Answer
Solutions of the provided equation in polar form are
$\begin{align}
& \sqrt[6]{1}\left[ \cos 30{}^\circ +i\sin 30{}^\circ \right],\sqrt[6]{1}\left[ \cos 90{}^\circ +i\sin 90{}^\circ \right],\sqrt[6]{1}\left[ \cos 150{}^\circ +i\sin 150{}^\circ \right], \\
& \sqrt[6]{1}\left[ \cos 210{}^\circ +i\sin 210{}^\circ \right],\sqrt[6]{1}\left[ \cos \left( 270{}^\circ \right)+i\sin \left( 270{}^\circ \right) \right],\sqrt[6]{1}\left[ \cos \left( 330{}^\circ \right)+i\sin \left( 330{}^\circ \right) \right] \\
\end{align}$
Solutions of the provided equation in rectangular form are
$\frac{\sqrt{3}}{2}+\frac{1}{2}i,0+1i,-\frac{\sqrt{3}}{2}+\frac{1}{2}i,-\frac{\sqrt{3}}{2}-\frac{1}{2}i,0-1i,\frac{\sqrt{3}}{2}-\frac{1}{2}i$
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 ${{n}^{th}}$ solutions and $\theta $ is in degrees.
There are exactly six solutions of the provided equation as the degree of the equation is $6$.
$\begin{align}
& {{x}^{6}}=-1 \\
& \sqrt[6]{{{x}^{6}}}=\sqrt[6]{-1} \\
& x=\sqrt[6]{-1} \\
& =\sqrt[6]{1\left( \cos 180{}^\circ +i\sin 180{}^\circ \right)}
\end{align}$
Apply DeMoivre’s theorem to find all solutions of the provided equation.
Here, $n=6$, as the number of solutions is six.
${{z}_{k}}=\sqrt[6]{1}\left[ \cos \left( \frac{180{}^\circ +360{}^\circ k}{6} \right)+i\sin \left( \frac{180{}^\circ +360{}^\circ k}{6} \right) \right]$
Here, $k=0,1,2,3,4,5$.
Insert the above values of $k$ to find six distinct solutions of the provided equation.
For $k=0$
$\begin{align}
& {{z}_{0}}=\sqrt[6]{1}\left[ \cos \left( \frac{180{}^\circ +360{}^\circ \cdot \left( 0 \right)}{6} \right)+i\sin \left( \frac{180{}^\circ +360{}^\circ \cdot \left( 0 \right)}{6} \right) \right] \\
& =\sqrt[6]{1}\left[ \cos 30{}^\circ +i\sin 30{}^\circ \right]
\end{align}$
The above solution is in the polar form. Substitute the value of $\cos 30{}^\circ $ and $\sin 30{}^\circ $ to find solution in rectangular form.
$\begin{align}
& {{z}_{0}}=\sqrt[6]{1}\left[ \frac{\sqrt{3}}{2}+i\frac{1}{2} \right] \\
& =\frac{\sqrt{3}}{2}+\frac{1}{2}i
\end{align}$
For $k=1$
$\begin{align}
& {{z}_{1}}=\sqrt[6]{1}\left[ \cos \left( \frac{180{}^\circ +360{}^\circ \cdot \left( 1 \right)}{6} \right)+i\sin \left( \frac{180{}^\circ +360{}^\circ \cdot \left( 1 \right)}{6} \right) \right] \\
& =\sqrt[6]{1}\left[ \cos 90{}^\circ +i\sin 90{}^\circ \right]
\end{align}$
The above solution is in the polar form. Substitute the value of $\cos \left( 90{}^\circ \right)$ and $\sin \left( 90{}^\circ \right)$ to find solution in rectangular form.
$\begin{align}
& {{z}_{1}}=\sqrt[6]{1}\left[ 0+i1 \right] \\
& =0+1i
\end{align}$
For $k=2$
$\begin{align}
& {{z}_{2}}=\sqrt[6]{1}\left[ \cos \left( \frac{180{}^\circ +360{}^\circ \cdot \left( 2 \right)}{6} \right)+i\sin \left( \frac{180{}^\circ +360{}^\circ \cdot \left( 2 \right)}{6} \right) \right] \\
& =\sqrt[6]{1}\left[ \cos 150{}^\circ +i\sin 150{}^\circ \right]
\end{align}$
The above solution is in the polar form. Substitute the value of $\cos \left( 150{}^\circ \right)$ and $\sin \left( 150{}^\circ \right)$ to find solution in rectangular form.
$\begin{align}
& {{z}_{2}}=\sqrt[6]{1}\left[ -\frac{\sqrt{3}}{2}+i\frac{1}{2} \right] \\
& =-\frac{\sqrt{3}}{2}+\frac{1}{2}i
\end{align}$
For $k=3$
$\begin{align}
& {{z}_{3}}=\sqrt[6]{1}\left[ \cos \left( \frac{180{}^\circ +360{}^\circ \cdot \left( 3 \right)}{6} \right)+i\sin \left( \frac{180{}^\circ +360{}^\circ \cdot \left( 3 \right)}{6} \right) \right] \\
& =\sqrt[6]{1}\left[ \cos 210{}^\circ +i\sin 210{}^\circ \right]
\end{align}$
The above solution is in the polar form. Substitute the value of $\cos \left( 210{}^\circ \right)$ and $\sin \left( 210{}^\circ \right)$ to find solution in rectangular form.
$\begin{align}
& {{z}_{3}}=\sqrt[6]{1}\left[ -\frac{\sqrt{3}}{2}+i\left( -\frac{1}{2} \right) \right] \\
& =-\frac{\sqrt{3}}{2}-\frac{1}{2}i
\end{align}$
For $k=4$
$\begin{align}
& {{z}_{4}}=\sqrt[6]{1}\left[ \cos \left( \frac{180{}^\circ +360{}^\circ \cdot \left( 4 \right)}{6} \right)+i\sin \left( \frac{180{}^\circ +360{}^\circ \cdot \left( 4 \right)}{6} \right) \right] \\
& =\sqrt[6]{1}\left[ \cos \left( 270{}^\circ \right)+i\sin \left( 270{}^\circ \right) \right]
\end{align}$
The above solution is in the polar form. Substitute the value of $\cos \left( 270{}^\circ \right)$ and $\sin \left( 270{}^\circ \right)$ to find solution in rectangular form.
$\begin{align}
& {{z}_{4}}=\sqrt[6]{1}\left[ 0+i\left( -1 \right) \right] \\
& =0-1i
\end{align}$
For $k=5$
$\begin{align}
& {{z}_{5}}=\sqrt[6]{1}\left[ \cos \left( \frac{180{}^\circ +360{}^\circ \cdot \left( 5 \right)}{6} \right)+i\sin \left( \frac{180{}^\circ +360{}^\circ \cdot \left( 5 \right)}{6} \right) \right] \\
& =\sqrt[6]{1}\left[ \cos \left( 330{}^\circ \right)+i\sin \left( 330{}^\circ \right) \right]
\end{align}$
The above solution is in the polar form. Substitute the value of $\cos \left( 330{}^\circ \right)$ and $\sin \left( 330{}^\circ \right)$ to find solution in rectangular form.
$\begin{align}
& {{z}_{5}}=\sqrt[6]{1}\left[ \frac{\sqrt{3}}{2}+i\left( -\frac{1}{2} \right) \right] \\
& =\frac{\sqrt{3}}{2}-\frac{1}{2}i
\end{align}$
