# 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:

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

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

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