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How do the roots of unity relate to De Moivre's Theorem? Others always pair the two up, but I do not understand why.

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If we let the $n$th root of unity be $$z=\cos(\theta)+i\cdot \sin(\theta)$$ then by De Moivre's Theorem$$z^n=(\cos(\theta)+i\cdot \sin(\theta))^n$$$$=\cos(n\cdot \theta)+i\cdot \sin(n\cdot \theta)$$where $n\in\mathbb{N}$. But:$$z^n=1=\cos(2\pi\cdot m)+i\cdot \sin(\pi+2\pi\cdot m)$$ Where $m\in\mathbb{Z}$. So, equating real parts,$$\cos(n\cdot \theta)=\cos(2\pi\cdot m)$$$$\therefore\theta=\frac{2\pi\cdot m}{n}$$ i.e. the $n$th roots of unity are $$z=\cos(\frac{2\pi\cdot m}{n})+i\cdot \sin(\frac{2\pi\cdot m}{n})$$Where $0\le m \le n-1$ as all other values of $m$ will produce the same values of $z$.

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  • $\begingroup$ In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root". $\endgroup$ Jan 5, 2019 at 0:27
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The roots of unity are all in the form $cos(x)+isin(x)$. Nth roots of unity, raised to the n power, will be 1. Thus, De Moivre's Theorem, which states that $cis(x)^n=cis(nx)$ also tells us that if $cis(x)$ is a nth root of unity, then $cis(x)^n=cis(nx)=1$. This process can also be reversed to find nth roots of unity, as, substituting in a n value, we then have a trig equation we can solve to find the values of x, and the nth roots of unity, $cis(x)$.

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