I'm having trouble trying to prove this:

Let $ m\in \mathbb Z$, m even and $w\in\mathbb C$ a primitive $2m$-th root of unity. Prove that $(w-1)^m$ is purely imaginary.

What I've tried to do so far is saying that if $(w-1)^m$ is purely imaginary then: $(w-1)^m= -(\overline {w-1)^m}$ and from here use the Binomial Theorem and try to manipulate both expressions to show they're equal. But still no luck.

Thanks for your help!


We have $w=e^{i k\pi/m}$ with $\gcd(k,2m)=1$. Now let's compute

$$\begin{align}w-1&=e^{i k\pi/m}-1\\&=e^{{i k\pi/2m}}\left(e^{{i k\pi/ 2m}}-e^{-{i k\pi/ 2m}}\right)\\&=2e^{{i k\pi/ 2m}}i \sin{{k\pi/2m}}\end{align}$$

And so

$$(w-1)^m=2^me^{{i k\pi/ 2}}i^m \sin^m{{k\pi\over 2m}}$$

Now $e^{{i k\pi/2}}=\pm i $ so $(w-1)^m$ is imaginary when $m$ is even.

  • $\begingroup$ Thanks you this is very helpful, but could you please explain a bit the third step ? $\endgroup$ – Ron Jun 15 '16 at 6:25
  • $\begingroup$ Just using $\sin{\theta}={e^{i \theta}-e^{-i\theta}\over 2i}$ $\endgroup$ – marwalix Jun 15 '16 at 6:28
  • $\begingroup$ Awesome! Thank you now I see it clear! :D $\endgroup$ – Ron Jun 15 '16 at 6:30

WLOG $$\omega=e^{i2\pi k/2m}=\cos\dfrac{\pi k}m+i\sin\dfrac{\pi k}m$$ where $(2k,2m)=1\iff(k,m)=1$

$\omega-1=2i\sin\dfrac{\pi k}{2m}\left(\cos\dfrac{\pi k}{2m}+i\sin\dfrac{\pi k}{2m}\right)$

$(\omega-1)^m=2^m(-1)^{m/2}\sin^m\dfrac{\pi k}{2m}\left(\cos\dfrac{\pi k}2+i\sin\dfrac{\pi k}2\right)$

As $m$ is even, $k$ is odd, $\cos\dfrac{\pi k}2$ will be zero

  • $\begingroup$ Sorry but I really don't understand the w-1=... step :s. Are you using the identity for: Cos (u)-Cos(v) ? $\endgroup$ – Ron Jun 15 '16 at 6:04
  • $\begingroup$ $$\cos2y+i\sin2y-1=2i\sin y\cos y-2\sin^2y=2i\sin y(\cos y+i\sin y)$$ $\endgroup$ – lab bhattacharjee Jun 15 '16 at 6:07
  • $\begingroup$ @labbhattacharjee Could you fix the really confusing first line! $\endgroup$ – almagest Jun 15 '16 at 6:13

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