[Math] Why is $ w^{2n} + w^n + 1 = 0 $

complex numbers

I know that $w^2 + w + 1 =0$ is true for the cube roots of unity, but why is $ w^{2n} + w^n + 1 = 0 $? I have seen this result in various websites but none of them has the derivation. The source is askiitians.com/iit-jee-algebra/complex-numbers/… its there under "Now, we consider some of the key results on nth roots of unity-". How is this derived? I tried substituting (-1-w) for w^2 to get $(-1-w)^{2n} + w^n + 1 = 0$ but that just complicates things.

Best Answer

Let $w=e^{\frac{2\pi i}{3}}$ and $\bar{w}=e^{-\frac{2\pi i}{3}}$ (which both are solutions to $x^2+x+1=0$), then consider $w^n$.

  • If $3\mid n$ then $n=3k$. Thus $w^n=e^{2\pi i k}=1$ and $w^{2n}+w^n+1=0$ is false.
  • If $n=3k\pm1$ then $w^n=e^{2\pi i k\pm\frac{2\pi i}{3}}=e^{\pm\frac{2\pi i}{3}}$. Thus if $w^n=w$, $w^{2n}=\bar w$, and vice versa. In either case you get $w+\bar w+1=0$ or $\bar w+w+1=0$, which are both true.

Another, perhaps easier, way to see why $$w^{2n}+w^{n}+1=0\tag{1}$$ is true when $3\not\mid n$ is to rewrite $(1)$ as $(w^n)^2+(w^n)^1+(w^n)^0$. Then it becomes clear that this is a geometric series which equals $$\frac{(w^n)^3-1}{w^n-1}=w^{2n}+w^n+1\tag{2}$$ Now $w$ is a third root of unity, so $(w^n)^3=1$, and $w^n\neq 1\iff3\not\mid n$. From this eq. $(2)$ gives the desired result.

Notice that this last approach generalizes the result:

Let $p$ be a prime number, and $\theta\in\mathbb{C}$ a $p$th root of unity, that is $\theta^p=1$. If $\theta\neq 1$ then $$\theta^{(p-1)n}+\ldots+\theta^2+\theta+1=0\iff p\not\mid n$$

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