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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 \(\eqref{eq:eq3}\).