# DeMoivre's Formula and roots of a complex number

The next theorem provides an important tool to calculate the $n$ -th power of a complex number:

Theorem (DeMoivre's Formula)

$\forall \,z\in \mathbb {C} ,\,n\in \mathbb {Z}$ we have that:

$z^{n}=(ze^{i\theta })^{n}=r^{n}(\cos(n\theta )+i\sin(n\theta ))\ .$ The computation of the $n$ -th roots of a complex number is a bit more difficult. That is, given $z\in \mathbb {C} {\mbox{ we are asking if }}\exists \,w\in \mathbb {C} \,\colon \,w^{n}=z{\mbox{ or, analagously, }}w=z^{\frac {1}{n}}.$ The number $w$ is called the n-th root of z.

By considering the polar form of a complex number we have:

$z=re^{i\theta }=re^{i(\theta +2k\pi )},\;k\in \mathbb {Z} {\mbox{ from what it follows that:}}$ $w=z^{\frac {1}{n}}=r^{\frac {1}{n}}e^{i\left({\frac {\theta +2k\pi }{n}}\right)}\;\forall \,k=0,1,\dots ,n-1\ .$ So, we have that every complex number admits $n$ distinct roots.

Example

We compute the $n$ -th roots of the unity. So, we consider $z=1\$ , and we want to find $w=1^{\frac {1}{n}}\ .$ $z=1=e^{i2k\pi }\Rightarrow w=e^{i{\frac {2k\pi }{n}}},{\mbox{ that is: }}$ ${\begin{cases}k=0&w_{0}=1\\k=1&w_{1}=e^{i{\frac {2\pi }{n}}}\\k=2&w_{2}=e^{i{\frac {4\pi }{n}}}=w_{1}^{2}\\\vdots &\vdots \\k=n-1&w_{n-1}=e^{i{\frac {2\pi (n-1)}{n}}}=w_{1}^{n-1}\end{cases}}$ From this example we see that the $n$ distinct roots of a complex number represent, in the plane $(Re(z),Im(z))$ defined via the isomorphism between $\mathbb {C}$ and $\mathbb {R} ^{2}$ , the $n$ vertices of a regular polygon.