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As stated in the Wikipedia article on roots of unity, the reciprocal of an $n$-th root of unity is its complex conjugate. They provide the following proof of this statement:

Let $z\in\mathbb{C}$ be a $n$-th root of unity $\Rightarrow$ $$\frac{1}{z}=z^{-1}=z^nz^{-1}=z^{n-1}=\overline{z}$$ I'm unable to figure out why the last equal sign holds.

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  • $\begingroup$ Does it help you see why this is the case if you write an $n$-th root of unity as $e^{2ki\pi/n}$? What's the multiplicative inverse of this number? $\endgroup$ – ah11950 Apr 19 '14 at 11:21
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It is always true that $z\overline z = |z|^2$, so when $|z| = 1$, we may conclude that $\overline z = 1/z$.

edit: and you know that roots of unity have $|z| = 1$ because $|z^k| = |z|^k$ (which you can prove by showing $|wz| = |w||z|$ and repeating) and hence any $n^\mathrm{th}$ root of unity has $|z|^n = 1$, which (since $|z|$ is a positive real number) means $|z| = 1$.

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  • $\begingroup$ How this is related to roots of unity? Suppose $z=re^{i\varphi}$. If $$1=z^n=r^ne^{in\varphi}$$ then $$|z|^n=1$$ since $\text{arg}(1)=0$. Thus, $z^n=1$ implies $|z|=1$. I think I've got it. $\endgroup$ – 0xbadf00d Apr 19 '14 at 11:59
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You should draw a picture, I think there is no way around that.

A suitable picture will be $S^1$ in the complex plane, each $n$-th root marked on it with a dot. $z^n=1$ is $(1,0)$ on the $x$-axis. From this point there are two equidistant points, $z$ and $z^{n-1}$, to its left and right on the circle. Since complex conjugation is the same as mirroring the complex number around the $x$-axis it should be clear that $\overline{z}=z^{n-1}$.

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$$zw=1$$ iff $$|z||w|=1$$ and $$\arg(z)+\arg(w)=0\ \ \ (\!\!\!\!\!\mod 2\pi).$$

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  • $\begingroup$ Maybe I am missing something, but what about $|1||-1|=1$ and $-1\cdot 1=-1$? $\endgroup$ – Phil-ZXX Apr 19 '14 at 11:33
  • $\begingroup$ $\arg(1)=0$, $\arg(-1)=\pi+2k\pi$. $\endgroup$ – Martín-Blas Pérez Pinilla Apr 19 '14 at 11:51

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