1
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.