Proving $(w-1)^m$ is purely imaginary.

Question

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.

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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!

Answer 1

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.

Answer 2

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