Answer
Solutions of the provided equation in polar form are
$\begin{align}
& \sqrt[6]{1}\left[ \cos 0{}^\circ +i\sin 0{}^\circ \right],\sqrt[6]{1}\left[ \cos 60{}^\circ +i\sin 60{}^\circ \right],\sqrt[6]{1}\left[ \cos 120{}^\circ +i\sin 120{}^\circ \right], \\
& \sqrt[6]{1}\left[ \cos 180{}^\circ +i\sin 180{}^\circ \right],\sqrt[6]{1}\left[ \cos \left( 240{}^\circ \right)+i\sin \left( 240{}^\circ \right) \right],\sqrt[6]{1}\left[ \cos \left( 300{}^\circ \right)+i\sin \left( 300{}^\circ \right) \right] \\
\end{align}$
Solutions of the provided equation in rectangular form are
$1.0,\frac{1}{2}+i\frac{\sqrt{3}}{2},-\frac{1}{2}+i\frac{\sqrt{3}}{2},-1,-\frac{1}{2}-\frac{\sqrt{3}}{2}i,\frac{1}{2}-\frac{\sqrt{3}}{2}i$
Work Step by Step
Solution of equation in complex number system can be found by application of DeMoivre’s theorem.
DeMoivre’s theorem for finding complex roots states that every complex number has two distinct complex square roots, three distinct complex cube roots, four distinct complex fourth roots, and so on. Each root has the same modulus.
$\begin{align}
& r=z_{k}^{n} \\
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +360{}^\circ k}{n} \right)+i\sin \left( \frac{\theta +360{}^\circ k}{n} \right) \right] \\
\end{align}$
Where $k$ is the number of distinct ${{n}^{th}}$ solution and $\theta $ is in degree.
There are exactly six solutions of the provided equation as degree of equation is $6$.
$\begin{align}
& {{x}^{6}}=1 \\
& \sqrt[6]{{{x}^{6}}}=\sqrt[6]{1} \\
& x=1
\end{align}$
The expression given above can be written in rectangular form as
$x=1+0i$
Convert the above expression into polar form as
$\begin{align}
& 1+0i=r\left( \cos \theta +i\sin \theta \right) \\
& =1\left( \cos 0{}^\circ +i\sin 0{}^\circ \right)
\end{align}$
Apply Demoivre’s theorem to find all solutions of the provided equation.
Here, $n=6$, as the number of solutions is six.
$\begin{align}
& {{z}_{k}}=\sqrt[6]{1}\left[ \cos \left( \frac{0+360{}^\circ \cdot k}{6} \right)+i\sin \left( \frac{0+360{}^\circ \cdot k}{6} \right) \right] \\
& =\sqrt[6]{1}\left[ \cos \left( \frac{0+360{}^\circ \cdot k}{6} \right)+i\sin \left( \frac{0+360{}^\circ \cdot k}{6} \right) \right]
\end{align}$
Here, $k=0,1,2,3,4,5$.
Insert the above values of $k$ to find six distinct solutions of the provided equation.
For $k=0$
$\begin{align}
& {{z}_{0}}=\sqrt[6]{1}\left[ \cos \left( \frac{0+360{}^\circ \cdot \left( 0 \right)}{6} \right)+i\sin \left( \frac{0+360{}^\circ \cdot \left( 0 \right)}{6} \right) \right] \\
& =\sqrt[6]{1}\left[ \cos 0{}^\circ +i\sin 0{}^\circ \right]
\end{align}$
Above solution is in the polar form. Substitute the value of $\cos 0{}^\circ $ and $\sin 0{}^\circ $ to find solution in rectangular form.
$\begin{align}
& {{z}_{0}}=\sqrt[6]{1}\left[ 1.00+i0.00 \right] \\
& =1.00+i0.00 \\
& =1.0
\end{align}$
For $k=1$
$\begin{align}
& {{z}_{1}}=\sqrt[6]{1}\left[ \cos \left( \frac{0+360{}^\circ \cdot \left( 1 \right)}{6} \right)+i\sin \left( \frac{0+360{}^\circ \cdot \left( 1 \right)}{6} \right) \right] \\
& =\sqrt[6]{1}\left[ \cos 60{}^\circ +i\sin 60{}^\circ \right]
\end{align}$
Above solution is in the polar form. Substitute the value of $\cos \left( 60{}^\circ \right)$ and $\sin \left( 60{}^\circ \right)$ to find solution in rectangular form.
$\begin{align}
& {{z}_{1}}=\sqrt[6]{1}\left[ \frac{1}{2}+i\frac{\sqrt{3}}{2} \right] \\
& =\frac{1}{2}+i\frac{\sqrt{3}}{2}
\end{align}$
For $k=2$
$\begin{align}
& {{z}_{2}}=\sqrt[6]{1}\left[ \cos \left( \frac{0+360{}^\circ \cdot \left( 2 \right)}{6} \right)+i\sin \left( \frac{0+360{}^\circ \cdot \left( 2 \right)}{6} \right) \right] \\
& =\sqrt[6]{1}\left[ \cos 120{}^\circ +i\sin 120{}^\circ \right]
\end{align}$
Above solution is in the polar form. Substitute the value of $\cos \left( 120{}^\circ \right)$ and $\sin \left( 120{}^\circ \right)$ to find solution in rectangular form.
$\begin{align}
& {{z}_{2}}=\sqrt[6]{1}\left[ -\frac{1}{2}+i\frac{\sqrt{3}}{2} \right] \\
& =-\frac{1}{2}+i\frac{\sqrt{3}}{2}
\end{align}$
For $k=3$
$\begin{align}
& {{z}_{3}}=\sqrt[6]{1}\left[ \cos \left( \frac{0+360{}^\circ \cdot \left( 3 \right)}{6} \right)+i\sin \left( \frac{0+360{}^\circ \cdot \left( 3 \right)}{6} \right) \right] \\
& =\sqrt[6]{1}\left[ \cos 180{}^\circ +i\sin 180{}^\circ \right]
\end{align}$
Above solution is in the polar form. Substitute the value of $\cos \left( 180{}^\circ \right)$ and $\sin \left( 180{}^\circ \right)$ to find solution in rectangular form.
$\begin{align}
& {{z}_{3}}=\sqrt[6]{1}\left[ -1+i0 \right] \\
& =-1
\end{align}$
For $k=4$
$\begin{align}
& {{z}_{4}}=\sqrt[6]{1}\left[ \cos \left( \frac{0+360{}^\circ \cdot \left( 4 \right)}{6} \right)+i\sin \left( \frac{0+360{}^\circ \cdot \left( 4 \right)}{6} \right) \right] \\
& =\sqrt[6]{1}\left[ \cos \left( 240{}^\circ \right)+i\sin \left( 240{}^\circ \right) \right]
\end{align}$
Above solution is in the polar form. Substitute the value of $\cos \left( 240{}^\circ \right)$ and $\sin \left( 240{}^\circ \right)$ to find solution in rectangular form.
$\begin{align}
& {{z}_{4}}=\sqrt[6]{1}\left[ -\frac{1}{2}+i\left( -\frac{\sqrt{3}}{2} \right) \right] \\
& =-\frac{1}{2}-\frac{\sqrt{3}}{2}i
\end{align}$
For $k=5$
$\begin{align}
& {{z}_{5}}=\sqrt[6]{1}\left[ \cos \left( \frac{0+360{}^\circ \cdot \left( 5 \right)}{6} \right)+i\sin \left( \frac{0+360{}^\circ \cdot \left( 5 \right)}{6} \right) \right] \\
& =\sqrt[6]{1}\left[ \cos \left( 300{}^\circ \right)+i\sin \left( 300{}^\circ \right) \right]
\end{align}$
Above solution is in the polar form. Substitute the value of $\cos \left( 300{}^\circ \right)$ and $\sin \left( 300{}^\circ \right)$ to find solution in rectangular form.
$\begin{align}
& {{z}_{5}}=\sqrt[6]{1}\left[ \frac{1}{2}+i\left( -\frac{\sqrt{3}}{2} \right) \right] \\
& =\frac{1}{2}-\frac{\sqrt{3}}{2}i
\end{align}$
