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: $$ |z|\cdot e^{i\alpha}$$ where $|z|$ corresponds to the norm and $\alpha$ 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

$$\sqrt[n]{|z|\cdot e^{i\alpha}}=\sqrt[n]{|z|}\cdot \sqrt[n]{e^{i\alpha}}= \sqrt[n]{|z|}\cdot e^{i\frac{\alpha+k360^\circ}{n}}$$

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 $e^{i\alpha}$ where $\alpha$ is it argument), then the $n$-th roots will be: $$e^{i\frac{\alpha+k360^\circ}{n}}$$ 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.

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: $$\displaystyle \begin{array}{ll} k=0 &\Rightarrow \ e^{i\frac{k 360^\circ}{n}}= e^0 = 1 \\ k=1 &\Rightarrow \ e^{i\frac{k 360^\circ}{n}}= e^{i\frac{360^\circ}{n}} \\ k=2 &\Rightarrow \ e^{i\frac{k 360^\circ}{n}}= e^{i\frac{2\cdot360^\circ}{n}} \\ \vdots & \vdots \\ k=n-1 &\Rightarrow \ e^{i\frac{k 360^\circ}{n}}= e^{i\frac{(n-1)\cdot360^\circ}{n}} \end{array} $$

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

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

$$k=1 \ \Rightarrow \ e^{i\frac{k 360^\circ}{3}}= e^{i\frac{360^\circ}{3}}= e^{i120^\circ} $$

$$k=2 \ \Rightarrow \ e^{i\frac{k 360^\circ}{3}}= e^{i\frac{2\cdot360^\circ}{3}} =e^{i240^\circ} $$

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

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

$$k=1 \ \Rightarrow \ e^{i\frac{k 360^\circ}{3}}= e^{i\frac{360^\circ}{3}}= e^{i120^\circ}= 1\cdot[\cos(120^\circ)+i \cdot\sin(120^\circ)] $$

$$k=2 \ \Rightarrow \ e^{i\frac{k 360^\circ}{3}}= e^{i\frac{2\cdot360^\circ}{3}} =e^{i240^\circ}=1\cdot[\cos(240^\circ)+i \cdot\sin(240^\circ)] $$