n-th roots

The trigonometric form can be expressed in the exponential form, that is, any complex number $z$ has a representation of the type: $| z | \cdot e^{i α}$ where $| z |$ corresponds to the norm and $α$ to the argument.

If we have the complex numbers of this trigonometric form it is very easy to calculate the $n$ -th roots, since we have

$\sqrt[n]{| z | \cdot e^{i α}} = \sqrt[n]{| z |} \cdot \sqrt[n]{e^{i α}} = \sqrt[n]{| z |} \cdot e^{i \frac{α + k 360^{\circ}}{n}}$

due to the properties of the roots and powers with rational exponents.

In the particular case of all the complex numbers whose norm is $1$ (they can be written as $e^{i α}$ where $α$ is it argument), then the $n$ -th roots will be: $e^{i \frac{α + k 360^{\circ}}{n}}$ For $k = 0$ we will have the first root, for $k = 1$ the second one, and successively up to coming to the $n$ -th root, which corresponds to $k = n - 1$ .

Therefore, we obtain $n$ different roots.

Example

Let's see a concrete example, we are going to find the $n$ -th roots of the unit, that is the number $1$ .

We want to determine the values $z$ such that $z^{n} = 1$ .

$1 = 1 \cdot [\cos (0^{\circ}) + i \cdot \sin (0^{\circ})] = 1 \cdot e^{i 0^{\circ}}$

Then, the $n$ roots of the unit are given by: $e^{i \frac{0^{\circ} + 360^{\circ} k}{n}} = e^{i \frac{360^{\circ} k}{n}}$

with $k = 0, 1, 2, \dots, (n - 1)$ .

Then they will be: $\begin{array}{ll} k = 0 & \Rightarrow e^{i \frac{k 360^{\circ}}{n}} = e^{0} = 1 \\ k = 1 & \Rightarrow e^{i \frac{k 360^{\circ}}{n}} = e^{i \frac{360^{\circ}}{n}} \\ k = 2 & \Rightarrow e^{i \frac{k 360^{\circ}}{n}} = e^{i \frac{2 \cdot 360^{\circ}}{n}} \\ ⋮ & ⋮ \\ k = n - 1 & \Rightarrow e^{i \frac{k 360^{\circ}}{n}} = e^{i \frac{(n - 1) \cdot 360^{\circ}}{n}} \end{array}$

In particular, for example, if $n = 3$ then the roots are:

$k = 0 \Rightarrow e^{i \frac{k 360^{\circ}}{3}} = e^{0} = 1$

$k = 1 \Rightarrow e^{i \frac{k 360^{\circ}}{3}} = e^{i \frac{360^{\circ}}{3}} = e^{i 120^{\circ}}$

$k = 2 \Rightarrow e^{i \frac{k 360^{\circ}}{3}} = e^{i \frac{2 \cdot 360^{\circ}}{3}} = e^{i 240^{\circ}}$

If we prefer to express it in trigonometric form we only need to do:

$k = 0 \Rightarrow e^{i \frac{k 360^{\circ}}{3}} = e^{0} = 1 \cdot [\cos (0^{\circ}) + i \cdot \sin (0^{\circ})]$

$k = 1 \Rightarrow e^{i \frac{k 360^{\circ}}{3}} = e^{i \frac{360^{\circ}}{3}} = e^{i 120^{\circ}} = 1 \cdot [\cos (120^{\circ}) + i \cdot \sin (120^{\circ})]$

$k = 2 \Rightarrow e^{i \frac{k 360^{\circ}}{3}} = e^{i \frac{2 \cdot 360^{\circ}}{3}} = e^{i 240^{\circ}} = 1 \cdot [\cos (240^{\circ}) + i \cdot \sin (240^{\circ})]$