Evaluate $\prod_{k=1}^{k=2n}[1+w^k]$, when $\omega$ is a cube root of unity

Question

My work:

I wrote it in this form $[(1+w)(1+w^2)(1+w^3)]^k$ where k is the number of repeat of this triplet in a product.

Every three things consecutive factors of $(1+w) ( 1+w^2) ( 1+w^3)$ count as one power to the exponent. I wrote $ k = \frac{q}{3}$ where $q$ is a number of terms in product, so I can group these products which give same value .

Here it is $q=2n$, so

$$ [ (1+w)(1+w^2)(1+w^3)]^{\frac{2n}{3}}$$

But this identity definitely doesn't seem to work for some powers like $n=2$, since

$$ (1+w)(1+w^2)(1+w^3)(1+w^4) = [(1+w)(1+w^2)^{\frac{1}{2}}(1+w^3)^{\frac{1}{2}}]^2$$ which doesn't match with derived identity

You've got to treat the cases $n=3s, 3s+1, 3s+2$ separately I reckon. I get (I've not checked) $2^{2s}, 2^{2s}, -2^{2s+1}\omega^2$. — ancient mathematician, May 20 '20 at 12:57

score 1 · Accepted Answer · answered May 20 '20 at 14:39

We have $$\begin{align} &(1+w)(1+w^2)(1+w^3)\\ =&(1+w)(1+\overline{w})(1+1)\\ =&2(1+w+\overline{w}+w\overline{w})\\ =&2(2+2\Re{w})\\ =&4\left(1+\cos{\frac{2\pi}{3}}\right)\\ =&2 \end{align} $$ Therefore, if $2n=3s+m,\ 0\leq m<3$ we have $$\prod_{k=1}^{2n}(1+w^k)=2^s\prod_{k=1}^{m}(1+w^k)=\begin{cases}2^s,&m\neq1\\2^{s-1}(1+i\sqrt3),&m=1 \end{cases}$$

Evaluate $\prod_{k=1}^{k=2n}[1+w^k]$, when $\omega$ is a cube root of unity

My work:

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