# DeMoivre's Formula and roots of a complex number

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

Theorem (DeMoivre's Formula)

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

${\displaystyle z^{n}=(ze^{i\theta })^{n}=r^{n}(\cos(n\theta )+i\sin(n\theta ))\ .}$

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

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

${\displaystyle z=re^{i\theta }=re^{i(\theta +2k\pi )},\;k\in \mathbb {Z} {\mbox{ from what it follows that:}}}$

${\displaystyle w=z^{\frac {1}{n}}=r^{\frac {1}{n}}e^{i\left({\frac {\theta +2k\pi }{n}}\right)}\;\forall \,k=0,1,\dots ,n-1\ .}$

So, we have that every complex number admits ${\displaystyle n}$ distinct roots.

Example

We compute the ${\displaystyle n}$-th roots of the unity. So, we consider ${\displaystyle z=1\ }$, and we want to find ${\displaystyle w=1^{\frac {1}{n}}\ .}$

${\displaystyle z=1=e^{i2k\pi }\Rightarrow w=e^{i{\frac {2k\pi }{n}}},{\mbox{ that is: }}}$

${\displaystyle {\begin{cases}k=0&w_{0}=1\\k=1&w_{1}=e^{i{\frac {2\pi }{n}}}\\k=2&w_{2}=e^{i{\frac {4\pi }{n}}}=w_{1}^{2}\\\vdots &\vdots \\k=n-1&w_{n-1}=e^{i{\frac {2\pi (n-1)}{n}}}=w_{1}^{n-1}\end{cases}}}$

From this example we see that the ${\displaystyle n}$ distinct roots of a complex number represent, in the plane ${\displaystyle (Re(z),Im(z))}$ defined via the isomorphism between ${\displaystyle \mathbb {C} }$ and ${\displaystyle \mathbb {R} ^{2}}$, the ${\displaystyle n}$ vertices of a regular polygon.