10.0 The ${n^{th}}$ root of unity

The equation ${x^n} = 1$ has $n$ roots which are called the ${n^{th}}$ roots of unity. $${x^n} = 1 = \cos 2k\pi + i\sin 2k\pi ...(1)$$ where $k$ is an integer. We can write equation $(1)$ as $$x = {\left( {\cos 2k\pi + i\sin 2k\pi } \right)^{\frac{1}{n}}}$$ Using De Moivre's theorem, $$x = \left( {\cos \frac{{2k\pi }}{n} + i\sin \frac{{2k\pi }}{n}} \right)\quad {\text{where }}k = 0,1,2,3,...,n - 1$$

Let $\alpha = \cos \frac{{2\pi }}{n} + i\sin \frac{{2\pi }}{n}$, then ${n^{th}}$ roots of unity are ${\alpha ^t}(t = 0,1,2,3,...,n - 1)$ i.e., the ${n^{th}}$ roots of unity are $1,\alpha ,{\alpha ^2},...,{\alpha ^{n - 1}}$ which are in Geometric Progression(G.P.).

Using G.P., the sum of ${n^{th}}$ roots of unity is $$1 + \alpha + {\alpha ^2} + {\alpha ^3} + ... + {\alpha ^{n - 1}} = \frac{{1 - {\alpha ^n}}}{{1 - \alpha }} = \frac{{1 - (\cos 2\pi + i\sin 2\pi )}}{{1 - \alpha }} = 0$$

The product of ${n^{th}}$ roots of unity is $$1.\alpha .{\alpha ^2}.{\alpha ^3}...{\alpha ^{n - 1}} = {\alpha ^{\frac{{n(n - 1)}}{2}}} = {\left( {\cos \frac{{2\pi }}{n} + i\sin \frac{{2\pi }}{n}} \right)^{\frac{{n(n - 1)}}{2}}} = \cos \left\{ {\pi (n - 1)} \right\} + i\sin \left\{ {\pi (n - 1)} \right\}$$