} =2^{2n}\omega^{(3m+1)}=2^{2n}\omega^{3m}\omega=2^{2n}\omega$

What have I done wrong here?

Answer 1

Hint : Evaluate first few factors and see. $$ 1st : (1-\omega+\omega^2)=-2\omega \\ 2nd : (1-\omega^2+\omega^4)=-2\omega^2 \\ 1st \times 2nd = \ldots \\ 3rd : (1-\omega^4+\omega^8)=-2\omega \\ 4th : (1-\omega^8+\omega^{16})= \ldots $$

Can you complete?

Answer 2

Since $$a^3+b^3=(a+b)(a^2-ab+b^2),$$ we obtain: $$\prod_{k=1}^{2n}(1-w^{2^{k-1}}+w^{2^{k}})=\prod_{k=1}^{2n}\frac{\left(1+w^{3\cdot2^{k-1}}\right)}{1+w^{2^{k-1}}}=\frac{2^{2n}}{((1+w)(1+w^2))^n}=2^{2n}.$$

Answer 3

Since $\omega$ is the third root of unity, and also since $2^{2n}-1\equiv 0 \mod{3}$

$$ \begin{align} 1+\omega^{2^{i}}+\omega^{2^{i+1}}&=0\\ \\ 1-\omega^{2^{i}}-\omega^{2^{i+1}}&=2\omega^{2^{i}}\\ \\ \prod_{i=0}^{2n-1}{\left(1-\omega^{2^{i}}+\omega^{2^{i+1}}\right)}&=\prod_{i=0}^{2n-1}{\left(2\omega^{2^{i}}\right)}\\ &=2^{2n}\omega^{2^{2n}-1}\\ &=2^{2n} \end{align} $$