For the calculation of the $$n$$-th power of a complex number expressed in trigonometric form let's see what the expression will be. We consider the product of $$n$$ complex numbers in trigonometric form:

$$$\big( |z|\cdot[\cos(\alpha)+i \cdot\sin(\alpha)] \big)^n=$$$

$$$= \big( |z|\cdot[\cos(\alpha)+i \cdot\sin(\alpha)] \big) \stackrel{(n)}{\cdots} \big( |z|\cdot[\cos(\alpha)+i \cdot\sin(\alpha)] \big)=$$$

$$$=|z|^n \cdot[\cos(n\alpha)+i \cdot\sin(n\alpha)]$$$ This formula is the one that gives the n-th power of a complex number in trigonometric form and that was given by Moivre.

Let's see an example: $$$ \displaystyle \begin{array}{rl} \big( 5\cdot[\cos(60^\circ)+i \cdot\sin(60^\circ)] \big)^3 =& 5^3 \cdot[\cos(3\cdot60^\circ)+i \cdot\sin(3\cdot60^\circ)] \\ =& 125 \cdot[\cos(180^\circ)+i \cdot\sin(180^\circ)] \end{array} $$$

Once the exponentiation is done, we can study the roots.

Given a complex number, any other complex number that is raised to the n-th exponent which gives an equal result to this first one, is said to be its $$n$$-th root.

Let's see whether, given any complex number whose module and argument are given by $$R$$ and $$\phi$$ respectively, it always has $$n$$-th roots.

From the definition the condition so that $$|z|\cdot[\cos(\alpha)+i \cdot\sin(\alpha)]$$ is an $$n$$-th root, is: $$$ \big( |z|\cdot[\cos(\alpha)+i \cdot\sin(\alpha)] \big)^n = R\cdot [\cos(\phi)+i \cdot\sin(\phi)]$$$

The two numbers represented by the first and second member of this equality have to be equal, that is, they will have to have the same norm and its arguments will have to differ in a multiple of $$360^\circ$$:

$$$ |z|^n=R \quad \text{ and } \quad n\alpha=\phi+k\cdot 360^\circ$$$

The norm $$|z|$$ is perfectly determined by the first of these equations, since it has to be a positive number and with its $$n$$-th power equal to $$R$$, whereby we conclude that:

$$$|z|=\sqrt[n]{R}$$$ This way, $$|z|$$ is the arithmetic $$n$$-th root of $$R$$.

As for the argument $$\alpha$$, the second equation gives us: $$$\alpha=\dfrac{\phi}{n}+\dfrac{k\cdot 360^\circ}{n}$$$

At first sight it might seem that it has infinite values, but in fact, for $$k = 0$$ to $$k = n-1$$ we will obtain $$n$$ different arguments, and from that we will obtain the same complex numbers that we had already found.

In short, any non zero complex number $$R\cdot [\cos(\phi)+i \cdot\sin(\phi)]$$ has $$n$$ different $$n$$-th roots, which norm is the same for all, (it is equal to the arithmetic $$n$$-th root of the norm $$R$$) and which arguments (except for multiples of $$360^\circ$$) are:

$$$ \dfrac{\phi}{n}, \ \dfrac{\phi+360^\circ}{n}, \ \dfrac{\phi+2\cdot 360^\circ}{n}, \ \dfrac{\phi+3\cdot 360^\circ}{n}, \ \dots \ , \ \dfrac{\phi+(n-1)\cdot 360^\circ}{n}$$$

Let's calculate the fourth roots of: $$$4\cdot[\cos(60^\circ)+i \cdot\sin(60^\circ)]$$$ We look for $$$ |z|^4=4 \ \Rightarrow \ |z|=\sqrt[4]{4}=\sqrt{2}$$$ And the arguments will be: $$$ \dfrac{60^\circ}{4}, \ \dfrac{60^\circ+360^\circ}{4}, \ \dfrac{60^\circ+2\cdot 360^\circ}{4}, \ \dfrac{60^\circ+3\cdot 360^\circ}{4}$$$

As we can see this method is very similar to the one used with the polar form of the complex numbers.