Tales of roots

The n ^th roots of unity ------------------------------------------

These numbers are aptly named: given a nonzero integer n, the *n^th roots of unity are all the complex numbers whose n ^th power equals 1. In other words, if we denote them by z, they satisfy the equation *z ⁿ* = 1.

For a given n, there are exactly n n^th roots of unity. A short calculation in ℂ shows that they can be expressed using the complex exponential: for 0 ≤ k ≤ n − 1, they are the numbers $z_k=e^{\frac{2ik\pi }{n}}=\cos \left(\frac{2k\pi }{n}\right)+i\sin \left(\frac{2k\pi }{n}\right)\mathrm{.}z\_k\; =\; e^\{\backslash frac\{2ik\backslash pi\}\{n\}\}\; =\; \backslash cos\; \backslash left(\; \backslash frac\{2k\backslash pi\}\{n\}\; \backslash right)\; +\; i\; \backslash sin\; \backslash left(\; \backslash frac\{2k\backslash pi\}\{n\}\; \backslash right).$

For example, if n = 2, both roots are real: unsurprisingly, we get 1 and −1.

For n = 3, they are known as the cube roots of unity: in the standard notation, they are 1, $j=e^{\frac{2i\pi }{3}}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}j\; =e^\{\backslash frac\{2i\backslash pi\}\{3\}\}\; =\; -\; \backslash frac\{1\}\{2\}\; +\; i\; \backslash frac\{\backslash sqrt\{3\}\}\{2\}$ and $j^2=e^{\frac{4i\pi }{3}}=-\frac{1}{2}-i\frac{\sqrt{3}}{2}j^2\; =\; e^\{\backslash frac\{4i\backslash pi\}\{3\}\}\; =\; -\; \backslash frac\{1\}\{2\}\; -\; i\; \backslash frac\{\backslash sqrt\{3\}\}\{2\}$ since j ² is the square of j.