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5. Powers and Roots

In this section we’re going to take a look at a really nice way of quickly computing integer powers and roots of complex numbers.

We’ll start with integer powers of $z = r{{\bf{e}}^{i\theta }}$ since they are easy enough. If $n$ is an integer then,

$$\begin{equation}{z^n} = {\left( {r{{\bf{e}}^{i\theta }}} \right)^n} = {r^n}{{\bf{e}}^{i\,\,n\theta }}\label{eq:eq1}\end{equation}$$

There really isn’t too much to do with powers other than working a quick example.

So, there really isn’t too much to integer powers of a complex number.

Note that if $r = 1$ then we have,

${z^n} = {\left( {{{\bf{e}}^{i\theta }}} \right)^n} = {{\bf{e}}^{i\,\,n\theta }}$

and if we take the last two terms and convert to polar form we arrive at a formula that is called de Moivre’s formula.

$${\left( {\cos \theta + i\sin \theta } \right)^n} = \cos \left( {n\,\theta } \right) + i\sin \left( {n\,\theta } \right) \hspace{0.25in} n = 0, \pm 1, \pm 2, \ldots$$

We now need to move onto computing roots of complex numbers. We’ll start this off “simple” by finding the n^th roots of unity. The n^th roots of unity for $n = 2,3, \ldots$ are the distinct solutions to the equation,

$${z^n} = 1$$

Clearly (hopefully) $z = 1$ is one of the solutions. We want to determine if there are any other solutions. To do this we will use the fact from the previous sections that states that ${z_1} = {z_2}$ if and only if

$${r_1} = {r_2} \hspace{0.25in} {\mbox{and}}\hspace{0.25in} {\theta _2} = {\theta _1} + 2\pi k\,\,\,{\mbox{for some integer }} \;k{\rm{ }}\left( {i.e.\,\,k = 0, \pm 1, \pm 2, \ldots } \right)$$

So, let’s start by converting both sides of the equation to complex form and then computing the power on the left side. Doing this gives,

$${\left( {r{{\bf{e}}^{i\theta }}} \right)^n} = 1\,{{\bf{e}}^{i\left( 0 \right)}} \hspace{0.5in} \Rightarrow \hspace{0.5in} {r^n}{{\bf{e}}^{i\,\,n\theta }} = 1\,{{\bf{e}}^{i\left( 0 \right)}}$$

So, according to the fact these will be equal provided,

$${r^n} = 1 \hspace{0.5in} n\theta = 0 + 2\pi k\, \hspace{0.25in} k = 0, \pm 1, \pm 2, \ldots$$

Now, $r$ is a positive integer (by assumption of the exponential/polar form) and so solving gives,

$$r = 1 \hspace{0.5in} \theta = \frac{{2\pi k}}{n}\, \hspace{0.25in} k = 0, \pm 1, \pm 2, \ldots$$

The solutions to the equation are then,

$$z = \exp \left( {i\,\,\frac{{2\pi k}}{n}} \right) \hspace{0.25in} k = 0, \pm 1, \pm 2, \ldots$$

Recall from our discussion on the polar form (and hence the exponential form) that these points will lie on the circle of radius $r$. So, our points will lie on the unit circle and they will be equally spaced on the unit circle at every $\frac{{2\pi }}{n}$ radians. Note this also tells us that there are $n$ distinct roots corresponding to $k = 0,1,2, \ldots ,n - 1$ since we will get back to where we started once we reach $k = n$

Therefore, there are $n$ n^th roots of unity and they are given by,

$$\begin{equation}\exp \left( {i\,\,\frac{{2\pi k}}{n}} \right) = \cos \left( {\frac{{2\pi k}}{n}} \right) + i\sin \left( {\frac{{2\pi k}}{n}} \right) \hspace{0.5in} k = 0,1,2, \ldots ,n - 1\label{eq:eq2}\end{equation}$$

There is a simpler notation that is often used to denote n^th roots of unity. First define,

$$\begin{equation}{\omega _n} = \exp \left( {i\,\,\frac{{2\pi }}{n}} \right)\label{eq:eq3}\end{equation}$$

then the n^th roots of unity are,

$$\omega _n^k = {\left( {\exp \left( {i\,\,\frac{{2\pi }}{n}} \right)} \right)^k} = \exp \left( {i\,\,\frac{{2\pi k}}{n}} \right) \hspace{0.5in} k = 0,1,2, \ldots n - 1$$

Or, more simply the n^th roots of unity are,

$$\begin{equation}1,{\omega _n},\omega _n^2, \ldots ,\omega _n^{n - 1}\label{eq:eq4}\end{equation}$$

where ${\omega _n}$ is defined in $\eqref{eq:eq3}$.

Now, let’s move on to more general roots. First let’s get some notation out of the way. We’ll define $z_{0}^{{1}/{n}\;}$ to be any number that will satisfy the equation

$$\begin{equation}{z^n} = {z_0}\label{eq:eq5}\end{equation}$$

To find the values of $z_{0}^{{1}/{n}\;}$ we’ll need to solve this equation and we can do that in the same way that we found the n^th roots of unity. So, if ${r_0} = \left| {{z_0}} \right|$ and ${\theta _0} = \arg {z_0}$ (note ${\theta _0}$ can be any value of the argument, but we usually use the principal value) we have,

$${\left( {r{{\bf{e}}^{i\theta }}} \right)^n} = {r_0}\,{{\bf{e}}^{i\,\,{\theta _0}}} \hspace{0.5in} \Rightarrow \hspace{0.5in} {r^n}{{\bf{e}}^{i\,\,n\theta }} = {r_0}\,{{\bf{e}}^{i\,\,{\theta _0}}}$$

So, this tells us that,

$$r = \sqrt[n]{{{r_0}}} \hspace{0.5in} \theta = \frac{{{\theta _0}}}{n} + \frac{{2\pi k}}{n} \hspace{0.25in} k = 0, \pm 1, \pm 2, \ldots$$

The distinct solutions to $\eqref{eq:eq5}$ are then,

$$\begin{equation}{a_k} = \sqrt[n]{{{r_0}}}\exp \left( {i\left( {\frac{{{\theta _0}}}{n} + \frac{{2\pi k}}{n}} \right)} \right) \hspace{0.25in} k = 0,1,2, \ldots ,n - 1\label{eq:eq6}\end{equation}$$

So, we can see that just as there were n n^th roots of unity there are also n n^th roots of ${z_0}$ .

Finally, we can again simplify the notation up a little. If $a$ is any of the n^th roots of ${z_0}$ then all the roots can be written as,

$$a,\,\,a{\omega _n},\,\,a\omega _n^2,\, \ldots ,,\,\,a\omega _n^{n - 1}$$

where ${\omega _n}$ is defined in $\eqref{eq:eq3}$.