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Paul

January 27, 2020

*i.e.*you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

## 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.

Of course, we could just do this by multiplying the number out, but this would be time consuming and prone to mistakes. Instead we can convert to exponential form and then use \(\eqref{eq:eq1}\) to quickly get the answer.

Here is the exponential form of \(3 + 3i\) .

\[r = \sqrt {9 + 9} = 3\sqrt 2 \hspace{0.5in} \tan \theta = \frac{3}{3} \hspace{0.25in} \Rightarrow \hspace{0.25in} {\mathop{\rm Arg}\nolimits} \,z = \frac{\pi }{4}\] \[3 + 3i = 3\sqrt 2 {{\bf{e}}^{i\frac{\pi }{4}}}\]Note that we used the principal value of the argument for the exponential form, although we didn’t have to.

Now, use \(\eqref{eq:eq1}\) to quickly do the computation.

\begin{align*}{\left( {3 + 3i} \right)^5} & = {\left( {3\sqrt 2 } \right)^5}{{\bf{e}}^{i\,\frac{{5\pi }}{4}}}\\ & = 972\sqrt 2 \,\left( {\cos \left( {\frac{{5\pi }}{4}} \right) + i\sin \left( {\frac{{5\pi }}{4}} \right)} \right)\\ & = 972\sqrt 2 \left( { - \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}i} \right)\\ & = - 972 - 972i\end{align*}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**.

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,

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

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,

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

then the n^{th} roots of unity are,

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

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

^{th}roots of unity for \(n\) = 2, 3, and 4.

We’ll start with \(n\) = 2.

\[{\omega _2} = \exp \left( {i\,\,\frac{{2\pi }}{2}} \right) = {{\bf{e}}^{i\,\pi }}\]This gives,

\begin{align*}1 = 1 \hspace{0.5in} {\rm{and}} \hspace{0.5in} {\omega _2} & = {{\bf{e}}^{i\,\pi }}\\ & = \cos \left( \pi \right) + i\sin \left( \pi \right)\\ & = - 1\end{align*}So, for \(n\) = 2 we have -1, and 1 as the n^{th} roots of unity. This should not be too surprising as all we were doing was solving the equation

and we all know that -1 and 1 are the two solutions.

While the result for \(n\) = 2 may not be that surprising that for \(n\) = 3 may be somewhat surprising. In this case we are really solving

\[{z^3} = 1\]and in the world of real numbers we know that the solution to this is \(z\) = 1. However, from the work above we know that there are 3 n^{th} roots of unity in this case. The problem here is that the remaining two are complex solutions and so are usually not thought about when solving for real solution to this equation which is generally what we wanted up to this point.

So, let’s go ahead and find the n^{th} roots of unity for \(n\) = 3.

This gives,

\begin{align*}1 = 1 \hspace{0.5in} {\omega _3} & = \exp \left( {i\,\,\frac{{2\pi }}{3}} \right) & \omega _3^2 & = \exp \left( {i\,\,\frac{{4\pi }}{3}} \right)\\ & = \cos \left( {\frac{{2\pi }}{3}} \right) + i\sin \left( {\frac{{2\pi }}{3}} \right) & \hspace{0.5in} & = \cos \left( {\frac{{4\pi }}{3}} \right) + i\sin \left( {\frac{{4\pi }}{3}} \right)\\ & = - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i & & = - \frac{1}{2} - \frac{{\sqrt 3 }}{2}i\end{align*}I’ll leave it to you to check that if you cube the last two values you will in fact get 1.

Finally, let’s go through \(n=4\). We’ll do this one much quicker than the previous cases.

\[{\omega _4} = \exp \left( {i\,\,\frac{{2\pi }}{4}} \right) = \exp \left( {i\,\,\frac{\pi }{2}} \right)\]This gives,

\begin{align*}1 = 1 \hspace{0.25in} {\omega _4} & = \exp \left( {i\,\,\frac{\pi }{2}} \right) \hspace{0.5in} & \omega _4^2 & = \exp \left( {i\,\,\pi } \right) \hspace{0.5in} & \omega _4^3 & = \exp \left( {i\,\,\frac{{3\pi }}{2}} \right)\\ & = i & & = - 1 & & = - i\end{align*}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,

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,

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

- \({\left( {2i} \right)^{\frac{1}{2}}}\)
- \({\left( {\sqrt 3 \, - i} \right)^{\frac{1}{3}}}\)

The first thing to do is write down the exponential form of the complex number we’re taking the root of.

\[2i = 2\exp \left( {i\,\frac{\pi }{2}} \right)\]So, if we use \({\theta _0} = \frac{\pi }{2}\) we can use \(\eqref{eq:eq6}\) to write down the roots.

\[{a_k} = \sqrt 2 \exp \left( {i\left( {\frac{\pi }{4} + \pi k} \right)} \right) \hspace{0.25in} k = 0,1\]Plugging in for \(k\) gives,

\begin{align*}{a_0} & = \sqrt 2 \exp \left( {i\frac{\pi }{4}} \right) \hspace{0.25in} & {a_1} & = \sqrt 2 \exp \left( {i\left( {\frac{{5\pi }}{4}} \right)} \right)\\ & = \sqrt 2 \left( {\cos \left( {\frac{\pi }{4}} \right) + i\sin \left( {\frac{\pi }{4}} \right)} \right) \hspace{0.25in} & & = \sqrt 2 \left( {\cos \left( {\frac{{5\pi }}{4}} \right) + i\sin \left( {\frac{{5\pi }}{4}} \right)} \right)\\ & = 1 + i & & = - 1 - i\end{align*}I’ll leave it to you to check that if you square both of these will get 2\(i\).

b Show Solution

Here’s the exponential form of the number,

\[\sqrt 3 - i = 2\exp \left( {i\,\left( { - \frac{\pi }{6}} \right)} \right)\]Using \(\eqref{eq:eq6}\) the roots are,

\[{a_k} = \sqrt[3]{2}\exp \left( {i\left( { - \frac{\pi }{{18}} + \frac{{2\pi k}}{3}} \right)} \right) \hspace{0.25in} k = 0,1,2\]Plugging in for \(k\) gives,

\begin{align*}{a_0} & = \sqrt[3]{2}\exp \left( {i\left( { - \frac{\pi }{{18}}} \right)} \right) = \sqrt[3]{2}\left( {\cos \left( { - \frac{\pi }{{18}}} \right) + i\sin \left( { - \frac{\pi }{{18}}} \right)} \right) = 1.24078 - 0.21878\,i\\ {a_1} & = \sqrt[3]{2}\exp \left( {i\frac{{11\pi }}{{18}}} \right) = \sqrt[3]{2}\left( {\cos \left( {\frac{{11\pi }}{{18}}} \right) + i\sin \left( {\frac{{11\pi }}{{18}}} \right)} \right) = - 0.43092 + 1.18394\,i\\ {a_2} & = \,\sqrt[3]{2}\exp \left( {i\frac{{23\pi }}{{18}}} \right) = \sqrt[3]{2}\left( {\cos \left( {\frac{{23\pi }}{{18}}} \right) + i\sin \left( {\frac{{23\pi }}{{18}}} \right)} \right) = - 0.80986 - 0.96516\,i\end{align*}As with the previous part I’ll leave it to you to check that if you cube each of these you will get \(\sqrt 3 - i\) .