\docname{sess6.7}
\doctitle{Finding $n$-th Roots}

\mylhead
\myrhead

\title{\doctitle}
\maketitle

To solve linear differential equations with constant coefficients, we need to be able to find the real and complex roots of polynomial equations. Though a lot of this is done today with calculators and computers, one still has to know how to do an important special case by hand: finding the roots of 
\[z^n=\alpha,\]
where $\alpha$ is a complex number, i.e., finding the $n$-th roots of $\alpha$. Polar representation will be a big help in this.

Let's begin with a special case: the {\bf $n$-th roots of unity}: the solutions to 
\[z^n=1.\]
To solve this equation, we use polar representation for both sides, setting $z=re^{i\theta}$ on the left, and using all possible polar angles on the right; using the exponential law to multiply, the above equation then becomes 
\[r^ne^{in\theta}=1\cdot e^{(2k\pi i)}, \qquad k=0,\pm 1, \pm 2,
\cdots .\]
Equating the absolute values and the polar angles of the two sides gives
\[
r^n=1, \qquad n\theta=2k\pi, \qquad k=0,\pm 1, \pm 2, \cdots,
\]
from which we conclude that
\begin{align}
\label{star} r&=1, \qquad &\theta&=\frac{2k\pi}{n}, \qquad k=0,1,\cdots,n-1.
\end{align}
In the above, we get only the value $r=1$, since $r$ must be real and non-negative. We don't need any integer values of $k$ other than $0,\cdots,n-1$, since they would not produce a complex number different from the above $n$ numbers. That is, if we add $an$, an integer multiple of $n$, to $k$, we get the same complex number:
\[
\theta'=\frac{2(k+an)\pi}n=\theta+2a\pi; \qquad\text{ and }\qquad e^{i\theta'}=e^{i\theta},\qquad\text{ since }e^{2a\pi i}=(e^{2\pi i})^a=1.\]
We conclude from (\ref{star}) therefore that
\begin{align}
\label{twentyone} \textit{the }n\textit{-th roots of 1 are the numbers
}\; e^{2k\pi i/n},\; k=0,\cdots,n-1.
\end{align}

This shows there are $n$ complex $n$-th roots of unity. They all lie
on the unit circle in the complex plane, since they have absolute
value 1; they are evenly spaced around the unit circle, starting with
1; the angle between two consecutive ones is $2\pi/n$. These facts are
illustrated in Figure~\ref{nthroots} for the case $n=6$.

\TODO
\begin{figure}[h]
\centering
%  \includegraphics[width=.5\textwidth]{figname}\\
  \caption{The six solutions to the equation $z^6 = 1$ lie on a unit circle in the
    complex plane.}\label{nthroots}
\end{figure}

From (\ref{twentyone}), we get another notation for the roots of unity ($\zeta$ is the Greek letter ``zeta''):
\begin{align}
\label{twentytwo}\textit{the }n\textit{-th roots of 1 are } 1,\zeta,
\zeta^2, \cdots, \zeta^{n-1}, \; \text{where } \zeta = e^{2\pi i/n}.
\end{align}

We now generalize the above to find the $n$-th roots of an arbitrary complex number $w$. We begin by writing $w$ in polar form: 
\begin{align*}
w&= re^{i\theta}; \hspace{1cm}\theta=\Arrg w,\; 0\leq \theta < 2\pi,
\end{align*}
i.e., $\theta$ is the principal value of the polar angle of $w$. Then the same reasoning as we used above shows that if $z$ is an $n$-th root of $w$, then
\begin{align}
\label{twentythree}z^n&=w=re^{i\theta} \quad \text{so} \quad
z=\sqrt[n]{r}e^{i(\theta+2k\pi)/n}, \quad k=0,1,\cdots,n-1.
\end{align}

Comparing this with (\ref{twentytwo}), we see that these $n$ roots can be written in the suggestive form 
\begin{align}
\label{twentyfour}\sqrt[n]{w}&= z_0,\, z_0\zeta,\, z_0\zeta^2, \cdots,z_0\zeta^{n-1},\hspace{1cm}\text{ where }z_0=\sqrt[n]{r}e^{i\theta/n}.
\end{align}
As a check, we see that all of the $n$ complex numbers in
(\ref{twentyfour}) satisfy $z^n = w$ :\\
\begin{center}
\begin{tabular}{rcll}
$(z_0 \zeta^i)^n $
&$=$& $z_0^n\zeta^{ni} = z_0^n\cdot 1^i$, \qquad
& since $\zeta^n=1$,  by (\ref{twentytwo});\\
&$=$& $w$, 
& by the definition (\ref{twentyfour}) of $z_0$ and (\ref{twentythree}).
\end{tabular}
\end{center}

\example Find in Cartesian form all values of \quad 
a)$\sqrt[3]{1}$ \quad b) $\sqrt[4]{1}$

\ans a) According to (\ref{twentytwo}), the cube roots of 1 are $1,\omega,$ and $\omega^2$, where 
\begin{align*}
\omega&=e^{2\pi i/3}=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}3=-\frac 12 + i\frac{\sqrt{3}}2\\
\omega^2&=e^{-2\pi i/3}=\cos\frac{-2\pi}{3}+i\sin\frac{-2\pi}3=-\frac 12-i\frac{\sqrt{3}}2.
\end{align*}

The greek letter $\omega$ (``omega'') is traditionally used for this cube root. Note that for the polar angle of $\omega^2$ we used $-2\pi/3$ rather than the equivalent angle $4\pi/3$, in order to take advantage of the identities
\[\cos(-x)=\cos x\hspace{1cm}\sin(-x)=-\sin x.\]
Note that $\omega^2=\overline{\omega}$. Another way to do this problem would be to draw the position of $\omega^2$ and $\omega$ on the unit circle and use geometry to figure out their coordinates.

b) To find $\sqrt[4]{i}$, we can use (\ref{twentyfour}). We know that $\sqrt[4]{1}=1,i,-1,-i$ (either by drawing the unit circle picture or by using (\ref{twentytwo})). Therefore by (\ref{twentyfour}), we get 
\begin{align*}
\sqrt[4]{i}&=z_0, z_0i, -z_0, -z_0i,&\text{where }z_0&=e^{\pi i/8}=\cos\frac{\pi}8+i\sin\frac{\pi}8;\\
&=a+ib, -b+ia, -a-ib, b-ia\; &\text{where }z_0&=a+ib=\cos\frac{\pi}8+i\sin\frac{\pi}8.
\end{align*}

\example Solve the equation $x^6-2x^3+2=0$.

\ans Treating this as a quadratic equation in $x^3$, we solve the quadratic by using the quadratic formula; the two roots are $1+i$ and $1-i$ (check this!), so the roots of the original equation satisfy either
\[x^3=1+i\quad \text{ or }\quad x^3=1-i .\]

This reduces the problem to finding the cube roots of the two complex numbers $1\pm i$. We begin by writing them in polar form:
\[1+i=\sqrt{2}e^{\pi i/4}, \qquad 1-i = \sqrt{2}e^{-\pi i/4} .\]
(Once again, note the use of the negative polar angle for $1-i$, which is more convenient for calculations.) The three cube roots of the first of these are (by (\ref{twentythree})),

\begin{align*}
\sqrt[6]{2}e^{\pi i/12}&=\sqrt[6]{2}\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)\\
\sqrt[6]{2}e^{3\pi
  i/4}&=\sqrt[6]{2}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right),
\quad \text{since }\frac{\pi}{12}+\frac{2\pi}3=\frac{3\pi}4;\\
\sqrt[6]{2}e^{-7\pi
  i/12}&=\sqrt[6]{2}\left(\cos\frac{7\pi}{12}-i\sin\frac{7\pi}{12}\right),
\quad \text{since }\frac{\pi}{12}-\frac{2\pi}3=-\frac{7\pi}{12}.
\end{align*}
The second cube root can also be written as $\displaystyle\sqrt[6]{2}\left(\frac{-1+i}{\sqrt{2}}\right)=\frac{-1+i}{\sqrt[3]{2}}$.

This gives three of the cube roots. The other three are the cube roots of $1-i$, which may be found by replacing $i$ by $-i$ everywhere (i.e., taking the complex conjugate).

The cube roots can also be described according to (\ref{twentyfour}) as
\[z_1,\, z_1\omega,\, z_1\omega^2\; \text{ and }\; z_2,\, z_2\omega,\,
z_2\omega^2
\; \text{ where }\; z_1=\sqrt[6]{2}e^{\pi i/12}, \; z_2=\sqrt[6]{2}e^{-\pi i/12}.\]

\hbcom{Should this have a concluding paragraph?  What about subsections?}