# How do the roots of unity relate to De Moivre's Theorem?

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

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 $$z$$.
• In the last line I would prefer that you said " an $n$th root" rather than " the $n$ root". Jan 5, 2019 at 0:27
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, $$cis(x)$$.